How to show that $ \int_0^\infty \frac{1}{\sqrt {t(t^2+21t+112)}} dt=\frac{1}{4\pi\sqrt {7}}\Gamma(\frac{1}{7})\Gamma(\frac{2}{7})\Gamma(\frac{4}{7})$ I've already known that
$$ \int \limits_0^\infty \frac{1}{\sqrt {t(t^2+21t+112)}} \mathrm{d}t=\frac{1}{4\pi\sqrt {7}}\Gamma(\frac{1}{7})\Gamma(\frac{2}{7})\Gamma(\frac{4}{7})$$
To get this answer, I let $\mathrm{d}u=\mathrm{d}\sqrt{t}$, then got
$$ \int \limits_0^\infty \frac{1}{\sqrt {t(t^2+21t+112)}} \mathrm{d}t=2\int \limits_0^\infty \frac{1}{\sqrt {u^4+21u^2+112}} \mathrm{d}u$$
$$2\int \limits_0^\infty \frac{1}{\sqrt {u^4+21u^2+112}} \mathrm{d}u=2\int \limits_0^\infty \frac{1}{\sqrt {u^2+\frac{21}{2}+\sqrt{\frac{7}{4}}i}{\sqrt {u^2+\frac{21}{2}-\sqrt{\frac{7}{4}}i}}} \mathrm{d}u$$
and it was pretty like Incomplete elliptic integral of the first kind. 
But, how to carry on?

As noted, this integral is reducible to an elliptic integral and the form of the proposed solution may suggest that its modulus could be a singular value. Indeed, from the form 
\begin{align}
 I&=2\int_0^\infty \frac{\mathrm{d}u}{\sqrt {u^4+21u^2+112}} \\
 &=2\int_{0}^\infty \frac{\mathrm{d}u}{\sqrt {u^2+\frac{21}{2}+\sqrt{\frac{7}{4}}i}{\sqrt {u^2+\frac{21}{2}-\sqrt{\frac{7}{4}}i}}} \\
 &=2\int_{0}^\infty \frac{\mathrm{d}u}{\sqrt {\left(  u^2+u_+^2\right)\left( u^2+u_-^2 \right)}} 
\end{align}
The opposite of the roots of the polynomial under the square root ($u_{\pm}^2=(21\pm\sqrt{7})/2$) are complex conjugate, as the coefficients of the polynomial are real. In the following, we first apply a Landen's transformation where these roots are replaced by their geometric and arithmetic means which are real. 
Defining $p=\sqrt{u_+u_-}$ and $q=(u_++u_-)/2$, we find from the coefficients of the polynomial 
\begin{align}
 p^2&=\sqrt{112}=4\sqrt{7}\\
 q^2&=\frac{1}{4}\left( u_+^2+u_-^2+2u_+u_- \right)\\
 &=\frac{1}{4}\left(21+2\sqrt{112} \right)=\frac{1}{4}\left(21+8\sqrt{7} \right)
\end{align}
With $u=x+\sqrt{x^2+p^2}$, remarking that 
\begin{equation}
\sqrt {\left(  u^2+u_+^2\right)\left(u^2+u_-^2 \right)}=2u\sqrt{x^2+q^2}
\end{equation} 
and that 
\begin{equation}
 du=\frac{u}{\sqrt{x^2+p^2}}
\end{equation} 
The integral becomes
\begin{equation}
 I=\int_{-\infty}^\infty\frac{dx}{\sqrt{\left( x^2+p^2 \right)\left( x^2+q^2 \right)}}
\end{equation} 
Now, changing $x=q\tan\theta$ and taking into account the parity of the function, one obtains an integral representation of the complete elliptic integral of the first kind:
\begin{align}
 I&=2\int_0^{\pi/2} \frac{d\theta}{\sqrt{p^2\cos^2\theta+q^2\sin^2\theta}}\\
 &=\frac{2}{p}\int_0^{\pi/2} \frac{d\theta}{\sqrt{1-\frac{p^2-q^2}{p^2}\sin^2\theta}}\\
 &=\frac{2}{p}\mathbf{K}\left( \sqrt{1-\frac{q^2}{p^2}} \right)%\\
 %&=\frac{2}{p}K'\left( \frac{q}{p} \right)
\end{align} 
Now, $\sqrt{1-\frac{q^2}{p^2}} =\frac{\sqrt{2}}{8}\left( 3-\sqrt{7} \right)=\lambda^*(7)$, where $\lambda^*(r)$ gives a singular value of the elliptic modulus $k_r$ for which $\mathbf{K}\left( \sqrt{1-k_r^2} \right)=\sqrt{r}\mathbf{K}(k_r)$ see here, here and here. Then, 
\begin{align}
 I&=\frac{2}{p}\mathbf{K}\left( k_7 \right)\\
 &=\frac{2}{2.7^{1/4}}\frac{\Gamma\left( \frac{1}{7} \right)\Gamma\left( \frac{2}{7} \right)\Gamma\left( \frac{4}{7} \right)}{4\pi 7^{1/4}}\\
 &=\frac{\Gamma\left( \frac{1}{7} \right)\Gamma\left( \frac{2}{7} \right)\Gamma\left( \frac{4}{7} \right)}{4\pi \sqrt{7}}
\end{align} 
as expected.
