difficult integral $\int_0^{\pi/2}\frac{x^2({1+\tan x})^2}{\sqrt{\tan x}({1-\tan x})}\sin{4x}dx$ [closed]

This is a complicated integral, the numerical value appears to me correct.Therefore how to prove this result?$$I=\int_0^{\pi/2}\frac{x^2({1+\tan x})^2}{\sqrt{\tan x}({1-\tan x})}\sin({4x})dx=\frac{\pi\sqrt{2}}{192}(35{\pi^2}-150+132\ln 2-84\ln^22)$$

$\displaystyle J=\int_0^{\tfrac{\pi}{2}}\dfrac{x^2({1+\tan x})^2}{\sqrt{\tan x}({1-\tan x})}\sin{(4x)}dx$

Perform the change of variable $y=\tan x$, one obtains:

$\displaystyle J=4\int_0^{+\infty} \sqrt{x}(\arctan x)^2\left(\dfrac{1+x}{1+x^2}\right)^3 dx$

Perform integration by parts:

$\displaystyle J=\dfrac{1}{2}\left[\left(\dfrac{\sqrt{x}(x-1)(7x^2+4x+7)}{(1+x^2)^2}+\dfrac{7}{\sqrt{2}}\left(\arctan(1+\sqrt{2x})-\arctan(1-\sqrt{2x})\right)\right)(\arctan x)^2\right]_0^{+\infty}-\int_0^{+\infty}\left(\dfrac{\sqrt{x}(x-1)(7x^2+4x+7)}{(1+x^2)^2}+\dfrac{7}{\sqrt{2}}\left(\arctan(1+\sqrt{2x})-\arctan(1-\sqrt{2x})\right)\right)\dfrac{\arctan x}{1+x^2}dx$

Therefore,

$\displaystyle J=\dfrac{7\pi^3}{8\sqrt{2}}-\int_0^{+\infty}\left(\dfrac{\sqrt{x}(x-1)(7x^2+4x+7)}{(1+x^2)^2}+\dfrac{7}{\sqrt{2}}\left(\arctan(1+\sqrt{2x})-\arctan(1-\sqrt{2x})\right)\right)\\\dfrac{\arctan x}{1+x^2}dx$

Let $\displaystyle A=\int_0^{+\infty}\dfrac{\sqrt{x}(x-1)(7x^2+4x+7)\arctan x}{(1+x^2)^3}dx$

Let $\displaystyle B=\int_0^{+\infty}\dfrac{\left(\arctan(1+\sqrt{2x})-\arctan(1-\sqrt{2x})\right)\arctan x}{1+x^2}dx$

Thus, $J=\dfrac{7\pi^3}{8\sqrt{2}}-A-\dfrac{7}{\sqrt{2}}B$

Perform integration by parts:

$\displaystyle A=-\dfrac{1}{4}\left[\left(\dfrac{11}{2\sqrt{2}}\Big(\log(x-\sqrt{2x}+1)-\log(x+\sqrt{2x}+1)\Big)+\dfrac{\sqrt{x}(x+1)(11x^2+4x+11)}{(1+x^2)^2}\right)\\\arctan x\right]_0^{+\infty}+\dfrac{1}{4}\times\int_0^{+\infty} \left(\dfrac{11}{2\sqrt{2}}\Big(\log(x-\sqrt{2x}+1)-\log(x+\sqrt{2x}+1)\Big)+\dfrac{\sqrt{x}(x+1)(11x^2+4x+11)}{(1+x^2)^2}\right)\\\dfrac{1}{1+x^2}dx$

Therefore,

$\displaystyle A=\dfrac{1}{4}\times\int_0^{+\infty}\left(\dfrac{11}{2\sqrt{2}}\Big(\log(x-\sqrt{2x}+1)-\log(x+\sqrt{2x}+1)\Big)+\dfrac{\sqrt{x}(x+1)(11x^2+4x+11)}{(1+x^2)^2}\right)\\\dfrac{1}{1+x^2}dx$

