$\int \frac{\mathrm{dx}}{{\cos }^3\mathrm{x}\sqrt{2\sin 2\mathrm{x}}}$ is equal to 

$\mathrm{Let} \begin{array}{l}\mathrm{I}=\int \frac{\mathrm{dx}}{{\cos }^3\mathrm{x}\sqrt{2\sin 2\mathrm{x}}}\\ =\int \frac{\mathrm{dx}}{{\cos }^3\mathrm{x}\sqrt{4\sin \mathrm{xcos}\mathrm{x}}}=\frac{1}{2}\int \frac{\mathrm{dx}}{{\cos }^{7/2}{\mathrm{xsin}}^{1/2}\mathrm{x}}\\ =\frac{1}{2}\int \frac{{\sec }^4\mathrm{x}}{\sqrt{\tan \mathrm{x}}}\mathrm{dx}=\frac{1}{2}\int \frac{\left(1+{\tan }^2\mathrm{x}\right){\sec }^2\mathrm{x}}{\sqrt{\tan \mathrm{x}}}\mathrm{dx}\end{array}$

$\begin{array}{l}\text{ Put }\tan \mathrm{x}=\mathrm{t}\Rightarrow {\sec }^2\mathrm{xdx}=\mathrm{dt}\\ \therefore \mathrm{I}=\frac{1}{2}\int \frac{1+{\mathrm{t}}^2}{\sqrt{\mathrm{t}}}\mathrm{dt}\\ =\frac{1}{2}\int {\mathrm{t}}^{-1/2}\mathrm{dt}+\frac{1}{2}\int {\mathrm{t}}^{3/2}\mathrm{dt}\\ ={\mathrm{t}}^{1/2}+\frac{{\mathrm{t}}^{5/2}}{5}+\mathrm{C}\\ =\sqrt{\tan \mathrm{x}}+\frac{1}{5}{\tan }^{5/2}\mathrm{x}+\mathrm{C}\end{array}$