Closed form of $\int_0^\infty \arctan^2 \left (\frac{2x}{1 + x^2} \right ) \, dx$ 
Can a closed form solution for the following integral be found:
$$\int_0^\infty \arctan^2 \left (\frac{2x}{1 + x^2} \right ) \, dx\,?$$

I have tried all the standard tricks such as integration by parts, various substitutions, and parametric differentiation (Feynman's trick), but all to no avail.
An attempt is letting
$$f(t):=\int_0^\infty\,\arctan^2\left(\frac{2tx}{1+x^2}\right)\,\text{d}x\,.$$
Therefore,
$$f'(t)=\int_0^\infty\,\frac{8x^2(x^2+1)}{\big(x^4+2(2t^2+1)x^2+1\big)^2}\,\left(1+x^2-4tx\arctan\left(\frac{2tx}{1+x^2}\right)^{\vphantom{a^2}}\right)\,\text{d}x\,.$$
This doesn't seem to go anywhere.  Help!

$$I=\int_0^\infty \arctan^2 \left (\frac{2x}{x^2 + 1} \right ) dx\overset{IBP}=4\int_0^\infty \frac{x(x^2-1)\arctan\left(\frac{2x}{x^2+1}\right)}{x^4+6x^2+1}dx$$
We have that: $$4\int\frac{x(x^2-1)}{x^4+6x^2+1}dx=(\sqrt 2 +1)\ln(x^2+(\sqrt 2+1)^2)-(\sqrt 2-1)\ln(x^2+(\sqrt 2-1)^2)$$
$$\frac{d}{dx}\arctan\left(\frac{2x}{x^2+1}\right)=\frac12\left(\frac{\sqrt 2+1}{x^2+(\sqrt 2+1)^2}-\frac{\sqrt 2-1}{x^2+(\sqrt 2-1)^2}\right)$$
Thus integrating by parts again and simplifying we obtain:
$$I=\int_0^\infty \frac{(\sqrt 2+1)^2 \ln(x^2+(\sqrt 2+1)^2)}{x^2+(\sqrt 2+1)^2}dx+\int_0^\infty \frac{(\sqrt 2-1)^2 \ln(x^2+(\sqrt 2-1)^2)}{x^2+(\sqrt 2-1)^2}dx$$
$$-\int_0^\infty \frac{\ln(x^2+(\sqrt 2-1)^2)}{x^2+(\sqrt 2+1)^2}dx-\int_0^\infty \frac{\ln(x^2+(\sqrt 2+1)^2)}{x^2+(\sqrt 2-1)^2)}dx$$
From here we have the following result:
$$\int_0^\infty \frac{\ln(x^2+a^2)}{x^2+b^2}dx=\frac{\pi}{b}\ln(a+b), \ a,b>0$$
So using this result and with some algebra everything simplifies to: 
$$\boxed{\int_0^\infty \arctan^2 \left (\frac{2x}{x^2 + 1} \right ) dx=2\pi \ln(1+\sqrt 2)-\sqrt 2\pi \ln 2}$$
