Tricky Integral -- $\int_0^1 \sqrt{x^2-4x+3} \arcsin(x)~dx$

TL;DR: I can't get a closed form for the integral below.

$$ \int_0^1 \sqrt{x^2-4x+3} \arcsin(x)~dx $$

I got an interesting question from a coworker a while ago:

Question:

The quantities $a$, $b$, and $c$ are chosen uniformly and independently from $[0, 1]$.

a) What is the probability a triangle can be constructed with $a$, $b$, and $c$ as side lengths?

b) Given we can form such a triangle, what is its expected area?

I can do a) pretty easily -- each constraint like $a < b + c$ cuts off a corner of the cube with area $1/6$, and the cut-off bits are disjoint, so the remaining area is $1/2$.

Part b) is where things get hairy. I can reduce the problem down to a single integral. I feel like it should be expressible in terms of known constants, though I admit I have no good reason to believe this.

$$ \frac{3}{40} \int_0^1 x \sqrt{3-4x+x^2} \left( \sqrt{1 - x^2} + \frac{\arcsin{x}}{x} \right)~dx $$

That can be split into two parts:

$$ \frac{3}{40} \int_0^1 x \sqrt{(3-4x+x^2)(1 - x^2)}~dx + \frac{3}{40} \int_0^1 \sqrt{3-4x+x^2} \arcsin(x)~dx $$

The first part can be solved exactly. $$ \begin{align*} \int_0^1 x \sqrt{(3-4x+x^2)(1 - x^2)}~dx &= \int_0^1 x \sqrt{(3-x)(1-x)(1-x)(1+x)}~dx \\ &= \int_0^1 x(1-x) \sqrt{(3-x)(1+x)}~dx \\ &= \frac{1}{12} (32 - 9 \sqrt{3} - 4\pi) \textrm{ by Mathematica} \end{align*} $$

The second part is still pretty stubborn.

Mathematica tells me the integral (without the 3/40 constant) is approximately 0.452854, but doesn't given an exact form. Does anyone have any ideas how to evaluate this further?

Solution 1:

WARNING: Incoming wall of math.

Let $\mathcal{I}$ denote the value of the definite integral,

$$\begin{align} \mathcal{I} &:=\int_{0}^{1}\mathrm{d}x\,\sqrt{x^{2}-4x+3}\arcsin{\left(x\right)}.\\ \end{align}$$

For our purposes here we may define the inverse sine function of a real argument via the usual integral representation

$$\arcsin{\left(z\right)}:=\int_{0}^{z}\mathrm{d}x\,\frac{1}{\sqrt{1-x^{2}}};~~~\small{-1\le z\le1}.$$

The integral definition of $\arcsin$ is particularly handy for deriving the following inverse trigonometric identity:

$$\forall z\in\left[0,1\right]:\arcsin{\left(1-2z^{2}\right)}=\frac{\pi}{2}-2\arcsin{\left(z\right)}.$$

Similarly, the inverse hyperbolic sine function of a real argument may be defined via the integral representation

$$\operatorname{arsinh}{\left(z\right)}:=\int_{0}^{z}\mathrm{d}x\,\frac{1}{\sqrt{1+x^{2}}};~~~\small{z\in\mathbb{R}},$$

and it can be verified through differentiation that the inverse hyperbolic sine may be expressed in the logarithmic form

$$\operatorname{arsinh}{\left(z\right)}=\ln{\left(z+\sqrt{1+z^{2}}\right)};~~~\small{z\in\mathbb{R}}.$$

Turning now to the main task of evaluating $\mathcal{I}$,

$$\begin{align} \mathcal{I} &=\int_{0}^{1}\mathrm{d}x\,\sqrt{x^{2}-4x+3}\arcsin{\left(x\right)}\\ &=\int_{0}^{1}\mathrm{d}x\,\sqrt{\left(3-x\right)\left(1-x\right)}\arcsin{\left(x\right)}\\ &=\int_{0}^{1}\mathrm{d}x\,\sqrt{\left(2+x\right)x}\arcsin{\left(1-x\right)};~~~\small{\left[x\mapsto1-x\right]}\\ &=4\int_{0}^{\frac12}\mathrm{d}x\,\sqrt{x\left(1+x\right)}\arcsin{\left(1-2x\right)};~~~\small{\left[x\mapsto2x\right]}\\ &=8\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,x^{2}\sqrt{1+x^{2}}\arcsin{\left(1-2x^{2}\right)};~~~\small{\left[x\mapsto x^{2}\right]}\\ &=8\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,x^{2}\sqrt{1+x^{2}}\left[\frac{\pi}{2}-2\arcsin{\left(x\right)}\right]\\ &=4\pi\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,x^{2}\sqrt{1+x^{2}}-16\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,x^{2}\sqrt{1+x^{2}}\arcsin{\left(x\right)}\\ &=\pi\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,\frac{x^{2}\left(4+4x^{2}\right)}{\sqrt{1+x^{2}}}-16\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,x^{2}\sqrt{1+x^{2}}\arcsin{\left(x\right)}\\ &=\pi\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,\frac{x^{2}\left(3+4x^{2}\right)}{\sqrt{1+x^{2}}}+\pi\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,\frac{x^{2}}{\sqrt{1+x^{2}}}\\ &~~~~~-16\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,x^{2}\sqrt{1+x^{2}}\arcsin{\left(x\right)}\\ &=\pi\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,\frac{\mathrm{d}}{\mathrm{d}x}\left[x^{3}\sqrt{1+x^{2}}\right]+\frac{\pi}{2}\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,\frac{2x^{2}+1-1}{\sqrt{1+x^{2}}}\\ &~~~~~-16\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,x^{2}\sqrt{1+x^{2}}\arcsin{\left(x\right)}\\ &=\frac{\pi}{2}\cdot\frac{\sqrt{3}}{2}\\ &~~~~~+\frac{\pi}{2}\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,\frac{2x^{2}+1}{\sqrt{1+x^{2}}}-\frac{\pi}{2}\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,\frac{1}{\sqrt{1+x^{2}}}\\ &~~~~~-16\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,x^{2}\sqrt{1+x^{2}}\arcsin{\left(x\right)}\\ &=\frac{\sqrt{3}\,\pi}{4}+\frac{\pi}{2}\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,\frac{\mathrm{d}}{\mathrm{d}x}\left[x\sqrt{1+x^{2}}\right]-\frac{\pi}{2}\operatorname{arsinh}{\left(\frac{1}{\sqrt{2}}\right)}\\ &~~~~~-16\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,x^{2}\sqrt{1+x^{2}}\arcsin{\left(x\right)}\\ &=\frac{\sqrt{3}\,\pi}{4}+\frac{\pi}{2}\cdot\frac{\sqrt{3}}{2}-\frac{\pi}{2}\operatorname{arsinh}{\left(\frac{1}{\sqrt{2}}\right)}\\ &~~~~~-16\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,x^{2}\sqrt{1+x^{2}}\arcsin{\left(x\right)}\\ &=\frac{\sqrt{3}\,\pi}{2}-\frac{\pi}{2}\operatorname{arsinh}{\left(\frac{1}{\sqrt{2}}\right)}-16\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,x^{2}\sqrt{1+x^{2}}\arcsin{\left(x\right)}.\\ \end{align}$$

