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?

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.
