Arctan and Log Integration Problem I have been trying to evaluate this integral: $\displaystyle \int_0^{\infty} \frac{\arctan^2 x \log^2 (1+x^2)}{x^2}\,dx$, but have failed despite all my attemps. I tried using the trig substitution $x = tan(a)$, and through a series of steps, was able to simplify the integral to $I(b)$ = $4\displaystyle \int_0^{\pi/2} \frac{(a\log(\cos(a))^b}{(\sin(a))^b}\,da$. I then tried to differentiate under the integral sign but failed. What should I do now?
Note: b = 2 in the latter integral. 
 A: Here is an idea:
From messing around with coefficients on OEIS A049034, it seems that
$$
\arctan^2(x)=\sum_{n=0}^\infty \frac{(-1)^n x^{2+2n}}{2n+2}\left(H_{n+\frac{1}{2}}+\log(4)\right)
$$
then
$$
I=\int_0^{\infty} \frac{\arctan^2 x \log^2 (1+x^2)}{x^2}\,dx
$$
$$
I=\int_0^{\infty} \frac{\sum_{n=0}^\infty \frac{(-1)^n x^{2+2n}}{2n+2}\left(H_{n+\frac{1}{2}}+\log(4)\right) \log^2 (1+x^2)}{x^2}\,dx
$$
$$
I=\int_0^{\infty} \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{2n+2}\left(H_{n+\frac{1}{2}}+\log(4)\right) \log^2 (1+x^2)\,dx
$$
$$
I=\log(4)\sum_{n=0}^\infty \frac{(-1)^n }{2n+2}\int_0^{\infty}  x^{2n}\log^2 (1+x^2)\,dx \\+ \sum_{n=0}^\infty \frac{(-1)^n H_{n+\frac{1}{2}}}{2n+2} \int_0^{\infty} x^{2n} \log^2 (1+x^2)\,dx 
$$
the Mellin transform of $\log(x^2+1)^2$ is
$$
\int_0^\infty x^{s-1} \log(x^2+1)\;dx = -\frac{2\pi}{s} \csc\left(\frac{\pi s}{2}\right)\left(\gamma + \psi\left(-\frac{s}{2}\right)\right)
$$
setting $s=2n+1$ then gives 
$$
\int_0^{\infty} x^{2n} \log^2 (1+x^2)\,dx  = -(-1)^n\frac{2\pi}{1+2n} H_{-n-\frac{3}{2}}
$$
$$
I=-\log(4)\sum_{n=0}^\infty \frac{1}{2n+2}\frac{2\pi}{1+2n} H_{-n-\frac{3}{2}} \\-\sum_{n=0}^\infty \frac{H_{n+\frac{1}{2}}}{2n+2}\frac{2\pi}{1+2n} H_{-n-\frac{3}{2}} 
$$
this gives
$$
I=-2\pi\sum_{n=0}^\infty \frac{H_{-n-\frac{3}{2}}\left(\log(4)+H_{n+\frac{1}{2}}\right)}{2+6n+4n^2} 
$$
but I don't trust the result yet because they don't appear to be numerically the same. There might be an error somewhere, but I'll leave it here and edit it if I work it out.