Define the auxiliary functions $f:\left[-1,1\right]\rightarrow\mathbb{R}_{>0}$ and $g:\left[-1,1\right]\rightarrow\mathbb{R}_{\ge0}$ via the respective expressions,

$$f{\left(x\right)}:=2\sqrt{1+x^{2}}\left[\sqrt{1-x^{2}}+x\arcsin{\left(x\right)}\right]$$

and

$$g{\left(x\right)}:=4x^{2}\sqrt{1+x^{2}}\left[\sqrt{1-x^{2}}+x\arcsin{\left(x\right)}\right].$$

We then obtain the following expression for the derivative of $f$ at $x\in\left(-1,1\right)$:

$$\begin{align} f^{\prime}{\left(x\right)} &=\frac{\mathrm{d}}{\mathrm{d}x}\bigg{[}2\sqrt{1+x^{2}}\left[\sqrt{1-x^{2}}+x\arcsin{\left(x\right)}\right]\bigg{]}\\ &=\frac{\mathrm{d}}{\mathrm{d}x}\left[2\sqrt{1+x^{2}}\right]\left[\sqrt{1-x^{2}}+x\arcsin{\left(x\right)}\right]\\ &~~~~~+2\sqrt{1+x^{2}}\frac{\mathrm{d}}{\mathrm{d}x}\left[\sqrt{1-x^{2}}+x\arcsin{\left(x\right)}\right]\\ &=\frac{2x}{\sqrt{1+x^{2}}}\left[\sqrt{1-x^{2}}+x\arcsin{\left(x\right)}\right]\\ &~~~~~+2\sqrt{1+x^{2}}\arcsin{\left(x\right)}\\ &=\frac{2x\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}}+\frac{2x^{2}}{\sqrt{1+x^{2}}}\arcsin{\left(x\right)}\\ &~~~~~+\frac{2+2x^{2}}{\sqrt{1+x^{2}}}\arcsin{\left(x\right)}\\ &=\frac{2x\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}}+\frac{2}{\sqrt{1+x^{2}}}\arcsin{\left(x\right)}+\frac{4x^{2}}{\sqrt{1+x^{2}}}\arcsin{\left(x\right)}.\\ \end{align}$$

In a similar fashion, we also obtain the following expression for the derivative of $g$ at $x\in\left(-1,1\right)$:

$$\begin{align} g^{\prime}{\left(x\right)} &=\frac{\mathrm{d}}{\mathrm{d}x}\bigg{[}4x^{2}\sqrt{1+x^{2}}\left[\sqrt{1-x^{2}}+x\arcsin{\left(x\right)}\right]\bigg{]}\\ &=\frac{\mathrm{d}}{\mathrm{d}x}\left[4x^{2}\sqrt{1+x^{2}}\right]\left[\sqrt{1-x^{2}}+x\arcsin{\left(x\right)}\right]\\ &~~~~~+4x^{2}\sqrt{1+x^{2}}\frac{\mathrm{d}}{\mathrm{d}x}\left[\sqrt{1-x^{2}}+x\arcsin{\left(x\right)}\right]\\ &=\left[\left(8x\sqrt{1+x^{2}}\right)+4x^{2}\left(\frac{2x}{2\sqrt{1+x^{2}}}\right)\right]\left[\sqrt{1-x^{2}}+x\arcsin{\left(x\right)}\right]\\ &~~~~~+4x^{2}\sqrt{1+x^{2}}\arcsin{\left(x\right)}\\ &=2x\left[6\sqrt{1+x^{2}}-\frac{2}{\sqrt{1+x^{2}}}\right]\left[\sqrt{1-x^{2}}+x\arcsin{\left(x\right)}\right]\\ &~~~~~+4x^{2}\sqrt{1+x^{2}}\arcsin{\left(x\right)}\\ &=2x\left[6\sqrt{1+x^{2}}-\frac{2}{\sqrt{1+x^{2}}}\right]\sqrt{1-x^{2}}\\ &~~~~~+2x\left[6\sqrt{1+x^{2}}-\frac{2}{\sqrt{1+x^{2}}}\right]x\arcsin{\left(x\right)}\\ &~~~~~+4x^{2}\sqrt{1+x^{2}}\arcsin{\left(x\right)}\\ &=2x\left[\frac{2\left(2+3x^{2}\right)\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}}\right]-\frac{4x^{2}}{\sqrt{1+x^{2}}}\arcsin{\left(x\right)}+16x^{2}\sqrt{1+x^{2}}\arcsin{\left(x\right)}.\\ \end{align}$$

