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

Theorem

 * $\ds \int \frac {\d x} {x^2 \paren {x^4 + a^4} } = \frac {-1} {a^4 x} - \frac {-1} {4 a^5 \sqrt 2} \map \ln {\frac {x^2 - a x \sqrt 2 + a^2} {x^2 + a x \sqrt 2 + a^2} } + \frac 1 {2 a^5 \sqrt 2} \paren {\map \arctan {1 - \frac {x \sqrt 2} a} - \map \arctan {1 + \frac {x \sqrt 2} a} }$