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

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

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

Proof
For $a > 0$:

Then:

Let $4 a c - b^2 > 0$.

Then:

Let $4 a c - b^2 < 0$.

Then:

Thus in both cases the same result applies, and so: