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

Theorem

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

Proof
Aiming for an expression in the form:
 * $\displaystyle \int u \frac {\mathrm d v} {\mathrm d x} \ \mathrm d x = u v - \int v \ \frac {\mathrm d u} {\mathrm d x} \ \mathrm d x$

in order to use the technique of Integration by Parts, let:

In order to make $u \dfrac {\mathrm d v} {\mathrm d x}$ equal to the integrand, let:

Select $s$ such that $m - s + n + 1 = 0$, and so $s = m + n + 1$:

Other instances of $s$ are left as they are, anticipating that they will cancel out later.

Thus:

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

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