There are some integrals that are impossible to express in terms of elementary function, for example, $ \int \frac{e^x}{x} dx $ is only expressible as a "special" function $Ei(x)$, the exponential integral, or $ \int \frac{\sin x}{x} dx = Si(x)$, the sine integral.
Is there a finite set of solutions $S$ of functions (and compositions of those functions) able to represent the set of solutions of a given finite set $C$ of functions?
What if we consider the set $S$, and ask if it is closed under indefinite integration? For example, if $C$ is the set of elementary functions, $S$ would have to include $Ei(x)$, and integrals of functions involving $Ei(x)$ and other functions in $S$ would be have to be closed under integration.
Could we iterate this process by making $S$ our new $C$, and always come up with a finite set as an answer? Will this process eventually stop producing new answers, making the set of all functions derivable from integrals on the set of compositions of elementary functions closed?
Your sets $C$ and $S$ cannot be finite if they are closed regarding composition because composition, e.g. iterated composition, is not finite.
If your set $C$ is finite, the set of the antiderivatives is finite.
The set of the elementary functions is infinite.
The set of functions you asked for are the Liouvillian functions. They are defined e.g. in section 1 of Davenport, J. H.: What Might "Understand a Function" Mean. In: Kauers, M.; Kerber, M., Miner, R.; Windsteiger, W.: Towards Mechanized Mathematical Assistants. Springer, Berlin/Heidelberg, 2007, page 55-65.
There is an uncountably infinite number of elementary functions non-integrable in the elementary functions. See my answer here.
In the answer to Why can't we define more elementary functions?, it is stated that the process of adding necessary new transcendents (antiderivatives) is infinite.