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!
 A: Samurai, this is for the second time you post problems that relate to me. First you posted this question here which is exactly similar with my rated problem on Brilliant.org. I have raised objection to mods but they can do nothing since your post doesn't violate any rules here. Okay, fine. I can accept their reason. Now you post this question which I believe it's taken from one of proposed problems in Brilliant Integration Contest - Season 1 that I held on Brilliant.org. The original problem was proposed by Jatin Yadav as PROBLEM 7 but a day later he deleted this problem and changed to another problem after no-one can solve it included himself. According to him, it's taken from here, on Math S.E. You may want to take a look there.
I tried to solve this problem for hours but no success. Here is my attempt:
Set $x=\tan y$, we get
\begin{align}
I&=\int_0^1\arctan^2x\,\ln\left(\frac{1+x^2}{2x^2}\right)\,dx\\
&=-\int_0^{\pi/4} \frac{y^2\ln\left(2\sin^2y\right)}{\cos^2y}\,dy\\
&=-2\int_0^{\pi/4} \frac{y^2\ln\left(1-\cos2y\right)}{1+\cos2y}\,dy\\
&=-\frac{1}{4}\int_0^{\pi/2} \frac{t^2\ln\left(1-\cos t\right)}{1+\cos t}\,dt\qquad\Rightarrow\qquad t=2y\\
\end{align}
Use integration by parts by taking $u=t^2$ and $dv=\dfrac{\ln\left(1-\cos t\right)}{1+\cos t}\,dt$, then
\begin{align}
v&=\int\frac{\ln\left(1-\cos t\right)}{1+\cos t}\,dt
\end{align}
Use integration by parts by taking $u=\ln\left(1-\cos t\right)$ and $dv=\dfrac{dt}{1+\cos t}$, by Weierstrass substitution: $x=\tan\left(\dfrac{t}{2}\right)$ then
\begin{align}
v=\int\frac{dt}{1+\cos t}=\int \,dx=\tan\left(\frac{t}{2}\right)=\frac{\sin t}{1+\cos t}
\end{align}
Hence
\begin{align}
\int\frac{\ln\left(1-\cos t\right)}{1+\cos t}\,dt
&=\frac{\sin t}{1+\cos t}\ln\left(1-\cos t\right)-\int\frac{\sin t}{1+\cos t}\cdot\frac{\sin t}{1-\cos t}\,dt\\
&=\frac{\sin t}{1+\cos t}\ln\left(1-\cos t\right)-t
\end{align}
and
\begin{align}
I&=-\frac{1}{4}\left[\frac{t^2\sin t}{1+\cos t}\ln\left(1-\cos t\right)-t^3\right]_0^{\pi/2}+\frac{1}{2}\int_0^{\pi/2}\left[\frac{t\sin t}{1+\cos t}\ln\left(1-\cos t\right)-t^2\right]\,dt\\
&=\frac{\pi^3}{32}+\frac{1}{2}\int_0^{\pi/2}\frac{t\sin t}{1+\cos t}\ln\left(1-\cos t\right)\,dt-\frac{\pi^3}{48}\\
&=\frac{\pi^3}{96}+\frac{1}{2}\int_0^{\pi/2}\frac{t\sin t}{1+\cos t}\ln\left(1-\cos t\right)\,dt
\end{align}
Consider
\begin{equation}
I(a)=\int_0^{\pi/2} \frac{t\sin t}{1+\cos t}\ln\left(1-a\cos t\right)\,dt
\end{equation}
so that $I(0)=0$ and
\begin{align}
I'(a)&=-\int_0^{\pi/2} \frac{t\sin t\cos t}{(1-a\cos t)(1+\cos t)}\,dt\\
&=\frac{1}{1+a}\int_0^{\pi/2} \left(\frac{t\sin t}{1+\cos t}-\frac{t\sin t}{1-a\cos t}\right)\,dt\\
\end{align}
