TITLE: Evaluation of $\int_0^1 \frac{\log(1+x)}{1+x}\log\left(\log\left(\frac{1}{x}\right)\right) \ dx$
QUESTION [28 upvotes]: I need some hints, clues for getting the closed form of 
$$\int_0^1 \frac{\log(1+x)}{1+x}\log\left(\log\left(\frac{1}{x}\right)\right) \ dx$$

REPLY [6 votes]: Substitute $x\mapsto e^{-x}$ and then $x\mapsto x/n$:
$$
\begin{align}\hspace{-1cm}
\int_0^1\frac{\log(1+x)}{1+x}\log(\log(1/x))\,\mathrm{d}x
&=\int_0^\infty\frac{\log(1+e^{-x})}{1+e^{-x}}\log(x)\,e^{-x}\,\mathrm{d}x\\
&=\sum_{n=2}^\infty\int_0^\infty(-1)^nH_{n-1}e^{-nx}\log(x)\,\mathrm{d}x\\
&=\sum_{n=2}^\infty\int_0^\infty(-1)^n\frac{H_{n-1}}ne^{-x}(\log(x)-\log(n))\,\mathrm{d}x\\
&=\sum_{n=2}^\infty(-1)^n\frac{H_{n-1}}n(-\gamma-\log(n))\tag{1}
\end{align}
$$
Using $(3)$ from this answer, we get
$$
\begin{align}
\sum_{n=2}^\infty(-1)^n\frac{H_{n-1}}n
&=\frac{\zeta(2)}{2}-\sum_{n=1}^\infty(-1)^{n-1}\frac{H_n}n\\
&=\frac12\log(2)^2\tag{2}
\end{align}
$$
Therefore, we get that $(1)$ is
$$
-\frac\gamma2\log(2)^2-\sum_{n=2}^\infty(-1)^n\frac{H_{n-1}}n\log(n)\tag{3}
$$

Comparison with Felix Marin's Answer
Using $(8)$ from this answer, we get
$$
\begin{align}
\sum_{n=2}^\infty(-1)^n\frac{H_{n-1}}n\log(n)
&=\sum_{n=2}^\infty(-1)^n\frac{\psi(n)+\gamma}n\log(n)\\
&=\sum_{n=2}^\infty(-1)^n\frac{\psi(n)}n\log(n)+\gamma^2\log(2)-\gamma\frac{\log(2)^2}{2}\tag{4}
\end{align}
$$
Thus, plugging $(4)$ into $(3)$, we see that $(1)$ is
$$
-\gamma^2\log(2)-\sum_{n=2}^\infty(-1)^n\frac{\psi(n)}n\log(n)\tag{5}
$$
which is the same as Felix Marin gets.

Further Investigation
Perhaps the next step is to consider the derivative of
$$
\sum_{n=2}^\infty(-1)^n\frac{H_{n-1}}{n^x}\tag{6}
$$