How to evaluate $\int_0^1 (\arctan x)^2 \ln(\frac{1+x^2}{2x^2}) dx$ 
Evaluate
  $$
\int_{0}^{1} \arctan^{2}\left(\, x\,\right)
\ln\left(\, 1 + x^{2} \over 2x^{2}\,\right)\,{\rm d}x
$$  

I substituted $x \equiv \tan\left(\,\theta\,\right)$ and got  
$$
-\int^{\pi/4}_{0}\theta^{2}\,{\ln\left(\, 2\sin^{2}\left(\,\theta\,\right)\,\right) \over \cos^{2}\left(\,\theta\,\right)}\,{\rm d}\theta
$$  
After this, I thought of using the Taylor Expansion of
$\ln\left(\, 2\sin^{2}\left(\,\theta\,\right)\,\right)$ near zero but that didn't do any good.  
Please Help!

By the way, there is a closed-form antiderivative (that could be proved by differentiation):
$$\int\arctan^2x\cdot
\ln\left(\frac{1+x^2}{2x^2}\right)\,dx=\\
\frac16\left[3 i \left\{\left(2 \operatorname{Li}_2(i
   x)-2 \operatorname{Li}_2(-i x)+\operatorname{Li}_2\left(\frac{2
   x}{x+i}\right)-\operatorname{Li}_2\left(\frac{2 x}{x-i}\right)\right)\cdot \ln \left(\frac{1+x^2}{x^2}\right)\\
+2 \left(2
   \operatorname{Li}_3\left(\frac{x}{x-i}\right)-2
   \operatorname{Li}_3\left(\frac{x}{x+i}\right)+\operatorname{Li}_3\left(\frac{2
   x}{x+i}\right)-\operatorname{Li}_3\left(\frac{2
   x}{x-i}\right)\right)\\
+\left(\operatorname{Li}_2\left(\frac{1}{2}-\frac{i
   x}{2}\right)-\operatorname{Li}_2\left(\frac{i
   x}{2}+\frac{1}{2}\right)\right)\cdot\ln2\right\}\\
+3 \left(2
   \operatorname{Li}_2\left(-x^2\right)+\ln ^2\left(1+x^2\right)+\ln
   \left(1+x^2\right)\cdot\ln2-2 \ln ^22\right)\cdot\arctan x\\
+6 x \ln \left(\frac{1+x^2}{2
   x^2}\right)\cdot\arctan^2x+4 \arctan^3x\right]\color{gray}{+C}$$
This enables us to evaluate a definite integral over any region.
