Solving Integral that contain exponential and Power

Question

I have an integral of this form:

$$\int_0^\infty e^{-\frac{x}{a}-\frac{z^2}{bx}-\frac{z}{bx}}\left(\frac{c}{c+x+z}\right)^K~dx$$

where $K$ is a positive integer.

$a$ , $b$ and $c$ are reals and $>0$

finally,$z$ is also reals.

My question is that; is it possible to find the solution of this integral?

Answer 1

Special case of $z=-c$ :

$\int_0^\infty e^{-\frac{x}{a}-\frac{c^2}{bx}+\frac{c}{bx}}\left(\dfrac{c}{x}\right)^K~dx$

$=c^K\int_0^\infty\dfrac{e^{-\frac{x}{a}-\frac{c^2-c}{bx}}}{x^K}dx$

$=c^K\int_0^\infty\dfrac{e^{-\frac{\sqrt{\frac{a(c^2-c)}{b}}x}{a}-\frac{c^2-c}{b\sqrt{\frac{a(c^2-c)}{b}}x}}}{\left(\sqrt{\dfrac{a(c^2-c)}{b}}x\right)^K}d\left(\sqrt{\dfrac{a(c^2-c)}{b}}x\right)$

$=\dfrac{b^\frac{K-1}{2}c^K}{a^\frac{K-1}{2}(c^2-c)^\frac{K-1}{2}}\int_0^\infty\dfrac{e^{-\sqrt{\frac{c^2-c}{ab}}\left(x+\frac{1}{x}\right)}}{x^K}dx$

$=\dfrac{b^\frac{K-1}{2}c^\frac{K+1}{2}}{a^\frac{K-1}{2}(c-1)^\frac{K-1}{2}}\int_{-\infty}^\infty\dfrac{e^{-\sqrt{\frac{c^2-c}{ab}}\left(e^x+\frac{1}{e^x}\right)}}{(e^x)^K}d(e^x)$

$=\dfrac{b^\frac{K-1}{2}c^\frac{K+1}{2}}{a^\frac{K-1}{2}(c-1)^\frac{K-1}{2}}\int_{-\infty}^\infty e^{-2\sqrt{\frac{c^2-c}{ab}}\cosh x}e^{(1-K)x}~dx$

$=\dfrac{b^\frac{K-1}{2}c^\frac{K+1}{2}}{a^\frac{K-1}{2}(c-1)^\frac{K-1}{2}}\left(\int_{-\infty}^0e^{-2\sqrt{\frac{c^2-c}{ab}}\cosh x}e^{(1-K)x}~dx+\int_0^\infty e^{-2\sqrt{\frac{c^2-c}{ab}}\cosh x}e^{(1-K)x}~dx\right)$

$=\dfrac{b^\frac{K-1}{2}c^\frac{K+1}{2}}{a^\frac{K-1}{2}(c-1)^\frac{K-1}{2}}\left(\int_\infty^0e^{-2\sqrt{\frac{c^2-c}{ab}}\cosh(-x)}e^{(1-K)(-x)}~d(-x)+\int_0^\infty e^{-2\sqrt{\frac{c^2-c}{ab}}\cosh x}e^{(1-K)x}~dx\right)$

$=\dfrac{b^\frac{K-1}{2}c^\frac{K+1}{2}}{a^\frac{K-1}{2}(c-1)^\frac{K-1}{2}}\left(\int_0^\infty e^{-2\sqrt{\frac{c^2-c}{ab}}\cosh x}e^{(K-1)x}~dx+\int_0^\infty e^{-2\sqrt{\frac{c^2-c}{ab}}\cosh x}e^{-(K-1)x}~dx\right)$

$=\dfrac{2b^\frac{K-1}{2}c^\frac{K+1}{2}}{a^\frac{K-1}{2}(c-1)^\frac{K-1}{2}}\int_0^\infty e^{-2\sqrt{\frac{c^2-c}{ab}}\cosh x}\cosh((K-1)x)~dx$

$=\dfrac{2b^\frac{K-1}{2}c^\frac{K+1}{2}}{a^\frac{K-1}{2}(c-1)^\frac{K-1}{2}}K_{K-1}\left(2\sqrt{\dfrac{c^2-c}{ab}}\right)$