Answer
$$\int \tan ^{5} \frac{x}{4} \sec ^{4} \frac{x}{4} d x = \frac{2}{3}\tan ^{6}(x/4)+\frac{1}{2}\tan ^{8}(x/4) +c$$
Work Step by Step
$$
\int \tan ^{5} \frac{x}{4} \sec ^{4} \frac{x}{4} d x
$$
Let $ u= \frac{x}{4}\ \ \to \ \ du=\frac{1}{4}du$
\begin{align*}
\int \tan ^{5} \frac{x}{4} \sec ^{4} \frac{x}{4} d x&=\frac{1}{4}\int \tan ^{5}u \sec ^{4} u du\\
&=4\int \tan ^{5}u (1+\tan^2u)\sec ^{2} u du\\
&=4\int[\tan ^{5}u\sec ^{2} u +\tan^7u\sec ^{2} u] du\\
&=4\left( \frac{1}{6}\tan ^{6}u+\frac{1}{8}\tan ^{8}u \right)+c\\
&= \frac{2}{3}\tan ^{6}(x/4)+\frac{1}{2}\tan ^{8}(x/4) +c
\end{align*}
