$$\boxed{\displaystyle\int_0^1\log\log\left(\frac1x+\sqrt{\frac1{x^2}-1}\right)\mathrm dx=-\gamma-2\ln\frac{2\Gamma(3/4)}{\Gamma(1/4)}}\tag{$\heartsuit$}$$
Derivation:
After the change of variables $x=\frac{1}{\cosh u}$ the integral becomes
$$\int_0^{\infty}\ln u \frac{\sinh u}{\cosh^2 u}du,$$
as was noticed above by Eric. We would like to integrate by parts to kill the logarithm but we get two divergent pieces. To go around this, let us consider another integral,
$$I(s)=\int_0^{\infty}u^s \frac{\sinh u}{\cosh^2 u}du,$$
with $s>0$. The integral we actually want to compute is equal to $I'(0)$, which will be later obtained in the limit.
Indeed, integrating once by parts one finds that
\begin{align}
I(s)&=s\int_0^{\infty}\frac{u^{s-1}du}{\cosh u}=s\cdot 2^{1-2 s}\Gamma(s)\left[\zeta\left(s,\frac14\right)-\zeta\left(s,\frac34\right)\right]=\\ &=2^{1-2 s}\Gamma(s+1)\left[\zeta\left(s,\frac14\right)-\zeta\left(s,\frac34\right)\right],
\end{align}
where $\zeta(s,a)=\sum_{n=0}^{\infty}(n+a)^{-s}$ denotes Hurwitz zeta function (in the way we have used its integral representaion (5) from here).
Now to get ($\heartsuit$), it suffices to use
\begin{align}
&\frac{\partial}{\partial s}\left[2^{1-2 s}\Gamma(s+1)\right]_{s=0}=-2\gamma-4\ln 2,\\
&\zeta\left(0,\frac14\right)-\zeta\left(0,\frac34\right)=\frac12,
\\
&\frac{\partial}{\partial s}\left[\zeta\left(s,\frac14\right)-\zeta\left(s,\frac34\right)\right]_{s=0}=-\ln\frac{\Gamma(\frac34)}{\Gamma(\frac14)}.
\end{align}
[See formulas (10) and (16) on the same page].
A closed form indeed exists for this integral:
$$\int_0^\infty\frac{\log\left(1+\frac{\pi^2}{4\,x}\right)}{e^{\sqrt{x}}-1}\mathrm dx=\pi^2\left(\log\frac{\pi\,A^9\sqrt{2}}{\Gamma(\frac{1}{4})^2}-\frac{9}{8}\right)+2\,\pi\,C,$$
where $A$ is the Glaisher-Kinkelin constant and $C$ is the Catalan constant.
A more general result:
$$\int_0^\infty\frac{\log\left(1+\frac{a}{x}\right)}{e^{\sqrt{x}}-1}\mathrm dx=8\,\pi^2\psi^{(-2)}\left(\frac{\sqrt{a}}{2\pi}\right)-\frac{a}2\left(1+\log\frac{4\pi^2}{a}\right)-2\pi\sqrt{a}\left(1+2\log\Gamma\left(\frac{\sqrt{a}}{2\pi}\right)\right),$$
where $\psi^{(-2)}(z)$ is the generalized polygamma function.
The proof is based on Binet's second formula, but I still need to sort out some details.
Best Answer
First, we transform the integral into a more computable form by using some substitutions.
$$\begin{align*}\displaystyle Q &= \int_0^1\sqrt{\frac{2-x}{(1-x)\,x}}\,\log\left(\frac{(2-x)\,x}{1-x}\right)dx\\ &=\int_0^1 \sqrt{\frac{1+u}{u(1-u)}}\log \left( \frac{(1+u)(1-u)}{u}\right)du \quad \color{blue}{\text{where }u=1-x}\\ &= \int_0^1 \frac{1+u}{\sqrt{u(1-u^2)}}\log \left( \frac{1-u^2}{u}\right)du \\ &= \frac{1}{2}\int_0^1 \frac{1+\sqrt{t}}{t^{\frac{3}{4}}\sqrt{1-t}}\log \left(\frac{1-t}{\sqrt{t}} \right)dt \quad \color{blue}{\text{where }t=u^2} \\ &= \frac{1}{2}\int_0^1 \frac{\log(1-t)}{t^{\frac{3}{4}}\sqrt{1-t}}dt-\frac{1}{4}\int_0^1 \frac{\log(t)}{t^{\frac{3}{4}}\sqrt{1-t}}dt \\ &\quad +\frac{1}{2}\int_0^1 \frac{\log(1-t)}{t^{\frac{1}{4}}\sqrt{1-t}}dt-\frac{1}{4}\int_0^1 \frac{\log(t)}{t^{\frac{1}{4}}\sqrt{1-t}}dt \tag{1} \end{align*}$$
These four integrals can be evaluated by calculating derivatives of beta function in terms of digamma function. For e.g.
$$ \begin{align*} \int_0^1 \frac{\log(1-t)}{t^{\frac{3}{4}}\sqrt{1-t}}dt &= \frac{d}{dz}\left\{ \int_0^1 t^{-\frac{3}{4}}(1-t)^{z-1} \; dt\right\}_{z=\frac{1}{2}}\\&= \frac{d}{dz}\left\{ \frac{\Gamma \left( \frac{1}{4}\right)\Gamma(z)}{\Gamma \left( \frac{1}{4}+z\right)} \right\}_{z=\frac{1}{2}}\\ &= \frac{\Gamma \left( \frac{1}{4}\right)\sqrt{\pi}}{\Gamma \left( \frac{3}{4}\right)}\left\{\psi_0 \left(\frac{1}{2} \right) -\psi_0 \left(\frac{3}{4} \right)\right\}\\ &= \pi^{3/2}\frac{\sqrt{2}}{\Gamma \left( \frac{3}{4}\right)^2}\left\{\log 2-\frac{\pi}{2} \right\} \tag{2} \\ \end{align*} $$ To get the last expression, I used the special values $$ \begin{align*} \psi_0 \left(\frac{3}{4}\right) &= -\gamma +\frac{\pi}{2}-3\log 2 \\ \psi_0 \left(\frac{1}{2}\right) &= -\gamma -2\log 2 \end{align*} $$ Using the same technique, the other three integrals can be evaluated: $$ \begin{align*} \int_0^1 \frac{\log(t)}{t^{\frac{3}{4}}\sqrt{1-t}}dt&= -\pi^{5/2}\frac{\sqrt2}{\Gamma \left( \frac{3}{4}\right)^2}\tag{3} \\ \int_0^1 \frac{\log(t)}{t^{\frac{1}{4}}\sqrt{1-t}}dt&=\frac{(4\pi-16)\Gamma \left( \frac{3}{4}\right)^2}{\sqrt{2\pi}} \tag{4} \\ \int_0^1 \frac{\log(1-t)}{t^{\frac{1}{4}}\sqrt{1-t}}dt&=\frac{2(-8+\pi+2\log 2)\Gamma \left( \frac{3}{4}\right)^2}{\sqrt{2\pi}} \tag{5} \end{align*} $$ Substituting the results of equations $(2),(3),(4)$ and $(5)$ in $(1)$ gives
$$Q=\frac{\Gamma\left(\frac34\right)^{-2}\pi^2\log2-\Gamma\left(\frac34\right)^2(4-2\log2)}{\sqrt{2\,\pi}}$$