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?