Proving that $\int_0^1 \frac{\arctan x}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx=\frac{\pi^3}{16}$ The following integral was proposed by Cornel Ioan Valean and appeared as Problem $12054$ in the American Mathematical Monthly earlier this year.

Prove $$\int_0^1 \frac{\arctan x}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx=\frac{\pi^3}{16}$$

I had small tries for it, such as writting:
$$I=\int_0^1 \frac{\arctan x}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx\overset{ x\to \tan \frac{x}{2}}=-\frac12 {\int_0^\frac{\pi}{2}\frac{x\ln(1-\sin x)}{\sin x} dx}$$
And with Feynman's trick we obtain:
$$J(t)=\int_0^\frac{\pi}{2} \frac{x\ln(1-t\sin x)}{\sin x}dx\Rightarrow J'(t)=\int_0^\frac{\pi}{2} \frac{x}{1-t\sin x}dx$$ But I don't see a way to obtain a closed from for the above integral.

Also from here we have the following relation:
$$\int_0^1 \frac{\arctan x \ln(1+x^2)}{x} dx =\frac23 \int_0^1 \frac{\arctan x \ln(1+x)}{x}dx$$
Thus we can rewrite the integral as:
$$I=\frac23 \int_0^1 \frac{\arctan x \ln(1+x)}{x}dx -2\int_0^1 \frac{\arctan x \ln(1-x)}{x}dx$$

Another option might be to rewrite:
$$\ln\left(\frac{1+x^2}{(1-x)^2}\right)= \ln\left(\frac{1+x}{1-x}\right)+\ln\left(\frac{1+x^2}{1-x^2}\right)$$
$$\Rightarrow I= \int_0^1 \frac{\arctan x}{x}\ln\left(\frac{1+x}{1-x}\right)dx+\int_0^1 \frac{\arctan x}{x}\ln\left(\frac{1+x^2}{1-x^2}\right)dx$$
And now to use the power expansion of the log functions to obtain:
$$\small I=\sum_{n=0}^\infty \frac{2}{2n+1}\int_0^1 \frac{\arctan x}{x} \, \left(x^{2n+1}+x^{4n+2}\right)dx=\sum_{n=0}^\infty \frac{2}{2n+1}\int_0^1\int_0^1 \frac{\left(x^{2n+1}+x^{4n+2}\right)}{1+y^2x^2}dydx$$

