Evaluating $\int_{-1}^{1}\frac{\arctan{x}}{1+x}\ln{\left(\frac{1+x^2}{2}\right)}dx$ This is a nice problem. I am trying to use nice methods to solve this integral, But I failed.
$$\int_{-1}^{1}\dfrac{\arctan{x}}{1+x}\ln{\left(\dfrac{1+x^2}{2}\right)}dx, $$
where  $\arctan{x}=\tan^{-1}{x}$
mark: this integral is my favorite one. Thanks to whoever has nice methods.
I have proved the following:
$$\int_{-1}^{1}\dfrac{\arctan{x}}{1+x}\ln{\left(\dfrac{1+x^2}{2}\right)}dx=\sum_{n=1}^{\infty}\dfrac{2^{n-1}H^2_{n-1}}{nC_{2n}^{n}}=\dfrac{\pi^3}{96}$$
where $$C_{m}^{n}=\dfrac{m}{(m-n)!n!},H_{n}=1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}$$
I also have got a few by-products 
$$\int_{-1}^{1}\dfrac{\arctan{x}}{1+x}\ln{\left(\dfrac{1+x^2}{2}\right)}dx=-I_{1}-2I_{2}$$
where $$I_{1}=\int_{0}^{1}\dfrac{\ln{(1-x^2)}}{1+x^2}\ln{\left(\dfrac{1+x^2}{2}\right)}dx=\dfrac{\pi}{4}\ln^2{2}+\dfrac{\pi^3}{32}-2K\times\ln{2}$$
and
$$I_{2}=\int_{0}^{1}\dfrac{x\arctan{x}}{1+x^2}\ln{(1-x^2)}dx=-\dfrac{\pi^3}{48}-\dfrac{\pi}{8}\ln^2{2}+K\times\ln{2}$$
and same methods,I have follow integral
$$\int_{0}^{1}\dfrac{\ln{(1-x^4)}\ln{x}}{1+x^2}dx=\dfrac{\pi^3}{16}-3K\times\ln{2}$$
where $ K $ denotes Catalan's Constant. 

FWIW, here's Maple:
> f:= arctan(x)/(1+x)*ln((1+x^2)/2);
> int(f, x=-1..1);

$$ {\frac {7}{64}}\,{\pi }^{3}-{\frac {5}{16}}\,\pi \, \left( \ln 
 \left( 2 \right)  \right) ^{2}-\ln  \left( 2 \right) {\it Catalan}+1/
2\, \left( \ln  \left( 1-i \right)  \right) ^{2}\pi -1/2\,i \left( 
\ln  \left( 1+i \right)  \right) ^{2}\ln  \left( 2 \right) +\ln 
 \left( 1+i \right) {\it Catalan}+1/2\, \left( \ln  \left( 1+i
 \right)  \right) ^{2}\pi -i{\it polylog} \left( 3,-i \right) +i{\it 
polylog} \left( 3,i \right) +1/2\,i \left( \ln  \left( 1-i \right) 
 \right) ^{2}\ln  \left( 2 \right) -1/2\,i \left( \ln  \left( 1-i
 \right)  \right) ^{3}-1/48\,i\ln  \left( 1+i \right) {\pi }^{2}+1/2\,
i \left( \ln  \left( 1+i \right)  \right) ^{3}+\ln  \left( 1-i
 \right) {\it Catalan}+1/48\,i\ln  \left( 1-i \right) {\pi }^{2}
$$
> simplify(%);

$$ \frac{\pi^3}{96}
$$
I don't know if that qualifies as "nice", but it's certainly easy.
