TITLE: Not the toughest integral, not the easiest one
QUESTION [6 upvotes]: Perhaps it's not amongst the toughest integrals, but it's interesting  to try to find an elegant
approach for the integral 
$$I_1=\int_0^1 \frac{\log (x)}{\sqrt{x (x+1)}} \, dx$$
$$=4 \text{Li}_2\left(-\sqrt{2}\right)-4 \text{Li}_2\left(-1-\sqrt{2}\right)+2 \log ^2\left(1+\sqrt{2}\right)-4 \log \left(2+\sqrt{2}\right) \log \left(1+\sqrt{2}\right)-\frac{\pi ^2}{3}$$
Do you see any such a way? Then I wonder if we can think of some elegant ways for the evaluation of the quadratic and cubic versions, that is 
$$I_2=\int_0^1 \frac{\log^2 (x)}{\sqrt{x (x+1)}} \, dx$$
$$I_3=\int_0^1 \frac{\log^3 (x)}{\sqrt{x (x+1)}} \, dx.$$
How far can we possibly go with the generalization such that we can get integrals in closed form?

REPLY [6 votes]: As an alternative, we may apply the substitution $x\mapsto\sinh^2{x}$.
\begin{align}
\int^1_0\frac{\ln{x}}{\sqrt{x(x+1)}}{\rm d}x
&=4\int^{-\ln(\sqrt{2}-1)}_0\ln(\sinh{x})\ {\rm d}x\\
&=4\int^{-\ln(\sqrt{2}-1)}_0\left(\color{green}{\ln\left(1-e^{-2x}\right)}-\ln{2}+x\right)\ {\rm d}x\\
&=-4\sum^\infty_{n=1}\frac{1}{n}\int^{-\ln(\sqrt{2}-1)}_0e^{-2nx}\ {\rm d}x+4\ln{2}\ln\left(\sqrt{2}-1\right)+2\ln^2\left(\sqrt{2}-1\right)\\
&=2\sum^\infty_{n=1}\frac{\left(3-2\sqrt{2}\right)^n-1}{n^2}+4\ln{2}\ln\left(\sqrt{2}-1\right)+2\ln^2\left(\sqrt{2}-1\right)\\
&=\color{red}{2{\rm Li}_2\left(3-2\sqrt{2}\right)-\frac{\pi^2}{3}+4\ln{2}\ln\left(\sqrt{2}-1\right)+2\ln^2\left(\sqrt{2}-1\right)}
\end{align}

Similarly, for the second integral,
\begin{align}
\int^1_0\frac{\ln^2{x}}{\sqrt{x(x+1)}}\ {\rm d}x
&=8\int^{-\ln(\sqrt{2}-1)}_0\ln^2(\sinh{x})\ {\rm d}x=8(A+B+C)
\end{align}
where
\begin{align}
A
&=\int^{-\ln(\sqrt{2}-1)}_0\left(x^2-2x\ln{2}-2\ln{2}\color{green}{\ln\left(1-e^{-2x}\right)}+\ln^2{2}\right)\ {\rm d}x\\
&=-{\rm Li}_2\left(3-2\sqrt{2}\right)\ln{2}+\frac{\pi^2}{6}\ln{2}-\ln^2{2}\ln\left(\sqrt{2}-1\right)-\ln{2}\ln^2\left(\sqrt{2}-1\right)-\frac{1}{3}\ln^3\left(\sqrt{2}-1\right)
\end{align}
and
\begin{align}
B
&=\int^{-\ln(\sqrt{2}-1)}_02x\ln\left(1-e^{-2x}\right)\ {\rm d}x\\
&=-2\sum^\infty_{n=1}\frac{1}{n}\int^{-\ln(\sqrt{2}-1)}_0xe^{-2nx}\ {\rm d}x\\
&=2\sum^\infty_{n=1}\frac{1}{n}\left[\frac{xe^{-2nx}}{2n}+\frac{e^{-2nx}}{4n^2}\right]^{-\ln(\sqrt{2}-1)}_0\\
&=-\ln\left(\sqrt{2}-1\right)\sum^\infty_{n=1}\frac{\left(3-2\sqrt{2}\right)^n}{n^2}+\frac{1}{2}\sum^\infty_{n=1}\frac{\left(3-2\sqrt{2}\right)^n}{n^3}-\frac{1}{2}\sum^\infty_{n=1}\frac{1}{n^3}\\
&=\frac{1}{2}{\rm Li}_3\left(3-2\sqrt{2}\right)-\frac{\zeta(3)}{2}-{\rm Li}_2\left(3-2\sqrt{2}\right)\ln\left(\sqrt{2}-1\right)
\end{align}
and
\begin{align}
C
&=\int^{-\ln(\sqrt{2}-1)}_0\ln^2\left(1-e^{-2x}\right)\ {\rm d}x=2\sum^\infty_{n=1}\frac{H_{n-1}}{n}\int^{-\ln(\sqrt{2}-1)}_0e^{-2nx}\ {\rm d}x\\
&=-\sum^\infty_{n=1}\frac{\left(H_n-\frac{1}{n}\right)\left(\left(3-2\sqrt{2}\right)^n-1\right)}{n^2}
\end{align}
Using the well-known generating function
$$\sum^\infty_{n=1}\frac{H_n}{n^2}z^n={\rm Li}_3(z)-{\rm Li}_3(1-z)+{\rm Li}_2(1-z)\ln(1-z)+\frac{1}{2}\ln{z}\ln^2(1-z)+\zeta(3)$$
we obtain
\begin{align}
C
&={\rm Li}_3\left(2\sqrt{2}-2\right)-{\rm Li}_2\left(2\sqrt{2}-2\right)\ln\left(2\sqrt{2}-2\right)-\ln\left(\sqrt{2}-1\right)\ln^2{2}-2\ln{2}\ln^2\left(\sqrt{2}-1\right)\\
&\ \ \ \ -\ln^3\left(\sqrt{2}-1\right)
\end{align}
Therefore,
\begin{align}
\int^1_0\frac{\ln^2{x}}{\sqrt{x(x+1)}}\ {\rm d}x
&=\color{blue}{4{\rm Li}_3\left(3-2\sqrt{2}\right)+8{\rm Li}_3\left(2\sqrt{2}-2\right)-4\zeta(3)-8{\rm Li}_2\left(3-2\sqrt{2}\right)\ln\left(2\sqrt{2}-2\right)}\\
&\ \ \ \ \color{blue}{-8{\rm Li}_2\left(2\sqrt{2}-2\right)\ln\left(2\sqrt{2}-2\right)+\frac{4\pi^2}{3}\ln{2}-24\ln{2}\ln^2\left(\sqrt{2}-1\right)}\\
&\ \ \ \ \color{blue}{-16\ln^2{2}\ln\left(\sqrt{2}-1\right)-\frac{32}{3}\ln^3\left(\sqrt{2}-1\right)}
\end{align}

Since the generating function of $\dfrac{H_n}{n^3}$ is known as well, I believe that one can work out the closed form of $\displaystyle\int^1_0\frac{\ln^3{x}}{\sqrt{x(x+1)}}\ {\rm d}x$ without too much difficulty. Nevertheless, the process will be immensely tedious.