I have a integral term that looks similar to $\int_0^\infty\exp(-u-ae^{-c_1u}-be^{-c_2u})\,du$ where the constants $a,b,c_1,c_2>0$. For the case where $b=0$ I can use the answer from: Integrating exponential of exponential function.
Using a similar approach where I let $t=\exp(-u)$, and taking the simpler case where a=b=1, I get: $$\int_0^1\exp(-(t^{c_1}+t^{c_2}))\,dt$$
Using this or otherwise, how would I progress?
In the event that there is no analytical solution, what would be an exponential lower-bound on the function, $\exp(-u-ae^{-c_1u}-be^{-c_2u})$ that takes into account all the constants, $a,b,c_1,c_2$
$\int_0^\infty e^{-u-ae^{-c_1u}-be^{-c_2u}}~du$
$=\int_\infty^0e^{-ae^{-c_1u}-be^{-c_2u}}~d(e^{-u})$
$=\int_0^1e^{-au^{c_1}-bu^{c_2}}~du$
$=\int_0^1\sum\limits_{n=0}^\infty\dfrac{(-1)^n(au^{c_1}+bu^{c_2})^n}{n!}du$
$=\int_0^1\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^nC_k^n(au^{c_1})^k(bu^{c_2})^{n-k}}{n!}du$
$=\int_0^1\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^na^kb^{n-k}u^{c_2n+(c_1-c_2)k}}{k!(n-k)!}du$
$=\left[\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^na^kb^{n-k}u^{c_2n+(c_1-c_2)k+1}}{k!(n-k)!(c_2n+(c_1-c_2)k+1)}\right]_0^1$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^na^kb^{n-k}}{k!(n-k)!(c_2n+(c_1-c_2)k+1)}$
Only in some very special cases this integral can have chance for expressable in terms of known special functions $\left(\text{e.g. when}\dfrac{c_1}{c_2}\text{or}\dfrac{c_2}{c_1}=1~\text{or}~2~\text{or}~3\right)$ .