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:
$$\int_0^{\infty } \frac{\arctan^2(x)}{x \left(x^2+1\right)^5} \, dx=\frac{55}{108}-\frac{1321}{12288}\pi^2+\frac{\pi^2}{4} \log (2)-\frac{7 }{8}\zeta (3)$$

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. 
