Evaluate the following indefinite integral ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}se{c}^4()ta{n}^2()d$

$\frac{1}{3}ta{n}^3()C$

$\frac{1}{3}ta{n}^3()\frac{1}{5}ta{n}^5()C$

$($C$)$

$\frac{1}{3}sec()tan()C$

$\frac{1}{3}se{c}^3()tan()tan()C$

$se{c}^3()ta{n}^5()C$