$$I:=\int_{0}^{\infty}\frac{\ln{(x)}}{\sqrt{x}\,\sqrt{x+1}\,\sqrt{2x+1}}\mathrm{d}x.$$
After first multiplying and dividing the integrand by 2, substitute $x=\frac{t}{2}$:
$$I=\int_{0}^{\infty}\frac{2\ln{(x)}}{\sqrt{2x}\,\sqrt{2x+2}\,\sqrt{2x+1}}\mathrm{d}x=\int_{0}^{\infty}\frac{\ln{\left(\frac{t}{2}\right)}}{\sqrt{t}\,\sqrt{t+2}\,\sqrt{t+1}}\mathrm{d}t.$$
Next, substituting $t=\frac{1}{u}$ yields:
$$\begin{align}
I
&=-\int_{0}^{\infty}\frac{\ln{(2u)}}{\sqrt{u}\sqrt{u+1}\sqrt{2u+1}}\mathrm{d}u\\
&=-\int_{0}^{\infty}\frac{\ln{(2)}}{\sqrt{u}\sqrt{u+1}\sqrt{2u+1}}\mathrm{d}u-\int_{0}^{\infty}\frac{\ln{(u)}}{\sqrt{u}\sqrt{u+1}\sqrt{2u+1}}\mathrm{d}u\\
&=-\int_{0}^{\infty}\frac{\ln{(2)}}{\sqrt{u}\sqrt{u+1}\sqrt{2u+1}}\mathrm{d}u-I\\
\implies I&=-\frac{\ln{(2)}}{2}\int_{0}^{\infty}\frac{\mathrm{d}x}{\sqrt{x}\sqrt{x+1}\sqrt{2x+1}}.
\end{align}$$
Making the sequence of substitutions $x=\frac{u-1}{2}$, then $u=\frac{1}{t}$, and finally $t=\sqrt{w}$, puts this integral into the form of a beta function:
$$\begin{align}
\int_{0}^{\infty}\frac{\mathrm{d}x}{\sqrt{x}\sqrt{x+1}\sqrt{2x+1}}
&=\int_{1}^{\infty}\frac{\mathrm{d}u}{\sqrt{u-1}\sqrt{u+1}\sqrt{u}}\\
&=\int_{1}^{\infty}\frac{\mathrm{d}u}{\sqrt{u^2-1}\sqrt{u}}\\
&=\int_{1}^{0}\frac{t^{3/2}}{\sqrt{1-t^2}}\frac{(-1)}{t^2}\mathrm{d}t\\
&=\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{t}\,\sqrt{1-t^2}}\\
&=\frac12\int_{0}^{1}\frac{\mathrm{d}w}{w^{3/4}\,\sqrt{1-w}}\\
&=\frac12\operatorname{B}{\left(\frac14,\frac12\right)}\\
&=\frac12\frac{\Gamma{\left(\frac12\right)}\Gamma{\left(\frac14\right)}}{\Gamma{\left(\frac34\right)}}\\
&=\frac{\pi^{3/2}}{2^{1/2}\Gamma^2{\left(\frac34\right)}}
\end{align}$$
Hence,
$$I=-\frac{\ln{(2)}}{2}\frac{\pi^{3/2}}{2^{1/2}\Gamma^2{\left(\frac34\right)}}=-\frac{\pi^{3/2}\,\ln{(2)}}{2^{3/2}\,\Gamma^2{\left(\frac34\right)}}.~~~\blacksquare$$
Possible Alternative: You could also derive the answer from the complete elliptic integral of the first kind instead of from the beta function by making the substitution $t=z^2$ instead of $t=\sqrt{w}$.
$$\begin{align}
\int_{0}^{\infty}\frac{\mathrm{d}x}{\sqrt{x}\sqrt{x+1}\sqrt{2x+1}}
&=\int_{1}^{\infty}\frac{\mathrm{d}u}{\sqrt{u-1}\sqrt{u+1}\sqrt{u}}\\
&=\int_{1}^{\infty}\frac{\mathrm{d}u}{\sqrt{u^2-1}\sqrt{u}}\\
&=\int_{1}^{0}\frac{t^{3/2}}{\sqrt{1-t^2}}\frac{(-1)}{t^2}\mathrm{d}t\\
&=\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{t}\,\sqrt{1-t^2}}\\
&=2\int_{0}^{1}\frac{\mathrm{d}z}{\sqrt{1-z^4}}\\
&=2\,K{(-1)}\\
&=\frac{\Gamma^2{\left(\frac14\right)}}{2\sqrt{2\pi}}\\
&=\frac{\pi^{3/2}}{2^{1/2}\Gamma^2{\left(\frac34\right)}}.
