About $ f(x) + c \space f(g(x)) = h(x) $ Let $g(x),h(x), f’(x) $ be functions that can be expressed by radicals , log and exp , but $f(x) $ can not.
Now consider functional equations like
$$ f(x) + c \space f(g(x)) = h(x) $$
Where $c^2 = 1$
The only solutions with h(x) not identically 0 I know are based on $g(x) = a + b x $ with solutions like $li(x)$.
How to find more solutions ? 
Is $ f(x) $ always of the form $ \int y(g(x)) t(x) dx $ ?
In particular I wonder about it when iterations of $ g (x) $ are cyclic.
—-
$ f(x) + f(-x) = 0 $ is not intresting ofcourse.
 A: Oh, I don't know... your last eqn might actually be pointing the way...
In any case, if you want "easy instances", consider  $g(x)=a/x$, so 
$$
f(x)\pm f(a/x)= h(x).
$$
It is evident that 
$$
h(a/x)=\pm h(x).
$$
You then see that 
$$
f(x)=\frac{1}{2} h(x) +k(x,a/x),
$$
where k is an antisymmetric (symmetric, respectively) function of its two arguments, solving the homogeneous equation. (Indeed, exponentiating the arguments of your trivial one gets one there...)
Similarly, for $g(x)=\sqrt{a^2-x^2}$,
$$
f(x)\pm f\left (\sqrt{a^2-x^2}\right ) = h(x), \qquad x\in          [0,a], \quad h(x)=\pm h\left ( \sqrt{a^2-x^2}\right ) .
$$
You then see that 
$$
f(x)=\frac{1}{2} h(x) +k\left ( x,\sqrt{a^2-x^2}\right ) ,
$$
where k is an antisymmetric (symmetric, respectively) function of its two arguments, solving the homogeneous equation.
Likewise, cycling around Babbage's equation solution, observe that
$$
f(x) + f\left ( \frac{a-x}{1+bx}  \right)=0,
$$
is solved by 
$$
k \left ( x, \frac{a-x}{1+bx}  \right),
$$
provided k be antisymmetric in its two arguments... and so on. I am only a physicist, so I don't have the complete beautiful machinery of Babbage's equation orbits under my belt...
So, the silly takeaway is to "stretch" the trivial cases by less trivial-looking transformations, conjugacies, etc... This was, in fact, the strategy Ernst Schroeder used 150 years ago on his eponymous equation solutions...
