Prove an action expansion's even-indexed terms have to be integrated, where the odd-indexed terms are only derivatives of the potential (WKB) After assuming a wavefunction of a form:
$$ \psi \approx A \exp{\left(i \frac{S(x)}{\hbar}\right)}$$
and letting
$$S = \hbar^0 S_0 + \hbar^1 S_1 + \hbar^2 S_2 +...$$ 
The odd-indexed terms of the action for a one-dimensional potential in the time-independent Schrodinger equation do not require integration if $p$ is known and differentiable. However the even-terms require non-trivial integration(which is extensively more computationally taxing). p is defined by:
$$ p = \sqrt{2 m ( E - V(x))} $$
The terms:
$$ S_0^\prime = p$$
$$ \boxed{S_0 = \int dx\, p =\pm \int dx \sqrt{2m(E-V(x))} } $$
$$ S_1^\prime = \frac{i}{2}\frac{1}{p}\frac{d p}{d x} $$
$$ \boxed{S_1 = \frac{i}{2} ln(p) } $$
$$ S_2^\prime = \frac{1}{8 p^3} \left(\frac{d p}{d x} \right)^2 - \frac{1}{4} \left(\frac{1}{p^2} \frac{d^2 p}{d x^2} - \frac{1}{p^3} \left(\frac{d p}{d x}\right)^2\right)$$
$$ \boxed{S_2 = \int dx \left(\frac{1}{4 p^2} \frac{d^2 p}{d x^2} + \frac{3}{8 p^3} \left( \frac{d p}{d x}\right)^2 \right)}\mathrm{\,requires\,\,integration} $$
Now for $S_3$:
$$ S_3^\prime = -\frac{i}{8 p^3} \frac{d^3 p}{d x^3} + \frac{3}{4} \left( \frac{i}{p^4} \frac{d p}{d x} \frac{d^2 p}{d x^2} - \frac{1}{p^5} \left( \frac{d p}{d x}\right)^3 \right)$$
After making an educated guess that:
$$ \frac{d}{dx} \left( -\frac{i}{8 p^3} \frac{d^2 p}{d x^2} + \frac{3 i}{16 p^4} \left(\frac{d p}{d x}\right)^2\right) = -\frac{i}{8 p^3} \frac{d^3 p}{d x^3} + \frac{3}{4} \left( \frac{i}{p^4} \frac{d p}{d x} \frac{d^2 p}{d x^2} - \frac{1}{p^5} \left( \frac{d p}{d x}\right)^3 \right)$$
Then
$$\int dx S_3^\prime = \int dx \frac{d}{dx} \left( -\frac{i}{8 p^3} \frac{d^2 p}{d x^2} + \frac{3 i}{16 p^4} \left(\frac{d p}{d x}\right)^2\right)$$
Clearly because of my guess
$$\boxed{S_3 = -\frac{i}{8 p^3} \frac{d^2 p}{d x^2} + \frac{3 i}{16 p^4} \left(\frac{d p}{d x}\right)^2} $$
How do we know that $S_2$ cannot be retrieved the same way? Surely I cannot try an ansatz for every possible function that could be a candidate for $S_2$. This problem has come up in general for different types of problems outside the scope of quantum mechanics. Is it only a consequence of this equation(1-D time-independent Schrodinger equation) that we have odd-indexed and even-indexed terms this way? Would it be different for 3-D?

The governing Schrödinger equation 
$$(e^{\frac{i}{\hbar}S})^{\prime\prime}~=~-k(x)^2 e^{\frac{i}{\hbar}S} ~\Leftrightarrow~ S^{\prime2} ~=~p(x)^2+i\hbar S^{\prime\prime} $$
can be turned into a fixed point equation 
$$S^{\prime} ~=~ \sqrt{p^2+i\hbar S^{\prime\prime}} ~=~ \sqrt{p^2+i\hbar \frac{d}{dx}\sqrt{p^2+i\hbar S^{\prime\prime}}} ~=~ \sqrt{p^2+i\hbar \frac{d}{dx}\sqrt{p^2+i\hbar  \frac{d}{dx}\sqrt{p^2+i\hbar S^{\prime\prime}}}} ~=~ \ldots $$
Expanding 
$$S^{\prime}~=~\sum_{n=0}^{\infty} (i\hbar)^{n}S^{\prime}_n,$$ 
one gets
$$S^{\prime}_0 ~=~ p,\qquad S^{\prime}_1 ~=~ \frac{p^{\prime}}{2p},\qquad S^{\prime}_2 ~=~ \frac{p^{\prime\prime}}{4p^2}-\frac{3(p^{\prime})^2}{8p^3}, $$
$$ S^{\prime}_3 ~=~ \frac{p^{\prime\prime\prime}}{8p^3}-\frac{3p^{\prime}p^{\prime\prime}}{4p^4}+\frac{3(p^{\prime})^3}{4p^5},\qquad, \ldots $$
OP then asks if there is a method to check if $f=S^{\prime}_n$ is a total derivative? Yes, one can check if the Euler-Lagrange operator
$$E(f)~=~\sum_{n=0}^{\infty} \left( - \frac{d}{dx}\right)^n \frac{\partial f}{\partial p^{(n)}}$$
vanishes on $f=S^{\prime}_n$. E.g.
$$E(S^{\prime}_0)~=~1~\neq~ 0, \qquad E(S^{\prime}_1)~=~ 0,\qquad E(S^{\prime}_2)~\neq~ 0,\qquad E(S^{\prime}_3)~=~ 0,\qquad\ldots.  $$
Here we are using the fact that a local functional can be written as a boundary term if and only if the Euler-Lagrange equation vanishes identically. See also this Phys.SE answer.
At least in this way it is possible to check operationally order by order in $n$ whether $S^{\prime}_n$ is a total derivative or not. We have not checked the calculations beyond OP's claims, nor investigated OP's conjectures.