\end{align}$$
Sub $x=\tanh{u}$, $dx = \operatorname{sech^2}{u} \, du$. Then the integral is
$$\int_{-\infty}^{\infty} du \, \frac{\operatorname{sech^2}{u}}{\pi^2+4 u^2} $$
Now, use Parseval. The Fourier transforms of the pieces of the integrand are
$$\int_{-\infty}^{\infty} du \, \frac{e^{i u k}}{\pi^2+4 u^2} = \frac14 \frac{\pi}{\pi/2} e^{-\pi |k|/2} $$
$$\int_{-\infty}^{\infty} du \, \operatorname{sech^2}{u} \, e^{i u k} = \pi k \operatorname{csch}{\left (\frac{\pi k}{2} \right )}$$
so by Parseval...
$$\int_{-\infty}^{\infty} du \, \frac{\operatorname{sech^2}{u}}{\pi^2+4 u^2} = \frac12 \frac{\pi}{2 \pi} \int_{-\infty}^{\infty} dk \, k \operatorname{csch}{\left (\frac{\pi k}{2} \right )} e^{-\pi |k|/2} = \int_0^{\infty} dk \frac{k \, e^{-\pi k/2}}{e^{\pi k/2}-e^{-\pi k/2}}$$
Expand the denominator:
$$\int_{-\infty}^{\infty} du \, \frac{\operatorname{sech^2}{u}}{\pi^2+4 u^2} = \sum_{m=0}^{\infty} \int_0^{\infty} dk \, k \, e^{-(1+m) \pi k} = \frac{1}{\pi^2} \sum_{m=0}^{\infty} \frac1{\left (1+m \right )^2} = \frac1{6}$$
Best Answer
$$I=-\int _0^1\frac{1-x^2}{1+x^2}\:\ln \left(\operatorname{arctanh}\left(x\right)\right)\:dx$$ $$=-\int _0^1\frac{1-x^2}{1+x^2}\ln \left(\ln \left(\frac{1+x}{1-x}\right)\right)\:dx-\ln \left(\frac{1}{2}\right)\int _0^1\frac{1-x^2}{1+x^2}\:dx$$ $$=-4\int _0^1\frac{t\ln \left(-\ln \left(t\right)\right)}{\left(1+t^2\right)\left(1+t\right)^2}\:dt-\ln \left(\frac{1}{2}\right)\left(\frac{\pi }{2}-1\right)$$ $$=-2\int _0^1\frac{\ln \left(-\ln \left(t\right)\right)}{1+t^2}\:dt+2\int _0^1\frac{\ln \left(-\ln \left(t\right)\right)}{\left(1+t\right)^2}\:dt-\ln \left(\frac{1}{2}\right)\left(\frac{\pi }{2}-1\right)$$ $$=-2\int _0^{\infty }\frac{\ln \left(x\right)}{1+e^{-2x}}\:e^{-x}\:dx+2\int _0^{\infty }\frac{\ln \left(x\right)}{\left(1+e^{-x}\right)^2}\:e^{-x}\:dx+\frac{\pi }{2}\ln \left(2\right)-\ln \left(2\right)$$ $$=-2\sum _{k=0}^{\infty }\left(-1\right)^k\:\int _0^{\infty }e^{-x\left(2k+1\right)}\ln \left(x\right)\:dx+2\int _0^{\infty }\frac{\ln \left(x\right)}{\left(1+e^{-x}\right)^2}\:e^{-x}\:dx+\frac{\pi }{2}\ln \left(2\right)-\ln \left(2\right)$$ $$=2\sum _{k=0}^{\infty }\left(-1\right)^k\:\left(\frac{\ln \left(2k+1\right)+\gamma }{2k+1}\right)+2\int _0^{\infty }\frac{\ln \left(x\right)}{\left(1+e^{-x}\right)^2}\:e^{-x}\:dx+\frac{\pi }{2}\ln \left(2\right)-\ln \left(2\right)$$ $$=2\sum _{k=0}^{\infty }\frac{\left(-1\right)^k\ln \left(2k+1\right)}{2k+1}+2\gamma \sum _{k=0}^{\infty }\frac{\left(-1\right)^k}{2k+1}+2\underbrace{\int _0^{\infty }\frac{\ln \left(x\right)}{\left(1+e^{-x}\right)^2}\:e^{-x}\:dx}_{K}+\frac{\pi }{2}\ln \left(2\right)-\ln \left(2\right)$$ $$\boxed{I=-2\beta '\left(1\right)+\frac{\gamma \pi }{2}+\ln \left(\pi \right)-\gamma +\frac{\pi }{2}\ln \left(2\right)-2\ln \left(2\right)}$$ Where $\displaystyle \beta '\left(s\right)$ is the derivative of the Dirichlet beta function.
Also see here for the integral $K$.