Let $\displaystyle C=\int_0^{+\infty}\dfrac{\left(\log(x-\sqrt{2x}+1)-\log(x+\sqrt{2x}+1)\right)}{1+x^2}dx$

Let $\displaystyle D=\int_0^{+\infty} \dfrac{\sqrt{x}(x+1)(11x^2+4x+11)}{(1+x^2)^3}dx$

Thus, $\displaystyle J=\dfrac{7\pi^3}{8\sqrt{2}}-\dfrac{11}{8\sqrt{2}}C-\dfrac{1}{4}D-\dfrac{7}{\sqrt{2}}B$

$\displaystyle D=\dfrac{1}{4}\times\left[\dfrac{\sqrt{x}(x-1)(25x^2+4x+25)}{(1+x^2)^2}+\dfrac{25}{\sqrt{2}}\Big(\arctan(1+\sqrt{2x})-\arctan(1-\sqrt{2x})\Big)\right]_0^{+\infty}$

Therefore,

$D=\dfrac{25\pi}{4\sqrt{2}}$

Define for $a\in [0,\sqrt{2}]$,

$\displaystyle F(a)=\int_0^{+\infty}\dfrac{\left(\log(x-a\sqrt{x}+1)-\log(x+a\sqrt{x}+1)\right)}{1+x^2}dx$

Note that,

$F(\sqrt{2})=C$ and $F(0)=0$

$\displaystyle F'(a)= \int_0^{+\infty} \dfrac{-2\sqrt{x}(1+x)}{(x-a\sqrt{x}+1)(x+a\sqrt{x}+1)(x^2+1)}dx$

$F'(a)=\left[\dfrac{-4\left( \mathrm{arctan}\left( \dfrac{2\sqrt{x}-a}{\sqrt{4-{{a}^{2}}}}\right) +\mathrm{arctan}\left( \dfrac{2\sqrt{x}+a}{\sqrt{4-{{a}^{2}}}}\right) \right) }{\sqrt{4-{{a}^{2}}}\cdot \left( {{a}^{2}}-2\right) }+\\\dfrac{2\sqrt{2} \Big( \mathrm{arctan}\left( \sqrt{2x}-1\right) +\mathrm{arctan}\left( \sqrt{2x}+1\right) \Big) }{{{a}^{2}}-2}\right]_0^{+\infty}$

Therefore,

$F'(a)=\dfrac{-4\pi}{(a^2-2)\sqrt{4-a^2}}+\dfrac{2\sqrt{2}\pi}{a^2-2}$

And then,

$\displaystyle C=F(\sqrt{2})=\int_0^{\sqrt{2}} F'(a)da=-\pi\left[\mathrm{log}\left( \dfrac{\left( \sqrt{2}+a\right)\left( \sqrt{4-{{a}^{2}}}-a\right) }{\left( \sqrt{2}-a\right)\left( \sqrt{4-{{a}^{2}}}+a\right) }\right)\right]_0^\sqrt{2}=-\pi\ln 2$

Define for $a\in [0,\sqrt{2}]$,

$\displaystyle G(a)=\int_0^{+\infty}\dfrac{\Big(\arctan(a\sqrt{x}+1)+\arctan(a\sqrt{x}-1)\Big)\arctan x}{1+x^2}dx$

Note that $G(\sqrt{2})=B$ and $G(0)=0$

$\displaystyle G'(a)=\int_0^{+\infty}\dfrac{\sqrt{x}}{1+x^2}\arctan(x)\left(\dfrac{1}{1+(a\sqrt{x}+1)^2}+\dfrac{1}{1+(a\sqrt{x}-1)^2}\right)dx$