Returning to the evaluation of $\mathcal{I}$,

$$\begin{align} \mathcal{I} &=\frac{\sqrt{3}\,\pi}{2}-\frac{\pi}{2}\operatorname{arsinh}{\left(\frac{1}{\sqrt{2}}\right)}-16\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,x^{2}\sqrt{1+x^{2}}\arcsin{\left(x\right)}\\ &=\frac{\sqrt{3}\,\pi}{2}-\frac{\pi}{2}\operatorname{arsinh}{\left(\frac{1}{\sqrt{2}}\right)}-\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,g^{\prime}{\left(x\right)}\\ &~~~~~+\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,2x\left[\frac{2\left(2+3x^{2}\right)\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}}\right]-\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,\frac{4x^{2}}{\sqrt{1+x^{2}}}\arcsin{\left(x\right)}\\ &=\frac{\sqrt{3}\,\pi}{2}-\frac{\pi}{2}\operatorname{arsinh}{\left(\frac{1}{\sqrt{2}}\right)}-\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,g^{\prime}{\left(x\right)}\\ &~~~~~+\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,\frac{2x\left(4+6x^{2}\right)\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}}-\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,f^{\prime}{\left(x\right)}\\ &~~~~~+\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,\frac{2x\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}}+\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,\frac{2}{\sqrt{1+x^{2}}}\arcsin{\left(x\right)}\\ &=\frac{\sqrt{3}\,\pi}{2}-\frac{\pi}{2}\operatorname{arsinh}{\left(\frac{1}{\sqrt{2}}\right)}-\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,g^{\prime}{\left(x\right)}-\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,f^{\prime}{\left(x\right)}\\ &~~~~~+\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,\frac{2x\left(5+6x^{2}\right)\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}}+2\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,\frac{\arcsin{\left(x\right)}}{\sqrt{1+x^{2}}}\\ &=\frac{\sqrt{3}\,\pi}{2}-\frac{\pi}{2}\operatorname{arsinh}{\left(\frac{1}{\sqrt{2}}\right)}\\ &~~~~~-\left[g{\left(\frac{1}{\sqrt{2}}\right)}-g{\left(0\right)}\right]-\left[f{\left(\frac{1}{\sqrt{2}}\right)}-f{\left(0\right)}\right]\\ &~~~~~+\int_{0}^{\frac12}\mathrm{d}y\,\frac{\left(5+6y\right)\sqrt{1-y}}{\sqrt{1+y}};~~~\small{\left[x=\sqrt{y}\right]}\\ &~~~~~+2\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,\frac{\arcsin{\left(x\right)}}{\sqrt{1+x^{2}}}\\ &=\frac{\sqrt{3}\,\pi}{2}-\frac{\pi}{2}\operatorname{arsinh}{\left(\frac{1}{\sqrt{2}}\right)}-\sqrt{3}\left[1+\arcsin{\left(\frac{1}{\sqrt{2}}\right)}\right]\\ &~~~~~-\sqrt{3}\left[1+\arcsin{\left(\frac{1}{\sqrt{2}}\right)}\right]+2\\ &~~~~~+\int_{0}^{\frac12}\mathrm{d}y\,\frac{\left(5+6y\right)\left(1-y\right)}{\sqrt{1-y^{2}}}\\ &~~~~~+2\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,\frac{\arcsin{\left(x\right)}}{\sqrt{1+x^{2}}}\\ &=2+\frac{\sqrt{3}\,\pi}{2}-\frac{\pi}{2}\operatorname{arsinh}{\left(\frac{1}{\sqrt{2}}\right)}-2\sqrt{3}\left[1+\arcsin{\left(\frac{1}{\sqrt{2}}\right)}\right]\\ &~~~~~+\int_{0}^{\frac12}\mathrm{d}y\,\frac{5+y-6y^{2}}{\sqrt{1-y^{2}}}\\ &~~~~~+2\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,\frac{\arcsin{\left(x\right)}}{\sqrt{1+x^{2}}}\\ &=2+\frac{\sqrt{3}\,\pi}{2}-\frac{\pi}{2}\operatorname{arsinh}{\left(\frac{1}{\sqrt{2}}\right)}-2\sqrt{3}\left[1+\arcsin{\left(\frac{1}{\sqrt{2}}\right)}\right]\\ &~~~~~+\int_{0}^{\frac12}\mathrm{d}y\,\frac{y}{\sqrt{1-y^{2}}}+\int_{0}^{\frac12}\mathrm{d}y\,\frac{3-6y^{2}}{\sqrt{1-y^{2}}}+\int_{0}^{\frac12}\mathrm{d}y\,\frac{2}{\sqrt{1-y^{2}}}\\ &~~~~~+2\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,\frac{\arcsin{\left(x\right)}}{\sqrt{1+x^{2}}}\\ &=2+\frac{\sqrt{3}\,\pi}{2}-\frac{\pi}{2}\operatorname{arsinh}{\left(\frac{1}{\sqrt{2}}\right)}-2\sqrt{3}\left[1+\arcsin{\left(\frac{1}{\sqrt{2}}\right)}\right]\\ &~~~~~+\int_{0}^{\frac{1}{4}}\mathrm{d}t\,\frac{1}{2\sqrt{1-t}};~~~\small{\left[y=\sqrt{t}\right]}\\ &~~~~~+\frac{3\sqrt{3}}{4}+2\arcsin{\left(\frac12\right)}\\ &~~~~~+2\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,\frac{\arcsin{\left(x\right)}}{\sqrt{1+x^{2}}}\\ &=2+\frac{\sqrt{3}\,\pi}{2}-\frac{\pi}{2}\operatorname{arsinh}{\left(\frac{1}{\sqrt{2}}\right)}-2\sqrt{3}\left[1+\arcsin{\left(\frac{1}{\sqrt{2}}\right)}\right]\\ &~~~~~-\frac{\sqrt{3}}{2}+1+\frac{3\sqrt{3}}{4}+\frac{\pi}{3}\\ &~~~~~+2\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,\frac{\arcsin{\left(x\right)}}{\sqrt{1+x^{2}}}\\ &=3-\frac{7\sqrt{3}}{4}+\frac{\pi}{3}-\frac{\pi}{2}\ln{\left(\frac{1+\sqrt{3}}{\sqrt{2}}\right)}+2\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,\frac{\arcsin{\left(x\right)}}{\sqrt{1+x^{2}}}.\\ \end{align}$$

