Answer
$$\frac{{{{\sec }^7}x}}{7} - \frac{{{{\sec }^5}x}}{5} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\sec }^5}x{{\tan }^3}x} dx \cr
& {\text{split }}{\sec ^5}x{\text{ and }}{\tan ^3}x \cr
& = \int {{{\sec }^4}x{{\tan }^2}x\sec x\tan x} dx \cr
& {\text{identity se}}{{\text{c}}^2}\theta = {\tan ^2}\theta + 1 \cr
& = \int {{{\sec }^4}x\left( {{{\sec }^2}x - 1} \right)\sec x\tan x} dx \cr
& = \int {\left( {{{\sec }^6}x - {{\sec }^4}x} \right)\sec x\tan x} dx \cr
& {\text{substitute }}u = \sec x,{\text{ }}du = \sec x\tan xdx \cr
& = \int {\left( {{u^6} - {u^4}} \right)} du \cr
& {\text{find the antiderivative by the power rule}} \cr
& = \frac{{{u^7}}}{7} - \frac{{{u^5}}}{5} + C \cr
& {\text{write in terms of }}x,{\text{ replace }}u = \sec x \cr
& = \frac{{{{\sec }^7}x}}{7} - \frac{{{{\sec }^5}x}}{5} + C \cr} $$
