Answer
$$\frac{1}{6} \tan \theta \sec ^{5} \theta-\frac{7}{24} \tan \theta \sec ^{3} \theta+\frac{1}{16} \tan \theta \sec \theta+\frac{1}{16} \ln |\sec \theta+\tan \theta|+C$$
Work Step by Step
Since
\begin{align*}
\int \sec ^{3} \theta \tan ^{4} \theta d \theta&=\int \sec ^{3} \theta(\sec ^{2}\theta-1) ^{2} d \theta\\
&=\int\left(\sec ^{7} \theta-2 \sec ^{5} \theta+\sec ^{3} \theta\right) d \theta
\end{align*}
Since
$$\int \sec ^{m} x d x=\frac{\tan x \sec ^{m-2} x}{m-1}+\frac{m-2}{m-1} \int \sec ^{m-2} x d x$$
then
\begin{aligned}
\int \sec ^{3} \theta d \theta
&=\frac{\tan \theta \sec \theta}{2}+\frac{1}{2} \int \sec \theta d \theta \\
&=\frac{1}{2} \tan \theta \sec \theta+\frac{1}{2} \ln |\sec \theta+\tan \theta|+C
\end{aligned}
and
\begin{aligned}
\int \sec ^{5} \theta d \theta
&=\frac{\tan \theta \sec ^{3} \theta}{4}+\frac{3}{4} \int \sec ^{3} \theta d \theta \\
&=\frac{1}{4} \tan \theta \sec ^{3} \theta+\frac{3}{4}\left[\frac{1}{2} \tan \theta \sec \theta+\frac{1}{2} \ln |\sec \theta+\tan \theta|\right]+C \\
&=\frac{1}{4} \tan \theta \sec ^{3} \theta+\frac{3}{8} \tan \theta \sec \theta+\frac{3}{8} \ln |\sec \theta+\tan \theta|+C
\end{aligned}
and
\begin{aligned}
\int \sec ^{7} \theta d \theta
&=\frac{\tan \theta \sec ^{5} \theta}{6}+\frac{5}{6} \int \sec ^{5} \theta d \theta \\
&=\frac{\tan \theta \sec ^{5} \theta}{6}+\frac{5}{6}\left[\frac{1}{4} \tan \theta \sec ^{3} \theta+\frac{3}{8} \tan \theta \sec \theta+\frac{3}{8} \ln |\sec \theta+\tan \theta|\right]+C \\
&=\frac{1}{6} \tan \theta \sec ^{5} \theta+\frac{5}{24} \tan \theta \sec ^{3} \theta+\frac{15}{48} \tan \theta \sec \theta+\frac{15}{48} \ln |\sec \theta+\tan \theta|+C
\end{aligned}
Hence
\begin{align*}
\int \sec ^{3} \theta \tan ^{4} \theta d \theta&=\int \sec ^{3} \theta(\sec ^{2}\theta-1) ^{2} d \theta\\
&=\int\left(\sec ^{7} \theta-2 \sec ^{5} \theta+\sec ^{3} \theta\right) d \theta\\
&= \frac{1}{6} \tan \theta \sec ^{5} \theta-\frac{7}{24} \tan \theta \sec ^{3} \theta+\frac{1}{16} \tan \theta \sec \theta+\frac{1}{16} \ln |\sec \theta+\tan \theta|+C
\end{align*}
