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

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Theorem

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


This needs considerable tedious hard slog to complete it.
In particular: This only takes on the case where $a > 0$. The case where $a < 0$ needs to be addressed.
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\(\ds \int \paren {a x^2 + b x + c}^{n + \frac 1 2} \rd x\) \(=\) \(\ds \int \paren {\frac {\paren {2 a x + b}^2 + 4 a c - b^2} {4 a} }^{n + \frac 1 2} \rd x\) Completing the Square
\(\ds \) \(=\) \(\ds \frac 1 {\paren {2 \sqrt a}^{2 n + 1} } \int \paren {\paren {2 a x + b}^2 + 4 a c - b^2}^{n + \frac 1 2} \rd x\) Linear Combination of Primitives


Let:

\(\ds z\) \(=\) \(\ds \paren {2 a x + b}^2\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d z} {\d x}\) \(=\) \(\ds 2 a \cdot 2 \paren {2 a x + b}\) Derivative of Power and Chain Rule for Derivatives
\(\ds \) \(=\) \(\ds 4 a \sqrt z\)


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


Then:

\(\ds \int \paren {a x^2 + b x + c}^{n + \frac 1 2} \rd x\) \(=\) \(\ds \frac 1 {\paren {2 \sqrt a}^{2 n + 1} } \int \frac {\paren {z + q}^{n + \frac 1 2} } {4 a \sqrt z} \rd z\) Integration by Substitution
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds \frac 1 {4 a \paren {2 \sqrt a}^{2 n + 1} } \int \frac {\paren {z + q}^{n + \frac 1 2} } {\sqrt z} \rd z\) Linear Combination of Primitives

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 $n := n + \dfrac 1 2$:

\(\ds \) \(\) \(\ds \int \paren {a x^2 + b x + c}^{n + \frac 1 2} \rd x\)
\(\ds \) \(=\) \(\ds \frac 1 {4 a \paren {2 \sqrt a}^{2 n + 1} } \int \frac {\paren {z + q}^{n + \frac 1 2} } {\sqrt z} \rd z\) from $(1)$
\(\ds \) \(=\) \(\ds \frac 1 {4 a \paren {2 \sqrt a}^{2 n + 1} } \paren {\frac {2 \paren {z + q}^{n + \frac 1 2} \sqrt z} {2 \paren {n + \frac 1 2} + 1} + \frac {2 \paren {n + \frac 1 2} q} {2 \paren {n + \frac 1 2} + 1} \int \frac {\paren {z + q}^{n - \frac 1 2} } {\sqrt z} \rd z}\) Primitive of $\dfrac {\paren {p x + q}^n} {\sqrt {a x + b} }$
\(\ds \) \(=\) \(\ds \frac 1 {4 a \paren {2 \sqrt a}^{2 n + 1} } \paren {\frac {\paren {z + q}^{n + \frac 1 2} \sqrt z} {n + 1} + \frac {\paren {2 n + 1} q} {2 \paren {n + 1} } \int \paren {z + q}^{n - \frac 1 2} \frac {\d z } {\sqrt z} }\) simplifying
\(\ds \) \(=\) \(\ds \frac 1 {4 a \paren {2 \sqrt a}^{2 n + 1} } \paren {\frac {\paren {\paren {2 a x + b}^2 + q}^{n + \frac 1 2} \paren {2 a x + b} } {n + 1} + \frac {\paren {2 n + 1} q} {2 \paren {n + 1} } \int \paren {\paren {2 a x + b}^2 + q}^{n - \frac 1 2} 4 a \rd x}\) substituting for $z$
\(\ds \) \(=\) \(\ds \frac 1 {4 a \paren {n + 1} } \frac {\paren {\paren {2 a x + b}^2 + 4 a c - b^2}^{n + \frac 1 2} \paren {2 a x + b} } {\paren {4 a}^{n + \frac 1 2} }\) substituting for $q$
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \frac {\paren {2 n + 1} \paren {4 a c - b^2} } {8 a \paren {n + 1} } \int \frac {\paren {\paren {2 a x + b}^2 + \paren {4 a c - b^2} }^{n - \frac 1 2} } {\paren {4 a}^{n - \frac 1 2} } \rd x\) and simplifying
\(\ds \) \(=\) \(\ds \frac {\paren {2 a x + b} \paren {a x^2 + b x + c}^{n + \frac 1 2} } {4 a \paren {n + 1} } + \frac {\paren {2 n + 1} \paren {4 a c - b^2} } {8 a \paren {n + 1} } \int \paren {a x^2 + b x + c}^{n - \frac 1 2} \rd x\) Completing the Square

$\blacksquare$


Sources

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