A closed form of $\int_0^1\log (- \log x)\log \left(\frac{1+x}{1-x}\right)\,dx$ Is it possible to obtain a closed form of the following integral?
$$\int_0^1\log (- \log x)\log \left(\frac{1+x}{1-x}\right)\,dx$$ I've made the change of variable $$t=\frac{1+x}{1-x} $$ but I feel like I'm turning in circles...
 A: We have the following closed form.

Proposition. $$
\int_0^1\log (- \log x)\log \left(\frac{1+x}{1-x}\right)\,dx=\gamma_1-2\ln^2 2-2\gamma \ln 2 -\gamma_1\Big({1,\small\frac12}\Big)\tag{$\star$}
$$ 

where $\gamma_1$ is the Stieltjes constant,
$$\gamma_1 = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log n}n-\int_1^N\frac{\log t}t\:dt\right)$$
 and where $\gamma_1(a,b)$ is the poly-Stieltjes constant, 
$$\gamma_1(a,b) = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log (n+a)}{n+b}-\int_1^N\frac{\log t}t\:dt\right)\!.$$
Proof. 
One may recall the classic integral representation of the Euler gamma function
$$
\frac{\Gamma(s)}{(a+1)^s}=\int_0^\infty t^{s-1} e^{-(a+1)t}\:dt, \qquad s>0,\, a>-1. \tag1
$$ By differentiating $(1)$ with respect to $s$, putting $s=1$ and making the change of variable $x=e^{-t}$, we get
$$
\int_0^1x^a\log\left(-\log x\right)\:dx=-\frac{\gamma+\log(a+1)}{a+1},\qquad a>-1, \tag2
$$ 
where $\displaystyle \gamma=\lim_{N\to+\infty}\left(\sum_{n=1}^N \frac1n-\int_1^N\frac{dt}t\right)$ is the Euler-Mascheroni constant.
From the standard Taylor series expansion,
$$
-\log (1-x)= \sum_{n=1}^{\infty} \frac{x^n}n, \qquad |x|<1,\tag3
$$ one gets
$$
\log (1+x)-\log (1-x)=2 \sum_{n=0}^{\infty} \frac{x^{2n+1}}{2n+1}, \qquad |x|<1.\tag4
$$ One may write the given integral as
$$
\int_0^1\log (- \log x)\log \left(\frac{1+x}{1-x}\right)\,dx
=\int_0^1\log (- \log x) \left(\log (1+x)-\log (1-x)\right)dx
$$
then, inserting $(4)$ into the latter integrand and using $(2)$, we obtain
$$
\begin{align}
\int_0^1\log (- \log x)\log \left(\frac{1+x}{1-x}\right)\,dx&=2\int_0^1\log (- \log x) \sum_{n=0}^{\infty} \frac{x^{2n+1}}{2n+1}\:dx\\
&=2\sum_{n=0}^{\infty} \frac1{2n+1}\int_0^1 x^{2n+1}\log (- \log x)\:dx\\
&=-2\sum_{n=0}^{\infty} \frac{\gamma+\log(2n+2)}{(2n+1)(2n+2)}\\
&=-\sum_{n=0}^{\infty} \frac{2\left(\gamma+\ln 2 \right)}{(2n+1)(2n+2)}-\sum_{n=1}^{\infty} \frac{2\log (n+1)}{(2n+1)(2n+2)}.\tag5
\end{align}
$$ 
On the one hand, using Abel's theorem and using $(3)$, one has
$$
\begin{align}
\sum_{n=0}^{\infty} \frac2{(2n+1)(2n+2)}&=\lim_{x \to 1^-}\sum_{n=0}^{\infty} \frac{2x^{2n+2}}{(2n+1)(2n+2)}\\
&=\lim_{x \to 1^-}\left( (1+x)\log(1+x)+(1-x)\log(1-x)\right)\\
&=2\ln2.\tag6
\end{align}
$$
On the other hand, using Theorem 2 here one has
$$
\begin{align}
\sum_{n=1}^{\infty} \frac{2\log (n+1)}{(2n+1)(2n+2)}=-\gamma_1+\gamma_1\Big({1,\small\frac12}\Big),\tag7
\end{align}
$$ since $\gamma_1(1,1)=\gamma_1$.
Finally, bringing all the steps together gives $(\star)$.
