Finding $\displaystyle \int\bigg(10\tan^3(x)+7\tan^2(x)+12\tan(x)+9\bigg)e^{x}dx$
What I tried
$$I=\int\bigg[10\tan(x)(1+\tan^2(x))+7(1+\tan^2(x))+2(\tan x+1)\bigg]e^x$$
$$I=\int \bigg[10\tan(x)\sec^2(x)+7\sec^2(x)+2(1+\tan x)\bigg]e^xdx$$
How do I solve it? Help me please.
The Ansatz $I=f(t)e^x+C,\,t:=\tan x$ gives $f(t)+(t^2+1)f^\prime(t)=10t^3+7t^2+12t+9$. Now seek a polynomial choice of $f$. It will be at most quadratic, say $f(t)=At^2+Bt+C$, so$$\begin{align}10t^3+7t^2+12t+9&=2At^3+(A+B)t^2+(2A+B)t+B+C\\\implies(A,\,B,\,C)&=(5,\,2,\,7).\end{align}$$
Hint: Integration by parts will be most helpful, i.e. $$I = uv - \int v \ du$$ by letting $u = 10\tan(x)\sec^2(x)+7\sec^2(x)+2(1+\tan x)$ and $dv = e^x dx$ for a start.
It might take you a few passes to get it to a reasonable form.
