Until now, I don't have a complete answer. The single parts of my answer are only hints that should be considered.
1). Complexity of Antidifferentiation
There are some approaches for complexity measures of antidifferentiation or integration. But I don't know if there are approaches we need here.
from Borel sets:
Dougherty, R.; Kechris, A. S.: The complexity of antidifferentiation. Adv. Math. 88 (1991) (2) 145-169
from computer algorithm theory:
Complexity of computing the antiderivative of a given function
Kawamura, A.: Computational Complexity in Analysis and Geometry. PhD thesis. University of Toronto, 2011
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Antiderivative means i.a. integration in closed form. Let's concentrate here on the elementary functions in closed form.
The elementary functions are analytic. Each elementary function can be splitted into bijective elementary functions by restricting its domain (that were the partial inverses).
2.) Non-Existence of Non-elementary Integrals
This section treats elementary functions that have a non-elementary integral.
Let $f$ be an elementary function, $f^{-1}$ the elementary inverse of $f$, and $c$ the integration constant.
The Integral of inverse functions gives
$$\int f(x)dx=xf(x)-\int f^{-1}(f(x))df(x)+c.$$
Because $f$ is elementary, $xf(x)$ is elementary. If $\int f(x)dx$ is (non)elementary, $\int f^{-1}(f(x))df(x)$ is therefore (non)elementary too. Marchisotto and Zakeri (see the reference below) give therefore the following Integrals of inverse functions theorem.
"If $f$ and $f^{-1}$ are elementary functions over some closed interval, then $\int f(x)dx$ is elementary if and only if $\int f^{-1}(x)dx$ is elementary."
That means, there are no elementary functions that are reasonably easy to antidifferentiate, and whose inverses are still elementary but have provably non-elementary antiderivatives.
That means, each elementary invertible elementary function with an elementary antiderivative has an inverse with an elementary antiderivative.
Marchisotto, E. A.; Zakeri, G.-A.: An invitation to integration in finite terms. College Math. J. 25 (1994) (4) 295-308
3.) The Structure of Elementary Invertible Elementary Functions
J. F. Ritt proved in [Ritt 1925] that if a bijective elementary function $f$ has an elementary inverse, then $f=f_n\circ\ ...\ \circ f_1$, where each $f_i$ is either an algebraic function of one variable or else $\exp$ or $\ln$.
R. H. Risch proved Ritt's result in [Risch 1979].
[Ritt 1925] Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90
[Ritt 1948] Ritt, J. F.: Integration in finite terms. Liouville's theory of elementary methods. Columbia Univbersity Press, New York, 1948
[Risch 1979] Risch, R. H.: Algebraic Properties of the Elementary Functions of Analysis. Amer. J. Math 101 (1979) (4) 743-759
That means: If a function $f$ and its inverse are elementary, $f$ is composited by $\exp$, $\ln$ and/or unary univalued algebraic functions. There exists a representation of $f$ as elementary function that doesn't contain multiary algebraic functions.
All the functions and inverse functions that are relevant for us must have this structure.
If the set of allowed basic functions in the compositions is the set {transcendental elementary standard functions, unary univalued algebraic functions}, the composition representation of $f$ may contain also multiary univalued algebraic functions with arguments that are algebraically dependent allowed basic functions of algebraically dependent functions (algebraic dependence over $\mathbb{C}$ is meant).
4.) The Structure of Elementary Integrable Elementary Functions
How can the structure mentioned in section 3 be combined with the structure from Liouville's theorem for integration in finite terms?
Until now, I cannot describe the elementary functions that are elementary invertible and elementary integrable. And I cannot describe the elementary functions that are elementary invertible and not elementary integrable.
But I can give some examples of the latter:
List of Functions Without Antiderivatives
Yadav D. K.: Six Conjectures on Indefinite Nonintegrable Functions or Nonelementary Functions. 2016, conjectures 4 and 5
5.) Complexity of Antidfifferentiation in Dependence of the Integrand's Class of Functions
Which classes of functions are easier to antidifferentiate?
The elementary standard functions? The rational functions? The explicit algebraic functions? The (explicit and implicit) algebraic functions?
And which of the algebraic functions? Antiderivatives of algebraic functions can be i.a. purely algebraic or purely logarithmic.
6.) Antiderivatives of the Elementary Standard Functions
Decide which antidifferentiation is easier.
