Integrating $\exp(-(\frac xa)^b-(\frac xc)^d)$

Question

For $a,b,c,d,x>0$, does $$ \int \exp\left(-\left(\frac{x}{a}\right)^b-\left(\frac{x}{c}\right)^d\right)\,\mathrm dx $$ have a closed form using erf, gammas, etc? Wolfram Alpha (standard) was not of help, so I am wondering if I am missing a standard transformation.

Answer 1

$\int e^{-\left(\frac{x}{a}\right)^b-\left(\frac{x}{c}\right)^d}~dx$

$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\left(\dfrac{x^b}{a^b}+\dfrac{x^d}{c^d}\right)^n}{n!}~dx$

$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^nC_k^n\dfrac{x^{bk}}{a^{bk}}\dfrac{x^{dn-dk}}{c^{dn-dk}}}{n!}~dx$

$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^nx^{dn+(b-d)k}}{k!(n-k)!a^{bk}c^{dn-dk}}~dx$

$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^nx^{dn+(b-d)k+1}}{k!(n-k)!a^{bk}c^{dn-dk}(dn+(b-d)k+1)}+C$

$=\sum\limits_{k=0}^\infty\sum\limits_{n=k}^\infty\dfrac{(-1)^nx^{dn+(b-d)k+1}}{k!(n-k)!a^{bk}c^{dn-dk}(dn+(b-d)k+1)}+C$

$=\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^{n+k}x^{dn+bk+1}}{k!n!a^{bk}c^{dn}(dn+bk+1)}+C$

Which relates to Srivastava-Daoust Function