Example equation which does not have a closed-form solution
Solution 1:
"Closed form" means you've given which symbols, functions and operations are allowed. Usually, "in closed form" means in finite terms of rational numbers, algebraic operations, Elementary functions and/or Special functions.
For indefinite integration of elementary functions by elementary functions, we have Liouville's theorem with Risch algorithm.
Solvability of differential equations in certain classes of functions (e.g. Elementary functions, Liouvillian functions, Special functions) are treated in Differential Galois theory and in Differential algebra.
A part of the answer is: equations that don't have a solution cannot be solved in closed form.
Let's restrict ourselves to equations in the complex numbers of one unknown and functions in the complex numbers.
The solutions of such equations are numbers. We have to look for closed-form numbers therefore. Among other things, the explicit algebraic numbers and the explicit elementary numbers are closed-form numbers (see below).
Consider that the function symbols in an equation actually mean functions and that certain functions have certain domains and certain ranges.
Consider that the function terms of each of the elementary standard functions (sin, cos, tan, cot, sec, csc, sinh, cosh, tanh, coth, sech, csch, arcsin, arccos, arctan, arccot, arcsec, arccsc, arcsinh, arccosh, arctanh, arccoth, arcsech, arccsch) can be brought to an algebraic-$\exp$-$\ln$-form.
So the function term of a trigonometric or hyperbolic function can be represented as a non-constant rational expression over the algebraic numbers of $e^{iz}$ or of $e^z$ respectively.
See e.g. the Wikipedia articles for the single functions or [Abramowitz/Stegun 1970].
[Abramowitz/Stegun 1970] Abramowitz, M.; Stegun, I.: Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. National Bureau of Standard 1970
1.) Range of functions
The existence of a solution can depend on the ranges of the functions in the equation.
Examples:
$$\sqrt{z}=-c\ \ \ \ (c>0)$$
$$\arcsin(z)=\frac{\pi}{2}+c\ \ \ \ (c>0)$$
Let $f$ a function of one variable. The situation doesn't change if you apply injective functions to the equation:
$$a+b\sqrt{z}=a-cb\ \ \ \ (c>0)$$
$$\ln(\sqrt{z})=\ln(-c)\ \ \ \ (c>0)$$
2.) Solutions by radicals
The numbers that can be represented by radical terms are called the explicit algebraic numbers. For solving algebraic equations by radicals, we have Abel-Ruffini theorem with Galois theory. Consider that each equation of the kind $A(x)=0$ ($A$ an algebraic function) leads to an algebraic equation.
There are criteria for determining whether a given quintic is solvable by radicals. see e.g. Wikipedia: Quintic function
Example:
$$1-z+z^5=0$$
Let $f$ a function of one variable. The situation doesn't change if you apply functions to $z$ and/or to the equation:
$$\cos(1-f(z)+f(z)^5)=1$$
3.) Solutions by algebraic numbers
Algebraic functions have algebraic values for each of its algebraic arguments. And certain transcendental functions have empty exceptional sets (see my answer at On the behavior of transcendental functions).
Let $A$ an algebraic function and $T$ a transcendental function. Equations of the kind
$$T(z)=A(z)\tag{1}$$
have algebraic solutions only at the exceptional points of $T$. If the exceptional set of $T$ is empty, the equation cannot have algebraic solutions therefore.
Example:
$$\cot(z)=z$$
Let $f$ a function of one variable. The situation doesn't change if you apply functions to $z$ and/or to the equation:
$$\sin(\cot(f(z)))=\sin(f(z))$$
An equation having no algebraic solutions doesn't have explicit algebraic solutions.
There are other closed-form representations for algebraic numbers (e.g. Bring radicals and theta functions). see e.g. Wikipedia: Quintic function - Beyond radicals.
4.) Solutions by elementary partial inverse functions
Let $E$ an elementary function. Solving an equation
$$E(z)=0\tag{2}$$
by rearranging it by operations we can read from the equation means applying the partial inverse functions of $E$. The incomprehensibly unfortunately hardly noticed theorem in [Ritt 1925], that is proved also in [Risch 1979], implies what kinds of elementary functions can have elementary partial inverses.
The elementary functions are generated from their complex function variable by applying finite numbers of $\exp$, $\ln$ and/or unary or multiary algebraic functions. The elementary functions having elementary partial inverse functions are generated from their complex function variable by applying finite numbers of $\exp$, $\ln$ and/or unary algebraic functions.
Let
$n\in\mathbb{N}_{>1}$,
$A$ a non-zero algebraic function of $n$ variables,
$f_1,...,f_n$ algebraically independent elementary functions of one variable.
Equations of the kind
$$A(f_1(z),...,f_n(z))=0\tag{3}$$
possibly don't have an elementary partial inverse function therefore. That means, such equations possibly cannot be rearranged by applying only elementary functions we can read from the equation.
Examples:
$$A(z,e^z)=0$$
$$A(z,\ln(z))=0$$
$$A(z,e^z,\ln(z))=0$$
$$A(z,\sin(z))=0$$
But consider the exceptional points of the functions on the left-hand sides of the equations.