Perform integration by parts,

$\displaystyle G'(a)=\\\left[\left(\dfrac{2a\mathrm{log}\left( \dfrac{a^2x-2a\sqrt{x}+2}{{{a}^{2}} x+2 a\sqrt{x}+2}\right) }{{{a}^{4}}-4}+\dfrac{\mathrm{log}\left( \dfrac{x+\sqrt{2x}+1}{x-\sqrt{2x}+1}\right) }{\sqrt{2}\left( {{a}^{2}}-2\right) }+\dfrac{\sqrt{2}\left( \mathrm{arctan}\left( \sqrt{2x}-1\right) +\mathrm{arctan}\left( \sqrt{2x}+1\right) \right) }{{{a}^{2}}+2}\right)\arctan(x)\right]_0^{+\infty}\\-\displaystyle \int_0^{+\infty}\left(\dfrac{2a\mathrm{log}\left( \dfrac{a^2x-2a\sqrt{x}+2}{{{a}^{2}} x+2 a\sqrt{x}+2}\right) }{{{a}^{4}}-4}+\\\dfrac{\mathrm{log}\left( \dfrac{x+\sqrt{2x}+1}{x-\sqrt{2x}+1}\right) }{\sqrt{2}\left( {{a}^{2}}-2\right) }+\dfrac{\sqrt{2}\left( \mathrm{arctan}\left( \sqrt{2x}-1\right) +\mathrm{arctan}\left( \sqrt{2x}+1\right) \right) }{{{a}^{2}}+2}\right)\dfrac{1}{1+x^2}dx$

Therefore,

$\displaystyle G'(a)=\dfrac{\pi^2}{\sqrt{2}(a^2+2)}-\dfrac{\pi\log 2}{\sqrt{2}(a^2-2)}+\int_{0}^{+\infty}\dfrac{2a\mathrm{log}\left( \dfrac{a^2x+2a\sqrt{x}+2}{{{a}^{2}} x-2a\sqrt{x}+2}\right) }{(a^4-4)(1+x^2)}dx-\int_{0}^{+\infty}\dfrac{\sqrt{2}\left( \mathrm{arctan}\left( \sqrt{2x}-1\right) +\mathrm{arctan}\left( \sqrt{2x}+1\right) \right) }{(a^2+2)(1+x^2)}dx$

Let $\displaystyle E=\int_{0}^{+\infty}\dfrac{\mathrm{arctan}\left( \sqrt{2x}-1\right) +\mathrm{arctan}\left( \sqrt{2x}+1\right)}{(1+x^2)}dx$

Define for $a\in [0,\sqrt{2}]$,

$\displaystyle H(a)=\int_{0}^{+\infty}\dfrac{\mathrm{log}\left( \dfrac{a^2x+2a\sqrt{x}+2}{{{a}^{2}} x-2a\sqrt{x}+2}\right) }{1+x^2}dx$

Note that $H(0)=0$

$\displaystyle H'(a)=\int_{0}^{+\infty}\dfrac{-4\sqrt{x}(a^2x-2)}{(x^2+1)(a^4x^2+4)}dx$

$\displaystyle H'(a)=\\\left[\dfrac{\sqrt{2}\mathrm{log}\left(\dfrac{x-\sqrt{2x}+1}{x+\sqrt{2x}+1}\right) }{{{a}^{2}}+2}+\dfrac{8a\Big( \mathrm{arctan}\left(a\sqrt{x}-1\right)+\mathrm{arctan}\left(a\sqrt{x}+1\right) \Big) }{{{a}^{4}}-4}-\dfrac{2\sqrt{2}\Big( \mathrm{arctan}\left(\sqrt{2x}-1\right) +\mathrm{arctan}\left( \sqrt{2x}+1\right) \Big) }{{{a}^{2}}-2}\right]_0^{+\infty}$

$\displaystyle H'(a)=\dfrac{8a\pi}{a^4-4}-\dfrac{2\sqrt{2}\pi}{a^2-2}$

$\displaystyle H(a)=\pi\log\left(\dfrac{(2-a^2)(\sqrt{2}+a)}{(2+a^2)(\sqrt{2}-a)}\right)=2\pi\log(a+\sqrt{2})-\pi\log(a^2+2)$

