Answer
$$\frac{{{{\sec }^4}\theta }}{4} - \frac{{{{\sec }^3}\theta }}{3} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\tan }^3}\theta {{\sec }^3}\theta d\theta } \cr
& {\text{split functions}} \cr
& = \int {{{\tan }^2}\theta {{\sec }^2}\theta \sec \theta \tan \theta d\theta } \cr
& {\text{use }}{\sec ^2}\theta - 1 = {\tan ^2}\theta \cr
& = \int {\left( {{{\sec }^2}\theta - 1} \right){{\sec }^2}\theta \sec \theta \tan \theta d\theta } \cr
& = \int {\left( {{{\sec }^3}\theta - {{\sec }^2}\theta } \right)\sec \theta \tan \theta d\theta } \cr
& u = \sec \theta ,{\text{ }}du = \sec \theta \tan \theta d\theta \cr
& = \int {\left( {{u^3} - {u^2}} \right)du} \cr
& {\text{find the antiderivative}} \cr
& = \frac{{{u^4}}}{4} - \frac{{{u^3}}}{3} + C \cr
& {\text{replacing }}u = \sec \theta \cr
& = \frac{{{{\sec }^4}\theta }}{4} - \frac{{{{\sec }^3}\theta }}{3} + C \cr} $$