This seems like an awesome integral and I would like to learn more so I am searching for more approaches.
Would any of you who also already solved it and submitted the answer to the AMM or know how to solve this integral kindly share the solution here?
Edit:
In the meantime I found a nice solution by Roberto Tauraso here and another impressive approach due to Yaghoub Sharifi here.
 A: By the identity
$$
\Im~{\log^2\left(\frac{1-x}{1+\text i x}\right)}=\arctan x\log\left(\frac{1+x^2}{(1-x)^2}\right)
$$
one easily has
$$
\begin{align}
&\int_0^1\frac{\arctan x}x\log\left(\frac{1+x^2}{(1-x)^2}\right)\text d x
\\=&\Im{\int_0^1\log^2\left(\frac{1-x}{1+\text ix}\right)\frac{\text d x}{x}}
\\=&\Im{\int_0^1\log^2y\left(\frac{1+y}{(1-y) (1+y^2)}+\text i~\frac{1}{1+y^2}\right)\text d y}
\\=&\int_0^1\frac{\log^2y}{1+y^2}\text d y
\\=&\frac{\pi^3}{16}
\end{align}
$$
with the obvious substitution $y=\dfrac{1-x}{1+\text ix}$. The last equality comes directly from series expansion.
A: Put
\begin{equation*}
I=\int_{0}^1\dfrac{\arctan x}{x}\ln\left(\dfrac{1+x^2}{(1-x)^2}\right)\, \mathrm{d}x.
\end{equation*}
Via the substitution $ x=\dfrac{z}{z+1}$ we get
\begin{equation*}
I = \int_{0}^{\infty}\dfrac{\arctan \frac{z}{z+1}\ln(2z^2+2z+1)}{z^2+z}\, \mathrm{d}z.
\end{equation*}
Put 
\begin{equation*}
\log z=\ln|z|+i\arg z, \quad -\pi<\arg z <\pi.
\end{equation*}
Then
\begin{equation*}
\arctan \frac{z}{z+1}\ln(2z^2+2z+1) = \text{Im}\left(\log^2(1+z+iz)\right).
\end{equation*}
Consequently
\begin{equation*}
I = \text{Im}\left(\int_{0}^{\infty}\dfrac{\log^2(1+z+iz)}{z^2+z}\right)\mathrm{d}z.
\end{equation*}
However, $ \log(z) $ is an analytic function in $ \text{Re} z>0 $. According to Cauchys integral theorem we get the same value if we integrate along the curve with the parametrization $ z=(1-i)s, s>0 $.
\begin{gather*}
I = \text{Im}\left(\int_{0}^{\infty}\dfrac{\ln^2(2s+1)}{s(s+1-is)}\, \mathrm{d}s\right) = \int_{0}^{\infty}\dfrac{\ln^2(2s+1)}{2s^2+2s+1}\, \mathrm{d}s = \\[2ex] \int_{0}^{\infty}\dfrac{2\ln^2(2s+1)}{(2s+1)^2+1}\, \mathrm{d}s = [t=2s+1] = \\[2ex] \int_{1}^{\infty}\dfrac{\ln^2(t)}{t^2+1}\, \mathrm{d}t =[u= 1/t] = \int_{0}^{1}\dfrac{\ln^2(u)}{u^2+1}\, \mathrm{d}u.
\end{gather*}
Thus
\begin{equation*}
2I = \int_{0}^{\infty}\dfrac{\ln^2(u)}{u^2+1}\, \mathrm{d}u
\end{equation*}
In order to evaluate this integral we integrate $ \dfrac{\log^3(z)}{z^2+1} $  along a keyhole contour and use residue calculus.
In this case $ \log z =\ln |z|+i\arg z, \quad 0<\arg z < 2\pi $.
We get
\begin{equation*}
I = \dfrac{\pi^3}{16}.
\end{equation*}
A: Starting with breaking the integral
$\displaystyle I=\int_0^1\frac{\arctan x}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)\ dx=\int_0^1\frac{\arctan x}{x}\ln(1+x^2)dx-2\int_0^1\frac{\arctan x}{x}\ln(1-x)dx$
then using the identity$\ \displaystyle\arctan x\ln(1+x^2)=-2\sum_{n=0}^{\infty}\frac{(-1)^n H_{2n}} {2n+1}x^{2n+1}$ for the first integral and series-expanding $\displaystyle\arctan x$ of the second integral, we get 
\begin{align*}
I&=-2\sum_{n=0}^{\infty}\frac{(-1)^n H_{2n}}{2n+1}\int_0^1x^{2n}\ dx-2\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\int_0^1x^{2n}\ln(1-x)\ dx\\
&=-2\sum_{n=0}^{\infty}\frac{(-1)^n H_{2n}}{(2n+1)^2}-2\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\left(-\frac{H_{2n+1}}{2n+1}\right)\\
&=-2\sum_{n=0}^{\infty}\frac{(-1)^n H_{2n}}{(2n+1)^2}-2\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\left(-\frac{H_{2n}}{2n+1}-\frac{1}{(2n+1)^2}\right)\\
&=2\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^3}=2\beta(3)=\frac{\pi^3}{16}
\end{align*}
where $\beta(3)=\frac{\pi^3}{32}$ is the Dirichlet beta function.
Note that we used the classical result  $\int_0^1 x^{n-1}\ln(1-x)dx=-\frac{H_n}{n}$ which can be proved as follows:
$$\int_0^1 x^{n-1}\ln(1-x)dx=-\sum_{k=1}^\infty\frac1k\int_0^1 x^{n+k-1}dx=-\sum_{k=1}^\infty\frac{1}{k(n+k)}\\=-\frac1n\sum_{k=1}^\infty\left(\frac1k-\frac1{n+k}\right)=-\frac1n\sum_{k=1}^n\frac1k=-\frac{H_n}{n}$$
A: We have $\log\left(\frac{1+x^2}{(1-x)^2}\right)=\log(1-x^4)-\log(1-x^2)-2\log(1-x)$, hence by integration by parts
$$ I_1 = \frac{3\pi^3}{32}-2\int_{0}^{1}\frac{1}{1+x^2}\sum_{n\geq 1}\frac{\chi(n) x^n}{n^2}\,dx\qquad \chi(n)=\left\{\begin{array}{rcl}1&\text{if}&n\equiv 1\pmod{2}\\ 2 & \text{if} & n\equiv 2\pmod{4}\\ 0 &\text{if}&n\equiv 0\pmod{4}\end{array}\right. $$
where
$$\int_{0}^{1}\frac{1}{1+x^2}\sum_{n\geq 1}\frac{\chi(n) x^n}{n^2}\,dx=\iint_{(0,1)^2}\frac{-\log(y)}{y(1+x^2)}\sum_{n\geq 1}\chi(n)(xy)^n\,dx\,dy=\iint_{(0,1)^2}\frac{-\log(y)}{y(1+x^2)}\cdot\frac{(1+xy)^2 xy}{1-(xy)^4}\,dx\,dy$$
equals
$$ \iint_{(0,1)^2}\frac{-\log(y)x(1+xy)}{(1+x^2)(1-x y)(1+x^2 y^2)}\,dx\,dy. $$
By performing a partial fraction decomposition this integral is reduced to four integrals in the $y$-variable, with two of them (namely $\int_{0}^{1}\frac{y\log y}{1-y^4}\,dy = -\frac{\pi^3}{32}$ and $\int_{0}^{1}\frac{\log(2)}{1+y^2}\,dy = \frac{\pi}{4}\log(2)$) being elementary and the remaining ones being
$$ J_1 = \int_{0}^{1}\frac{\arctan(y)\log(y)}{1-y^2}\,dy,\qquad J_2=\int_{0}^{1}\frac{\log(y)\log(1-y)}{1+y^2}\,dy. $$
$J_2$ can be tackled by performing the substitution $y=\tan\theta$ and exploiting the Fourier series of $\log \sin$ and $\log\cos$. By performing the substitution $y\mapsto\frac{1+y}{1-y}$ $J_1$ is reduced to
$$ \int_{0}^{1}\frac{\arctan(x)\operatorname{arctanh}(x)}{x}\,dx $$
then, by computing $\int_{0}^{1}\frac{x^{2k}}{2k+1}\operatorname{arctanh}(x)\,dx$, to the series (also appearing here)
$$ \sum_{k\geq 0}\frac{(-1)^k H_k}{(2k+1)^2}=\int_{0}^{1}\frac{\log(1+z^2)\log(z)}{1+z^2}\,dx. $$
Via $z\to\tan\theta$ we have that both $J_1$ and $J_2$ are reduced to the integrals $\int_{0}^{\pi/4}\log^2(\sin\theta)\,d\theta$ and $\int_{0}^{\pi/4}\log(\sin\theta)\log(\cos\theta)\,d\theta$, which are well-known and related to Euler sums with weight $3$. Luckily the contributions related to $\pi\log^2(2),\pi^2\log(2),K\log(2)$ and $\text{Im}\,\text{Li}_3\left(\frac{1+i}{2}\right)$ cancel out and only leave a rational multiple of $\pi^3$.