Define for $a\in [0,\sqrt{2}]$,

$\displaystyle K(a)=\int_{0}^{+\infty}\dfrac{\mathrm{arctan}\left( a\sqrt{x}-1\right) +\mathrm{arctan}\left( a\sqrt{x}+1\right)}{(1+x^2)}dx$

Note that $K(\sqrt{2})=E$ and $K(0)=0$

$\displaystyle K'(a)=\int_{0}^{+\infty} \dfrac{2\sqrt{x}\left( 2+{{a}^{2}}x\right) }{\left( {{a}^{2}}x-2a\sqrt{x}+2\right)\left( {{a}^{2}}x+2a\sqrt{x}+2\right)\left( {{x}^{2}}+1\right) }dx$

$\displaystyle K'(a)=\\\left[\dfrac{2a\mathrm{log}\left( \dfrac{{{a}^{2}}x-2a\sqrt{x}+2}{{{a}^{2}} x+2a\sqrt{x}+2}\right) }{{{a}^{4}}-4}+\dfrac{\mathrm{log}\left( \dfrac{x+\sqrt{2x}+1}{x-\sqrt{2x}+1}\right) }{\sqrt{2}\left( {{a}^{2}}-2\right) }+\dfrac{\sqrt{2}\left( \mathrm{arctan}\left( \sqrt{2x}-1\right) +\mathrm{arctan}\left( \sqrt{2x}+1\right) \right) }{{{a}^{2}}+2}\right]_{0}^{+\infty}$

$\displaystyle K'(a)=\dfrac{\sqrt{2}\pi}{a^2+2}$

$\displaystyle E=K(\sqrt{2})=\int_0^{\sqrt{2}}\dfrac{\sqrt{2}\pi}{a^2+2}da=\pi\left[\arctan\left(\dfrac{a}{\sqrt{2}}\right)\right]_0^{\sqrt{2}}=\dfrac{\pi^2}{4}$

Thus,

$\displaystyle G'(a)=\dfrac{\pi^2}{\sqrt{2}(a^2+2)}-\dfrac{\pi\log 2}{\sqrt{2}(a^2-2)}+\dfrac{2a\pi}{a^4-4}\Big(2\log(a+\sqrt{2})-\log(a^2+2)\Big)-\dfrac{\pi^2}{2\sqrt{2}(a^2+2)}$

$\displaystyle G'(a)=\dfrac{\pi^2}{2\sqrt{2}(a^2+2)}+\dfrac{\pi a \log(a^2+2)}{2(a^2+2)}-\dfrac{\pi a\log 2}{2(a^2+2)}-\dfrac{\pi a \log\left(\dfrac{a}{\sqrt{2}}+1\right)}{2\left(\left(\dfrac{a}{\sqrt{2}}\right)^2+1\right)}+\dfrac{\pi}{2(a^2-2)}\Big(2a\log(a+\sqrt{2})-\sqrt{2}\log 2-a\log(a^2+2)\Big)$

$\displaystyle \int_0^{\sqrt{2}}\dfrac{\pi^2}{2\sqrt{2}(a^2+2)}da=\pi^2\left[\dfrac{1}{4}\arctan\left(\dfrac{x}{\sqrt{2}}\right)\right]_0^{\sqrt{2}}=\dfrac{\pi^3}{16}$

$\displaystyle \int_0^{\sqrt{2}}\dfrac{\pi a\log(a^2+2)}{2(a^2+2)}da=\dfrac{\pi}{8}\left[\left(\log(a^2+2)\right)^2\right]_0^{\sqrt{2}}=\dfrac{3\pi(\log 2)^2}{8}$

