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