<s>[INST] Evaluating $\int (\tan^3x+\tan^4x) dx $ using substitution solving $$\int (\tan^3x+\tan^4x) dx $$ using substitution $$t = \tan x$$
My approach has led me to $ \int (1+t)t\sin^2xdt$ which has an $x$ too much and isn't easily solvable for me. If I remove the $x$ I get $\sin^2(\arctan(t))$ and that's not too nice to work with...
 [/INST] 
unfortunately $\sec$ is not used in my country

If you feel unconfortable with $\sec $ you may proceed from your last integral
\begin{equation*}
I=\int (1+t)t\sin ^{2}\left( \arctan t\right) \,dt
\end{equation*}
by using the following trigonometric identity
\begin{equation*}
\sin ^{2}x=\frac{\tan ^{2}x}{1+\tan ^{2}x},\qquad x=\arctan t,
\end{equation*}
which you can derive from the fundamental identity $\sin^2 x+\cos^2 x=1$ to obtain 
\begin{equation*}
\sin ^{2}\left( \arctan t\right) =\frac{t^{2}}{1+t^{2}}.
\end{equation*}
Consequently
\begin{equation*}
I=\int (1+t)t\frac{t^{2}}{1+t^{2}}\,dt=\int \frac{t^{3}+t^{4}}{1+t^{2}}\,dt,
\end{equation*}
which is integrable by partial fractions. By long division we compute 
\begin{equation*}
\frac{t^{4}+t^{3}}{1+t^{2}}=t^{2}+t-1+\frac{-t+1}{1+t^{2}}.
\end{equation*}
So
\begin{equation*}
I=\int \left( t^{2}+t-1\right) dt+\int \frac{-t+1}{1+t^{2}}dt.
\end{equation*}
Since 
\begin{eqnarray*}
\int \frac{-t+1}{1+t^{2}}dt &=&-\frac{1}{2}\int \frac{2t}{1+t^{2}}dt+\int 
\frac{1}{1+t^{2}}dt \\
&=&-\frac{1}{2}\ln \left( 1+t^{2}\right) +\arctan t+C,
\end{eqnarray*}
we thus have
\begin{eqnarray*}
I &=&\frac{1}{3}t^{3}+\frac{1}{2}t^{2}-t-\frac{1}{2}\ln \left(
1+t^{2}\right) +\arctan t+C \\
&=&\frac{1}{3}\tan ^{3}x+\frac{1}{2}\tan ^{2}x-\tan x-\frac{1}{2}\ln \left(
1+\tan ^{2}x\right) +x+C.
\end{eqnarray*}
</s>