$\displaystyle \int_0^{\sqrt{2}}\dfrac{-\pi a\log 2}{2(a^2+2)}da=-\dfrac{\pi \log 2}{4}\Big[\log(a^2+2)\Big]_0^{\sqrt{2}}=-\dfrac{\pi (\log 2)^2}{4}$

$\displaystyle \int_0^{\sqrt{2}} \dfrac{-\pi a \log\left(\dfrac{a}{\sqrt{2}}+1\right)}{2\left(\left(\dfrac{a}{\sqrt{2}}\right)^2+1\right)}da=-\pi\int_0^1\dfrac{a\log(a+1)}{a^2+1}da$

Let $\displaystyle L=\int_0^1\dfrac{a\log(a+1)}{a^2+1}da$

Define for $t\in[0,1]$,

$\displaystyle M(t)=\int_0^1\dfrac{a\log(at+1)}{a^2+1}da$

Note that $M(1)=L$ and $M(0)=0$.

$\displaystyle M'(t)=\int_0^1\dfrac{a^2}{(a^2+1)(at+1)}da$

$\displaystyle M'(t)=\left[\dfrac{\log(at+1)}{t^3+t}+\dfrac{t\log(a^2+1)}{2(t^2+1)}-\dfrac{\arctan(a)}{t^2+1}\right]_0^1=\dfrac{\log(1+t)}{t}-\dfrac{t\log(1+t)}{1+t^2}-\dfrac{\pi}{4(t^2+1)}+\dfrac{t\log 2}{2(t^2+1)}$

$\displaystyle\int_0^1\dfrac{\log(1+t)}{t}dt=\sum_{n=0}^{+\infty}\int_0^1 \left(\dfrac{(-1)^n t^{n+1}}{(n+1)t}\right)dt=-\sum_{n=1}^{+\infty} \dfrac{(-1)^n}{n^2}=\sum_{n=0}^{+\infty} \dfrac{1}{(2n+1)^2}-\sum_{n=1}^{+\infty} \dfrac{1}{(2n)^2}=\left(\zeta(2)-\sum_{n=1}^{+\infty} \dfrac{1}{(2n)^2}\right)-\sum_{n=1}^{+\infty} \dfrac{1}{(2n)^2}=\dfrac{1}{2}\zeta(2)=\dfrac{\pi^2}{12}$

$\displaystyle\int_0^1 \dfrac{-\pi}{4(t^2+1)}dt=-\dfrac{\pi}{4}\Big[\arctan t\Big]_0^1=-\dfrac{\pi^2}{16}$

$\displaystyle\int_0^1\dfrac{t\log 2}{2(t^2+1)}dt=\dfrac{\log 2}{4}\Big[\log(t^2+1)\Big]_0^1=\dfrac{(\log 2)^2}{4}$

$\displaystyle L=\int_0^1 M'(t)dt=\dfrac{\pi^2}{12}-\dfrac{\pi^2}{16}+\dfrac{(\log 2)^2}{4}-L$

Thus,

$L=\dfrac{\pi^2}{96}+\dfrac{(\log 2)^2}{8}$

And therefore,

$\displaystyle B=G\left(\sqrt{2}\right)=\int_0^{\sqrt{2}}G'(a)da=\dfrac{5\pi^3}{96}+\dfrac{\pi}{2}\int_0^{\sqrt{2}}\dfrac{2a\log(a+\sqrt{2})-\sqrt{2}\log 2-a\log(a^2+2)}{a^2-2}da$

Let $\displaystyle P=\int_0^{\sqrt{2}}\dfrac{2a\log(a+\sqrt{2})-\sqrt{2}\log 2-a\log(a^2+2)}{a^2-2}da$