Let $f$ an elementary function of one variable. The situation doesn't change if you apply elementary functions to $z$ and/or functions to the equation:
$$a+bA(f(z),\sin(f(z)))=a$$
$$\cos(A(f(z),\sin(f(z))))=1$$
The method presented in [Khovanskii 2014] and articles of A. Khovanskii and Y. Burda can prove for individual given elementary functions whether they have elementary partial inverse functions or not. [Belov-Kanel/Malistov/Zaytsev 2020] proves that the equation
$$\tan(z)-z=a\ \ \ \ \ (a\in\overline{\mathbb{Q}})$$
cannot be solved by rearranging it by applying only elementary functions readable from the equation.
The algebraic functions are subdivided into the explicit algebraic functions and the implicit algebraic functions. The elementary functions are therefore subdivided into the explicit elementary functions and the implicit elementary functions. An elementary function having no elementary partial inverse function don't have explicit elementary partial inverse functions.
[Ritt 1925] Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90
[Risch 1979] Risch, R. H.: Algebraic Properties of the Elementary Functions of Analysis. Amer. J. Math. 101 (1979) (4) 743-759
[Khovanskii 2014] Khovanskii, A.: Topological Galois Theory. Solvability and Unsolvability of Equations in Finite Terms. Springer 2014
[Belov-Kanel/Malistov/Zaytsev 2020] Belov-Kanel, A.; Malistov, A.; Zaytsev, R.: Solvability of equations in elementary functions. Journal of Knot Theory and Its Ramifications 29 (2020) (2) 204-205
5.) Solutions by elementary numbers
The Elementary numbers (see also) are generated by applying a finite number of elementary functions (see above) to the rational numbers.
For the exponential polynomials, we have the main theorem of [Lin 1983]:
If Schanuel's conjecture is true, an equation $P(x,e^x)=0$ with $P(X,Y)\in\overline{\mathbb{Q}}(X,Y)$ an irreducible polynomial involving $X$ and $Y$ cannot have a solution $x\neq 0$ that is an elementary number.
Example:
$$z+\sin(z)=0$$
Let $f$ an elementary function of one variable. The situation doesn't change if you apply elementary functions to $z$ or functions to the equation:
$$a+bA(f(z),\sin(f(z)))=a$$
$$\cos(A(f(z),\sin(f(z))))=1$$
An equation having no elementary solutions don't have explicit elementary solutions.
In A conjecture about the solvability of rational equations of transcendental functions by elementary numbers, I present an extension of Lin's theorem that I meanwhile have proven.
[Chow 1999] presents a method for this problem for the explicit elementary numbers.
[Lin 1983] Ferng-Ching Lin: Schanuel's Conjecture Implies Ritt's Conjectures. Chin. J. Math. 11 (1983) (1) 41-50
[Chow 1999] Chow, T.: What is a closed-form number. Am. Math. Monthly 106 (1999) (5) 440-448
6.) Solutions by Lambert W and elementary functions
Let $f$ an elementary function of one variable. According to Lin's theorem (see above), irreducible algebraic equations over $\overline{\mathbb{Q}}$ of simultaneously $f(z)$ and $e^{f(z)}$ don't have an elementary solution except $f(z)=0$.
The non-elementarity of Lambert W has been proven already by Liouville. Lambert W is the inverse relation of the function $F\colon\mathbb{C}\to\mathbb{C},z\mapsto ze^z$. Lambert W can therefore be used for solving equations that lead to equations of the kind $f_0(f_1(z)e^{f_2(z)})=f_0(y)$, wherein $f_0,f_1,f_2$ are functions of one variable, $f_1=f_2$ and $y$ is a complex variable or a complex constant.
Let
$a,b,c$ complex constants,
$f$ a function of one variable with appropriate partial inverse functions in closed-form,
$y$ a complex variable or complex constant.
Equations of the kind $(a+bf(z))^ce^{f(z)}=y$ can be solved by only Lambert W, elementary functions and appropriate partial inverse functions of $f$. Equations that lead to algebraic equations over $\overline{\mathbb{Q}}$ of simultaneously $f(z)$ and $e^{f(z)}$ that aren't of the above kind of equations are candidates for equations not solvable using only Lambert W, elementary functions and appropriate partial inverse functions of $f$.
Example:
$$(1+z+z^2)e^z=y$$
Equations where $f_1$ is a non-constant non-polynomial rational function of $z$ and $f_2$ is a non-constant polynomial function of $z$ or vice versa are such candidates.
Example:
$$\frac{z}{1+z}e^z=y$$
The situations don't change if you apply functions to $z$ and/or to the equations.
Equations of the kind
$$f_0(R_1(f(z))e^{R_2(f(z))})=f_0(y),$$
wherein $R_1(f(z)),R_2(f(z))$ are non-constant rational expressions of $f(z)$ can be solved in closed form by only generalized Lambert W ([Mezö/Baricz 2017]), elementary functions and appropriate partial inverse functions of $f$.
[Mezö/Baricz 2017] Mezö, I.; Baricz, A.: On the generalization of the Lambert W function. Transact. Amer. Math. Soc. 369 (2017) (11) 7917–7934
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See also Trigonometric/polynomial equations and the algebraic nature of trig functions.