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

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

Then:
 * $\displaystyle \int \frac {\mathrm d x} {\left({a x^2 + b x + c}\right)^{n + \frac 1 2} } = \frac {2 \left({2 a x + b}\right)} {\left({2 n - 1}\right) \left({4 a c - b^2}\right) \left({a x^2 + b x + c}\right)^{n - \frac 1 2} } + \frac {8 a \left({n - 1}\right)} {\left({2 n - 1}\right) \left({4 a c - b^2}\right)} \int \frac {\mathrm d x} {\left({a x^2 + b x + c}\right)^{n - \frac 1 2} }$