Integral $\int_0^\infty\frac{\ln\left(\sqrt{x+1\vphantom{x^0}}-1\right)\,\ln\left(\sqrt{x^{-1}+1}+1\right)}{(x+1)^{3/2}}dx$ Another integral similar to my previous question:
$$\int_0^\infty\frac{\ln\left(\sqrt{x+1\vphantom{x^0}}-1\right)\,\ln\left(\sqrt{x^{-1}+1}+1\right)}{(x+1)^{3/2}}dx$$
Can someone suggest how to evaluate it? Is there a closed form?
 A: To evaluate
$$
\int^{\infty}_{0}\frac{\log\left(\sqrt{x+1}-1\right)\log\left(\sqrt{1/x+1}+1\right)}{(x+1)^{3/2}}dx.\tag 1
$$
I will evaluate first
$$
I(x)=\int\frac{\log\left(\sqrt{x+1}-1\right)\log\left(\sqrt{1/x+1}+1\right)}{(x+1)^{3/2}}dx\tag 2
$$
and then use limits to find the answer. The evaluation is based on Mathematica program (see Wolfram alpha...etc).
First write
$$
I(x)=-2\int\log\left(\sqrt{x+1}-1\right)\log\left(\sqrt{1/x+1}+1\right)\frac{d}{dx}\left(\frac{1}{\sqrt{x+1}}\right)dx.
$$
Then using integration by parts we find
$$
I(x)=-\frac{2}{\sqrt{x+1}}\log\left(\sqrt{x+1}-1\right)\log\left(\sqrt{1/x+1}+1\right)+
$$
$$
+\int\frac{\log\left(1+\sqrt{1+\frac{1}{x}}\right)}{(1+x)(-1+\sqrt{1+x})}dx+\int\frac{(\sqrt{x^2+x}-x-1)\log\left(-1+\sqrt{1+x}\right)}{x(1+x)^{3/2}}dx=
$$
$$
=-\frac{2}{\sqrt{x+1}}\log\left(\sqrt{x+1}-1\right)\log\left(\sqrt{1/x+1}+1\right)+
$$
$$
+\int\frac{\log\left(1+\sqrt{1+\frac{1}{x}}\right)}{(1+x)(-1+\sqrt{1+x})}dx
+\int\frac{\log\left(-1+\sqrt{1+x}\right)}{(1+x)\sqrt{x}}dx
-\int\frac{\log\left(-1+\sqrt{1+x}\right)}{x\sqrt{1+x}}dx\tag 3
$$
But all three integrals can evaluated with Mathematica and we write for $x>0$
$$
I_1(x)=\int\frac{\log\left(-1+\sqrt{1+x}\right)}{(1+x)\sqrt{x}}dx
-\int\frac{\log\left(-1+\sqrt{1+x}\right)}{x\sqrt{1+x}}dx=
2\pi i\cot^{(-1)}(\sqrt{x})-
$$
$$
-\frac{1}{2}\log\left(-1+\sqrt{1+x}\right)\left(\log 4+\log\left(-1+\sqrt{1+x}\right)-2\log\left(1+\sqrt{1+x}\right)\right)-
$$
$$
-4i\cdot\textrm{Li}\left(2,\frac{1-i\sqrt{x}}{\sqrt{1+x}}\right)+i\cdot\textrm{Li}\left(2,1-\frac{2}{1+i\sqrt{x}}\right)+\textrm{Li}\left(2,\frac{1}{2}\left(1-\sqrt{1+x}\right)\right)+C_1.\tag 4
$$
Also if $y=1/x$, then
$$
I_2(x)=\int\frac{\log\left(1+\sqrt{1+\frac{1}{x}}\right)}{(1+x)(-1+\sqrt{1+x})}dx=-\int\frac{\log\left(1+\sqrt{1+y}\right)}{y+1}dy-\int\frac{\log(1+\sqrt{1+y})}{\sqrt{y(y+1)}}dy=
$$
$$
=-4i\log( x)\arcsin\left(\sqrt{\frac{1}{2}\left(1+\sqrt{1+x^{-1}}\right)}\right)-
$$
$$
-4\arcsin\left(\frac{1}{2}\left(1+\sqrt{1+x^{-1}}\right)\right)^2+\log(x)\log\left(1+\sqrt{1+x^{-1}}\right)-
$$
$$
-2\pi i \log\left(1+\sqrt{1+x^{-1}}\right)-2\log\left(1+\sqrt{1+x^{-1}}\right)\log\left(1+\sqrt{1+x}\right)+
$$
$$
8i\arcsin\left(\sqrt{\frac{1}{2}\left(1+\sqrt{1+x^{-1}}\right)}\right)\log\left(-1+\sqrt{x}+\sqrt{1+x}\right)+
$$
$$
+2\textrm{Li}\left(2,-\sqrt{1+x^{-1}}\right)-4\textrm{Li}\left(2,-\frac{1}{\sqrt{x}}\left(-1+\sqrt{1+x}\right)\right)
$$
Hence
$$
I(x)=I_1(x)+I_2(x)
$$
and we can see easily using Mathematica that
$$
\lim_{x\rightarrow\infty}I(x)=-4G-\left(1-\frac{i}{4}\right)\pi^2+2\pi i \log 2-\frac{1}{2}\log^22
$$
Also
$$
\lim_{x\rightarrow0}I(x)=\left(-\frac{4}{3}+\frac{i}{4}\right)\pi^2+2\pi i\log 2+\frac{1}{2}\log^22
$$
Hence
$$
\int^{\infty}_{0}\frac{\log\left(\sqrt{x+1}-1\right)\log\left(\sqrt{1/x+1}+1\right)}{(x+1)^{3/2}}dx=-4G+\frac{\pi^2}{3}-\log^22
$$
QED
A: Yes, there is a closed form:
$$\frac{\pi^2}3-\ln^22-4\,G,$$
where $G$ is the Catalan constant:
$$G=-\int_0^1\frac{\ln x}{x^2+1}dx.$$
