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?
[Math] Integral $\int_0^1\frac{\log(x)\log(1+x)}{\sqrt{1-x}}\,dx$
calculusdefinite integralshypergeometric functionintegrationlogarithms
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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$\ds{I \equiv \int_{0}^{\infty}x^{2}\expo{-x^{2}}{\rm erf}\pars{x}\ln\pars{x} \,\dd x}$. Let's $\ds{{\cal I}\pars{\mu} \equiv \int_{0}^{\infty}x^{\mu}\expo{-x^{2}}{\rm erf}\pars{x}\,\dd x}$ such that $\ds{I = \lim_{\mu \to 2}\totald{{\cal I}\pars{\mu}}{\mu}}$
Since $\ds{{\rm erf}\pars{x} \stackrel{{\rm def.}}{=}{2 \over \root{\pi}}\int_{0}^{x}\expo{-y^{2}}\,\dd y}$: \begin{align} {\cal I}\pars{\mu}&=\int_{0}^{\infty}x^{\mu}\expo{-x^{2}} {2 \over \root{\pi}}\int_{0}^{\infty}\Theta\pars{x - y}\expo{-y^{2}}\,\dd y\,\dd x \\[3mm]&= {2 \over \root{\pi}}\int_{0}^{\pi/2}\dd\theta\,\cos^{\mu}\pars{\theta} \Theta\pars{\cos\pars{\theta} - \sin\pars{\theta}} \overbrace{\int_{0}^{\infty}\dd r\,r^{\mu + 1}\expo{-r^{2}}} ^{\Gamma\pars{1 + \mu/2}/2} \\[3mm]&={1 \over \root{\pi}}\,\Gamma\pars{1 + {\mu \over 2}} \int_{0}^{\pi/4}\dd\theta\,\cos^{\mu}\pars{\theta} \end{align}
\begin{align} I&=\lim_{\mu \to 2}\totald{{\cal I}\pars{\mu}}{\mu}= {1 \over 2\root{\pi}}\,\overbrace{\Psi\pars{2}}^{\ds{1 - \gamma}}\ \overbrace{\int_{0}^{\pi/4}\dd\theta\,\cos^{2}\pars{\theta}}^{\ds{\pars{\pi + 2}/8}} \\[3mm]&+ {1 \over \root{\pi}}\,\overbrace{\Gamma\pars{2}}^{\ds{1}} \overbrace{% \int_{0}^{\pi/4}\dd\theta\,\cos^{2}\pars{\theta}\ln\pars{\cos\pars{\theta}}} ^{\ds{\braces{4G + \pi - 2\bracks{1 + \ln\pars{2}} - \pi\ln\pars{4}}/16}} \end{align} $\Gamma\pars{z}$ and $\Psi\pars{z}$ are the $\it Gamma$ and $\it Digamma$ functions, respectively. $\gamma$ and $G$ are the $\it Euler-Mascheroni$ and Catalan constants, respectively.
