A log arctan integral $\int_0^1 \log x \arctan^2 x \, dx$ Here is an integral that arose while solving another problem. Can we express
$$\mathcal{J}=\int_0^1 \log x \arctan^2 x \, {\rm d}x$$
in terms of known mathematical constants or special functions? I had a couple of ideas for this one. For example begin by parts , that is:
\begin{align*}
\int_{0}^{1} \log x \arctan^2 x\, {\rm d}x &= \left [ \left ( x \log x - x \right )\arctan^2 x \right ]_0^1 - \int_{0}^{1}\left ( x \log x - x \right )\arctan^2 x \, {\rm d}x \\ 
 &=-\frac{\pi^2}{16} - \int_{0}^{1}x \log x \arctan^2 x \, {\rm d}x + \int_{0}^{1}x \arctan^2 x \, {\rm d}x \\ 
 &= - \frac{\pi^2}{16} - \int_{0}^{1}x \log x \arctan^2 x \, {\rm d}x -\frac{\pi}{4}+\frac{\pi^2}{16} +\frac{\log 256}{16}
\end{align*}
since the RHS integral is trivial. We can even find an elementary antiderivative. The problem is with the second. An idea that pumped to me while writing down my thoughts is that probably the easiest way of computing is by trying to evaluate the integral
$$\int_{0}^{1} t^{s-1} \arctan^2 x \, {\rm d}x$$
Of course $s$ is imposed on restrictions that are yet unknown to me. Playing around I see that this method is fruitless since:
$$\int_{0}^{1}x^{s-1} \arctan^2 x \, {\rm d}x = \left [ \frac{x^s}{s} \arctan^2 x  \right ]_0^1  - \frac{2}{s}\int_{0}^{1} \frac{x^{s-1} \arctan x}{x^2+1} \, {\rm d}x$$
and if someone tries to apply parts again at the second integral beginning with the rational function then he encounters hypergeometrics. I don't know any of that. So, any help is welcome!
 A: We have
$$\frac{d}{dx}\arctan^2(x) = 2\cdot\frac{\arctan x}{1+x^2}\tag{1}$$
hence it follows that the Taylor series of $\arctan(x)^2$ is given by
$$\arctan^2(x)=2\sum_{n\geq 0}(-1)^n \frac{x^{2n+2}}{2n+2} \sum_{m=0}^{n}\frac{1}{2m+1}\tag{2} $$
and:
$$ \mathcal{J} = \color{red}{\sum_{n\geq 1}\frac{(-1)^{n}\left(2H_{2n}-H_{n}\right)}{(2n)(2n+1)^2}}.\tag{3}$$
Now Mathematica provides:
$$ \sum_{n\geq 1}\frac{H_n}{(2n)(2n+1)^2}=\frac{2\pi^2\log(2)+8\log^2(2)-14\,\zeta(3)}{8}\tag{4}$$
but the alternating sign in $(3)$ gives an extra level of complexity. Anyway,
$$ \frac{1}{(2n)(2n+1)^2}=\frac{1}{2n}-\frac{1}{2n+1}-\frac{1}{(2n+1)^2}\tag{5}$$
hence the crucial value to compute is $\color{red}{2\,S_6-S_3}$, where:

$$ S_1 = \sum_{n\geq 1}\frac{(-1)^n H_n}{2n}=\frac{\log^2(2)-\zeta(2)}{4},\qquad S_2 = \sum_{n\geq 1}\frac{(-1)^n H_n}{2n+1}=K-\frac{\pi}{2}\log 2$$
  $$ \color{red}{S_3} = \sum_{n\geq 1}\frac{(-1)^n H_n}{(2n+1)^2}=\,\color{red}{???}\qquad  S_4=\sum_{n\geq 1}\frac{(-1)^n H_{2n}}{2n}=\frac{\log^2(2)}{8}-\frac{5\pi^2}{96} $$
  $$ S_5=\sum_{n\geq 1}\frac{(-1)^n H_{2n}}{2n+1}=-\frac{\pi}{8}\log(2),\qquad \color{red}{S_6}=\sum_{n\geq 1}\frac{(-1)^n H_{2n}}{(2n+1)^2}=\,\color{red}{???} $$

We have:
$$ S_3 = \int_{0}^{1}\frac{\log(x)\log(1+x^2)}{1+x^2}\,dx $$
that Mathematica is able to convert into a complicated expression involving $\text{Im}\,\text{Li}_3\left(\frac{1+i}{2}\right)$ (besides $\pi^3,K\log(2)$ and $\pi\log^2(2)$) but I am stil clueless about $S_6$.
