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

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

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

Proof
Let:

Also let $q = 4 a c - b^2$.

Then:

From Primitive of $\dfrac 1 {\left({p x + q}\right)^n \sqrt{a x + b} }$:
 * $\displaystyle \int \frac {\mathrm d x} {\left({p x + q}\right)^n \sqrt{a x + b} } = \frac {\sqrt{a x + b} } {\left({n - 1}\right) \left({a q - b p}\right) \left({p x + q}\right)^{n-1} } + \frac {\left({2 n - 3}\right) a} {2 \left({n - 1}\right) \left({a q - b p}\right)} \int \frac {\mathrm d x} {\left({p x + q}\right)^{n-1} \sqrt{a x + b} }$

Here $p = 1, a = 1, b = 0$ and $n := n + \dfrac 1 2$: