there is a closed form solution for $I_n$ for $$I_n = \int_{0}^{\infty} e^{-\Lambda x} x^n e^{i \tau ( c_1 - c_2 e^{-c_3 x} ) } dx$$ for integer $n$, for example, for $n=0$:
$$I_0=(\lambda (c_3+\lambda))^{-1}$$$$I_0=\frac{1}{\lambda (c_3+\lambda)}e^{i c_1 \tau}$$ $$\qquad\times\left[e^{i c_1 \tau} \left((c_3+\lambda) \, _1F_2\left(\frac{\lambda}{2 c_3};\frac{1}{2},\frac{\lambda}{2 c_3}+1;-\frac{1}{4} c_2^2 \tau^2\right)-i c_2 \lambda \tau \, _1F_2\left(\frac{\lambda}{2 c_3}+\frac{1}{2};\frac{3}{2},\frac{\lambda}{2 c_3}+\frac{3}{2};-\frac{1}{4} c_2^2 \tau^2\right)\right)\right].$$$$\qquad\times \left[(c_3+\lambda) \, _1F_2\left(\frac{\lambda}{2 c_3};\tfrac{1}{2},\tfrac{\lambda}{2 c_3}+1;-\tfrac{1}{4} c_2^2 \tau^2\right)-i c_2 \lambda \tau \, _1F_2\left(\tfrac{\lambda}{2 c_3}+\tfrac{1}{2};\tfrac{3}{2},\tfrac{\lambda}{2 c_3}+\tfrac{3}{2};-\tfrac{1}{4} c_2^2 \tau^2\right)\right].$$
for larger integer $n$ the expressions are similar, in terms of hypergeometric functions, but much longer.