= Solution
Use <integration by substitution> $t=\tan x$, so $dt=dx/\cos^2x$. The <indefinite integral> becomes $\int dt/[t(t^2-1)]$. Its <partial fraction decomposition> is
$$
\frac1{t(t^2-1)}=-\frac1t+\frac12\frac1{t-1}+\frac12\frac1{t+1}.
$$
Integrating gives $-\ln|t|+\tfrac12\ln|t^2-1|+C$. Hence
$$
\boxed{\int\frac{dx}{\cos^2x(\tan^3x-\tan x)}=\frac12\ln\left|1-\frac1{\tan^2x}\right|+C.}
$$
The formula applies on each interval where the printed integrand is defined: $\cos x\ne0$ and $\tan x\notin\{0,1,-1\}$. The <constant of integration> may differ between such intervals.
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