Hello Bob,
We are given to integrate:
$$\int\frac{\tan^3\left(\frac{1}{z} \right)}{z^2}\,dz$$
Let's use the substitution:
$$u=\frac{1}{z}\,\therefore\,du=-\frac{1}{z^2}\,dz$$
and the integral becomes:
$$-\int\tan^3(u)\,du$$
Next, let's employ the Pythagorean identity $$\tan^2(x)+1=\sec^2(x)$$, and rewrite the integral as:
$$-\int\tan(u)\cdot\tan^2(u)\,du=-\int\tan(u)\left(\sec^2(u)-1 \right)\,du=-\int\tan(u)\sec^2(u)-\tan(u)\,du=$$
$$\int\frac{\sin(u)}{\cos(u)}\,du-\int\tan(u)\sec^2(u)\,du$$
On the first integral, use the substitution:
$$v=\cos(u)\,\therefore\,dv=-\sin(u)\,du$$
On the second integral, use the substitution:
$$w=\tan(u)\,\therefore\,dw=\sec^2(u)\,du$$
And now we have:
$$-\int\frac{1}{v}\,dv-\int w\,dw=-\ln|v|-\frac{w^2}{2}+C$$
Back-substituting for $v$ and $w$, we have:
$$\ln|\sec(u)|-\frac{1}{2}\tan^2(u)+C$$
Back-substituting for $u$, we have:
$$\ln\left|\sec\left(\frac{1}{z} \right) \right|-\frac{1}{2}\tan^2\left(\frac{1}{z} \right)+C$$
Hence, we may state:
$$\int\frac{\tan^3\left(\frac{1}{z} \right)}{z^2}\,dz=\ln\left|\sec\left(\frac{1}{z} \right) \right|-\frac{1}{2}\tan^2\left(\frac{1}{z} \right)+C$$