$f(x)\rightarrow \int f(x)dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f^{-1}(x)\rightarrow \int f^{-1}(x)dx$
$x^n\rightarrow \frac{x^{n+1}}{n+1}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x^{\frac{1}{n}}\rightarrow \frac{n{x}^{\frac{n+1}{n}}}{n+1}$
$a^x\rightarrow \frac{a^x}{\ln(a)}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \log_a(x)\rightarrow \frac{x(\ln(x)-1)}{\ln(a)}$
$e^x\rightarrow e^x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ln(x)\rightarrow x\ (\ln(x)-1)$
$\sin(x)\rightarrow -\cos(x)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \arcsin(x)\rightarrow x\ \arcsin(x)+\sqrt{1-x^2}$
$\cos(x)\rightarrow \sin(x)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \arccos(x)\rightarrow x\ \arccos(x)-\sqrt{1-x^2}$
$\tan(x)\rightarrow -\ln(\cos(x))\ \ \ \ \ \ \ \ \ \ \ \arctan(x)\rightarrow x\ \arctan(x)-\frac{1}{2}\ln(1+x^2)$
$\cot(x)\rightarrow \ln(\sin(x))\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ arccot(x)\rightarrow \frac{1}{2}x\pi-x\ \arctan(x)+\frac{1}{2}\ln(1+x^2)$
$\sec(x)\rightarrow \ln\left(\frac{1+\sin(x)}{\cos(x)}\right)\ \ \ \ \ \ \ \ \ \ \ \ \ arcsec(x)\rightarrow x\ arcsec(x)-\ln\left(x\left(1+\sqrt{{\frac{-1+x^2}{x^2}}}\right)\right)$
$\csc(x)\rightarrow -\ln\left(\frac{1+\cos(x)}{\sin(x)}\right)\ \ \ \ \ \ \ \ \ \ arccsc(x)\rightarrow x\ arccsc(x)+\ln\left(x\left(1+\sqrt{\frac{-1+x^2}{x^2}}\right)\right)$
$\sinh(x)\rightarrow \cosh(x)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ arcsinh(x)\rightarrow x\ arcsinh(x)-\sqrt{1+x^2}$
$\cosh(x)\rightarrow \sinh(x)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ arccosh(x)\rightarrow x\ arccosh(x)-\sqrt{x-1}\sqrt{x+1}$
$\tanh(x)\rightarrow \ln(\cosh(x))\ \ \ \ \ \ \ \ \ \ \ \ arctanh(x)\rightarrow x\ arctanh(x)+\frac{1}{2}\ln(-(x-1)(x+1))$
$\coth(x)\rightarrow \ln(\sinh(x))\ \ \ \ \ \ \ \ \ \ \ \ \ arccoth(x)\rightarrow x\ arccoth(x)+\frac{1}{2}\ln((x-1)(x+1))$
$sech(x)\rightarrow \arctan(\sinh(x))\ \ \ \ \ \ arcsech(x)\rightarrow x\ arcsech(x)-\arctan\left(\sqrt{-\frac{x-1}{x}}\sqrt{\frac{x+1}{x}}\right)$
$csch(x)\rightarrow \ln(\tanh(\frac{x}{2}))\ \ \ \ \ \ \ \ \ \ \ \ arccsch(x)\rightarrow x\ arccsch(x)+\ln\left(x\left(1+\sqrt{\frac{1+x^2}{x^2}}\right)\right)$
7.) Complexity of the Integrands from Integration Rules
The common integration rules beside Change of Variable are Integration by Substitution and Partial Integration. The complexity of the resulting integrands depends on the standard functions in the integrand. See the following reference for the form of the integrands.
Produktregel, Quotientenregel, Reziprokenregel, Kettenregel und Umkehrregel für die Integration
Here are techniques for Integration By Parts:
Alcantara, E.: Integrals of composite functions through tabular integration by parts. Asia Pacific Higher Educ. Res. J. 2 (2015) (1)
Alcantara, E.: On the Derivation of Some Reduction Formula through Tabular Integration by Parts. Asia Pacific J. Multidisciplinary Res 3 (2015) (1) 80-84
Mardeli Jandja, M.; Lutfi, M.: The Five Columns Rule in Solving Definite Integration by Parts Through Transformation of Integral Limits. J. Phys.: Conf. Ser. 1028 (2018) 012109
Best Answer
There are an infinite number of functions out there, and you can put an integral sign in front of any of them. Some of those functions are pretty strange and/or ugly. There is no reason why such an integral should have a representation in the very limited set of elementary functions. In fact when you get such a representation, you could consider yourself lucky.
For practical purposes, any convergent integral can be evaluated to any desired degree of accuracy. There are numerous numeric methods available with which to do that. And even if an integral should have some elementary function representation, that most often is a lot harder to evaluate than just running a numeric method on the integral. And you wouldn't be more accurate if the answer were cos(28.34) and you had to evaluate that.
Similarly, if you are going to use the integral in a further computation, it might be easiest to leave it as an integral, rather than tangling yourself up in things that have a lot of terms like sec(log^{-1}(x^{2})) -- or much worse.
So why do calculus classes teach you to find anti-derivatives? First, if the problem is going to resolve itself into an easy anti-derivative you might as well use it. Second, to familiarize you better with the underlying concepts, such as the chain rule and the product rule of derivatives. Third to give you some experience with actual answers so you have some idea what the answer ought to look like. If your numeric method evaluates to 2034.86, and you know the answer can't be larger than 80, then you know you made a mistake in your computation.
Of course, nowadays, all those integrals can be accurately evaluated online; but still, you should have some knowledge of what kind of answer to expect and what it all means.