Now consider
\begin{equation}
I(b)=\int_0^{\pi/2} \frac{t\sin t}{1+b\cos t}\,dt
\end{equation}
Use integration by parts by taking $u=t$ and $dv=\dfrac{\sin t}{1+b\cos t}\,dt$, then 
\begin{align}
I(b)&= \frac{t\ln(1+b\cos t)}{b}\bigg|_0^{\pi/2}-\frac{1}{b}\int_0^{\pi/2}\ln(1+b\cos t)\,dt\\
&=-\frac{1}{b}\int_0^{\pi/2}\ln(1+b\cos t)\,dt
\end{align}
Consider
\begin{equation}
J(b)=\int_0^{\pi/2}\ln(1+b\cos t)\,dt
\end{equation}
so that $J(0)=0$ and
\begin{align}
J'(b)&=\int_0^{\pi/2} \frac{\cos t}{1+b\cos t}\,dt\\
&=\frac{1}{b}\int_0^{\pi/2} \left(1-\frac{1}{1+b\cos t}\right)\,dt\\
&=\frac{\pi}{2b}-\int_0^{\pi/2} \frac{dt}{1+b\cos t}\qquad\Rightarrow\qquad x=\tan\left(\frac{t}{2}\right)\\
&=\frac{\pi}{2b}-\int_0^{1} \frac{2}{1+b+(1-b)x^2}\,dx\qquad\Rightarrow\qquad x=\sqrt{\frac{1+b}{1-b}}\tan z\\
&=\frac{\pi}{2b}-\frac{2}{\sqrt{1-b^2}}\arctan\left(\sqrt{\frac{1-b}{1+b}}\right)\\
J(b)&=\frac{\pi}{2}\ln b-\int\frac{2}{\sqrt{1-b^2}}\arctan\left(\sqrt{\frac{1-b}{1+b}}\right)\,db\\
\end{align}
Again we use integration by parts by taking $u=\arctan\left(\sqrt{\frac{1-b}{1+b}}\right)$ and $dv=\dfrac{2}{\sqrt{1-b^2}}$, we have
\begin{align}
J(b)&=\frac{\pi}{2}\ln b-2\arctan\left(\sqrt{\frac{1-b}{1+b}}\right)\arcsin b-\int\frac{\arcsin b}{1-b}\sqrt{\frac{1-b}{1+b}}\,db\\
&=\frac{\pi}{2}\ln b-2\arctan\left(\sqrt{\frac{1-b}{1+b}}\right)\arcsin b-\int\frac{\arcsin b}{\sqrt{1-b^2}}\,db\\
&=\frac{\pi}{2}\ln b-2\arctan\left(\sqrt{\frac{1-b}{1+b}}\right)\arcsin b-\frac{\arcsin^2 b}{2}
\end{align}
Therefore
\begin{align}
I'(a)&=\frac{\pi^2}{8(1+a)}-\frac{1}{a(1+a)}\left[\frac{\pi}{2}\ln (-a)-2\arctan\left(\sqrt{\frac{1+a}{1-a}}\right)\arcsin (-a)-\frac{\arcsin^2(-a)}{2}\right]\\
\end{align}
From this step, I give up. Perhaps someone else want to continue it. Be my guest...
A: This is not an answer, but my approach suggests that the answer is
$$ I := \int_{0}^{1} \arctan^{2} x \log \left( \frac{x^{2}+1}{2x^{2}} \right) \, dx
= \frac{19\pi^{3}}{192} + \frac{5\pi}{16}\log^{2}2 - 6 \Im \mathrm{Li}_{3}\left( \frac{1+i}{2} \right). $$

My approach is to write
$$ I = \int_{0}^{1} \arctan^{2} x \log \left( \frac{x^{2}+1}{2} \right) \, dx - 2\int_{0}^{1} \arctan^{2} x \log x \, dx, $$
introduce function $f(z) = \log \left( \frac{1+iz}{\sqrt{2}} \right)$ and write
$$ \arctan^{2} x \log \left( \frac{x^{2}+1}{2} \right) = -\frac{1}{4} ( f(x) - f(-x))^{2}(f(x) + f(-x)). $$
This allows to write, with a bit help of complex analysis,
$$ \int_{0}^{1} \arctan^{2} x \log \left( \frac{x^{2}+1}{2} \right) \, dx = - \frac{\sqrt{2}}{4} \int_{-\pi/4}^{\pi/4} \theta^{2}\log(\sqrt{2}e^{i\theta} - 1) e^{i\theta} \, d\theta, $$
which seems more tractable than the original one. But I was stuck here.
A: 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.
A: I dont'have (yet) a complete solution for now.
Let $I=\displaystyle \int_0^1 (\arctan(x))^2 \ln\Big(\dfrac{1+x^2}{2x^2}\Big)dx$
$I=\displaystyle \int_0^1 (\arctan(x))^2\ln(1+x^2)dx-2\int_0^1 (\arctan(x))^2\ln(x)dx-\ln(2)\int_0^1 (\arctan(x))^2dx$
I know how to compute the last one ;)
Let $J=\displaystyle \int_0^{\frac{\pi}{2}} \log(\sin x)dx$
The value of $J$ is well known to be $-\dfrac{\pi}{2}\ln 2$
Perform integration by parts:
$J=\displaystyle \Big[x\log(\sin x)\Big]_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}}\dfrac{x}{\tan x}dx=-\int_0^{\frac{\pi}{2}}\dfrac{x}{\tan x}dx$
Let $K=-J$
Perform the change of variable $u=\tan x$
$K=\displaystyle \int_0^{+\infty}\dfrac{\arctan x}{x(1+x^2)}dx$
$K\displaystyle =\int_0^{1}\dfrac{\arctan x}{x(1+x^2)}dx+\int_1^{+\infty}\dfrac{\arctan x}{x(1+x^2)}dx$
In the second integral, in the right member perform the change of variable $u=\dfrac{1}{x}$:
$K= \displaystyle\int_0^1 \dfrac{\arctan x }{x(1+x^2)}dx+ \int_0^1 \dfrac{x\arctan \Big(\dfrac{1}{x}\Big) }{1+x^2}dx$
For $x>0$, $\arctan \Big(\dfrac{1}{x}\Big)+\arctan x=\dfrac{\pi}{2}$
Therefore:
$K=\displaystyle\int_0^1 \dfrac{\arctan x }{x(1+x^2)}dx+\dfrac{\pi}{2}\int_0^1\dfrac{x}{1+x^2}dx-\int_0^1\dfrac{x\arctan x}{1+x^2}dx$
For $x\neq 0$ ,$\dfrac{1}{x(1+x^2)}=\dfrac{1}{x}-\dfrac{x}{1+x^2}$
$K=\displaystyle\int_0^1 \dfrac{\arctan x }{x}dx-2\int_0^1 \dfrac{x\arctan x }{1+x^2}dx+\dfrac{\pi}{2}\int_0^1\dfrac{x}{1+x^2}dx$
The derivative of $(\arctan x)^2$ is $\dfrac{2\arctan x}{1+x^2}$
Hence:
$\displaystyle \int_0^1 \dfrac{2x\arctan x }{1+x^2}dx=\Big[x(\arctan x)^2\Big]_0^1-\int_0^1 (\arctan x)^2dx=\dfrac{\pi^2}{16}-\int_0^1 (\arctan x)^2dx$
Therefore:
$K=\displaystyle\int_0^1 \dfrac{\arctan x }{x}dx-\dfrac{\pi^2}{16}+\int_0^1 (\arctan x)^2dx+\dfrac{\pi}{4}\Big[\log(1+x^2)\Big]_0^1$
Recall $K=\dfrac{\pi}{2}\ln 2$
Therefore:
$\displaystyle\int_0^1 (\arctan x)^2 dx=\dfrac{\pi^2}{16}-G+\dfrac{\pi}{4}\ln 2$
Where $G=\displaystyle\int_0^1 \dfrac{\arctan x }{x}dx$ is the Catalan's constant.
