Chapter 8 - Techniques of Integration - 8.2 Trigonometric Integrals - Exercises - Page 403: 36

$$\frac{1}{5}\sec^5\theta -\frac{1}{3}\sec^3\theta+c$$

\begin{align*} \int \tan ^{3} \theta \sec ^{3} \theta d \theta&=\int\left(\sec ^{2} \theta-1\right) \sec ^{2} \theta\sec \theta \tan \theta d \theta\\ &=\int\left(\sec ^{4} \theta-\sec ^{2} \theta\right) \sec \theta \tan \theta d \theta\\ &=\frac{1}{5}\sec^5\theta -\frac{1}{3}\sec^3\theta+c \end{align*}