An arctan integral $\int_0^{\infty } \frac{\arctan(x)}{x \left(x^2+1\right)^5} \, dx$

Here is another solution: Let $(I_n)$ by

$$ I_n = \int_{0}^{\infty} \frac{\arctan x}{x(1+x^2)^n} \, dx = \int_{0}^{\frac{\pi}{2}} \frac{\theta}{\sin\theta} \cos^{2n-1}\theta \, d\theta. $$

Then by a simple calculation,

$$ I_n - I_{n+1} = \int_{0}^{\frac{\pi}{2}} \theta \sin\theta \cos^{2n-1}\theta \, d\theta = \frac{1}{2n} \int_{0}^{\frac{\pi}{2}} \cos^{2n}\theta \, d\theta. $$

Since $I_n \to 0$ as $n \to \infty$, we find that

$$ I_n = \sum_{k=n}^{\infty} \frac{1}{2k} \int_{0}^{\frac{\pi}{2}} \cos^{2k}\theta \, d\theta. $$

Splitting the summation as $\sum_{k=n}^{\infty} = \sum_{k=1}^{\infty} - \sum_{k=1}^{n-1}$, we find that

$$ I_n = \frac{\pi}{2}\left( \log 2 - \sum_{k=1}^{n-1} \frac{1}{2k} \frac{(2k-1)!!}{(2k)!!} \right), $$

where $n!!$ denotes the double factorial.

Edit 1. In general, we have

$$ \int_{0}^{\infty} \frac{\arctan^s x}{x(1+x^2)^{n+1}} \, dx = \int_{0}^{\frac{\pi}{2}} \theta^s \cot \theta \, d\theta - \sum_{k=1}^{n} \int_{0}^{\frac{\pi}{2}} \theta^s \sin\theta \cos^{2k-1}\theta \, d\theta. \tag{1} $$

Currently I have no idea how to obtain a simple formula for the following integral

$$ \int_{0}^{\frac{\pi}{2}} \theta^s \sin\theta \cos^{2k-1}\theta \, d\theta, \tag{2} $$

even when $s = 2$. On the other hand, for any $s > 0$ and $N \geq \lfloor s/2 \rfloor$ we have

\begin{align*} \int_{0}^{\frac{\pi}{2}} \theta^s \cot \theta \, d\theta &= 2^{-s}\cos\left(\frac{\pi s}{2}\right)\Gamma(1+s)\zeta(1+s) \\ &\quad + \left(\frac{\pi}{2}\right)^s \sum_{k=0}^{N} (-1)^k \pi^{-2k} \frac{\Gamma(2k-s)}{\Gamma(-s)} \eta(2k+1) \\ &\quad + \frac{(-1)^{N+1}}{2^s \Gamma(-s)} \int_{0}^{\infty} \frac{t^{2N+1-s}}{1+t^2} \left( \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{1+s}} e^{-\pi n t} \right) \, dt, \end{align*}

where $\eta(s)$ denotes the Dirichlet eta function. (My solution is somewhat involved, so I will post later if it seems useful to our problem.) In particular, when $s$ is a positive integer, then the integral part vanishes and the formula becomes much simpler. Thus the formula (*) gives a closed form as long as we can figure out the integral (2).

Example 1. For example, when $s = 2$ then we can use $N = 1$ and then \begin{align*} \int_{0}^{\frac{\pi}{2}} \theta^2 \cot \theta \, d\theta &= -\frac{1}{2}\zeta(3) + \frac{\pi^2}{4} \log 2 - \frac{1}{2}\eta(3) \\ &= \frac{\pi^2}{4}\log 2 - \frac{7}{8}\zeta(3). \end{align*} Since we can figure out the integral (2) for $s = 2$ and $k = 1, \cdots, 4$, we easily obtain OP's last identity.

Here is another example:

Example 2. Using the formula with $s = 6$, we can check that \begin{align*} \int_{0}^{\infty} \frac{\arctan^6 x}{x(1+x^2)^3} \, dx &= \frac{\pi^6}{64} \log 2 -\frac{45 \pi^4}{128} \zeta(3) + \frac{675 \pi^2}{128} \zeta(5) -\frac{5715}{256} \zeta(7) \\ &\quad - \frac{11 \pi^6}{2048} + \frac{705 \pi^4}{4096} - \frac{8595 \pi^2}{4096} + \frac{135}{16} \\ &\approx 0.0349464822054751922142122595622\cdots. \end{align*}

What about Feynman's way? If we take: $$ f(a) = \int_{0}^{+\infty}\frac{\arctan(ax)}{x(1+x^2)^5}\,dx $$ we have $f(0)=0$ and: $$ f'(a)=\int_{0}^{+\infty}\frac{dx}{(1+x^2)^5 (1+a^2 x^2)} $$ that is a manageable integral through partial fraction decomposition / the residue theorem.

We have: $$\begin{eqnarray*} f'(a) &=& \frac{\pi}{256(1+a)^5}\left(35+175 a+345 a^2+325 a^3+128 a^4\right)\\&=&\frac{\pi}{256(1+a)^5}\left(8-52(a+1)+138 (a+1)^2-187(a+1)^3+128(a+1)^4\right)\end{eqnarray*} $$ and it is not difficult to integrate such expression over $(0,1)$.

I'm not sure if this is nice and simple, but here is one way:

Substitute $y = \arctan x$ to rewrite as $$\int_0^{\pi/2} \frac{y}{\tan y \sec^{8} y} \, dy$$ Now write this as an integral in $y$ and $\sec y$: $$\int_0^{\pi/2} y \cdot \frac{\sec y \tan y}{(\sec^2 y - 1) \sec^{9} y} \, dy$$

You can integrate this by parts. Finding the antiderivative of $\dfrac{\sec y \tan y}{(\sec^2 y - 1) \sec^{9} y}$ amounts (after a change of variable) to finding the antiderivative of $$\frac{1}{t^{11} - t^{9}} = \frac{1}{t^{9}(t-1)(t+1)}.$$ This has an elementary partial fractions decomposition which I don't have the energy to carry out.

Does it look any nicer? $$\int\frac{\arctan x}{x\,\left(x^2+1\right)^5}\, dx=\frac12\,\Im\operatorname{Li}_2\left(e^{2\,i\arctan x}\right)\\-\frac x{9216\,\left(x^2+1\right)^4}\left(3963x^6+12995x^4+14525x^2+5637\right)\\+\left[\ln\frac{2x}{\sqrt{x^2+1}}+\frac{12x^6+42x^4+52x^2+25}{24\,\left(x^2+1\right)^4}-\frac{1321}{3072}\right]\cdot\arctan x$$