integral of the product of a trigonometric and an exponential function Since tan has an odd power I would normally aim to sub $u=\sec(x)$, but I cant get rid of the $2^x$.
$$\int 2^x \tan^9(x^2)\sec(x^2)dx$$
I also tried integrating by parts but it got more complicated. 
 A: This is not an answer to the posted question (the $2^x$ should probably be $2x$ as discussed in the comments). However, I hope my solution will be still be instructive to those who happen to come across this question anyway:
$$\int 2x\tan^9(x^2)\sec(x^2) \, dx$$ We first start by simplifying the argument in the tangent and secant terms. Let's make a $u$-substitution where $u = x^2 \implies du = 2x \, dx$. So we have
$$ \int \tan^9(u)\sec(u) \, du$$
It takes some practice to see what another substitution should be. It has to do with the parity [even-ness vs. odd-ness] of both the powers of the tangent and secant. Click here for Pauls Online Notes, or here for some more general approaches to these types of integration problems along with examples. Anyway, the appropriate substitution is $v=\sec(u) \implies dv = \sec(u)\tan(u) \, du$ and we're left with $$\int \tan^9(u)\sec(u)\, du \\ = \int \tan(u)\tan^8(u)\sec(u)\,du \\ = \int \tan^8(u)\sec(u)\tan(u)\,du \\ = \int [\tan^2(u)]^4\sec(u)\tan(u)\,du \\ = \int [\sec^2(u)-1]^4 \sec(u)\tan(u)\,du \\ = \int [v^2-1]^4 \, dv$$
At this point it's just expanding into a polynomial, which is always welcome when it comes to integration:
$$ \int [v^2-1]^4 \, dv \\ = \int [(v^2-1)(v^2-1)]^2\,dv \\ = \int [v^4-2v^2+1]^2\, dv \\ = \int (v^4-2v^2+1)(v^4-2v^2+1)\,dv \\ = \int (v^8 -2v^6+v^4-2v^6+4v^4-2v^2+v^4-2v^2+1)\, dv \\ = \int (v^8 -4v^6 + 6v^4-4v^2+1)\,dv \\ = \frac{1}{9}v^9-\frac{4}{7}v^7+\frac{6}{5}v^5-\frac{4}{3}v^3+v + C$$
But we know $v=\sec(u)$, and that $u=x^2$ from earlier. So we know $v=\sec(x^2)$. Let's back-substitute and replace all the $v$'s with $\sec(x^2)$ to get:
$$ \frac{1}{9}\sec^9(x^2)-\frac{4}{7}\sec^7(x^2)+\frac{6}{5}\sec^5(x^2)-\frac{4}{3}\sec^3(x^2)+\sec(x^2)+C$$
