Table of Integrals, Series, and Products, 8th Edition

$\begin{array}{ccc}{\displaystyle \int \frac{\text{d}x}{z^m{t}^n\sqrt{z}}}\hfill & =\hfill & \frac{\sqrt{z}}{z^m}[\frac{-1}{\left(n-1\right)\Delta }\frac{1}{t^{n-1}}\hfill \\ +{\displaystyle \sum _{k=2}^{n-1}{\left(-1\right)}^k\frac{\left(2n+2m-3\right)\left(2n+2m-5\right)\dots \left(2n+2m-2k+1\right)b^{k-1}}{2^{k-1}\left(n-1\right)\left(n-2\right)\dots \left(n-k\right){\Delta }^k}\cdot \frac{1}{t^{n-k}}]}\hfill \\ +{\left(-1\right)}^{n-1}\frac{\left(2n+2m-3\right)\left(2n+2m-5\right)\dots \left(-2m+3\right)\left(-2m+1\right)b^{n-1}}{2^{n-1}\left(n-1\right)!{\Delta }^{n-1}}{\displaystyle \int \frac{\text{d}x}{t{z}^m\sqrt{z}}}\hfill \\ \text{For}n=1\hfill \\ {\displaystyle \int \frac{\text{d}x}{z^{m}t\sqrt{z}}}\hfill & =\hfill & \frac{2}{\left(2m-1\right)\Delta }\frac{1}{z^{m-1}\sqrt{z}}+\frac{\beta }{\Delta }{\displaystyle \int \frac{\text{d}x}{t{z}^{m-1}\sqrt{z}}}\hfill \end{array}$

2.25 Forms containing $\sqrt{a+bx+c{x}^2}$

Integration techniques

2.251 It is possible to rationalize the integrand in integrals of the form ${\displaystyle \int R(}$