Theorem: Primitive of Half Integer Power of a x squared plus b x plus c

Let $a \in \R_{\ne 0}$.
Then:
:$\ds \int \paren {a x^2 + b x + c}^{n + \frac 1 2} \rd x = \frac {\paren {2 a x + b} \paren {a x^2 + b x + c}^{n + \frac 1 2} } {4 a \paren {n + 1} } + \frac {\paren {2 n + 1} \paren {4 a c - b^2} } {8 a \paren {n + 1} } \int \paren {a x^2 + b x + c}^{n - \frac 1 2} \rd x$

Proof:
{{finish|This only takes on the case where $a > 0$. The case where $a < 0$ needs to be addressed.}}
{{begin-eqn}}
{{eqn | l = \int \paren {a x^2 + b x + c}^{n + \frac 1 2} \rd x
      | r = \int \paren {\frac {\paren {2 a x + b}^2 + 4 a c - b^2} {4 a} }^{n + \frac 1 2} \rd x
      | c = Completing the Square
}}
{{eqn | r = \frac 1 {\paren {2 \sqrt a}^{2 n + 1} } \int \paren {\paren {2 a x + b}^2 + 4 a c - b^2}^{n + \frac 1 2} \rd x
      | c = Linear Combination of Primitives
}}
{{end-eqn}}
Let:
{{begin-eqn}}
{{eqn | l = z
      | r = \paren {2 a x + b}^2
}}
{{eqn | ll= \leadsto
      | l = \frac {\d z} {\d x}
      | r = 2 a \cdot 2 \paren {2 a x + b}
      | c = Derivative of Power and Chain Rule for Derivatives
}}
{{eqn | r = 4 a \sqrt z
      | c = 
}}
{{end-eqn}}
Also let $q = 4 a c - b^2$.
Then:
{{begin-eqn}}
{{eqn | l = \int \paren {a x^2 + b x + c}^{n + \frac 1 2} \rd x
      | r = \frac 1 {\paren {2 \sqrt a}^{2 n + 1} } \int \frac {\paren {z + q}^{n + \frac 1 2} } {4 a \sqrt z} \rd z
      | c = Integration by Substitution
}}
{{eqn | n = 1
      | r = \frac 1 {4 a \paren {2 \sqrt a}^{2 n + 1} } \int \frac {\paren {z + q}^{n + \frac 1 2} } {\sqrt z} \rd z
      | c = Linear Combination of Primitives
}}
{{end-eqn}}
From Primitive of $\dfrac {\paren {p x + q}^n} {\sqrt {a x + b} }$:
:$\ds \int \frac {\paren {p x + q}^n} {\sqrt {a x + b} } \rd x = \frac {2 \paren {p x + q}^n \sqrt {a x + b} } {\paren {2 n + 1} a} + \frac {2 n \paren {a q - b p} } {\paren {2 n + 1} a} \int \frac {\paren {p x + q}^{n - 1} } {\sqrt {a x + b} } \rd x$
Here $p = 1, a = 1, b = 0$ and $n := n + \dfrac 1 2$:
{{begin-eqn}}
{{eqn | o = 
      | r = \int \paren {a x^2 + b x + c}^{n + \frac 1 2} \rd x
      | c = 
}}
{{eqn | r = \frac 1 {4 a \paren {2 \sqrt a}^{2 n + 1} } \int \frac {\paren {z + q}^{n + \frac 1 2} } {\sqrt z} \rd z
      | c = from $(1)$
}}
{{eqn | r = \frac 1 {4 a \paren {2 \sqrt a}^{2 n + 1} } \paren {\frac {2 \paren {z + q}^{n + \frac 1 2} \sqrt z} {2 \paren {n + \frac 1 2} + 1} + \frac {2 \paren {n + \frac 1 2} q} {2 \paren {n + \frac 1 2} + 1} \int \frac {\paren {z + q}^{n - \frac 1 2} } {\sqrt z} \rd z}
      | c = Primitive of $\dfrac {\paren {p x + q}^n} {\sqrt {a x + b} }$
}}
{{eqn | r = \frac 1 {4 a \paren {2 \sqrt a}^{2 n + 1} } \paren {\frac {\paren {z + q}^{n + \frac 1 2} \sqrt z} {n + 1} + \frac {\paren {2 n + 1} q} {2 \paren {n + 1} } \int \paren {z + q}^{n - \frac 1 2} \frac {\d z } {\sqrt z} }
      | c = simplifying
}}
{{eqn | r = \frac 1 {4 a \paren {2 \sqrt a}^{2 n + 1} } \paren {\frac {\paren {\paren {2 a x + b}^2 + q}^{n + \frac 1 2} \paren {2 a x + b} } {n + 1} + \frac {\paren {2 n + 1} q} {2 \paren {n + 1} } \int \paren {\paren {2 a x + b}^2 + q}^{n - \frac 1 2} 4 a \rd x}
      | c = substituting for $z$
}}
{{eqn | r = \frac 1 {4 a \paren {n + 1} } \frac {\paren {\paren {2 a x + b}^2 + 4 a c - b^2}^{n + \frac 1 2} \paren {2 a x + b} } {\paren {4 a}^{n + \frac 1 2} }
      | c = substituting for $q$
}}
{{eqn | o = 
      | ro= +
      | r = \frac {\paren {2 n + 1} \paren {4 a c - b^2} } {8 a \paren {n + 1} } \int \frac {\paren {\paren {2 a x + b}^2 + \paren {4 a c - b^2} }^{n - \frac 1 2} } {\paren {4 a}^{n - \frac 1 2} } \rd x
      | c = and simplifying
}}
{{eqn | r = \frac {\paren {2 a x + b} \paren {a x^2 + b x + c}^{n + \frac 1 2} } {4 a \paren {n + 1} } + \frac {\paren {2 n + 1} \paren {4 a c - b^2} } {8 a \paren {n + 1} } \int \paren {a x^2 + b x + c}^{n - \frac 1 2} \rd x
      | c = Completing the Square
}}
{{end-eqn}}
{{qed}}