$$ \begin{array}{|l|}\hline \mbox{}\\ \quad{\displaystyle\int_{0}^{\infty}x^{2}\expo{-x^{2}} \,\mathrm{erf}\pars{x}\ln\pars{x}\,\dd x}\quad \\[2mm] = \quad{{\displaystyle\quad\pars{\pi + 2}\pars{1 - \gamma} + 4G + \pi - 2\bracks{1 + \ln\pars{2}} - \pi\ln\pars{4}\quad} \over {\displaystyle 16\root{\pi}}}\quad \\ \mbox{}\\ \hline \end{array} \approx 0.0436462 $$
Edited for a more concise derivation:
Define $\mathcal{I}$ to be the value of the definite integral,
$$\mathcal{I}:=\int_{0}^{1}\frac{\ln^{2}{\left(x\right)}}{\sqrt{x^{2}-x+1}}\,\mathrm{d}x.\tag{1}$$
The definite integral $\mathcal{I}$ is found to have as an approximate numerical value
$$\mathcal{I}\approx2.10029.\tag{2}$$
We begin by transforming the integral via an Euler substitution of the first kind:
$$\sqrt{x^{2}-x+1}=x+t\implies x=\frac{1-t^{2}}{1+2t},\tag{3}$$
$$\implies\mathrm{d}x=\frac{\left(-2\right)\left(1+t+t^{2}\right)}{\left(1+2t\right)^{2}}\,\mathrm{d}t,$$
$$\implies\sqrt{x^{2}-x+1}=\frac{1-t^{2}}{1+2t}+t=\frac{1+t+t^{2}}{1+2t}.$$
Under the transformation $(3)$, the integral $\mathcal{I}$ becomes
$$\begin{align} \mathcal{I} &=\int_{0}^{1}\frac{\ln^{2}{\left(x\right)}}{\sqrt{x^{2}-x+1}}\,\mathrm{d}x\\ &=2\int_{0}^{1}\frac{\ln^{2}{\left(\frac{1-t^{2}}{1+2t}\right)}}{1+2t}\,\mathrm{d}t;~~~\small{\left[\sqrt{x^{2}-x+1}=x+t\right]}\\ &=2\int_{0}^{1}\frac{\left[\ln{\left(1-t\right)}+\ln{\left(1+t\right)}-\ln{\left(1+2t\right)}\right]^{2}}{1+2t}\,\mathrm{d}t.\tag{4}\\ \end{align}$$
Next, using the algebraic identity
$$\left(a+b-c\right)^{2}=a^{2}+2b^{2}-\left(a-b\right)^{2}+\left(a-c\right)^{2}-2bc,$$
with $a=\ln{\left(1-t\right)}\land b=\ln{\left(1+t\right)}\land c=\ln{\left(1+2t\right)}$, we may expand the integral $\mathcal{I}$ as a sum of five simpler logarithmic integrals:
$$\begin{align} \mathcal{I} &=2\int_{0}^{1}\frac{\left[\ln{\left(1-t\right)}+\ln{\left(1+t\right)}-\ln{\left(1+2t\right)}\right]^{2}}{1+2t}\,\mathrm{d}t\\ &=2\int_{0}^{1}\frac{\ln^{2}{\left(1-t\right)}}{1+2t}\,\mathrm{d}t+4\int_{0}^{1}\frac{\ln^{2}{\left(1+t\right)}}{1+2t}\,\mathrm{d}t\\ &~~~~~-2\int_{0}^{1}\frac{\left[\ln{\left(1-t\right)}-\ln{\left(1+t\right)}\right]^{2}}{1+2t}\,\mathrm{d}t\\ &~~~~~+2\int_{0}^{1}\frac{\left[\ln{\left(1-t\right)}-\ln{\left(1+2t\right)}\right]^{2}}{1+2t}\,\mathrm{d}t\\ &~~~~~-4\int_{0}^{1}\frac{\ln{\left(1+t\right)}\ln{\left(1+2t\right)}}{1+2t}\,\mathrm{d}t\\ &=2\int_{0}^{1}\frac{\ln^{2}{\left(1-t\right)}}{1+2t}\,\mathrm{d}t+4\int_{0}^{1}\frac{\ln^{2}{\left(1+t\right)}}{1+2t}\,\mathrm{d}t\\ &~~~~~-2\int_{0}^{1}\frac{\ln^{2}{\left(\frac{1-t}{1+t}\right)}}{1+2t}\,\mathrm{d}t+2\int_{0}^{1}\frac{\ln^{2}{\left(\frac{1-t}{1+2t}\right)}}{1+2t}\,\mathrm{d}t\\ &~~~~~-4\int_{0}^{1}\frac{\ln{\left(1+t\right)}\ln{\left(1+2t\right)}}{1+2t}\,\mathrm{d}t.\tag{5}\\ \end{align}$$
Now, we'll find it convenient to introduce the following two auxiliary functions:
$$J{\left(z\right)}:=\int_{0}^{1}\frac{\ln^{2}{\left(y\right)}}{1+zy}\,\mathrm{d}y;~~~\small{z\ge-1},\tag{6a}$$
and
$$H{\left(a,c\right)}:=\int_{0}^{1}\frac{\ln^{2}{\left(1+ay\right)}}{1+cy}\,\mathrm{d}y;~~~\small{a\ge-1\land c>-1}.\tag{6b}$$
As we shall see, the integral $\mathcal{I}$ can be expressed entirely in terms of the auxiliary functions $J$ and $H$, and hence, entirely in terms of elementary functions and the standard polylogarithms:
$$\begin{align} \mathcal{I} &=2\int_{0}^{1}\frac{\ln^{2}{\left(1-t\right)}}{1+2t}\,\mathrm{d}t+4\int_{0}^{1}\frac{\ln^{2}{\left(1+t\right)}}{1+2t}\,\mathrm{d}t\\ &~~~~~-2\int_{0}^{1}\frac{\ln^{2}{\left(\frac{1-t}{1+t}\right)}}{1+2t}\,\mathrm{d}t+2\int_{0}^{1}\frac{\ln^{2}{\left(\frac{1-t}{1+2t}\right)}}{1+2t}\,\mathrm{d}t\\ &~~~~~-\int_{0}^{1}\frac{4\ln{\left(1+t\right)}\ln{\left(1+2t\right)}}{1+2t}\,\mathrm{d}t\\ &=2\int_{0}^{1}\frac{\ln^{2}{\left(u\right)}}{3-2u}\,\mathrm{d}u;~~~\small{\left[1-t=u\right]}\\ &~~~~~+4\,H{\left(1,2\right)}\\ &~~~~~-\int_{0}^{1}\frac{4\ln^{2}{\left(v\right)}}{\left(3-v\right)\left(1+v\right)}\,\mathrm{d}v;~~~\small{\left[\frac{1-t}{1+t}=v\right]}\\ &~~~~~+2\int_{0}^{1}\frac{\ln^{2}{\left(w\right)}}{1+2w}\,\mathrm{d}w;~~~\small{\left[\frac{1-t}{1+2t}=w\right]}\\ &~~~~~\small{-\left(\left[\ln{\left(1+t\right)}\ln^{2}{\left(1+2t\right)}\right]_{0}^{1}-\int_{0}^{1}\frac{\ln^{2}{\left(1+2t\right)}}{1+t}\,\mathrm{d}t\right)};~~~\small{I.B.P.s}\\ &=\frac23\int_{0}^{1}\frac{\ln^{2}{\left(u\right)}}{1-\frac23u}\,\mathrm{d}u+4\,H{\left(1,2\right)}\\ &~~~~~-\frac13\int_{0}^{1}\frac{\ln^{2}{\left(v\right)}}{1-\frac13v}\,\mathrm{d}v-\int_{0}^{1}\frac{\ln^{2}{\left(v\right)}}{1+v}\,\mathrm{d}v+2\,J{\left(2\right)}\\ &~~~~~-\ln{(2)}\ln^{2}{(3)}+H{\left(2,1\right)}\\ &=\frac23\,J{\left(-\frac23\right)}+4\,H{\left(1,2\right)}-\frac13\,J{\left(-\frac13\right)}\\ &~~~~~-J{\left(1\right)}+2\,J{\left(2\right)}-\ln{(2)}\ln^{2}{(3)}+H{\left(2,1\right)}.\tag{7}\\ \end{align}$$
