Conjecture $\int_0^1\ln\ln\left(\frac{1+x}{1-x}\right)\frac{\ln x}{1-x^2}\,dx\stackrel?=\frac{\pi^2}{24}\,\ln\left(\frac{A^{36}}{16\,\pi^3}\right)$

I did some numeric experiments with integrals involving double logarithms (because they received much interest both on this site and in published papers, sometimes under names of Malmsten—Vardi—Adamchik integrals).

It appears that $${\large\int}_0^1\ln\ln\left(\frac{1+x}{1-x}\right)\cdot\frac{\ln x}{1-x^2}\,dx\stackrel{\color{gray}?}=\frac{\pi^2}{24}\,\ln\left(\frac{A^{36}}{16\,\pi^3}\right),$$ where $A=\exp\left(\frac1{12}-\zeta'(-1)\right)$ is the Glaisher—Kinkelin constant (I have more than $1000$ decimal digits confirming this conjecture). How can we prove it?

Using the substitution suggested by @AccidentalFourierTransform , and the integral $\displaystyle \,\,\, \int \frac{1}{\sinh x} dx=\ln\tanh\left(\frac{x}{2}\right)+C$, we have

$$\begin{align} -2\int_0^1 \ln \ln \left(\frac{1+x}{1-x}\right) \frac{\ln x}{1-x^2}dx \\&= -\int_0^{\infty}\ln x \,\,\ln \tanh\left(\frac{x}{2}\right)dx \\&=\int_0^{\infty}\ln x \int_x^{\infty}\frac1{\sinh t} dt dx \\&=\int_0^{\infty} \int_0^{t}\ln x \frac1{\sinh t} dx dt \\&=\int_0^{\infty} \frac{x\ln x-x}{\sinh x} dx \\&=\int_0^{\infty} (x\ln x -x) \frac{2 e^{-x}}{1-e^{-2x}}\,dx \\&=\int_0^{\infty} (x\ln x-x)2\sum_{n=0}^{\infty} e^{-x(2n+1)}dx \\&=2\sum_{n=0}^{\infty} \left(\frac{d}{ds} \frac{\Gamma(s)}{(2n+1)^s} \Bigg{|}_{s=2}-\frac1{(2n+1)^2}\right) \\&=2\sum_{n=0}^{\infty} \left(\frac{\Gamma'(2)-1}{(2n+1)^2}-\frac{\ln(2n+1)}{(2n+1)^2}\right) \\&=2(\Gamma'(2)-1)\frac{\pi^2}{8}-2\left(\sum_{n=1}^{\infty}\frac{\ln(n)}{n^2}-\sum_{n=1}^{\infty} \frac{\ln(2n)}{(2n)^2} \right) \\&=2(\Gamma'(2)-1)\frac{\pi^2}{8}+2\zeta'(2)+\frac12\left(\ln2 \frac{\pi^2}{6}-\zeta'(2)\right) \\&=-\frac{\pi^2}{12}\ln\left(\frac{A^{36}}{16\pi^3}\right) \end{align}$$

Hint. One may observe that, by the change of variable $u=\dfrac{1-x}{1+x}$ one gets $$ I={\int}_0^1\log\log\left(\frac{1+x}{1-x}\right)\cdot\frac{\log x}{1-x^2}\,dx=\int_0^1\log\left(-\log u\right)\cdot\frac{\log (1-u)-\log (1+u)}{2u}\,du. \tag1 $$ Then by a standard Taylor expansion, we have $$ \frac{\log (1-u)-\log (1+u)}{2u}= -\sum_{n=0}^{\infty} \frac{u^{2n}}{2n+1}, \qquad |u|<1,\tag2 $$ giving $$ I=-\sum_{n=0}^{\infty} \frac{1}{2n+1}\int_0^1u^{2n}\log\left(-\log u\right)\:du.\tag3 $$ The latter integral is easily obtained using the well-known integral representation of the Euler gamma function $$ \frac{\Gamma(s)}{(a+1)^s}=\int_0^\infty t^{s-1} e^{-(a+1)t}\:dt. \tag4 $$ By differentiating $(4)$ with respect to $s$ and putting $s=1$ we produce $$ \int_0^1u^a\log\left(-\log u\right)\:du=-\frac{\gamma+\log(a+1)}{a+1} \tag5 $$ leading to $$ I=\sum_{n=0}^{\infty} \frac{\gamma+\log(2n+1)}{(2n+1)^2}=\left.\left(\gamma-\frac{d}{ds} \right)\left(\left(1-2^{-s}\right)\zeta(s)\right)\right|_{s=2}=\frac{\pi^2}{24}\,\ln\left(\frac{A^{36}}{16\,\pi^3}\right)\tag6 $$ as announced.

These integrals have been studied by Adamchik, Vardi, Moll and many others. One may have a look at this interesting paper.