An arctan integral $\int_0^{\infty } \frac{\arctan(x)}{x \left(x^2+1\right)^5} \, dx$ 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:
$$\int_0^{\infty } \frac{\arctan^2(x)}{x \left(x^2+1\right)^5} \, dx=\frac{55}{108}-\frac{1321}{12288}\pi^2+\frac{\pi^2}{4} \log (2)-\frac{7 }{8}\zeta (3)$$
 A: 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. 
A: Does it look any nicer?
$$\int\frac{\arctan x}{x\,\left(x^2+1\right)^5}\, dx=\frac12\,\Im\operatorname{Li}_2\left(e^{2\,i\arctan x}\right)\\-\frac x{9216\,\left(x^2+1\right)^4}\left(3963x^6+12995x^4+14525x^2+5637\right)\\+\left[\ln\frac{2x}{\sqrt{x^2+1}}+\frac{12x^6+42x^4+52x^2+25}{24\,\left(x^2+1\right)^4}-\frac{1321}{3072}\right]\cdot\arctan x$$
A: 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 $(0,1)$.