The function $H$ can be expressed in terms of $J$, dilogarithms, and elementary functions. Define $\gamma:=\frac{a-c}{c}$. For $0<a\land0<c$, we have $-1<\gamma$ and
$$\begin{align} H{\left(a,c\right)} &=\int_{0}^{1}\frac{\ln^{2}{\left(1+ay\right)}}{1+cy}\,\mathrm{d}y\\ &=\int_{\frac{1}{1+a}}^{1}\frac{\ln^{2}{\left(x\right)}}{x\left[c+\left(a-c\right)x\right]}\,\mathrm{d}x;~~~\small{\left[\frac{1}{1+ay}=x\right]}\\ &=\frac{1}{c}\int_{\frac{1}{1+a}}^{1}\frac{\ln^{2}{\left(x\right)}}{x}\,\mathrm{d}x-\frac{a-c}{c}\int_{\frac{1}{1+a}}^{1}\frac{\ln^{2}{\left(x\right)}}{c+\left(a-c\right)x}\,\mathrm{d}x\\ &=\frac{1}{3c}\ln^{3}{\left(1+a\right)}-\frac{a-c}{c^{2}}\int_{\frac{1}{1+a}}^{1}\frac{\ln^{2}{\left(x\right)}}{1+\left(\frac{a-c}{c}\right)x}\,\mathrm{d}x\\ &=\frac{1}{3c}\ln^{3}{\left(1+a\right)}-\frac{\gamma}{c}\int_{\frac{1}{1+a}}^{1}\frac{\ln^{2}{\left(x\right)}}{1+\gamma x}\,\mathrm{d}x\\ &=\frac{1}{3c}\ln^{3}{\left(1+a\right)}-\frac{\gamma}{c}\int_{0}^{1}\frac{\ln^{2}{\left(x\right)}}{1+\gamma x}\,\mathrm{d}x\\ &~~~~~+\frac{\gamma}{c}\int_{0}^{\frac{1}{1+a}}\frac{\ln^{2}{\left(x\right)}}{1+\gamma x}\,\mathrm{d}x\\ &=\frac{1}{3c}\ln^{3}{\left(1+a\right)}-\frac{\gamma}{c}J{\left(\gamma\right)}\\ &~~~~~+\frac{\gamma}{c\left(1+a\right)}\int_{0}^{1}\frac{\ln^{2}{\left(\frac{w}{1+a}\right)}}{1+\left(\frac{\gamma}{1+a}\right)w}\,\mathrm{d}w;~~~\small{\left[\left(1+a\right)x=w\right]}\\ &=\frac{1}{3c}\ln^{3}{\left(1+a\right)}-\left(\frac{a-c}{c^{2}}\right)J{\left(\frac{a-c}{c}\right)}\\ &~~~~~\small{+\frac{\gamma}{c\left(1+a\right)}\int_{0}^{1}\frac{\ln^{2}{\left(w\right)}-2\ln{\left(w\right)}\ln{\left(1+a\right)}+\ln^{2}{\left(1+a\right)}}{1+\left(\frac{\gamma}{1+a}\right)w}\,\mathrm{d}w}\\ &=\frac{1}{3c}\ln^{3}{\left(1+a\right)}-\left(\frac{a-c}{c^{2}}\right)J{\left(\frac{a-c}{c}\right)}\\ &~~~~~+\frac{\gamma}{c\left(1+a\right)}\int_{0}^{1}\frac{\ln^{2}{\left(w\right)}}{1+\left(\frac{\gamma}{1+a}\right)w}\,\mathrm{d}w\\ &~~~~~-\frac{2}{c}\ln{\left(1+a\right)}\int_{0}^{1}\frac{\left(\frac{\gamma}{1+a}\right)\ln{\left(w\right)}}{1+\left(\frac{\gamma}{1+a}\right)w}\,\mathrm{d}w\\ &~~~~~+\frac{1}{c}\ln^{2}{\left(1+a\right)}\int_{0}^{1}\frac{\left(\frac{\gamma}{1+a}\right)}{1+\left(\frac{\gamma}{1+a}\right)w}\,\mathrm{d}w\\ &=\frac{1}{3c}\ln^{3}{\left(1+a\right)}-\left(\frac{a-c}{c^{2}}\right)J{\left(\frac{a-c}{c}\right)}\\ &~~~~~+\frac{\gamma}{c\left(1+a\right)}\,J{\left(\frac{\gamma}{1+a}\right)}\\ &~~~~~+\frac{2}{c}\ln{\left(1+a\right)}\int_{0}^{1}\frac{\ln{\left(1+\left(\frac{\gamma}{1+a}\right)w\right)}}{w}\,\mathrm{d}w\\ &~~~~~+\frac{1}{c}\ln^{2}{\left(1+a\right)}\ln{\left(1+\left(\frac{\gamma}{1+a}\right)\right)}\\ &=\frac{1}{3c}\ln^{3}{\left(1+a\right)}-\left(\frac{a-c}{c^{2}}\right)J{\left(\frac{a-c}{c}\right)}\\ &~~~~~+\frac{a-c}{c^{2}\left(1+a\right)}\,J{\left(\frac{a-c}{c\left(1+a\right)}\right)}\\ &~~~~~-\frac{2}{c}\ln{\left(1+a\right)}\operatorname{Li}_{2}{\left(-\left(\frac{\gamma}{1+a}\right)\right)}\\ &~~~~~+\frac{1}{c}\ln^{2}{\left(1+a\right)}\ln{\left(\frac{a\left(1+c\right)}{\left(1+a\right)c}\right)}\\ &=\frac{a-c}{c^{2}\left(1+a\right)}\,J{\left(\frac{a-c}{c\left(1+a\right)}\right)}-\frac{a-c}{c^{2}}\,J{\left(\frac{a-c}{c}\right)}\\ &~~~~~-\frac{2}{c}\ln{\left(1+a\right)}\operatorname{Li}_{2}{\left(\frac{c-a}{c\left(1+a\right)}\right)}\\ &~~~~~+\frac{1}{3c}\ln^{3}{\left(1+a\right)}+\frac{1}{c}\ln^{2}{\left(1+a\right)}\ln{\left(\frac{a\left(1+c\right)}{\left(1+a\right)c}\right)}.\tag{8}\\ \end{align}$$
The function $J$ reduces to the trilogarithm. For $-1\le z\land z\neq0$,
$$\begin{align} J{\left(z\right)} &=\int_{0}^{1}\frac{\ln^{2}{\left(y\right)}}{1+zy}\,\mathrm{d}y\\ &=-\frac{2}{z}\operatorname{Li}_{3}{\left(-z\right)}.\tag{9}\\ \end{align}$$
Thus, continuing from where we left off at the last line of $(7)$,
$$\begin{align} \mathcal{I} &=\frac23\,J{\left(-\frac23\right)}-\frac13\,J{\left(-\frac13\right)}-J{\left(1\right)}+2\,J{\left(2\right)}\\ &~~~~~+4\,H{\left(1,2\right)}+H{\left(2,1\right)}-\ln{(2)}\ln^{2}{(3)}\\ &=2\operatorname{Li}_{3}{\left(\frac23\right)}-2\operatorname{Li}_{3}{\left(\frac13\right)}+2\operatorname{Li}_{3}{\left(-1\right)}-2\operatorname{Li}_{3}{\left(-2\right)}\\ &~~~~~\small{+4\left[-\operatorname{Li}_{3}{\left(\frac14\right)}+\operatorname{Li}_{3}{\left(\frac12\right)}-\ln{(2)}\operatorname{Li}_{2}{\left(\frac14\right)}-\frac56\ln^{3}{(2)}+\frac12\ln^{2}{(2)}\ln{(3)}\right]}\\ &~~~~~\small{+\left[2\operatorname{Li}_{3}{\left(-1\right)}-2\operatorname{Li}_{3}{\left(-\frac13\right)}-\ln{(9)}\operatorname{Li}_{2}{\left(-\frac13\right)}-\frac23\ln^{3}{(3)}+\ln{(4)}\ln^{2}{(3)}\right]}\\ &~~~~~-\ln{(2)}\ln^{2}{(3)}\\ &=2\operatorname{Li}_{3}{\left(\frac23\right)}-2\operatorname{Li}_{3}{\left(\frac13\right)}-2\operatorname{Li}_{3}{\left(-\frac13\right)}-2\operatorname{Li}_{3}{\left(-2\right)}\\ &~~~~~-4\operatorname{Li}_{3}{\left(\frac14\right)}+4\operatorname{Li}_{3}{\left(\frac12\right)}+4\operatorname{Li}_{3}{\left(-1\right)}\\ &~~~~~-4\ln{(2)}\operatorname{Li}_{2}{\left(\frac14\right)}-2\ln{(3)}\operatorname{Li}_{2}{\left(-\frac13\right)}\\ &~~~~~-\frac{10}{3}\ln^{3}{(2)}+2\ln^{2}{(2)}\ln{(3)}+\ln{(2)}\ln^{2}{(3)}-\frac23\ln^{3}{(3)}.\blacksquare\\ \end{align}$$
For $z\notin[1,\infty)$,
$$\small{\operatorname{Li}_{3}{\left(z\right)}=-\operatorname{Li}_{3}{\left(\frac{z}{z-1}\right)}-\operatorname{Li}_{3}{\left(1-z\right)}+\frac{\ln^3{\left(1-z\right)}}{6}-\frac{\ln{\left(z\right)}\ln^2{\left(1-z\right)}}{2}+\frac{\pi^2}{6}\ln{\left(1-z\right)}+\zeta{(3)}}.$$
For $0<z<1$,
$$\small{\operatorname{Li}_{3}{\left(z\right)}+\operatorname{Li}_{3}{\left(1-z\right)}+\operatorname{Li}_{3}{\left(\frac{z}{z-1}\right)}=\frac{\ln^3{\left(1-z\right)}}{6}-\frac{\ln{\left(z\right)}\ln^2{\left(1-z\right)}}{2}+\frac{\pi^2}{6}\ln{\left(1-z\right)}+\zeta{(3)}}.$$
Setting $z=\frac13$,
$$\small{\operatorname{Li}_{3}{\left(\frac13\right)}+\operatorname{Li}_{3}{\left(\frac23\right)}+\operatorname{Li}_{3}{\left(-\frac12\right)}=\frac{\ln^3{\left(\frac23\right)}}{6}-\frac{\ln{\left(\frac13\right)}\ln^2{\left(\frac23\right)}}{2}+\zeta{(2)}\ln{\left(\frac23\right)}+\zeta{(3)}}.$$
$$\small{\operatorname{Li}_{3}{\left(\frac13\right)}+\operatorname{Li}_{3}{\left(\frac23\right)}+\operatorname{Li}_{3}{\left(-\frac12\right)}=\frac{\ln^{3}{(2)}-3\ln{(2)}\ln^{2}{(3)}+2\ln^{3}{(3)}}{6}+\zeta{(2)}\ln{\left(\frac23\right)}+\zeta{(3)}}$$
$$\small{\operatorname{Li}_{3}{\left(\frac13\right)}+\operatorname{Li}_{3}{\left(\frac23\right)}+\operatorname{Li}_{3}{\left(-\frac12\right)}=\frac{\ln^{2}{\left(\frac32\right)}\ln{\left(18\right)}}{6}-\zeta{(2)}\ln{\left(\frac32\right)}+\zeta{(3)}}$$
Best Answer
$$\int_0^1\frac{\ln(1+x)\ln x}{\sqrt{1-x}}dx=16-8\ln2+4\ln^2\left(1+\sqrt2\right)\\+\sqrt2\left[2\ln^22+8\left(\ln2-1\right)\ln\left(1+\sqrt2\right)-\frac{7\pi^2}3+16\operatorname{Li}_2\!\left(\frac1{\sqrt{2}}\right)\right].$$