$$\boxed{\displaystyle\int_0^1\log\log\left(\frac1x+\sqrt{\frac1{x^2}-1}\right)\mathrm dx=-\gamma-2\ln\frac{2\Gamma(3/4)}{\Gamma(1/4)}}\tag{$\heartsuit$}$$
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].
Using
$$ K(ik) =
\frac{1}{\sqrt{1+k^2}}K\left(\sqrt{\frac{k^2}{k^2+1}}\right) $$
and a substitution $t^2 = \frac{k^2}{1+k^2}$, rewrite the integral as
$$ \int_0^\infty K(i k)^3\,dk = \int_0^1 K(t)^3\,dt. $$
There is a paper "Moments of elliptic integrals and critical L-values"
by Rogers, Wan and Zucker (http://arxiv.org/abs/1303.2259; also one of
the authors' earlier papers: http://arxiv.org/abs/1101.1132), and the
authors, by relating this integral to an L-series of a modular form (their theorems 1 and 2),
show that
$$ \int_0^1 K(k)^3\,dk = \frac{3}{5}K(1/\sqrt{2})^4 =
\frac{3\Gamma(\frac14)^8}{1280\pi^2}, $$
using $K(1/\sqrt{2}) = \frac14 \pi^{-1/2}\Gamma(\frac14)^2$.
Best Answer
This hypergeometric function is not an elementary function, but its inverse is - see Bring radical.
\begin{align} I&= \int_0^1\arctan{_4F_3}\left(\frac15,\frac25,\frac35,\frac45;\frac12,\frac34,\frac54;\frac{x}{64}\right)\,dx \\ &=\frac{3125}{48}\left(5+3\pi+6\ln2-3\alpha^4+4\alpha^3+6\alpha^2-12\alpha\\-12\left(\alpha^5-\alpha^4+1\right)\arctan\frac1\alpha-6\ln\left(1+\alpha^2\right)\right)\\ &=0.7857194\dots \end{align} where $\alpha$ is the positive root of the polynomial $625\alpha^4-500\alpha^3-100\alpha^2-20\alpha-4$. It can be expressed in radicals as follows:
$$\alpha=\frac15+\sqrt\beta+\sqrt{\frac15-\beta +\frac1{25\sqrt\beta}},$$ where $$\beta=\frac1{30}\left(\frac\gamma5-\frac4\gamma+2\right),$$ where $$\gamma=\sqrt[3]{15\sqrt{105}-125}.$$