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?
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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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.
A: $$\int_0^1\frac{\ln(1+x)\ln x}{\sqrt{1-x}}dx=16-8\ln2+4\ln^2\left(1+\sqrt2\right)\\+\sqrt2\left[2\ln^22+8\left(\ln2-1\right)\ln\left(1+\sqrt2\right)-\frac{7\pi^2}3+16\operatorname{Li}_2\!\left(\frac1{\sqrt{2}}\right)\right].$$
A: The integral may readily be decomposed into a sum of integrals of products of log-linear terms:
$$\begin{align}
\mathcal{I}
&=\int_{0}^{1}\frac{\ln{\left(x\right)}\ln{\left(1+x\right)}}{\sqrt{1-x}}\,\mathrm{d}x\\
&=2\int_{0}^{1}\ln{\left(1-y^2\right)}\ln{\left(2-y^2\right)}\,\mathrm{d}y;~~~\small{\left[\sqrt{1-x}=y\right]}\\
&=2\int_{0}^{1}\ln{\left(1-y\right)}\ln{\left(\sqrt{2}-y\right)}\,\mathrm{d}y\\
&~~~~~+2\int_{0}^{1}\ln{\left(1-y\right)}\ln{\left(\sqrt{2}+y\right)}\,\mathrm{d}y\\
&~~~~~+2\int_{0}^{1}\ln{\left(1+y\right)}\ln{\left(\sqrt{2}-y\right)}\,\mathrm{d}y\\
&~~~~~+2\int_{0}^{1}\ln{\left(1+y\right)}\ln{\left(\sqrt{2}+y\right)}\,\mathrm{d}y.\\
\end{align}$$
Each of these four integrals can be resolved in terms of dilogarithms in a systematic manner, for instance by using the general closed forms for two integrals I derive below. Since a final result has already been provided in another response, I leave the plugging-and-chugging step as an exercise to the fearless reader.

Suppose $0<a\land0<a+b$. Then we find:
$$\begin{align}
J{(a,b)}
&=\int_{0}^{1}\ln{\left(1-y\right)}\ln{\left(a+by\right)}\,\mathrm{d}y\\
&=\int_{0}^{1}\ln{\left(w\right)}\ln{\left(a+b-bw\right)}\,\mathrm{d}w;~~~\small{\left[1-y=w\right]}\\
&=\small{-\int_{0}^{1}\frac{1}{w}\left[\frac{\left(a+b\right)\ln{\left(a+b\right)}}{b}-\frac{bw+\left(a+b-bw\right)\ln{\left(a+b-bw\right)}}{b}\right]\,\mathrm{d}w}\\
&=\int_{0}^{1}\frac{bw-bw\ln{\left(a+b\right)}+\left(a+b-bw\right)\ln{\left(1-\frac{b}{a+b}w\right)}}{bw}\,\mathrm{d}w\\
&=1-\ln{\left(a+b\right)}+\int_{0}^{1}\frac{\left(a+b-bw\right)\ln{\left(1-\frac{b}{a+b}w\right)}}{bw}\,\mathrm{d}w\\
&=1-\ln{\left(a+b\right)}+\int_{0}^{1}\frac{\left(1-cw\right)\ln{\left(1-cw\right)}}{cw}\,\mathrm{d}w;~~~\small{\left[c:=\frac{b}{a+b}\right]}\\
&=\small{1-\ln{\left(a+b\right)}-\int_{0}^{1}\ln{\left(1-cw\right)}\,\mathrm{d}w+\frac{1}{c}\int_{0}^{1}\frac{\ln{\left(1-cw\right)}}{w}\,\mathrm{d}w}\\
&=1-\ln{\left(a+b\right)}+1+\frac{1-c}{c}\ln{\left(1-c\right)}-\frac{1}{c}\,\operatorname{Li}_{2}{\left(c\right)}\\
&=2-\ln{\left(a+b\right)}+\frac{1-c}{c}\ln{\left(1-c\right)}-\frac{1}{c}\,\operatorname{Li}_{2}{\left(c\right)}\\
&=2+\frac{a\ln{\left(a\right)}-\left(a+b\right)\ln{\left(a+b\right)}-\left(a+b\right)\operatorname{Li}_{2}{\left(\frac{b}{a+b}\right)}}{b}\\
\end{align}$$

