Originally Posted by

**Matt Westwood** I might have an angle on this:

a) Complete the square to get the integral in the form $\displaystyle \int (\sqrt{z^2 + q})^n$

b) Substitute $\displaystyle u = z^2$ to get it in the form $\displaystyle \int \frac {(\sqrt{u + q})^n} {\sqrt u}$

c) Use the standard result $\displaystyle \int \frac {\left({p x + q}\right)^n} {\sqrt{a x + b} } \ \mathrm d x = \frac {2 \left({p x + q}\right)^n \sqrt{a x + b} } {\left({2 n + 1}\right) a} + \frac {2 n \left({a q - b p}\right)} {\left({2 n + 1}\right) a} \int \frac {\left({p x + q}\right)^{n-1} } {\sqrt{a x + b} } \ \mathrm d x$

but it's late and I need to sleep on it.