$$\boxed{\displaystyle\int_0^1\log\log\left(\frac1x+\sqrt{\frac1{x^2}-1}\right)\mathrm dx=-\gamma-2\ln\frac{2\Gamma(3/4)}{\Gamma(1/4)}}\tag{$\heartsuit$}$$
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
For $a > 0$, let $b = \frac12 + \frac1a$ and $I(a)$ be the integral $$I(a) = \int_0^1 \frac{\log(a+x)}{\sqrt{x(1-x)(2-x)}}dx$$ Substitute $x$ by $\frac{1}{p+\frac12}$, it is easy to check we can rewrite $I(a)$ as
$$ I(a) = -\sqrt{2}\int_\infty^{\frac12}\frac{\log\left[a (p + b)/(p + \frac12)\right]}{\sqrt{4p^3 - p}} dp $$ Let $\wp(z), \zeta(z)$ and $\sigma(z)$ be the Weierstrass elliptic, zeta and sigma functions associated with the ODE:
$$\wp'(z)^2 = 4\wp(z)^3 - g_2 \wp(z) - g_3\quad\text{ for }\quad g_2 = 1 \;\text{ and }\; g_3 = 0.$$
In terms of $\wp(z)$, we can express $I(a)$ as
$$I(a) = \sqrt{2}\int_0^\omega \log\left[a \left(\frac{\wp(z) + b}{\wp(z) + \frac12}\right)\right] dz = \frac{1}{\sqrt{2}}\int_{-\omega}^\omega \log\left[a \left(\frac{\wp(z) + b}{\wp(z) + \frac12}\right)\right] dz $$
where $\;\displaystyle \omega = \int_\frac12^\infty \frac{dp}{\sqrt{4p^3 - p}} = \frac{\pi^{3/2}}{2\Gamma\left(\frac34\right)^2}\;$ is the half period for $\wp(z)$ lying on real axis. Since $g_3 = 0$, the double poles of $\wp(z)$ lies on a square lattice $\mathbb{L} = \{\; 2\omega ( m + i n ) : m, n \in \mathbb{Z} \;\}$ and and we can pick the other half period $\;\omega'$ as $\;i\omega$.
Notice $\wp(\pm i \omega) = -\frac12$. If we pick $u \in (0,\omega)$ such that $\wp(\pm i u) = -b$, the function inside the square brackets in above integral is an ellitpic function with zeros at $\pm i u + \mathbb{L}$ and poles at $\pm i \omega + \mathbb{L}$. We can express $I(a)$ in terms of $\sigma(z)$ as
$$I(a) = \frac{1}{\sqrt{2}}\int_{-\omega}^\omega \log\left[ C\frac{\sigma(z-iu)\sigma(z+iu)}{\sigma(z-i\omega)\sigma(z+i\omega)}\right] dz \quad\text{ where }\quad C = a\left(\frac{\sigma(-i\omega)\sigma(i\omega)}{\sigma(-iu)\sigma(iu)}\right). $$
Let $\varphi_{\pm}(\tau)$ be the integral $\displaystyle \int_{-\omega}^\omega \log\sigma(z+\tau) dz$ for $\Im(\tau) > 0$ and $< 0$ respectively. Notice $\sigma(z)$ has a simple zero at $z = 0$. We will choose the branch cut of $\log \sigma(z)$ there to be the ray along the negative real axis.
