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

Theorem

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

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

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

In order to make $u \dfrac {\d v} {\d x}$ 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:
 * $\ds \int \paren {a x + b}^m \paren {p x + q}^n \rd x = \frac {\paren {a x + b}^m \paren {p x + q}^{n + 1} } {\paren {m + n + 1} p} + \frac {m \paren {b p - a q} } {\paren {m + n + 1} p} \int \paren {a x + b}^{m - 1} \paren {p x + q}^n \rd x$

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