Quantum tunneling effect in a potential of the kind $V(x)=A\frac{x^2}{1+x^4}$ Given a potential: $$V(x)=A\frac{x^2}{1+x^4}$$ with $A\gt 1$ and a quantum particle inside the well around the point $x=0$. I'm stuck on the calculation of the transmission and reflection coefficients for this particle vs. its energy.
 A: Here's a not-so-clever answer.
The plot of the function is shown below. 

The red line denotes the energy of the particle being tunneled which expressed in terms of A. The black line denotes the max value of the potential which is A/2.
The task is to evaluate the transmission coefficient of the particle through one of the bumps of the potential. 
According to the WKB approximation the tunneling transmission coefficient across a given barrier is given by. 

To evaluate the integral, taylor expand the square root in equation 1 around the point x = 1. And one would arrive at (for 0 < c < 0.5).

Now, the limits of the integral are determined by the points at which the line U(x) = cA (Energy of the particle) intercepts the bumps of the curve. The integral of the square root in equation 1 must be evaluated between these points because the square root will give rise to imaginary numbers at all other points. To obtain the values of x at which the line U(x) = cA intercepts the bumps, one must solve the 4th power polynomial equation.

The four roots are given by 

Two of these roots/intercepts are on LHS bump and the other two are on RHS bump. Since we are only interested in the intercepts on one of the bumps we select only the positive roots which correspond to the intercept of U(x) on the RHS bump. 

The the above values in equation 5 become the limits of the integral in eqtn (1). 
Now to complete the problem one must integrate all the terms in equation 2 with respect to x and plug in the limits of the integral given in equation 5, which is a routine (and yet tedious) task. The result can be substituted in equation 1 to obtain the transmission coefficient. 
I believe the process becomes easier if c is known. The general equation for all values of c (c < 0.5) becomes rather large and messy.
References:
1. A. Messiah (1991), "Quantenmechanik 1", Degruyter, 1991.


*G. Squires, (1995). "Problems in quantum mechanics", Cambridge University Press, Cambridge, UK.   

A: In order to compute the transmission coefficient, we can use the first correction in the WKB approximation. Ignoring constants that we can pull outside the integral, we essentially are faced with the integration problem,
$$I(x) = \int \mathrm dx \, \sqrt{V(x)-E}.$$
In the case of your potential, we thus have,
$$I(x) = \sqrt{E}\int \mathrm dx \, \sqrt{\frac{Cx^2}{1+x^4}-1} $$
where $C := A/E$. We can now employ the generalised binomial theorem to expand the square root, using the Pochhammer symbol ${}_rP_k$, obtaining,
$$\sqrt{\frac{Cx^2}{1+x^4}-1} = \sum_{k=0}^\infty \frac{(-1)^k {}_{1/2}P_k}{k!}C^{1/2-k} \left(\frac{x^2}{1+x^4} \right)^{1/2-k}.$$
We can now integrate a general term over $x$, which yields a hypergeometric function. There are further simplifications for the cases $x>0$ and $x<0$. These lead to,
$$I(x) = \mathrm{sgn}(-x)\frac{\sqrt{E}}{4}\sum_{k=0}^\infty i^{k+1}\frac{(-1)^k {}_{1/2}P_k}{k!}C^{1/2-k} B \left(-x^4; \frac{1-k}{2},\frac{1+k}{2} \right)$$
where $B(z;a,b)$ is the incomplete beta function. If $x_1$ and $x_2$ denote the two classical turning points, we then have that,
$$T = \exp \left[ -2 \frac{\sqrt{2m}}{\hbar}[I(x_2)-I(x_1)]\right] \left( 1 + \frac14 \exp \left[ -2 \frac{\sqrt{2m}}{\hbar}[I(x_2)-I(x_1)]\right]\right)^{-2}.$$