$\displaystyle P=\int_0^{\sqrt{2}}\dfrac{a\log(a^2+2)}{2\sqrt{2}(a+\sqrt{2})}da-\int_0^{\sqrt{2}}\dfrac{a\log(a+\sqrt{2})}{\sqrt{2}(a+\sqrt{2})}da+\\\displaystyle \int_0^{\sqrt{2}}\dfrac{\log 2}{2(a+\sqrt{2})}da+\dfrac{1}{2\sqrt{2}}\int_0^{\sqrt{2}}\dfrac{2a\log(a+\sqrt{2})-\sqrt{2}\log 2-a\log(a^2+2)}{a-\sqrt{2}}da$

$\displaystyle \int_0^{\sqrt{2}}\dfrac{-a\log(a+\sqrt{2})}{\sqrt{2}(a+\sqrt{2})}da=\\\left[\sqrt{2}\log(a+\sqrt{2})+\dfrac{\sqrt{2}}{2}\Big(\log(a+\sqrt{2})\Big)^2+\log(a+\sqrt{2})\Big (a-\sqrt{2}\log(a+\sqrt{2})\Big)-a\right]_0^{\sqrt{2}}=(\log 2)^2-\dfrac{5}{2}\log 2+1$

$\displaystyle \int_0^{\sqrt{2}}\dfrac{\log 2}{2(a+\sqrt{2})}da=\left[\dfrac{1}{2}\log(a+\sqrt{2})\log 2\right]_0^{\sqrt{2}}=\dfrac{1}{2}\Big(\log(2)\Big)^2$

Perform change of variable $x=\dfrac{a}{\sqrt{2}}$

$\displaystyle \int_0^{\sqrt{2}}\dfrac{a\log(a^2+2)}{2\sqrt{2}(a+\sqrt{2})}da=\\\displaystyle\dfrac{1}{2}\log 2\int_0^1\dfrac{x}{x+1}dx+\dfrac{1}{2}\int_0^1\dfrac{x\log(1+x^2)}{x+1}dx=\dfrac{1}{2}\log 2\Big[x-\log(1+x)\Big]_0^1+\dfrac{1}{2}\Big[(x-\log(1+x))\log(1+x^2)\Big]_0^1-\dfrac{1}{2}\int_0^1 \dfrac{2x(x-\log(1+x))}{1+x^2}da=\dfrac{1}{2}\log 2\Big[x-\log(1+x)\Big]_0^1+\dfrac{1}{2}\Big[(x-\log(1+x))\log(1+x^2)\Big]_0^1-\int_0^1 \dfrac{x^2}{1+x^2}dx+L=\dfrac{1}{2}\log 2\Big[x-\log(1+x)\Big]_0^1+\dfrac{1}{2}\Big[(x-\log(1+x))\log(1+x^2)\Big]_0^1-\Big[x-\arctan x\Big]_0^1+L=\\-\dfrac{7}{8}\Big(\log 2\Big)^2+\log 2+\dfrac{\pi^2}{96}+\dfrac{\pi}{4}-1$

Let $\displaystyle Q=\dfrac{1}{2\sqrt{2}}\int_0^{\sqrt{2}}\dfrac{2a\log(a+\sqrt{2})-\sqrt{2}\log 2-a\log(a^2+2)}{a-\sqrt{2}}da$

Perform change of variable $x=1-\dfrac{a}{\sqrt{2}}$

$\displaystyle Q=\int_0^1 \left(\dfrac{\log 2}{2x}-\dfrac{\log(2-x)}{x}+\log(2-x)+\dfrac{1}{2}\left(\dfrac{1}{x}-1\right)\log(x^2-2x+2)\right)dx$

$\displaystyle Q=\int_0^1 \log(2-x)dx-\dfrac{1}{2}\int_0^1\log(x^2-2x+2)dx+\int_0^1 \dfrac{1}{2x}\log\left(\dfrac{x^2}{2}-x+1\right)dx-\int_0^1 \dfrac{\log\left(1-\dfrac{x}{2}\right)}{x}dx$

