A closed form for $\int_0^1\frac{\ln(-\ln x)\ \operatorname{li}^2x}{x}dx$

I will start from Vladimir's result and my calculation. Let

$$ F(s) = \int_{0}^{\infty} z^{s-1} \mathrm{Ei}(-z)^{2} \, dz = \frac{2\Gamma(s)}{s} \int_{0}^{\frac{1}{2}} \frac{u^{s-1}}{1 - u} \, du. $$

Then we have

\begin{align*} \int_{0}^{\infty} z^{s-1} \mathrm{Ei}(-z)^{2} \log z \, dz &= F'(s) \\ &= F(s) \left( \psi_{0}(s) - \frac{1}{s} \right) + \frac{2\Gamma(s)}{s} \int_{0}^{\frac{1}{2}} \frac{u^{s-1}\log u}{1-u} \, du. \end{align*}

Plugging $s = 1$ to both sides,

\begin{align*} I &= F(1) \left( \psi_{0}(1) - 1 \right) + 2 \int_{0}^{\frac{1}{2}} \frac{\log u}{1-u} \, du \\ &= -2(\gamma+1)\log 2 + 2 \left[ \mathrm{Li}_{2}(1-u) \right]_{0}^{\frac{1}{2}} \\ &= -\zeta(2) - \log^{2} 2 - 2(\gamma + 1) \log 2 \end{align*}

as desired. Here,

\begin{align*} \mathrm{Li}_{2}(z) = - \int_{0}^{z} \frac{\log(1 - t)}{t} \, dt \end{align*}

is the dilogarithm and we exploited the special values

\begin{align*} \mathrm{Li}_{2}(1) &= \zeta(2), \\ \mathrm{Li}_{2}\left(\tfrac{1}{2}\right) &= \frac{1}{2}\zeta(2) - \frac{1}{2}\log^{2} 2. \end{align*}

Change the integration variable $z=\ln x$, then apply the formula $(12)$. The integral becomes $$I=\int_{-\infty}^0\ln(-z)\,\operatorname{Ei}^2z\ dz,\tag1$$ where $\operatorname{Ei}z$ is the exponential integral: $$\operatorname{Ei}z=-\int_{-z}^\infty\frac{e^{-t}}t dt.\tag2$$

In this form the integral can be evaluated by Mathematica: $$I=-\frac{\pi ^2}{6}-\ln^2 2-2\,(1+\gamma)\ln2,\tag3$$ where $\gamma$ is the Euler-Mascheroni Constant.