Answer
$$\frac{{{{\tan }^8}x}}{8} + \frac{{{{\tan }^6}x}}{6} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\tan }^5}x{{\sec }^4}x} dx \cr
& {\text{split }}{\sec ^4}x \cr
& = \int {{{\tan }^5}x{{\sec }^2}x{{\sec }^2}x} dx \cr
& {\text{identity se}}{{\text{c}}^2}x = {\tan ^2}x + 1 \cr
& = \int {{{\tan }^5}x\left( {{{\tan }^2}x + 1} \right){{\sec }^2}x} dx \cr
& = \int {\left( {{{\tan }^7}x + {{\tan }^5}x} \right){{\sec }^2}x} dx \cr
& {\text{substitute }}u = \tan x,{\text{ }}du = {\sec ^2}xdx \cr
& = \int {\left( {{u^7} + {u^5}} \right)du} \cr
& {\text{find the antiderivative by the power rule}} \cr
& = \frac{{{u^8}}}{8} + \frac{{{u^6}}}{6} + C \cr
& {\text{write in terms of }}x,{\text{ replace }}u = \tan x \cr
& = \frac{{{{\tan }^8}x}}{8} + \frac{{{{\tan }^6}x}}{6} + C \cr} $$
