Is there some close form solution available for following integral? I need to solve the following integral $$\int_d^{\infty}\left(\frac{a}{bx-c}\right)^{p}x^f e^{-x}dx$$ where $a,b,c,d$ are all positive values and $p,f$ are positive integers. I have checked in the book "Integral series and products" by Ryzhik but I am unable to find an expression that matches with mine. I will be very thankful if somebody guides me how to solve the above integral. Thanks in advance.
 A: Partial Hint
Assuming all your parameters are positive (the fact they also be integers is another question that does not stress the following idea), you can think about integrating by parts, choosing for example
$$f'(x) = \left(\frac{a}{bx-c}\right)^p ~~~~~~~~~~~ f(x) = \frac{(c-b x) \left(\frac{a}{b x-c}\right)^p}{b (p-1)}$$
$$g(x) = x^f e^{-x} ~~~~~~~~~~~ g'(x) = e^{-x}x^{\ f-1}(f-x)$$
Hence you'd get
$$\frac{(c-b x) \left(\frac{a}{b x-c}\right)^p}{b (p-1)}x^f\ e^{-x}\bigg|_d^{+\infty}  - \int_d^{+\infty} \frac{(c-b x) \left(\frac{a}{b x-c}\right)^p}{b (p-1)} e^{-x}x^{\ f-1}(f-x)\ \text{d}x$$
The first term gives you 
$$\frac{e^{-d} d^f (c-b d) \left(\frac{a}{b d-c}\right)^p}{b (p-1)}$$
The second term, the integral, can be arranged to be written in this way:
$$\frac{a^p}{b(p-1)}\int_d^{+\infty} \frac{e^{-x} x^{\ f-1} (f-x)}{(bx-c)^{p-1}}\ \text{d}x$$
Which can be split into two pieces which are
$$f \frac{e^{-x}x^{\ f-1}}{(bx-c)^{p-1}} ~~~~~~~~~~~~~ (1)$$
and
$$\frac{e^{-x} x^{\ f}}{(bx-c)^{p-1}} ~~~~~~~~~~~~~ (2)$$
Which both have to be integrated.
Hence the "new" problem is to deal with those two integrals. I'll think more about. I'll add details later!
