Primitive of Reciprocal of x cubed by x fourth plus a fourth

Theorem

 * $\ds \int \frac {\d x} {x^3 \paren {x^4 + a^4} } = \frac {-1} {2 a^4 x^2} - \frac 1 {2 a^6} \arctan \frac {x^2} {a^2}$