This looks to me like the Lagrangian for an exponential central potential of the form
U(x) = k (exp[-2a(x - c)] - 2 exp[-a(x - c)]) with total energy E>0 and k, a, c > 0, while b is proportional to the angular momentum. At x=c, there is a stable equilibrium. Since x=r/√E, the integral runs from 0 to r0/√E.
Can you confirm all of this?
The difference of the exponentials in U goes to zero quickly as x→∞, so it makes sense to do a Taylor series expansion about x=c:
U(c) = -k
U'(c) = 0
U ≈ k(-1+a²(x-c)²)
This is a harmonic oscillator potential. Substitute this into your Lagrangian and you get the well-known solutions for a harmonic oscillator.