# Prove that primitives of $\frac{x^3}{{\rm e}^x - 1}$ have no closed form in terms of elementary functions

It is known the following indefinite integral $$\int \frac{x^3}{{\rm e}^x - 1} dx$$ cannot be evaluated in closed form in terms of any of the elementary functions of mathematics. A proof of this can be found here. The proof given there expresses the integral in terms of four infinite series, one of which is the dilogarithm function. This term is then shown it cannot be expressed in closed form in terms of elementary functions using the Risch algorithm.

My question is, using the Risch algorithm, is it possible to shown directly from the form of the integral given above that it cannot be expressed in closed form in terms of any of the elementary functions of mathematics?

I must confess my experience in working with the Risch algorithm is rather limited (and yes I do understand a 100 plus page document exists somewhere which completely describes its implementation) so any outline of a possible proof would be greatly appreciated.

Let $\theta = \exp(x)$. Then your problem becomes $$\int \frac{x^3}{\exp(x)-1}dx = \int \frac{p(\theta)}{q(\theta)}dx$$ where $p(\theta) = x^3 \in \mathbb{Q}(x)[\theta]$ and $q(\theta) = \theta-1 \in \mathbb{Q}(x)[\theta]$.
Because $\deg_\theta(p(\theta)) = 0 < 1 = \deg_\theta(q(\theta))$ and $q$ is a monic, square-free polynomial and $\gcd(p,q) = \gcd(q,\theta) = 1$, we may apply the Rothstein-Trager method: \begin{align} R(z) = \operatorname{res}_\theta(p(\theta)-zq'(\theta),q(\theta)) &= \operatorname{res}_\theta(x^3-z,\theta-1) = x^3-z \end{align} The root of $R(z)$ is $z = x^3$. Since $x^3 \not \in \mathbb{\bar{Q}}$, the integral is not elementary.