When we move $\tau$ around, as long as we don't cross the real axis, the line segment $[\tau-\omega,\tau+\omega]$ won't touch the branch cut and everything will be well behaved. We have
$$\begin{align} & \varphi_{\pm}(\tau)''' = -\wp(\tau+\omega) + \wp(\tau-\omega) = 0\\ \implies & \varphi_{\pm}(\tau)'' = \zeta(\tau+\omega) - \zeta(\tau-\omega) \quad\text{ is a constant}\\ \implies & \varphi_{\pm}(\tau)'' = 2 \zeta(\omega)\\ \implies & \varphi_{\pm}(\tau) = \zeta(\omega) \tau^2 + A_{\pm} \tau + B_{\pm} \quad\text{ for some constants } A_{\pm}, B_{\pm} \end{align} $$
Let $\eta = \zeta(\omega)$ and $\eta' = \zeta(\omega')$. For elliptic functions with general $g_2, g_3$, there is always an identity $$\eta \omega' - \omega \eta' = \frac{\pi i}{2}$$ as long as $\omega'$ is chosen to satisfy $\Im(\frac{\omega'}{\omega}) > 0$. In our case, $\omega' = i\omega$ and the symmetric of $\mathbb{L}$ forces $\eta = \frac{\pi}{4\omega}$. This implies
$$\varphi_{\pm}(\tau) = \frac{\pi}{4\omega}\tau^2 + A_{\pm}\tau + B_{\pm}$$
Because of the branch cut, $A_{+} \ne A_{-}$ and $B_{+} \ne B_{+}$. In fact, we can evaluate their differences as
$$\begin{align} A_{+} - A_{-} &= \lim_{\epsilon\to 0} \left( -\log\sigma(i\epsilon-\omega) + \log\sigma(-i\epsilon-\omega) \right) = - 2 \pi i\\ B_{+} - B_{-} &= \lim_{\epsilon\to 0} \int_{-\omega}^0 \left( \log\sigma(i\epsilon+z) - \log\sigma(-i\epsilon+z) \right) dz = 2\pi i\omega \end{align} $$ Apply this to our expression of $I(a)$, we get
$$\begin{align} I(a) &= \frac{1}{\sqrt{2}}\left(2\omega\log C + \varphi_{-}(-iu)+\varphi_{+}(iu)-\varphi_{-}(-i\omega)-\varphi_{+}(i\omega)\right)\\ &= \frac{1}{\sqrt{2}}\left\{ 2\omega\log\left[a\left(\frac{\sigma(-i\omega)\sigma(i\omega)}{\sigma(-iu)\sigma(iu)}\right)\right] + \frac{\pi}{2\omega}(\omega^2 - u^2) + 2\pi(u-\omega) \right\} \end{align} $$
Back to our original problem where $a = \sqrt{2} \iff b = \frac{1+\sqrt{2}}{2}$. One can use the duplication formula for $\wp(z)$ to vertify $u = \frac{\omega}{2}$. From this, we find: $$I(\sqrt{2}) = \sqrt{2}\omega\left\{ \log\left[\sqrt{2}\left(\frac{\sigma(-i\omega)\sigma(i\omega)}{\sigma(-i\frac{\omega}{2})\sigma(i\frac{\omega}{2})}\right)\right] - \frac{5\pi}{16}\right\} $$
It is known that $| \sigma(\pm i\omega) | = e^{\pi/8}\sqrt[4]{2}$. Furthermore, we have the identity:
$$\wp'(z) = - \frac{\sigma(2z)}{\sigma(z)^4} \quad\implies\quad \left|\sigma\left( \pm i\frac{\omega}{2} \right)\right| = \left|\frac{\sigma(\pm i \omega)}{\wp'\left(\pm i\frac{\omega}{2}\right)}\right|^{1/4} = \left(\frac{\sigma(\omega)}{1+\sqrt{2}}\right)^{1/4} $$ Combine all these, we get a result matching other answer.
$$\begin{align} I(\sqrt{2}) &= \sqrt{2}\omega\left\{\log\left[\sqrt{2}\sigma(\omega)^{3/2}\sqrt{1+\sqrt{2}}\right] - \frac{5\pi}{16}\right\}\\ &= \frac{\pi^{3/2}}{\sqrt{2}\Gamma\left(\frac34\right)^2}\left\{\frac78\log 2 + \frac12\log(\sqrt{2}+1) - \frac{\pi}{8} \right\} \end{align}$$