Integral $\int_0^1\frac{\log(x)\log(1+x)}{\sqrt{1-x}}\,dx$ I'm trying to evaluate this definite integral:
$$\int_0^1\frac{\log(x) \log(1+x)}{\sqrt{1-x}} dx$$
It's clear that the result can be expressed in terms of derivatives of a hypergeometric function with respect to its parameters. I obtained the following form:
$$4 \left(1 - \log 2\right){_2F_1}^{(0,1,0,0)}\left(1, 0; \tfrac{3}{2}; -1\right) - 2 {_2F_1}^{(1,1,0,0)}\left(1, 0; \tfrac{3}{2}; -1\right) - 2{_2F_1}^{(0,1,1,0)}\left(1, 0; \tfrac{3}{2}; -1\right)$$
Is it possible to expand these derivatives to some explicit form and further simplify this result? Or maybe you could suggest a different way to evaluate this integral that gives a simpler result without going through hypergeometric functions?

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\begin{align}
&\bbox[10px,#ffd]{\ds{\int_{0}^{1}
{\ln\pars{x}\ln\pars{1 + x} \over \root{1 - x}}\,\dd x}}
\,\,\,\stackrel{x\ =\ 1 - t^{2}}{=}\,\,\,
\int_{-1}^{1}\ln\pars{1 - t^{2}}\ln\pars{2 - t^{2}}\,\dd t
\\[5mm] = &\
2\int_{-1}^{1}\ln\pars{1 - t}\ln\pars{\root{2} - t}\,\dd t +
2\int_{-1}^{1}\ln\pars{1 - t}\ln\pars{\root{2} + t}\,\dd t
\\[1cm] = &\
\bracks{\int_{-1}^{1}\ln^{2}\pars{1 - t}\,\dd t +
\int_{-1}^{1}\ln^{2}\pars{\root{2} - t}\,\dd t -
\int_{-1}^{1}\ln^{2}\pars{1 - t \over \root{2} - t}\,\dd t}
\\[5mm] + &\
\bracks{\int_{-1}^{1}\ln^{2}\pars{1 - t}\,\dd t +
\int_{-1}^{1}\ln^{2}\pars{\root{2} + t}\,\dd t -
\int_{-1}^{1}\ln^{2}\pars{1 - t \over \root{2} + t}\,\dd t}
\\[1cm] = &\
2\int_{-1}^{1}\ln^{2}\pars{1 - t}\,\dd t +
2\int_{-1}^{1}\ln^{2}\pars{\root{2} - t}\,\dd t
\\[5mm] - &\
\int_{-1}^{1}\ln^{2}\pars{1 - t \over \root{2} - t}\,\dd t -
\int_{-1}^{1}\ln^{2}\pars{1 - t \over \root{2} + t}\,\dd t
\label{1}\tag{1}
\end{align}
The first and the second integral, in \eqref{1}, are quite trivial:
\begin{align}
\int_{-1}^{1}\ln^{2}\pars{1 - t}\,\dd t & = 4 - 4\ln\pars{2} + 2\ln^{2}\pars{2}
\\[5mm]
\int_{-1}^{1}\ln^{2}\pars{\root{2} - t}\,\dd t & =
4 - 4\root{2}\ln\pars{1 + \root{2}} + 2\ln^{2}\pars{1 + \root{2}}
\end{align}
With the change of variables
$\ds{{1 - t \over \root{2} \pm t} = x}$, the third and fourth integral, in \eqref{1}, can be rewritten as
\begin{align}
\int_{-1}^{1}\ln^{2}\pars{1 - t \over \root{2} - t}\,\dd t & = 
\pars{\root{2} - 1}\int_{0}^{2\root{2} - 2}
{\ln^{2}\pars{x} \over \pars{1 - x}^{2}}\,\dd x
\\[5mm]
\int_{-1}^{1}\ln^{2}\pars{1 - t \over \root{2} + t}\,\dd t & = 
\pars{\root{2} + 1}\int_{0}^{2\root{2} + 2}
{\ln^{2}\pars{x} \over \pars{1 + x}^{2}}\,\dd x
\end{align}
Both integrals are straightforward evaluated by successive integration by parts.