Let $\mathcal{J}$ denote the value of the definite integral,

$$\begin{align} \mathcal{J} &:=\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,\frac{2\arcsin{\left(x\right)}}{\sqrt{1+x^{2}}}.\\ \end{align}$$

Integrating by parts and applying a certain Euler substitution, we find

$$\begin{align} \mathcal{J} &=\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,\frac{2\arcsin{\left(x\right)}}{\sqrt{1+x^{2}}}\\ &=2\arcsin{\left(\frac{1}{\sqrt{2}}\right)}\operatorname{arsinh}{\left(\frac{1}{\sqrt{2}}\right)}-\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,\frac{2\operatorname{arsinh}{\left(x\right)}}{\sqrt{1-x^{2}}};~~~\small{I.B.P.s}\\ &=\frac{\pi}{2}\ln{\left(\frac{1}{\sqrt{2}}+\sqrt{1+\frac12}\right)}-2\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,\frac{\ln{\left(x+\sqrt{1+x^{2}}\right)}}{\sqrt{1-x^{2}}}\\ &=\frac{\pi}{2}\ln{\left(\frac{1+\sqrt{3}}{\sqrt{2}}\right)}+2\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,\frac{\ln{\left(\frac{1}{x+\sqrt{1+x^{2}}}\right)}}{\sqrt{1-x^{2}}}\\ &=\frac{\pi}{2}\ln{\left(\frac{1+\sqrt{3}}{\sqrt{2}}\right)}+2\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,\frac{\ln{\left(-x+\sqrt{1+x^{2}}\right)}}{\sqrt{1-x^{2}}}\\ &=\frac{\pi}{2}\ln{\left(\frac{1+\sqrt{3}}{\sqrt{2}}\right)}\\ &~~~~~+2\int_{1}^{-\frac{1}{\sqrt{2}}+\sqrt{\frac32}}\mathrm{d}y\,\frac{\left(-1\right)\left(1+y^{2}\right)}{2y^{2}}\cdot\frac{\ln{\left(y\right)}}{\sqrt{1-\left(\frac{1-y^{2}}{2y}\right)^{2}}};~~~\small{\left[\sqrt{1+x^{2}}=x+y\right]}\\ &=\frac{\pi}{2}\ln{\left(\frac{1+\sqrt{3}}{\sqrt{2}}\right)}+\int_{\frac{\sqrt{3}-1}{\sqrt{2}}}^{1}\mathrm{d}y\,\frac{2\left(1+y^{2}\right)\ln{\left(y\right)}}{y\sqrt{4y^{2}-\left(1-y^{2}\right)^{2}}}\\ &=\frac{\pi}{2}\ln{\left(\frac{1+\sqrt{3}}{\sqrt{2}}\right)}+\int_{\frac{\sqrt{3}-1}{\sqrt{2}}}^{1}\mathrm{d}y\,\frac{2\left(1+y^{2}\right)\ln{\left(y\right)}}{y\sqrt{-1+6y^{2}-y^{4}}}\\ &=\frac{\pi}{2}\ln{\left(\frac{1+\sqrt{3}}{\sqrt{2}}\right)}\\ &~~~~~+\int_{\frac{\sqrt{3}-1}{\sqrt{2}}}^{1}\mathrm{d}y\,\frac{2\ln{\left(y\right)}}{y\sqrt{-1+6y^{2}-y^{4}}}+\int_{\frac{\sqrt{3}-1}{\sqrt{2}}}^{1}\mathrm{d}y\,\frac{2y^{2}\ln{\left(y\right)}}{y\sqrt{-1+6y^{2}-y^{4}}}\\ &=\frac{\pi}{2}\ln{\left(\frac{1+\sqrt{3}}{\sqrt{2}}\right)}\\ &~~~~~+\int_{\frac{\sqrt{2}}{\sqrt{3}-1}}^{1}\mathrm{d}t\,\frac{\left(-t^{-2}\right)2\ln{\left(\frac{1}{t}\right)}}{t^{-1}\sqrt{-1+6t^{-2}-t^{-4}}};~~~\small{\left[y=\frac{1}{t}\right]}\\ &~~~~~+\int_{\left(\frac{\sqrt{3}-1}{\sqrt{2}}\right)^{2}}^{1}\mathrm{d}u\,\frac{\ln{\left(\sqrt{u}\right)}}{\sqrt{-1+6u-u^{2}}};~~~\small{\left[y=\sqrt{u}\right]}\\ &=\frac{\pi}{2}\ln{\left(\frac{1+\sqrt{3}}{\sqrt{2}}\right)}\\ &~~~~~-\int_{1}^{\frac{1+\sqrt{3}}{\sqrt{2}}}\mathrm{d}t\,\frac{2t\ln{\left(t\right)}}{\sqrt{-t^{4}+6t^{2}-1}}\\ &~~~~~+\int_{2-\sqrt{3}}^{1}\mathrm{d}u\,\frac{\ln{\left(u\right)}}{2\sqrt{-1+6u-u^{2}}}\\ &=\frac{\pi}{2}\ln{\left(\frac{1+\sqrt{3}}{\sqrt{2}}\right)}\\ &~~~~~-\int_{1}^{2+\sqrt{3}}\mathrm{d}u\,\frac{\ln{\left(u\right)}}{2\sqrt{-u^{2}+6u-1}};~~~\small{\left[t=\sqrt{u}\right]}\\ &~~~~~+\int_{2-\sqrt{3}}^{1}\mathrm{d}u\,\frac{\ln{\left(u\right)}}{2\sqrt{-1+6u-u^{2}}}\\ &=\frac{\pi}{2}\ln{\left(\frac{1+\sqrt{3}}{\sqrt{2}}\right)}+\int_{2-\sqrt{3}}^{1}\mathrm{d}x\,\frac{\ln{\left(x\right)}}{2\sqrt{8-\left(x-3\right)^{2}}}-\int_{1}^{2+\sqrt{3}}\mathrm{d}x\,\frac{\ln{\left(x\right)}}{2\sqrt{8-\left(x-3\right)^{2}}}\\ &=\frac{\pi}{2}\ln{\left(\frac{1+\sqrt{3}}{\sqrt{2}}\right)}+\frac12\int_{2-\sqrt{3}}^{1}\mathrm{d}x\,\frac{\ln{\left(x\right)}}{\sqrt{\left(3+2\sqrt{2}-x\right)\left(x-3+2\sqrt{2}\right)}}\\ &~~~~~-\frac12\int_{1}^{2+\sqrt{3}}\mathrm{d}x\,\frac{\ln{\left(x\right)}}{\sqrt{\left(3+2\sqrt{2}-x\right)\left(x-3+2\sqrt{2}\right)}},\\ \end{align}$$

