Primitive of Reciprocal of x by Half Integer Power of a x squared plus b x plus c

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

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

Proof
Now take:

Let:

Thus:

So: