According to Mathematica, we have that
$$\int_0^{\infty } \frac{\arctan(x)}{x \left(x^2+1\right)^5} \, dx=\pi \left(\frac{\log (2)}{2}-\frac{1321}{6144}\right)$$
that frankly speaking looks pretty nice.
However Mathematica shows that
$$\int \frac{\arctan(x)}{x \left(x^2+1\right)^5} \, dx$$
$$=-\frac{1}{2} i \text{Li}_2\left(e^{2 i \tan ^{-1}(x)}\right)-\frac{1}{2} i \tan ^{-1}(x)^2+\tan ^{-1}(x) \log \left(1-e^{2 i \tan ^{-1}(x)}\right)-\frac{65}{256} \sin \left(2 \tan ^{-1}(x)\right)-\frac{23 \sin \left(4 \tan ^{-1}(x)\right)}{1024}-\frac{5 \sin \left(6 \tan ^{-1}(x)\right)}{2304}-\frac{\sin \left(8 \tan ^{-1}(x)\right)}{8192}+\frac{65}{128} \tan ^{-1}(x) \cos \left(2 \tan ^{-1}(x)\right)+\frac{23}{256} \tan ^{-1}(x) \cos \left(4 \tan ^{-1}(x)\right)+\frac{5}{384} \tan ^{-1}(x) \cos \left(6 \tan ^{-1}(x)\right)+\frac{\tan ^{-1}(x) \cos \left(8 \tan ^{-1}(x)\right)}{1024}$$
and this form doesn't look that nice.
Having given the nice form of the closed form I wonder if we can find a very nice and simple way of getting the answer. What do you think?
A supplementary question:
$$\int_0^{\infty } \frac{\arctan^2(x)}{x \left(x^2+1\right)^5} \, dx=\frac{55}{108}-\frac{1321}{12288}\pi^2+\frac{\pi^2}{4} \log (2)-\frac{7 }{8}\zeta (3)$$
Best Answer
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).
Here is another example: