If $\int {\tan }^{7}x \mathrm{d}x=f\left(x\right)+C$ then 

Putting $\tan x=t$ we have $\mathrm{d}x=\frac{\mathrm{d}t}{1+t^2},$ So

$\int {\tan }^{7}x \mathrm{d}x=\int \frac{t^7}{1+t^2} \mathrm{d}t$

$=\int \left(t^5-t^3+t-\frac{t}{1+t^2}\right)\mathrm{d}t$

$=\frac{t^6}{6}-\frac{t^4}{4}+\frac{t^2}{2}-\frac{1}{2}\log \left(1+t^2\right)+C$

$=\frac{1}{6}{\tan }^{6}x-\frac{1}{4}{\tan }^{4}x+\frac{1}{2}{\tan }^{2}x$

$-\frac{1}{2}{\mathrm{logsec}}^{2}x+C$

$\begin{array}{r}=\frac{1}{6}{\tan }^{6}x-\frac{1}{4}{\tan }^{4}x+\frac{1}{2}{\tan }^{2}x\\ +\log |\cos x|+C.\end{array}$