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:

 $\displaystyle v$ $=$ $\displaystyle \left({a x + b}\right)^s$ $\displaystyle \implies \ \$ $\displaystyle \frac {\mathrm d v} {\mathrm d x}$ $=$ $\displaystyle a s \left({a x + b}\right)^{s-1}$ Derivative of Power and Derivative of Function of Constant Multiple: Corollary

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

 $\displaystyle u$ $=$ $\displaystyle \frac {\left({a x + b}\right)^{m - s + 1} } {a s} \left({p x + q}\right)^n$ $\displaystyle \implies \ \$ $\displaystyle \frac {\mathrm d u} {\mathrm d x}$ $=$ $\displaystyle \frac {a \left({m - s + 1}\right) \left({a x + b}\right)^{m - s} \left({p x + q}\right)^n + p n \left({a x + b}\right)^{m - s + 1} \left({p x + q}\right)^{n-1} } {a s}$ Product Rule for Derivatives and above $\displaystyle$ $=$ $\displaystyle \frac {\left({a x + b}\right)^{m - s} \left({p x + q}\right)^{n-1} } {a s} \left({a \left({m - s + 1}\right) \left({p x + q}\right) + p n \left({a x + b}\right) }\right)$ extracting common factor $\displaystyle$ $=$ $\displaystyle \frac {\left({a x + b}\right)^{m - s} \left({p x + q}\right)^{n-1} } {a s} \left({a p x \left({m - s + 1 + n}\right) + a q \left({m - s + 1}\right) + p b n}\right)$ separating out terms in $x$

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

 $\displaystyle \frac {\mathrm d u} {\mathrm d x}$ $=$ $\displaystyle \frac {\left({a x + b}\right)^{m - s} \left({p x + q}\right)^{n-1} } {a \left({m + n + 1}\right)} \left({a q \left({m - \left({m + n + 1}\right) + 1}\right) + p b n}\right)$ term in $x$ vanishes $\displaystyle$ $=$ $\displaystyle \frac {\left({a x + b}\right)^{m - s} \left({p x + q}\right)^{n-1} } {a \left({m + n + 1}\right)} \left({n \left({p b - a q}\right)}\right)$ simplifying

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

Thus:

 $\displaystyle$  $\displaystyle \int \left({a x + b}\right)^m \left({p x + q}\right)^n \ \mathrm d x$ $\displaystyle$ $=$ $\displaystyle \int \frac {\left({a x + b}\right)^{m - s + 1} } {a s} \left({p x + q}\right)^n a s \left({a x + b}\right)^{s-1} \ \mathrm d x$ in the form $\displaystyle \int u \frac {\mathrm d v} {\mathrm d x} \ \mathrm d x$ $\displaystyle$ $=$ $\displaystyle \frac {\left({a x + b}\right)^{m - s + 1} } {a s} \left({p x + q}\right)^n \left({a x + b}\right)^s$ in the form $\displaystyle u v - \int v \frac {\mathrm d u} {\mathrm d x} \ \mathrm d x$ $\displaystyle$  $\, \displaystyle - \,$ $\displaystyle \int \left({a x + b}\right)^s \frac {\left({a x + b}\right)^{m - s} \left({p x + q}\right)^{n-1} } {a \left({m + n + 1}\right)} \left({n \left({p b - a q}\right)}\right) \ \mathrm d x$ $\displaystyle$ $=$ $\displaystyle \frac {\left({a x + b}\right)^{m+1} \left({p x + q}\right)^n} {\left({m + n + 1}\right) a} - \frac {n \left({b p - a q}\right)} {\left({m + n + 1}\right) a} \int \left({a x + b}\right)^m \left({p x + q}\right)^{n-1} \ \mathrm d x$ Primitive of Constant Multiple of Function

$\blacksquare$

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 $n$.