and then,

$$\begin{align} \mathcal{J} &=\frac{\pi}{2}\ln{\left(\frac{1+\sqrt{3}}{\sqrt{2}}\right)}+\frac12\int_{2-\sqrt{3}}^{1}\mathrm{d}x\,\frac{\ln{\left(x\right)}}{\sqrt{\left(3+2\sqrt{2}-x\right)\left(x-3+2\sqrt{2}\right)}}\\ &~~~~~-\frac12\int_{1}^{2+\sqrt{3}}\mathrm{d}x\,\frac{\ln{\left(x\right)}}{\sqrt{\left(3+2\sqrt{2}-x\right)\left(x-3+2\sqrt{2}\right)}}\\ &=\frac{\pi}{2}\ln{\left(\frac{1+\sqrt{3}}{\sqrt{2}}\right)}+\frac12\int_{\frac{3+2\sqrt{2}-2+\sqrt{3}}{2-\sqrt{3}-3+2\sqrt{2}}}^{\frac{3+2\sqrt{2}-1}{1-3+2\sqrt{2}}}\mathrm{d}y\,\frac{\left(-1\right)\ln{\left(\frac{\left(3+2\sqrt{2}\right)+\left(3-2\sqrt{2}\right)y}{1+y}\right)}}{\left(1+y\right)\sqrt{y}}\\ &~~~~~-\frac12\int_{\frac{3+2\sqrt{2}-1}{1-3+2\sqrt{2}}}^{\frac{3+2\sqrt{2}-2-\sqrt{3}}{2+\sqrt{3}-3+2\sqrt{2}}}\mathrm{d}y\,\frac{\left(-1\right)\ln{\left(\frac{\left(3+2\sqrt{2}\right)+\left(3-2\sqrt{2}\right)y}{1+y}\right)}}{\left(1+y\right)\sqrt{y}};~~~\small{\left[\frac{3+2\sqrt{2}-x}{x-3+2\sqrt{2}}=y\right]}\\ &=\frac{\pi}{2}\ln{\left(\frac{1+\sqrt{3}}{\sqrt{2}}\right)}+\frac12\int_{\frac{1+\sqrt{2}}{-1+\sqrt{2}}}^{\frac{1+2\sqrt{2}+\sqrt{3}}{-1+2\sqrt{2}-\sqrt{3}}}\mathrm{d}y\,\frac{\ln{\left(\frac{\left(3+2\sqrt{2}\right)+\left(3-2\sqrt{2}\right)y}{1+y}\right)}}{\left(1+y\right)\sqrt{y}}\\ &~~~~~-\frac12\int_{\frac{1+2\sqrt{2}-\sqrt{3}}{-1+2\sqrt{2}+\sqrt{3}}}^{\frac{1+\sqrt{2}}{-1+\sqrt{2}}}\mathrm{d}y\,\frac{\ln{\left(\frac{\left(3+2\sqrt{2}\right)+\left(3-2\sqrt{2}\right)y}{1+y}\right)}}{\left(1+y\right)\sqrt{y}}\\ &=\frac{\pi}{2}\ln{\left(\frac{1+\sqrt{3}}{\sqrt{2}}\right)}+\frac12\int_{\left(1+\sqrt{2}\right)^{2}}^{\left(\sqrt{6}+\sqrt{3}+\sqrt{2}+2\right)^{2}}\mathrm{d}y\,\frac{\ln{\left(\frac{\left(3+2\sqrt{2}\right)+\left(3-2\sqrt{2}\right)y}{1+y}\right)}}{\left(1+y\right)\sqrt{y}}\\ &~~~~~-\frac12\int_{\left(\sqrt{6}+\sqrt{3}-\sqrt{2}-2\right)^{2}}^{\left(1+\sqrt{2}\right)^{2}}\mathrm{d}y\,\frac{\ln{\left(\frac{\left(3+2\sqrt{2}\right)+\left(3-2\sqrt{2}\right)y}{1+y}\right)}}{\left(1+y\right)\sqrt{y}}\\ &=\frac{\pi}{2}\ln{\left(\frac{1+\sqrt{3}}{\sqrt{2}}\right)}+\int_{1+\sqrt{2}}^{\sqrt{6}+\sqrt{3}+\sqrt{2}+2}\mathrm{d}t\,\frac{\ln{\left(\frac{\left(3+2\sqrt{2}\right)+\left(3-2\sqrt{2}\right)t^{2}}{1+t^{2}}\right)}}{\left(1+t^{2}\right)}\\ &~~~~~-\int_{\sqrt{6}+\sqrt{3}-\sqrt{2}-2}^{1+\sqrt{2}}\mathrm{d}t\,\frac{\ln{\left(\frac{\left(3+2\sqrt{2}\right)+\left(3-2\sqrt{2}\right)t^{2}}{1+t^{2}}\right)}}{\left(1+t^{2}\right)};~~~\small{\left[y=t^{2}\right]}\\ &=\frac{\pi}{2}\ln{\left(\frac{1+\sqrt{3}}{\sqrt{2}}\right)}+\int_{\tan{\left(\frac{3\pi}{8}\right)}}^{\tan{\left(\frac{11\pi}{24}\right)}}\mathrm{d}t\,\frac{\ln{\left(\frac{\left(3+2\sqrt{2}\right)+\left(3-2\sqrt{2}\right)t^{2}}{1+t^{2}}\right)}}{1+t^{2}}\\ &~~~~~-\int_{\tan{\left(\frac{5\pi}{24}\right)}}^{\tan{\left(\frac{3\pi}{8}\right)}}\mathrm{d}t\,\frac{\ln{\left(\frac{\left(3+2\sqrt{2}\right)+\left(3-2\sqrt{2}\right)t^{2}}{1+t^{2}}\right)}}{1+t^{2}}\\ &=\frac{\pi}{2}\ln{\left(\frac{1+\sqrt{3}}{\sqrt{2}}\right)}+\int_{\frac{3\pi}{4}}^{\frac{11\pi}{12}}\mathrm{d}\varphi\,\frac{\sec^{2}{\left(\frac{\varphi}{2}\right)}}{2}\cdot\frac{\ln{\left(\frac{\left(3+2\sqrt{2}\right)+\left(3-2\sqrt{2}\right)\tan^{2}{\left(\frac{\varphi}{2}\right)}}{1+\tan^{2}{\left(\frac{\varphi}{2}\right)}}\right)}}{1+\tan^{2}{\left(\frac{\varphi}{2}\right)}}\\ &~~~~~-\int_{\frac{5\pi}{12}}^{\frac{3\pi}{4}}\mathrm{d}\varphi\,\frac{\sec^{2}{\left(\frac{\varphi}{2}\right)}}{2}\cdot\frac{\ln{\left(\frac{\left(3+2\sqrt{2}\right)+\left(3-2\sqrt{2}\right)\tan^{2}{\left(\frac{\varphi}{2}\right)}}{1+\tan^{2}{\left(\frac{\varphi}{2}\right)}}\right)}}{1+\tan^{2}{\left(\frac{\varphi}{2}\right)}};~~~\small{\left[t=\tan{\left(\frac{\varphi}{2}\right)}\right]}\\ &=\frac{\pi}{2}\ln{\left(\frac{1+\sqrt{3}}{\sqrt{2}}\right)}+\frac12\int_{\frac{3\pi}{4}}^{\frac{11\pi}{12}}\mathrm{d}\varphi\,\ln{\left(\frac{\left(3+2\sqrt{2}\right)+\left(3-2\sqrt{2}\right)\tan^{2}{\left(\frac{\varphi}{2}\right)}}{\sec^{2}{\left(\frac{\varphi}{2}\right)}}\right)}\\ &~~~~~-\frac12\int_{\frac{5\pi}{12}}^{\frac{3\pi}{4}}\mathrm{d}\varphi\,\ln{\left(\frac{\left(3+2\sqrt{2}\right)+\left(3-2\sqrt{2}\right)\tan^{2}{\left(\frac{\varphi}{2}\right)}}{\sec^{2}{\left(\frac{\varphi}{2}\right)}}\right)}\\ &=\frac{\pi}{2}\ln{\left(\frac{1+\sqrt{3}}{\sqrt{2}}\right)}+\frac12\int_{\frac{3\pi}{4}}^{\frac{11\pi}{12}}\mathrm{d}\varphi\,\ln{\left(3+2\sqrt{2}\cos{\left(\varphi\right)}\right)}\\ &~~~~~-\frac12\int_{\frac{5\pi}{12}}^{\frac{3\pi}{4}}\mathrm{d}\varphi\,\ln{\left(3+2\sqrt{2}\cos{\left(\varphi\right)}\right)},\\ \end{align}$$

and then after setting $\alpha:=\arcsin{\left(\frac{2\sqrt{2}}{3}\right)}\in\left(0,\frac{\pi}{2}\right)$,

