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.