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

Theorem

 * $\displaystyle \int \left({a x + b}\right)^m \left({p x + q}\right)^n \ \mathrm d x = \frac 1 {\left({n + 1}\right) \left({b p - a q}\right)} \left({\left({a x + b}\right)^{m+1} \left({p x + q}\right)^{n+1} - a \left({m + n + 2}\right) \int \left({a x + b}\right)^m \left({p x + q}\right)^{n+1} \ \mathrm d x}\right)$

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


 * $\displaystyle \int \left({a x + b}\right)^m \left({p x + q}\right)^n \ \mathrm d x = \frac {\left({a x + b}\right)^{m+1} \left({p x + q}\right)^n} {\left({m + n + 1}\right) a} - \frac {n \left({b p - a q}\right)} {\left({m + n + 1}\right) a} \int \left({a x + b}\right)^m \left({p x + q}\right)^{n-1} \ \mathrm d x$

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

Also defined as
This can also be reported as:
 * $\displaystyle \int \left({a x + b}\right)^m \left({p x + q}\right)^n \ \mathrm d x = \frac 1 {\left({m + 1}\right) \left({a q - b p}\right)} \left({\left({a x + b}\right)^{m+1} \left({p x + q}\right)^{n+1} - p \left({m + n + 2}\right) \int \left({a x + b}\right)^{m+1} \left({p x + q}\right)^n \ \mathrm d x}\right)$

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