Answer
CAS:$\displaystyle \quad\tan x+\frac{1}{3}\tan^{3}x+\mathrm{C}$
Table:$\displaystyle \quad=\frac{1}{3}\tan x\sec^{2}x+\frac{2}{3}\tan x+C$.
(equivalent, after applying $\tan^{2}x=\sec^{2}x-1$)
Work Step by Step
Using Table of integrals:
$\color{blue}{77. \displaystyle \quad\int\sec^{n}udu=\frac{1}{n-1}\tan u\sec^{n-2}u+\frac{n-2}{n-1}\int\sec^{n-2}udu }$
$\displaystyle \int\sec^{4}xdx=\frac{1}{3}\tan x\sec^{2}x+\frac{2}{3}\int\sec^{2}xdx$
$\color{blue}{8. \quad \displaystyle \int\sec^{2}udu=\tan u+C }$
$=\displaystyle \frac{1}{3}\tan x\sec^{2}x+\frac{2}{3}\tan x+C$.
Geogebra CAS gives $\displaystyle \int\sec^{4}xdx=\tan x+\frac{1}{3}\tan^{3}x+\mathrm{C}$
Applying the identity
$\tan^{2}x=\sec^{2}x-1$
$\displaystyle \tan x+\frac{1}{3}\tan^{3}x=\\=\tan x+\frac{1}{3}\tan x(\sec^{2}x-1)=\frac{2}{3}\tan x+\frac{1}{3}\tan x\sec^{2}x,$
so the results are equivalent.
