Closed form solution of $\int \exp(-a (b-x)^{3/2}-cx)\text dx$

Question

Does following integral have a closed form solution (a, b, and c are constants) $$ \int \exp(-a (b-x)^{3/2}-cx)\text dx $$ If not possible, what about a function with close behavior. $$ \int \exp(-a (b-x)^{2}-cx)\text dx $$

Answer 1

For $\int e^{-a(b-x)^\frac{3}{2}-cx}~dx$ ,

Let $u=(b-x)^\frac{1}{2}$ ,

Then $x=b-u^2$

$dx=-2u~du$

$\therefore\int e^{-a(b-x)^\frac{3}{2}-cx}~dx$

$=\int-2ue^{-au^3-c(b-u^2)}~du$

$=-2e^{-bc}\int ue^{-au^3+cu^2}~du$

$=-2e^{-bc}\int ue^{-u^2(au-c)}~du$

$=-2e^{-bc}\int\sum\limits_{n=0}^\infty\dfrac{(-1)^nu^{2n+1}(au-c)^n}{n!}du$

$=-2e^{-bc}\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^nu^{2n+1}C_k^na^k(-c)^{n-k}u^k}{n!}du$

$=-2e^{-bc}\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^ka^kc^{n-k}u^{2n+k+1}}{k!(n-k)!}du$

$=-2e^{-bc}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^ka^kc^{n-k}u^{2n+k+2}}{k!(n-k)!(2n+k+2)}+C$

$=-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{2(-1)^ka^kc^{n-k}(b-x)^{n+\frac{k}{2}+1}}{k!(n-k)!(2n+k+2)e^{bc}}+C$