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

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

Then:
 * $\ds \int \frac {\d x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} } = \frac {2 \paren {2 a x + b} } {\paren {2 n - 1} \paren {4 a c - b^2} \paren {a x^2 + b x + c}^{n - \frac 1 2} } + \frac {8 a \paren {n - 1} } {\paren {2 n - 1} \paren {4 a c - b^2} } \int \frac {\d x} {\paren {a x^2 + b x + c}^{n - \frac 1 2} }$

Proof
Let:

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

Then:

From Primitive of $\dfrac 1 {\paren {p x + q}^n \sqrt {a x + b} }$:
 * $\ds \int \frac {\d x} {\paren {p x + q}^n \sqrt {a x + b} } = \frac {\sqrt {a x + b} } {\paren {n - 1} \paren {a q - b p} \paren {p x + q}^{n - 1} } + \frac {\paren {2 n - 3} a} {2 \paren {n - 1} \paren {a q - b p} } \int \frac {\d x} {\paren {p x + q}^{n - 1} \sqrt {a x + b} }$

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