<s>[INST] Find $\int _0^1\frac{12\arctan ^2 x\ln (\frac{(1-x)^2}{1+x^2})-\ln ^3(\frac{(1-x)^2}{1+x^2})}{x}\:dx$ I want to find and prove that: $$\int _0^1\frac{12\arctan ^2\left(x\right)\ln \left(\frac{\left(1-x\right)^2}{1+x^2}\right)-\ln ^3\left(\frac{\left(1-x\right)^2}{1+x^2}\right)}{x}\:dx=\frac{9 \pi ^4}{16}$$
I tried to split the integral:
$$=12\int _0^1\frac{\arctan ^2\left(x\right)\ln \left(\frac{\left(1-x\right)^2}{1+x^2}\right)}{x}\:dx+12\int _0^1\frac{\ln ^2\left(1-x\right)\ln \left(1+x^2\right)}{x}\:dx$$
$$-6\int _0^1\frac{\ln \left(1-x\right)\ln ^2\left(1+x^2\right)}{x}\:dx+\int _0^1\frac{\ln ^3\left(1+x^2\right)}{x}\:dx-8\int _0^1\frac{\ln ^3\left(1-x\right)}{x}\:dx$$
But the first $3$ integrals are very tough, expanding some terms yield very complicated series.
I also attempted to integrate by parts but it was rather messy and I'd prefer not to put it here. I have faith that there must be a trick for finding this one because of its closed forms simplicity.
Note $2$: I found this integral in a certain mathematics group, I've ask for hints but received none.
 [/INST] This is not an answer.
I had a look at
$$\int _0^1\frac{\arctan ^2\left(x\right)\log \left(\frac{\left(1-x\right)^2}{1+x^2}\right)}{x}\,dx$$ and used
$$\log \left(\frac{\left(1-x\right)^2}{1+x^2}\right)=-4\sum_{n=1}^\infty \frac{  \sin ^2\left(\frac{n\pi  }{4}\right)}{n}\,x^n$$ I have not been able to compute
$$I_n=\int_0^1 x^{n-1}\,\arctan ^2\left(x\right)\,dx$$ for general $n$ but they look interesting
$$\left(
\begin{array}{cc}
n & I_n \\
 1 & -C+\frac{\pi ^2}{16}+\frac{1}{4} \pi  \log (2) \\
 2 & -\frac{\pi }{4}+\frac{\pi ^2}{16}+\frac{\log (2)}{2} \\
 3 & \frac{1}{3}+\frac{C}{3}-\frac{\pi }{6}+\frac{\pi ^2}{48}-\frac{1}{12} \pi  \log
   (2) \\
 4 & \frac{1}{12}+\frac{\pi }{12}-\frac{\log (2)}{3} \\
 5 & -\frac{4}{15}-\frac{C}{5}+\frac{\pi }{10}+\frac{\pi ^2}{80}+\frac{1}{20} \pi 
   \log (2) \\
 6 & -\frac{13}{180}-\frac{13 \pi }{180}+\frac{\pi ^2}{48}+\frac{23 \log (2)}{90} \\
 7 & \frac{73}{315}+\frac{C}{7}-\frac{2 \pi }{21}+\frac{\pi ^2}{112}-\frac{1}{28} \pi
    \log (2) \\
 8 & \frac{29}{420}+\frac{19 \pi }{420}-\frac{22 \log (2)}{105}
\end{array}
\right)$$ Hard to find a pattern.
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