Does there exist a general closed form solution (in terms of elementary or special functions) to the differential equation:
$$ \frac{df(x)}{dx} = \frac{P(f(x))}{P(x)} $$
when $P(x)$ is a polynomial of degree higher than 3? (excluding the trivial case $f(x)=x$).
Context:
I'm trying to find the action of a certain class of composition operators
$$C_f(x,\frac{d}{dx}) = e^{P(x) \frac{d}{dx}} $$
where $P(x)$ is a polynomial in $\mathbb{C}$ of degree $n \geq 3$, such that for a complex function $g$
$$C_f(g) = g \circ f$$
After some manipulations, one arrives at the Abel equation
$$ f(x) = \alpha^{-1}(\alpha(x) + 1) $$
where
$$ \alpha(x) = \int^x \frac{dt}{P(t)} $$
Differentiating this last expression, one obtains a differential equation that all the family of iterations of $f$ (even fractional ones) must satisfy:
$$ \frac{df(x)}{dx} = \frac{P(f(x))}{P(x)} $$
I already know the basic properties of this function, and I know how to calculate it numerically. What I'm trying to find is whether there exists a general closed form expression for $f$ when $\deg P \geq 3$ (in the case $n \leq 2$, $f(x)$ is a Möbius transformation).
There are some special cases I've checked manually, such as the case $P(x) = ax^n$, whose solution is a combination of a rational function and radicals, but I don't know if this holds in general, or how to prove it.
After revisiting my question I think I found an answer. Let $P(x)$ be a monic polynomial of degree $3$ for simplicity, and furthermore, let two of its roots be $0$ and $1$, and the third one be real. This polynomial can thus be put in the form:
$$P(x)=x(x-1)(x-r)$$
where $r\neq 0, 1$ is the third root.
The problem then reduces to finding the explicit form of an inverse function of
$$\int^x \frac{dt}{P(t)} = \frac1{r(r-1)} \left( (r-1) \log x - r \log(1-x) + \log(x-r)\right) = \alpha(x)$$
Between two of the logarithmic singularities the function (regarded as a real function) is one-to-one, so an inverse exists. Grouping the logarithms and simplifying, we obtain
$$\alpha(x) = \frac1{r(1-r)} \log \left( \frac{(\frac1x-1)^r} {1-\frac rx} \right) $$
To find the inverse, we must solve (setting $z = \frac1x$):
$$y = \frac{(z-1)^r} {1- rz}$$
for $z$ in terms of $y$. If $r$ is a natural number this is equivalent to solving a polynomial equation of degree $r$, and we know from Galois theory that there isn't a general closed form algebraic solution except when $r<5$. If $r$ is irrational the situation is even worse.
So it looks like the answer to my question is no, except in some special cases. There is still the possibility that although the inverse function $\alpha^{-1}(x)$ isn't expressible in closed form, its composition with $\alpha(x) + c$, where $c\neq0$ is an appropriate constant, leads to a closed form expression, but it seems unlikely and I don't know how to explore that possibility anyways.