What is the integral of $\coth(\sqrt{\tan(x)+1}*\cos(x\sin(x)))$? I don't know where to start. Trying to take a Taylor series, the first derivative is:$$\frac{d}{dx}(\coth(\sqrt{\tan(x) + 1}\cos(x\sin(x)))) = \frac{\sec^2(x)\cos(x\sin(x))}{(2\sqrt{\tan(x) + 1}}-\sin(x \sin(x))\sqrt{\tan(x) + 1}(\sin(x)+x\cos(x)))(-csch^2(\sqrt{\tan(x) + 1}\cos(x\sin(x))))$$Which is very complicated, so there is no use trying to get a Taylor Series. Please help!
 A: The limit of your imagination seems to very very close to $\infty$ (which is not bad).
Considering the monster
$$f(x)=\coth(\cos(x\sin(x))\sqrt{\tan(x)+1})$$ it has vertical asymptotes at $x=-\frac \pi 4$ and $x=\frac \pi 2$. However, the function
$$g(x)=\left(\frac{\pi }{2}-x\right) \left(x+\frac{\pi }{4}\right)f(x)$$ is not so bad and does not any discontinuity or asymptote.
So, suppose that you find a good polynomial which fits $g(x)$, that is to say
$$g(x)=\sum_{n=0}^p a_n \,x^n$$ which makes that
$$\int f(x)\,dx=\sum_{n=0}^p a_n\int\frac{x^n}{ \left(x+\frac{\pi }{4}\right)\left(\frac{\pi }{2}-x\right)}\,dx$$ Each ot these integrals
$$I_n=\int\frac{x^n}{ \left(x+\frac{\pi }{4}\right)\left(\frac{\pi }{2}-x\right)}\,dx$$ is simple since
$$\frac{3 \pi  n }{4}I_n=\frac{2 x^{n+1}}{\pi -2
   x} \, _2F_1\left(1,1;1-n;\frac{\pi }{\pi -2 x}\right)+$$ $$\left(x+\frac{\pi }{4}\right)^n \, _2F_1\left(-n,-n;1-n;\frac{\pi }{4 x+\pi }\right)$$
Now, my question : are you sure that you will not prefer numerical integration ?
