Reduction Formula for Primitive of Power of a x + b by Power of p x + q/Increment of Power

Theorem

 * $\ds \int \paren {a x + b}^m \paren {p x + q}^n \rd x = \frac 1 {\paren {n + 1} \paren {b p - a q} } \paren {\paren {a x + b}^{m + 1} \paren {p x + q}^{n + 1} - a \paren {m + n + 2} \int \paren {a x + b}^m \paren {p x + q}^{n + 1} \rd x}$

Proof
From Reduction Formula for Primitive of Power of $a x + b$ by Power of $p x + q$: Decrement of Power:


 * $\ds \int \paren {a x + b}^m \paren {p x + q}^n \rd x = \frac {\paren {a x + b}^{m + 1} \paren {p x + q}^n} {\paren {m + n + 1} a} - \frac {n \paren {b p - a q} } {\paren {m + n + 1} a} \int \paren {a x + b}^m \paren {p x + q}^{n - 1} \rd x$

Putting $n + 1$ for $n$, this yields:

Also defined as
This can also be reported as:
 * $\ds \int \paren {a x + b}^m \paren {p x + q}^n \rd x = \frac 1 {\paren {m + 1} \paren {a q - b p} } \paren {\paren {a x + b}^{m + 1} \paren {p x + q}^{n + 1} - p \paren {m + n + 2} \int \paren {a x + b}^{m + 1} \paren {p x + q}^n \rd x}$

by interchanging the roles of $m$ and $n$.