Integration of $1/(x^4 \sin x +2x)$ $$\int\frac{1}{x^4 \sin x +2x} dx\ $$  How to evaluate this integral. How to go about evaluating these integrals?
 A: Maple does not find a closed form for the antiderivative.  Since this is the transcendental case, for which Maple's implementation of the Risch algorithm is
supposed to be complete, that should indicate that there is no elementary antiderivative.
Wolfram Alpha also finds no closed form.
A: As Robert Israel answered, there is no closed form for the antiderivative.
I you need to compute the integral over a short range (far away from $\pi$), you could expand the integrand as a Taylor series built at $x=0$. What you would get is $$\frac{1}{x^4 \sin x +2x}=\frac{1}{2 x}-\frac{x^3}{4}+\frac{x^5}{24}+\frac{59 x^7}{480}-\frac{839
   x^9}{20160}-\frac{82657 x^{11}}{1451520}+O\left(x^{13}\right) $$ and so, for the antiderivative $$\int\frac{dx}{x^4 \sin x +2x}=\frac{\log (x)}{2}-\frac{x^4}{16}+\frac{x^6}{144}+\frac{59 x^8}{3840}-\frac{839
   x^{10}}{201600}-\frac{82657 x^{12}}{17418240}+O\left(x^{14}\right)$$ You could observe that the plots of the integrand and the expansion are extremely close up to $x=0.5$
A: Hint:
$\int\dfrac{1}{x^4\sin x+2x}dx$
$=\int\dfrac{1}{2x\left(1+\dfrac{x^3\sin x}{2}\right)}dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{3n-1}\sin^nx}{2^{n+1}}dx$
$=\int\left(\dfrac{1}{2x}+\sum\limits_{n=1}^\infty\dfrac{(-1)^nx^{3n-1}\sin^nx}{2^{n+1}}\right)dx$
But this approach is only suitable for the very limited ranges of $x$ : $\left|\dfrac{x^3\sin x}{2}\right|\leq1$
