Answer
$$\frac{1}{4} \tan x \sec ^{3} x-\frac{5}{8} \tan x \sec x+\frac{3}{8} \ln |\sec x+\tan x|+C$$
Work Step by Step
Use
$$\int \sec ^{m} x d x=\frac{\tan x \sec ^{m-2} x}{m-1}+\frac{m-2}{m-1} \int \sec ^{m-2} x d x$$
Then
\begin{align*}
\int \tan ^{4} x \sec x d x&=\int\left(\sec ^{2} x-1\right)^{2} \sec x d x\\
&=\int \sec ^{5} x d x-2 \int \sec ^{3} x d x+\int \sec x d x\\
&=\left(\frac{1}{4} \tan x \sec ^{3} x+\frac{3}{4} \int \sec ^{3} x d x\right)-2 \int \sec ^{3} x d x+\int \sec x d x\\
&=\frac{1}{4} \tan x \sec ^{3} x-\frac{5}{4} \int \sec ^{3} x d x+\int \sec x d x\\
&=\frac{1}{4} \tan x \sec ^{3} x-\frac{5}{4}\left(\frac{1}{2} \tan x \sec x+\frac{1}{2} \int \sec x d x\right)+\int \sec x d x\\
&=\frac{1}{4} \tan x \sec ^{3} x-\frac{5}{8} \tan x \sec x+\frac{3}{8} \int \sec x d x\\
&=\frac{1}{4} \tan x \sec ^{3} x-\frac{5}{8} \tan x \sec x+\frac{3}{8} \ln |\sec x+\tan x|+C
\end{align*}