$$\begin{align} \mathcal{J} &=\frac{\pi}{2}\ln{\left(\frac{1+\sqrt{3}}{\sqrt{2}}\right)}+\frac12\int_{\frac{3\pi}{4}}^{\frac{11\pi}{12}}\mathrm{d}\varphi\,\ln{\left(3+2\sqrt{2}\cos{\left(\varphi\right)}\right)}\\ &~~~~~-\frac12\int_{\frac{5\pi}{12}}^{\frac{3\pi}{4}}\mathrm{d}\varphi\,\ln{\left(3+2\sqrt{2}\cos{\left(\varphi\right)}\right)}\\ &=\frac{\pi}{2}\ln{\left(\frac{1+\sqrt{3}}{\sqrt{2}}\right)}+\frac12\int_{\frac{\pi}{12}}^{\frac{\pi}{4}}\mathrm{d}\varphi\,\ln{\left(3-2\sqrt{2}\cos{\left(\varphi\right)}\right)}\\ &~~~~~-\frac12\int_{\frac{\pi}{4}}^{\frac{7\pi}{12}}\mathrm{d}\varphi\,\ln{\left(3-2\sqrt{2}\cos{\left(\varphi\right)}\right)};~~~\small{\left[\varphi\mapsto\pi-\varphi\right]}\\ &=\frac{\pi}{2}\ln{\left(\frac{1+\sqrt{3}}{\sqrt{2}}\right)}\\ &~~~~~+\frac12\int_{\frac{\pi}{12}}^{\frac{\pi}{4}}\mathrm{d}\varphi\,\ln{\left(3\right)}+\frac12\int_{\frac{\pi}{12}}^{\frac{\pi}{4}}\mathrm{d}\varphi\,\ln{\left(1-\frac{2\sqrt{2}}{3}\cos{\left(\varphi\right)}\right)}\\ &~~~~~-\frac12\int_{\frac{\pi}{4}}^{\frac{7\pi}{12}}\mathrm{d}\varphi\,\ln{\left(3\right)}-\frac12\int_{\frac{\pi}{4}}^{\frac{7\pi}{12}}\mathrm{d}\varphi\,\ln{\left(1-\frac{2\sqrt{2}}{3}\cos{\left(\varphi\right)}\right)}\\ &=\frac{\pi}{2}\ln{\left(\frac{1+\sqrt{3}}{\sqrt{2}}\right)}-\frac{\pi\ln{\left(3\right)}}{12}\\ &~~~~~+\int_{0}^{\frac{\pi}{4}}\mathrm{d}\varphi\,\ln{\left(1-\frac{2\sqrt{2}}{3}\cos{\left(\varphi\right)}\right)}\\ &~~~~~-\frac12\int_{0}^{\frac{\pi}{12}}\mathrm{d}\varphi\,\ln{\left(1-\frac{2\sqrt{2}}{3}\cos{\left(\varphi\right)}\right)}\\ &~~~~~-\frac12\int_{0}^{\frac{7\pi}{12}}\mathrm{d}\varphi\,\ln{\left(1-\frac{2\sqrt{2}}{3}\cos{\left(\varphi\right)}\right)}\\ &=\frac{\pi}{2}\ln{\left(\frac{1+\sqrt{3}}{\sqrt{2}}\right)}-\frac{\pi\ln{\left(3\right)}}{12}\\ &~~~~~+\int_{0}^{\frac{\pi}{4}}\mathrm{d}\varphi\,\ln{\left(1-\sin{\left(\alpha\right)}\cos{\left(\varphi\right)}\right)}\\ &~~~~~-\frac12\int_{0}^{\frac{\pi}{12}}\mathrm{d}\varphi\,\ln{\left(1-\sin{\left(\alpha\right)}\cos{\left(\varphi\right)}\right)}\\ &~~~~~-\frac12\int_{0}^{\frac{7\pi}{12}}\mathrm{d}\varphi\,\ln{\left(1-\sin{\left(\alpha\right)}\cos{\left(\varphi\right)}\right)}.\\ \end{align}$$

The remaining logarithmic integrals can be evaluated in terms of Clausen functions using the following integration formula, which holds for any $\left(\alpha,\vartheta\right)\in\left(0,\frac{\pi}{2}\right)\times\mathbb{R}$:

$$\begin{align} \int_{0}^{\theta}\mathrm{d}\varphi\,\ln{\left(1-\sin{\left(\alpha\right)}\cos{\left(\varphi\right)}\right)} &=\operatorname{Cl}_{2}{\left(2\theta+2\omega\right)}-\operatorname{Cl}_{2}{\left(2\theta\right)}-\operatorname{Cl}_{2}{\left(2\omega\right)}\\ &~~~~~-\theta\ln{\left(\sec^{2}{\left(\frac{\alpha}{2}\right)}\right)}-\omega\ln{\left(\tan^{2}{\left(\frac{\alpha}{2}\right)}\right)}\\ \end{align}$$

where

$$\omega:=\arctan{\left(\frac{\tan{\left(\frac{\alpha}{2}\right)}\sin{\left(\vartheta\right)}}{1-\tan{\left(\frac{\alpha}{2}\right)}\cos{\left(\vartheta\right)}}\right)}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right).$$

Recall that the Clausen function may be defined for real arguments via the integral representation,

$$\operatorname{Cl}_{2}{\left(\theta\right)}:=-\int_{0}^{\theta}\mathrm{d}\varphi\,\ln{\left(\left|2\sin{\left(\frac{\varphi}{2}\right)}\right|\right)};~~~\small{\theta\in\mathbb{R}}.$$

