Primitive of Reciprocal of x squared by Cube of Root of a x squared plus b x plus c

Theorem
Let $a \in \R_{\ne 0}$.

Then:
 * $\displaystyle \int \frac {\d x} {x^2 \paren {\sqrt {a x^2 + b x + c} }^3} = -\frac {a x^2 + 2 b x + c} {c^2 x \sqrt {a x^2 + b x + c} } + \frac {b^2 - 2 a c} {2 c^2} \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3} - \frac {3 b} {2 c^2} \int \frac {\d x} {x \sqrt {a x^2 + b x + c} }$