Are there basic techniques for proving that no elementary antiderivative exists?

Question

Background: It seems there is an algorithm called the full Risch algorithm that can decide whether a given function has an antiderivative that can be expressed in terms of elementary functions. There is also an older "(classical) Risch algorithm," which attempts to do the same, but without guaranteed success. The distinction between these algorithms, and whether they work, and whether they have been fully implemented, is apparently controversial. (See also)

Question: Given a particular function $f$, are there any "basic" ways that one might be able to show that $f$ has no elementary antiderivative? Do there exist standard techniques for this? Are there any easy examples?

Edit: Relaxing the question slightly, rather than seeking general methods that might apply to many functions, even special cases are welcome. For example, "It is known that $\dots$ has no elementary antiderivative because $\dots$". I'm looking for functions that have a well-behaved antiderivative, by the way, but that cannot be expressed with elementary functions.

Edit 2: Joako's answer made me realize I should really add the restriction that $f$ be an elementary function to begin with. Otherwise, the question is almost equivalent to asking "What functions can't be expressed as elementary functions, but have a derivative?"

Edit 3: Just found a duplicate with lots of answers.

Answer 1

The most elementary result I know is Chebyshev's theorem of differential binomials. It gives you a simple test for a certain class of functions.

Chebyshev's theorem: If $a, b \in\mathbb{R} $ and $m,p,n \in\mathbb{Q}$ then the (indefinite) integral of $x^m\left(a+bx^n\right)^p $ can be written in terms of elementary functions if and only if one of $p, (m+1)/n, p+ (m+1)/n $ is an integer!

Example: $\displaystyle \int \frac{3x}{ \sqrt{x^3+ 9 } }\, dx $

Answer: No elementary antiderivative exists because $-\frac{1}{2}\not\in\mathbb{Z}$, $\frac{2}{3}\not\in\mathbb{Z}$ and $ \frac{1}{6} \not\in\mathbb{Z}$.

Example: Does $\displaystyle \int \frac{1}{1+x^{n}}\, dx $ have an elementary anti-derivative?

Answer: Remarkably, yes! Take $m=0$, $a=b=1$ and $p = -1$. Since $p$ is an integer, there is an elementary anti-derivative! (Here for an explicit calculation of this).

Answer 2

This is a comment to long for the comment section

Maybe the derivative $R'(x)$ of the Rvachëv function works as an example of what you ask in the last part of your edit:

It fulfills $$R'(x)=2R(2x+1)-2R(2x-1)$$ in the whole real line, but the function is defined as: $$R(x)=\begin{cases} 0,\quad |x|\geq 1\\ F(x+1),\ \text{otherwise} \end{cases}$$ with $F(x)$ the Fabius function, which is "an example of an infinitely differentiable function that is nowhere analytic".

So you have a "well behaved function" since is a smooth bump function, but which closed-form formula is unattainable (I believe).

In this answer there is a clever way to approximate the function through piecewise polynomials, but as a "gross approach" of how it looks like (for having a general idea), it looks pretty close to $$q(x)=\begin{cases} 1,\quad x=0\\ 0,\quad |x|\geq 1\\ \dfrac{1}{1+\exp\left(\dfrac{1-2|x|}{x^2-|x|}\right)},\ \text{otherwise}\end{cases}$$ both here in Desmos.

and if it don't work, since $q(x)$ is also a smooth function, I don't believe there is going to be a closed form for $$h(x)=\int\limits_{-\infty}^{x}q(y)\ dy$$ which be also a well-behaved smooth transition function, which could be aproximated through a power series, but a non-piecewise power series will never match it since it has a constant value in a non-zero measure interval (it will violate the Identity theorem for power series).