$\displaystyle\int_0^1 \log(2-x)dx=\Big[(x-2)\log(2-x)-x\Big]_0^1=2\log 2-1$

$\displaystyle -\dfrac{1}{2}\int_0^1\log(x^2-2x+2)dx=\left[-\dfrac{1}{2}x\log(x^2-2x+2)+\dfrac{1}{2}\log(x^2-2x+2)-\arctan(x-1)+x\right]_0^1=\\ -\dfrac{1}{2}\log 2-\dfrac{\pi}{4}+1$

$\displaystyle \int_0^1 \dfrac{-\log\left(1-\dfrac{x}{2}\right)}{x}dx= \\ \displaystyle\int_0^1\left(\dfrac{1}{x}\sum_{n=1}^{+\infty} \dfrac{1}{n}\left(\dfrac{x}{2}\right)^n\right)dx=\displaystyle \sum_{n=1}^{+\infty} \dfrac{1}{2^n n}\int_0^1 x^{n-1}dx=\sum_{n=1}^{+\infty} \dfrac{1}{2^n n^2}=Li_2\left(\dfrac{1}{2}\right)=\dfrac{\pi^2}{12}-\dfrac{\Big(\log 2\Big)^2}{2}$

Let $R=\displaystyle\int_0^1 \dfrac{\log\left(\dfrac{1}{2}x^2-x+1\right)}{2x}dx$

Define for $a\in [0,1]$,

$\displaystyle S(a)=\int_0^1 \dfrac{\log(ax^2-2ax+1)}{2x}dx$

Note that $S\left(\dfrac{1}{2}\right)=R$ and $S(0)=0$

$\displaystyle S'(a)=\int_0^1 \dfrac{x-2}{2(ax^2-2ax+1)}dx$

$\displaystyle S'(a)=\int_0^1 \dfrac{x-2}{2(ax^2-2ax+1)}dx$

$\displaystyle S'(a)=\left[\dfrac{\log(ax^2-2ax+1}{4a}-\dfrac{1}{2\sqrt{a(1-a)}}\arctan\left(\sqrt{\dfrac{a}{1-a}}(x-1)\right)\right]_0^1$

$\displaystyle S'(a)=\dfrac{\log(1-a)}{4a}-\dfrac{1}{2\sqrt{a(1-a)}}\arctan\left(\sqrt{\dfrac{a}{1-a}}\right)$

Perform the change of variable $x=2a$,

$\displaystyle \int_0^{\tfrac{1}{2}} \dfrac{\log(1-a)}{4a} da=\int_0^1 \dfrac{\log\left(1-\dfrac{x}{2}\right)}{4x}dx=\dfrac{\Big(\log 2\Big)^2}{8}-\dfrac{\pi^2}{48}$

$\displaystyle \int_0^{\tfrac{1}{2}} \dfrac{-1}{2\sqrt{a(1-a)}}\arctan\left(\sqrt{\dfrac{a}{1-a}}\right)da=-\left[\dfrac{1}{2}\left(\arctan\left(\sqrt{\dfrac{a}{1-a}}\right)\right)^2\right]_0^{\tfrac{1}{2}}=-\dfrac{\pi^2}{32}$

Thus,

$R=\dfrac{\Big(\log 2\Big)^2}{8}-\dfrac{5\pi^2}{96}$

$Q=\dfrac{\pi^2}{32}-\dfrac{\pi}{4}+\dfrac{3\log 2}{2}-\dfrac{3\Big(\log 2\Big)^2}{8}$

$P=\dfrac{\Big(\log 2\Big)^2}{4}+\dfrac{\pi^2}{24}$

$B=\dfrac{\pi\Big(\log 2\Big)^2}{8}+\dfrac{7\pi^3}{96}$

And finally,

$J=\dfrac{11\pi\log 2}{8\sqrt{2}}-\dfrac{7\pi\Big(\log 2\Big)^2}{8\sqrt{2}}+\dfrac{35\pi^3}{96\sqrt{2}}-\dfrac{25\pi}{16\sqrt{2}}$