Integral ${\large\int}_0^\infty\frac{dx}{\sqrt[4]{7+\cosh x}}$ How to prove the following conjectured identity?
$$\int_0^\infty\frac{dx}{\sqrt[4]{7+\cosh x}}\stackrel{\color{#a0a0a0}?}=\frac{\sqrt[4]6}{3\sqrt\pi}\Gamma^2\big(\tfrac14\big)\tag1$$
It holds numerically with precision of at least $1000$ decimal digits.
Are there any other integers under the radical except $7$ and $1$ that result in a nice closed form?
 A: This is a partial answer to the second question. Mathematica could evaluate
$$\int_0^\infty\frac{dx}{\sqrt[4]{a+\cosh x}},$$
in term of the following Appell function:
$$
F_1\left(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{5}{4},\sqrt{a^2-1}-a,\frac{1}{\sqrt{a^2-1}-a}\right).
$$
For $a=0$ and $a=1$ there is a closed-form of this Appell function, so we get
$$\int_0^\infty\frac{dx}{\sqrt[4]{\cosh x}} = \frac{4\sqrt{\pi}\,\Gamma\left(\frac{9}{8}\right)}{\Gamma\left(\frac{5}{8}\right)}$$
and
$$\int_0^\infty\frac{dx}{\sqrt[4]{1+\cosh x}} = \frac{\Gamma^2\left(\frac{1}{4}\right)}{2^{3/4}\sqrt{\pi}}.$$
Numerically I've got your conjectured form for $a=7$ too:
$$\int_0^\infty\frac{dx}{\sqrt[4]{7+\cosh x}}\stackrel{\color{#a0a0a0}?}=\frac{\sqrt[4]6}{3\sqrt\pi}\Gamma^2\left(\tfrac14\right),$$
or in term of elliptic $K$ function:
$$\int_0^\infty\frac{dx}{\sqrt[4]{7+\cosh x}}\stackrel{\color{#a0a0a0}?}=\frac{4\sqrt[4]2\sqrt[4]3}{3}K\left(\tfrac{\sqrt{2}}{2}\right).$$
A related, somehow generalized question is here.
A: 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.
A: 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.
A: Take the integral in the form $$I=\frac{1}{2^{1/4}}\int_{0}^{1}\frac{du}{\left(3u^{4}+u^{2}\right)^{1/4}\left(1-u^{2}\right)^{1/2}}=\frac{1}{2^{1/4}}\int_{0}^{1}\frac{du}{\left(3u^{2}+1\right)^{1/4}\left(1-u^{2}\right)^{1/2}\left(u^{2}\right)^{1/4}}
 $$ then put $u^{2}=s
 $ $$=\frac{1}{2^{5/4}}\int_{0}^{1}\frac{ds}{\left(3s+1\right)^{1/4}\left(1-s\right)^{1/2}s^{3/4}}
 $$ and now put $s=1-t
 $ $$=\frac{1}{2^{7/4}}\int_{0}^{1}\frac{dt}{\left(1-3t/4\right)^{1/4}\left(1-t\right)^{3/4}t^{1/2}}.
 $$ Now recalling the identity $$\,_{2}F_{1}\left(a,b;c;z\right)=\frac{\Gamma\left(c\right)}{\Gamma\left(b\right)\Gamma\left(c-b\right)}\int_{0}^{1}\frac{t^{b-1}\left(1-t\right)^{c-b-1}}{\left(1-tz\right)^{a}}dt
 $$ we have $$I=\frac{1}{2^{7/4}}\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{4}\right)}{\Gamma\left(\frac{3}{4}\right)}\,_{2}F_{1}\left(\frac{1}{4},\frac{1}{2};\frac{3}{4};\frac{3}{4}\right)
 $$ and in this case it is possible calculate the exact value of the hypergeometric function (see the update in the Jack D'Aurizio's answer for reference) $$\,_{2}F_{1}\left(\frac{1}{4},\frac{1}{2};\frac{3}{4};\frac{3}{4}\right)=\frac{2\sqrt{2}}{3^{3/4}}
 $$ and so $$I=\frac{\Gamma^{2}\left(\frac{1}{4}\right)}{6^{3/4}\sqrt{\pi}}.
 $$ The result is not equal to $\sqrt[4]{6}\Gamma^{2}\left(\frac{1}{4}\right)/\left(2\sqrt{\pi}\right)
 $ but I haven't found a mistake in my calculations.
