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

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

Then:
 * $\ds \int \frac {\d x} {x \paren {\sqrt {a x^2 + b x + c} }^3} = \frac 1 {c \sqrt {a x^2 + b x + c} } + \frac 1 c \int \frac {\d x} {x \sqrt {a x^2 + b x + c} } - \frac b {2 c} \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3}$