Integral $\int_0^1\frac{\log(x)\log(1+x)}{\sqrt{1-x}}\,dx$
$$\int_0^1\frac{\ln(1+x)\ln x}{\sqrt{1-x}}dx=16-8\ln2+4\ln^2\left(1+\sqrt2\right)\\+\sqrt2\left[2\ln^22+8\left(\ln2-1\right)\ln\left(1+\sqrt2\right)-\frac{7\pi^2}3+16\operatorname{Li}_2\!\left(\frac1{\sqrt{2}}\right)\right].$$
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}$$
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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.