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?

Yes, there is a closed form: $$\frac{\pi^2}3-\ln^22-4\,G,$$ where $G$ is the Catalan constant: $$G=-\int_0^1\frac{\ln x}{x^2+1}dx.$$

Is known that

\begin{cases} \int\limits_0^1 \dfrac {y\,\mathrm dy}{(1+y^2)^2} = \dfrac14\\ \int\limits_0^1 \ln y \dfrac {y\,\mathrm dy}{(1+y^2)^2} = -\dfrac{\ln2}4\\ \int\limits_0^1 \ln (1-y^2) \dfrac {y\,\mathrm dy}{(1+y^2)^2} = -\dfrac{\ln2}4\\ \int\limits_0^1 \ln (1+y) \dfrac {y\,\mathrm dy}{(1+y^2)^2} = \dfrac{\pi-2\ln2}{16}. \tag1\end{cases}

Substitution $$x=\dfrac{4y^2}{(1-y^2)^2},\quad \sqrt{x+1\mathstrut} = \dfrac{1+y^2}{1-y^2},\quad \sqrt{x^{-1}+1\mathstrut} = \dfrac{1+y^2}{2y},\quad y\in(0,1)$$

allows to write \begin{align} &I=\int_0^\infty\frac{\ln\left(\sqrt{x+1\large\mathstrut}-1\right)\,\ln\left(\sqrt{x^{-1}+1\large\mathstrut}+1\right)}{(x+1){\Large\!^{^{\normalsize\,^3/_2}}}}\mathrm dx\\[8pt] &=\int_0^1\ln\dfrac{2y^2}{1-y^2}\,\ln\dfrac{(1+y)^2}{2y}\,\dfrac{(1-y^2)^3}{(1+y^2)^3}\ \dfrac{8y(1-y^2)+16y^3}{(1-y^2)^3}\,\mathrm dy\\[8pt] &=8\int_0^1\left(\ln2+\ln\dfrac{y^2}{1-y^2}\right)\,\left(\ln2+\ln\dfrac{(1+y)^2}{4y}\right)\,\dfrac{y\ \mathrm dy}{(1+y^2)^2}\\ &=8\ln2\int_0^1\left(-\ln2+\ln y+2\ln(1+y)-\ln(1-y^2)\Large\mathstrut\right)\,\dfrac{y\ \mathrm dy}{(1+y^2)^2}\\ &+8\int_0^1\ln\dfrac{y^2}{1-y^2}\,\ln\dfrac{(1+y)^2}{4y}\,\dfrac{y\ \mathrm dy}{(1+y^2)^2}\\ &=(-2\ln2-2\ln2 +\pi-2\ln2+2\ln2)\ln2+8\int_0^1\ln\dfrac{y^2}{1-y^2}\,\ln\dfrac{(1+y)^2}{4y}\,\dfrac{y\ \mathrm dy}{(1+y^2)^2}, \end{align} $$I=(\pi-4\ln2)\ln2+I_1,\quad\text{where}\quad I_1=8\int_0^1\ln\dfrac{y^2}{1-y^2}\,\ln\dfrac{(1+y)^2}{4y}\,\dfrac{y\ \mathrm dy}{(1+y^2)^2}.\tag2$$

By parts: \begin{align} &I_1=8\int_0^1\,\ln\dfrac{(1+y)^2}{4y}\,\mathrm d\int_0^y\ln\dfrac{y^2}{1-y^2}\,\dfrac{y\ \mathrm dy}{(1+y^2)^2}\\[8pt] &=8\left.\left(\ln\dfrac{(1+y)^2}{4y}\,\int_0^y\ln\dfrac{y^2}{1-y^2}\,\dfrac{y\ \mathrm dy}{(1+y^2)^2}\right)\right|_{\,0}^{\,1}\\[8pt] &-8\int_0^1\,\left(\ln\dfrac{(1+y)^2}{4y}\right)'\left(\int_0^y\ln\dfrac{y^2}{1-y^2}\,\dfrac{y\mathrm dy}{(1+y^2)^2}\right)\mathrm dy\\[8pt] &= 0 - 2\int_0^1\,\left(\dfrac2{1+y}-\dfrac1y\right)\left(\dfrac{4y^2}{1+y^2}\ln y +\dfrac{1-y^2}{1+y^2}\ln(1-y^2) - \ln(1+y^2)\right)\,\mathrm dy\\[8pt] &=2\int_0^1\left(\dfrac{1-y}{y(1+y)}\left(\dfrac{4y^2}{1+y^2}\ln y +\dfrac{1-y^2}{1+y^2}\ln(1-y^2)\right)-\left(\dfrac2{1+y}-\dfrac1y\right) \ln(1+y^2)\right)\mathrm dy\\[8pt] &= 8\int_0^1\,\dfrac{y(1-y)}{(1+y)(1+y^2)}\ln y\,\mathrm dy +2\int_0^1\,\dfrac{(1-y)^2}{y(1+y^2)}\ln(1-y)\,\mathrm dy\\[8pt] &+2\int_0^1\,\dfrac{(1-y)^2}{y(1+y^2)}\ln(1+y)\,\mathrm dy -2\int_0^1\,\left(\dfrac1y-\dfrac2{1+y}\right)\ln(1+y^2)\,\mathrm dy \end{align} (see also Wolfram Alpha undefined integral).

At last - calculating of the remaining integrals and final summation: \begin{align} &I=\pi\ln2-4\ln^22 + 8\left(\dfrac{\pi^2}{12}-G\right) +2\left(2G-\dfrac\pi{12}(2\pi+3\ln2)\right)\\[4pt] &+2\,\dfrac\pi{12}(\pi-3\ln2) -2\left(\dfrac{\pi^2}{24} -2\left(\dfrac34\ln^2 2 - \dfrac{\pi^2}{48}\right)\right),\\[8pt] &\color{brown}{\boxed{\ \mathbf{I=\dfrac{\pi^2}3 - \ln^2 2 -4G \approx -0.85444\,72569.}\ \large\mathstrut}}\\ \end{align} (see also Wolfram Alpha 1st integral, 2nd integral, 3rd integral, 4th integral, 5th integral)

Obtained result corresponds with the exact value.