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?

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.