Suppose $0<a\land0<a+b\land0<a-b$. Then we find:
$$\begin{align}
K{(a,b)}
&=\int_{0}^{1}\ln{\left(1+y\right)}\ln{\left(a+by\right)}\,\mathrm{d}y\\
&=\small{\ln{(2)}\ln{\left(a+b\right)}-\int_{0}^{1}y\left[\frac{b\ln{\left(1+y\right)}}{a+by}+\frac{\ln{\left(a+by\right)}}{1+y}\right]\,\mathrm{d}y}\\
&=\ln{(2)}\ln{\left(a+b\right)}\\
&~~~~~\small{-\int_{0}^{1}\left[\ln{\left(1+y\right)}-\frac{a\ln{\left(1+y\right)}}{a+by}+\ln{\left(a+by\right)}-\frac{\ln{\left(a+by\right)}}{1+y}\right]\,\mathrm{d}y}\\
&=\small{\ln{(2)}\ln{\left(a+b\right)}-\left(2\ln{(2)}-1\right)+\frac{b+a\ln{\left(a\right)}-\left(a+b\right)\ln{\left(a+b\right)}}{b}}\\
&~~~~~\small{+\int_{0}^{1}\frac{a\ln{\left(1+y\right)}}{a+by}\,\mathrm{d}y+\int_{0}^{1}\frac{\ln{\left(a+by\right)}}{1+y}\,\mathrm{d}y}\\
&=\small{2-2\ln{(2)}+\ln{(2)}\ln{\left(a+b\right)}+\frac{a\ln{\left(a\right)}-\left(a+b\right)\ln{\left(a+b\right)}}{b}}\\
&~~~~~\small{+\int_{0}^{1}\frac{a\ln{\left(1+y\right)}}{a+by}\,\mathrm{d}y+\ln{(2)}\ln{\left(a+b\right)}-\int_{0}^{1}\frac{b\ln{\left(1+y\right)}}{a+by}\,\mathrm{d}y}\\
&=\small{2-2\ln{(2)}+2\ln{(2)}\ln{\left(a+b\right)}+\frac{a\ln{\left(a\right)}-\left(a+b\right)\ln{\left(a+b\right)}}{b}}\\
&~~~~~\small{+\left(a-b\right)\int_{0}^{1}\frac{\ln{\left(1+y\right)}}{a+by}\,\mathrm{d}y}\\
&=\small{2-2\ln{(2)}+2\ln{(2)}\ln{\left(a+b\right)}+\frac{a\ln{\left(a\right)}-\left(a+b\right)\ln{\left(a+b\right)}}{b}}\\
&~~~~~\small{+\left(a-b\right)\int_{1}^{2}\frac{\ln{\left(w\right)}}{a-b+bw}\,\mathrm{d}w};~~~\small{\left[1+y=w\right]}\\
&=\small{2-2\ln{(2)}+2\ln{(2)}\ln{\left(a+b\right)}+\frac{a\ln{\left(a\right)}-\left(a+b\right)\ln{\left(a+b\right)}}{b}}\\
&~~~~~\small{+\int_{0}^{2}\frac{\ln{\left(w\right)}}{1+\frac{b}{a-b}w}\,\mathrm{d}w-\int_{0}^{1}\frac{\ln{\left(w\right)}}{1+\frac{b}{a-b}w}\,\mathrm{d}w}\\
&=\small{2-2\ln{(2)}+2\ln{(2)}\ln{\left(a+b\right)}+\frac{a\ln{\left(a\right)}-\left(a+b\right)\ln{\left(a+b\right)}}{b}}\\
&~~~~~\small{+2\int_{0}^{1}\frac{\ln{\left(2u\right)}}{1+\frac{2b}{a-b}u}\,\mathrm{d}u-\frac{a-b}{b}\operatorname{Li}_{2}{\left(-\frac{b}{a-b}\right)}};~~~\small{\left[w=2u\right]}\\
&=\small{2-2\ln{(2)}+2\ln{(2)}\ln{\left(a+b\right)}+\frac{a\ln{\left(a\right)}-\left(a+b\right)\ln{\left(a+b\right)}}{b}}\\
&~~~~~+2\ln{(2)}\int_{0}^{1}\frac{\mathrm{d}u}{1+\frac{2b}{a-b}u}+2\int_{0}^{1}\frac{\ln{\left(u\right)}}{1+\frac{2b}{a-b}u}\,\mathrm{d}u\\
&~~~~~-\frac{a-b}{b}\operatorname{Li}_{2}{\left(-\frac{b}{a-b}\right)}\\
&=2-2\ln{(2)}+2\ln{(2)}\ln{\left(a+b\right)}\\
&~~~~~+\frac{a\ln{\left(a\right)}-\left(a+b\right)\ln{\left(a+b\right)}+\left(a-b\right)\ln{(2)}\ln{\left(\frac{a+b}{a-b}\right)}}{b}\\
&~~~~~+\frac{a-b}{b}\left[\operatorname{Li}_{2}{\left(-\frac{2b}{a-b}\right)}-\operatorname{Li}_{2}{\left(-\frac{b}{a-b}\right)}\right].\\
\end{align}$$
