$\int \tan x \tan 2 x \tan 3 x ~d x$ is equal to

$\frac{1}{3} \log |\sec 3 x|-\frac{1}{2} \log |\sec 2 x|-\log |\sec x|+C$

We have,

$\tan 3 x=\tan (2 x+x)$

$\Rightarrow \tan 3 x=\frac{\tan 2 x+\tan x}{1-\tan x \tan 2 x}$

$\Rightarrow \tan x \tan 2 x \tan 3 x=\tan 3 x-\tan 2 x-\tan x$

∴  $\int \tan x \tan 2 x \tan 3 x d x$

$=\int(\tan 3 x-\tan 2 x-\tan x) d x$

$=\frac{1}{3} \log |\sec 3 x|-\frac{1}{2} \log |\sec 2 x|-\log |\sec x|+C$