Having obtained explicit expressions for each of the integrals comprising $\mathcal{J}$, a little algebraic elbow-grease yields a greatly simplified final value:

$$\begin{align} \mathcal{J} &=\frac{\pi}{2}\ln{\left(\frac{1+\sqrt{3}}{\sqrt{2}}\right)}-\frac{\pi\ln{\left(3\right)}}{12}\\ &~~~~~+\int_{0}^{\frac{\pi}{4}}\mathrm{d}\varphi\,\ln{\left(1-\sin{\left(\alpha\right)}\cos{\left(\varphi\right)}\right)}\\ &~~~~~-\frac12\int_{0}^{\frac{\pi}{12}}\mathrm{d}\varphi\,\ln{\left(1-\sin{\left(\alpha\right)}\cos{\left(\varphi\right)}\right)}\\ &~~~~~-\frac12\int_{0}^{\frac{7\pi}{12}}\mathrm{d}\varphi\,\ln{\left(1-\sin{\left(\alpha\right)}\cos{\left(\varphi\right)}\right)}\\ &=\frac{\pi}{2}\ln{\left(\frac{1+\sqrt{3}}{\sqrt{2}}\right)}-\frac{\pi\ln{\left(3\right)}}{12}\\ &~~~~~+\operatorname{Cl}_{2}{\left(\pi\right)}-2\operatorname{Cl}_{2}{\left(\frac{\pi}{2}\right)}-\frac{\pi}{4}\ln{\left(\sec^{2}{\left(\frac{\alpha}{2}\right)}\right)}-\frac{\pi}{4}\ln{\left(\tan^{2}{\left(\frac{\alpha}{2}\right)}\right)}\\ &~~~~~-\frac12\operatorname{Cl}_{2}{\left(\frac{\pi}{2}\right)}+\frac12\operatorname{Cl}_{2}{\left(\frac{\pi}{6}\right)}+\frac12\operatorname{Cl}_{2}{\left(\frac{\pi}{3}\right)}+\frac{\pi}{24}\ln{\left(\sec^{2}{\left(\frac{\alpha}{2}\right)}\right)}+\frac{\pi}{12}\ln{\left(\tan^{2}{\left(\frac{\alpha}{2}\right)}\right)}\\ &~~~~~-\frac12\operatorname{Cl}_{2}{\left(\frac{3\pi}{2}\right)}+\frac12\operatorname{Cl}_{2}{\left(\frac{7\pi}{6}\right)}+\frac12\operatorname{Cl}_{2}{\left(\frac{\pi}{3}\right)}+\frac{7\pi}{24}\ln{\left(\sec^{2}{\left(\frac{\alpha}{2}\right)}\right)}+\frac{\pi}{12}\ln{\left(\tan^{2}{\left(\frac{\alpha}{2}\right)}\right)}\\ &=\frac{\pi}{2}\ln{\left(\frac{1+\sqrt{3}}{\sqrt{2}}\right)}-\frac{\pi\ln{\left(3\right)}}{12}\\ &~~~~~+\frac{\pi}{12}\ln{\left(\sec^{2}{\left(\frac{\alpha}{2}\right)}\right)}-\frac{\pi}{12}\ln{\left(\tan^{2}{\left(\frac{\alpha}{2}\right)}\right)}\\ &~~~~~+\frac12\operatorname{Cl}_{2}{\left(\frac{7\pi}{6}\right)}+\frac12\operatorname{Cl}_{2}{\left(\frac{\pi}{6}\right)}+\operatorname{Cl}_{2}{\left(\frac{\pi}{3}\right)}-2\operatorname{Cl}_{2}{\left(\frac{\pi}{2}\right)}\\ &=\frac{\pi}{2}\ln{\left(\frac{1+\sqrt{3}}{\sqrt{2}}\right)}-\frac{\pi\ln{\left(3\right)}}{12}\\ &~~~~~+\frac{\pi}{12}\ln{\left(\frac32\right)}-\frac{\pi}{12}\ln{\left(\frac12\right)}\\ &~~~~~-\frac12\operatorname{Cl}_{2}{\left(\frac{5\pi}{6}\right)}+\frac12\operatorname{Cl}_{2}{\left(\frac{\pi}{6}\right)}+\operatorname{Cl}_{2}{\left(\frac{\pi}{3}\right)}-2\operatorname{Cl}_{2}{\left(\frac{\pi}{2}\right)}\\ &=\frac54\operatorname{Cl}_{2}{\left(\frac{\pi}{3}\right)}-2C+\frac{\pi}{2}\ln{\left(\frac{1+\sqrt{3}}{\sqrt{2}}\right)}.\\ \end{align}$$

Finally(!), we arrive at our desired result:

$$\begin{align} \mathcal{I} &=3-\frac{7\sqrt{3}}{4}+\frac{\pi}{3}+\frac{\sqrt{3}\,\pi}{2}-\frac{\pi}{2}\operatorname{arsinh}{\left(\frac{1}{\sqrt{2}}\right)}-2\sqrt{3}\arcsin{\left(\frac{1}{\sqrt{2}}\right)}+\mathcal{J}\\ &=3-\frac{7\sqrt{3}}{4}+\frac{\pi}{3}-2C+\frac54\operatorname{Cl}_{2}{\left(\frac{\pi}{3}\right)}.\\ \end{align}$$

Solution 2:

This is not an answer.

We could use $$\sqrt{x^2-4x+3}=\sum_{n=0}^\infty a_n\, x^n$$ with $$a_n=\frac{2(2 n-3)\, a_{n-1}-(n-3)\, a_{n-2}}{3 n} \qquad \text{where}\qquad a_0=\sqrt{3}\qquad a_1=-\frac{2}{\sqrt{3}}$$ and $$\int_0^1 x^n\arcsin(x)\,dx=\frac{\pi }{2( n+1)}-\frac{\sqrt{\pi }\,\,\Gamma \left(\frac{n}{2}+1\right)}{(n+1)^2 \,\, \Gamma \left(\frac{n+1}{2}\right)}$$ but the convergence is very slow.