A: Let $I$ be our integral 
Since $\displaystyle \small\int \ln\left(\frac{1+x^2}{2x^2}\right)\ dx=x\ln\left(\frac{1+x^2}{2x^2}\right)+2\arctan(x)$, so by integration by parts we have 
$$I=2\left(\frac{\pi}{4}\right)^3-4\underbrace{\int_0^1\frac{\arctan^2(x)}{1+x^2}\ dx}_{\frac13\left(\frac{\pi}{4}\right)^3}-\underbrace{\int_0^1\frac{2x}{1+x^2}\ln\left(\frac{1+x^2}{2x^2}\right)\arctan(x)\ dx}_{IBP}$$
$$=\frac{\pi^3}{96}+\int_0^1\frac{\ln(1+x^2)}{1+x^2}\ln\left(\frac{1+x^2}{2x^2}\right)\ dx-\int_0^1\frac{2\ln(1+x^2)\arctan(x)}{x(1+x^2)}\ dx$$
$$=\frac{\pi^3}{96}+A-B\tag1$$

For $A$, use $x=\tan\theta$
$$A=2\ln2\int_0^{\pi/4}\ln(\cos\theta)\ d\theta+4\int_0^{\pi/4}\ln(\cos\theta)\ln(\sin\theta)\ d\theta$$
$$A=2\ln2\underbrace{\int_0^{\pi/4}\ln(\cos\theta)\ d\theta}_{\text{common integral}}+2\underbrace{\int_0^{\pi/2}\ln(\cos\theta)\ln(\sin\theta)\ d\theta}_{\text{beta function}}$$
$$=2\ln2\left(\frac{G}{2}-\frac{\pi}{2}\ln^22\right)+2\left(\frac{\pi}{2}\ln^22-\frac{\pi^3}{48}\right)$$
$$\boxed{A=G\ln2+\frac{\pi}{2}\ln^22-\frac{\pi^3}{24}}$$

For $B$, write $\frac{2}{x(1+x^2)}=\frac{2}{x}-\frac{2x}{1+x^2}$
$$B=2\int_0^1\frac{\ln(1+x^2)\arctan(x)}{x}\ dx-\int_0^1\frac{2x\ln(1+x^2)\arctan(x)}{1+x^2}\ dx$$
The first integral is already evaluated here
$$\int_0^1\frac{\ln(1+x^2)\arctan(x)}{x}\ dx=\frac{\pi^3}{16}+\frac{\pi}{8}\ln^22+G\ln2+2\Im\operatorname{Li}_3(1-i)$$
For the second one, apply integration by parts
$$\int_0^1\frac{2x\ln(1+x^2)\arctan(x)}{1+x^2}\ dx=\frac{\pi}{8}\ln^22-\frac12\int_0^1\frac{\ln^2(1+x^2)}{1+x^2}\ dx$$
$$=\frac{\pi}{8}\ln^22-2\int_0^{\pi/4}\ln^2(\cos\theta)\ d\theta$$
we proved here
$$\int_0^{\pi/4}\ln^2(\cos\theta)\ d\theta=\frac{7\pi^3}{192}+\frac{5\pi}{16}\ln^22-\frac{G}{2}\ln2+\Im\operatorname{Li_3}(1-i)$$
Collecting the results we have 
$$\boxed{B=\frac{19\pi^3}{96}+\frac{3\pi}{4}\ln^22+G\ln2+6\Im\operatorname{Li_3}(1-i)}$$
Now plug the boxed results in $(1)$ we get
$$I=-\frac{11\pi^3}{48}-\frac{\pi}{4}\ln^22-6\Im\operatorname{Li_3}(1-i)$$
