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I want to know if the Laplace transform of

$x^\alpha (1+ax)^\beta$

has any closed form?

I really appreciate your help.

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    Yes, but are you alright with the Kummer confluent hypergeometric function as a "closed form"?2011-08-07

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Yes it does, but requires special functions known as Tricomi confluent hypergeometric function (see functions.wolfram.com). The integral reduces to the one stated in the linked page after change of variables $x = \frac{y}{a}$:

$ LT_s(x^\alpha (1+a x)^\beta) = a^{\alpha - 1} \Gamma(1 + \alpha) U(1 + \alpha, 2 + \alpha + \beta, \frac{s}{a}) $

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    ...and [Whittaker functions and confluent hypergeometric functions are more or less the same thing](http://dlmf.nist.gov/13.14.i). :D2011-08-08