Primitive of x over Cube of Root of a x squared plus b x plus c

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

Then:
 * $\ds \int \frac {x \rd x} {\paren {\sqrt {a x^2 + b x + c} }^3} = \frac {2 \paren {b x + 2 c} } {\paren {b^2 - 4 a c} \sqrt {a x^2 + b x + c} } + C$

Proof
First:

Then: