Integral ${\large\int}_0^\infty\frac{dx}{\sqrt[4]{7+\cosh x}}$

I will follow @user15302's idea. In this answer, I showed that

$$ \int_{0}^{\infty} \frac{dx}{(a + \cosh x)^{s}} \, dx = \frac{1}{(a+1)^{s}} \int_{0}^{1} \frac{v^{s-1}}{\sqrt{(1-v)(1-bv)}} \, dv, $$

where $b = \frac{a-1}{a+1}$. Now let $I$ denote the Vladimir's integral and set $s = \frac{1}{4}$ and $a = 7$. Then we have $b = \frac{3}{4}$ and

$$ I = 2^{-3/4} \int_{0}^{1} \frac{1}{v^{3/4}\sqrt{(1-v)(1-\frac{3}{4}v)}} \, dv. $$

The reason why the case $b = \frac{3}{4}$ is special is that, if we plug $v = \operatorname{sech}^2 t$ then we can utilize the triple angle formula to get the following surprisingly neat integral

$$ I = 2^{5/4} \int_{0}^{\infty} \frac{\cosh t}{\sqrt{\cosh 3t}} \, dt. $$

Now using the substitution $x = e^{-6t}$, we easily find that

$$ I = \frac{1}{3 \sqrt[4]{2}} \int_{0}^{1} \frac{x^{-11/12} + u^{-7/12}}{\sqrt{1+x}} \, dx = \frac{1}{3 \sqrt[4]{2}} \int_{0}^{\infty} \frac{dx}{x^{11/12}\sqrt{1+x}}. $$

The last integral can be easily calculated by the following formula

$$ \int_{0}^{\infty} \frac{x^{a-1}}{(1+x)^{a+b}} \, dx = \beta(a, b) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}. $$

Therefore we obtain the following closed form

$$ I = \frac{\Gamma(\frac{1}{12})\Gamma(\frac{5}{12})}{3 \sqrt[4]{2}\sqrt{\pi}}. $$

In order to verify that this is exactly the same as Vladimir's result, We utilize the Legendre multiplication formula and the reflection formula to find that

$$ \Gamma(\tfrac{1}{12})\Gamma(\tfrac{5}{12}) = \frac{\Gamma(\tfrac{1}{12})\Gamma(\tfrac{5}{12})\Gamma(\tfrac{9}{12})}{\Gamma(\tfrac{3}{4})} = \frac{2 \pi \cdot 3^{1/4} \Gamma(\frac{1}{4})}{\Gamma(\tfrac{3}{4})} = 2^{1/2} 3^{1/4} \Gamma(\tfrac{1}{4})^2. $$

This completes the proof.

By replacing $x$ with $4u$, then $\cosh u$ with $\frac{1}{t}$, we have:

$$ I = \frac{1}{2^{3/4}}\int_{0}^{1}\frac{dt}{(1-t^2+t^4)^{1/4}(1-t^2)^{1/2}}=\frac{1}{2^{3/4}}\int_{0}^{1/2}\frac{dt}{(1-t(1-t))^{1/4}(t(1-t))^{1/2}} $$ Next, by replacing $t(1-t)$ with $v/4$, $$ I=\frac{1}{2^{11/4}}\int_{0}^{1}\frac{dv}{(1-v/4)^{1/4}(v(1-v))^{1/2}}$$ then, by setting $v=4-3z$, $$ I = \frac{3^{1/4}}{2^{9/4}}\int_{1}^{4/3}\frac{dz}{z^{1/4}((4-3z)(1+z))^{1/2}}=\frac{3^{1/4}}{2^{5/4}}\int_{1}^{2/\sqrt{3}}\sqrt{\frac{z}{(4-3z^2)(1+z^2)}}\,dz$$ that, at least, looks manageable. We also have: $$ I = \frac{1}{2^{1/4}}\int_{0}^{1}\frac{du}{(3u^4+u^2)^{1/4}(1-u^2)^{1/2}}\tag{1}$$ that Mathematica gladly evaluates to: $$ I = \frac{2^{1/4}\,\Gamma\left(\frac{1}{4}\right)^2}{3^{3/4}\sqrt{\pi}}. $$ Now we just need to understand how.

I think this problem can be solved by invoking the theory of $j$-invariants for (hyper?)-elliptic curves, but I am not so confident in the topic to find the right change of variables that brings our integral into a complete elliptic integral. I think that Noam Elkies would solve this problem in a few seconds, so I am asking his help.

Update. Found. Our claim was proven by Zucker and Joyce in Special values of the hypergeometric series II, it is the result $(7\!\cdot\! 6)$. It is derived through standard hypergeometric manipulations, by starting with the elliptic modulus $k$ for which: $$\frac{K'(k)}{K(k)}=3.$$ The modular function to be considered for regarding our integral as a period is so the elliptic lambda function.