Answer
$$\frac{\sec ^{5} x}{5}-\frac{\sec ^{3} x}{3}+C $$
Work Step by Step
We evaluate the trigonometric integral as follows:
\begin{align*}
\int \tan ^{3} x \sec ^{3} x d x&=\int\left(\sec ^{2} x-1\right) \sec ^{2} x \cdot \sec x \cdot \tan x d x\\
&=\int \sec ^{4} x \cdot \sec x \cdot \tan x d x-\int \sec ^{2} x \cdot \sec x \cdot \tan x d x\\
&=\frac{\sec ^{5} x}{5}-\frac{\sec ^{3} x}{3}+C
\end{align*}
Where we used the fact that $\sec^2 x = \tan^2 x +1$