I found the following integral and I am stuck with the procedure to find the given closed form in terms of the complete Gamma function
$$\displaystyle I= \int_{0}^{\infty} \frac{dx}{\sqrt{112+21x^2+x^4}} = \frac{\Gamma\left(\frac{1}{7}\right)\Gamma\left(\frac{2}{7}\right)\Gamma\left(\frac{4}{7}\right)}{8\pi\sqrt{7}}.$$
I know that this integral is related to the complete elliptic integral of the first kind throught the following integral representation:
$$K(p) = \int_{0}^{\infty} \frac{dt}{\sqrt{(1+t^2)(1+q^2t^2)}}$$
where $p$ is the modulus, $q$ is the complementary modulus:
$$p^2+q^2=1.$$
However, this is not a normal elliptic integral as we will see.
If we factorize the polynomial in the radicand we found that
$$112+21x^2+x^4 = \left(x^2+\frac{21}{2}+i\frac{\sqrt{7}}{2}\right)\left(x^2+\frac{21}{2}-i\frac{\sqrt{7}}{2}\right).$$
For simplicity, if we denote
$$ a_{1} = \frac{21}{2}+i\frac{\sqrt{7}}{2} \quad a_{2}=\frac{21}{2}-i\frac{\sqrt{7}}{2}$$
then \begin{align*}I & = \int_{0}^{\infty} \frac{dx}{\sqrt{112+21x^2+x^4}} = \int_{0}^{\infty} \frac{dx}{\sqrt{(x^2+a_{1})(x^2+a_{2})}} \\ & = \frac{1}{\sqrt{a_{1}a_{2}}} \int_{0}^{\infty} \frac{dx}{\sqrt{\left(\frac{x^2}{a_{1}}+1\right)(\frac{x^2}{a_{2}}+1)}}.\end{align*}
Making the substitution $\displaystyle w^2 = \frac{x^2}{a_{1}}$
$$ I = \frac{1}{\sqrt{a_{1}a_{2}}} \int_{0}^{\infty} \frac{dw}{\sqrt{\left(w^2+1\right)(\frac{a_{1}}{a_{2}}w^2+1)}}.$$
If now we make the substitution $\displaystyle \theta = \arctan(w)$ and putting the values of $a_{1}$ and $a_{2}$ we recover the Legendre form of the complete elliptic integral of the first kind:
\begin{align*} I & = \frac{\sqrt{\frac{21}{2}+i\frac{\sqrt{7}}{2}}}{4\sqrt{7}} \int_{0}^{\frac{\pi}{2}} \frac{d\theta}{\sqrt{1-\left(\frac{1}{32}-i\frac{3\sqrt{7}}{32}\right)\sin^2\theta}} \\ & = \frac{1}{4}\sqrt{\frac{1}{7}\left(\frac{21}{2}+i\frac{\sqrt{7}}{2}\right)}K\left(\sqrt{\frac{1}{32}-i\frac{3\sqrt{7}}{32}}\right).\end{align*}
The expression in the right hand is numerically identical to the solution in terms of the Gamma function. So, I tried using the hypergeometric representation of the elliptic integral to try to obtain the closed form in terms of Gamma functions:
$$K\left(\sqrt{\frac{1}{32}-i\frac{3\sqrt{7}}{32}}\right) = \frac{\pi}{2}{}_{2}F_{1}\left(\frac{1}{2},\frac{1}{2};1;\frac{1}{32}-i\frac{3\sqrt{7}}{32}\right).$$
Using the folowing quadratic transformation
$$F\left({a,b\atop 2b};z\right)=\left(1-z\right)^{-\frac{a}{2}}F\left({\frac{1% }{2}a,b-\tfrac{1}{2}a\atop b+\tfrac{1}{2}};\frac{z^{2}}{4z-4}\right),$$
I get rid of the imaginary unit in the argument of ${}_{2}F_{1}$: \begin{align*} I & = \frac{1}{4}\sqrt{\frac{1}{7}\left(\frac{21}{2}+i\frac{\sqrt{7}}{2}\right)}\frac{\pi}{2}{}_{2}F_{1}\left(\frac{1}{2},\frac{1}{2};1;\frac{1}{32}-i\frac{3\sqrt{7}}{32}\right) \\ & =\frac{\pi}{8}\frac{\sqrt{\frac{1}{7}\left(\frac{21}{2}+i\frac{\sqrt{7}}{2}\right)}}{\left(\frac{1}{32}-i\frac{3\sqrt{7}}{32}\right)^{\frac{1}{4}}} {}_{2}F_{1}\left(\frac{1}{4},\frac{1}{4};1;\frac{1}{64}\right).\end{align*}
Wolfram evaluates
$$ {}_{2}F_{1}\left(\frac{1}{4},\frac{1}{4};1;\frac{1}{64}\right) = \frac{7^{\frac{3}{4}}\Gamma\left(\frac{2}{7}\right)\Gamma\left(\frac{4}{7}\right)\Gamma\left(\frac{8}{7}\right)}{2\pi^2}.$$
But I do not know how to obtain this expression nor how to proceed further. I will be very grateful for your help.
