$I = \int_{0}^{u} z^{a} \exp(-b z) (c z + d)^{-g}\, dz$ where $u, a, b, c, d, g \in \mathbb R^{+}$

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I have the following integral, $$I = \int_{0}^{u} z^{a} \exp(-b z) (c z + d)^{-g}\, dz$$ where $u, a, b, c, d, g \in \mathbb R^{+}$. I need a closed form expression for the integral above.

Since, the exponent of the term $(cz + d)$ is negative, using binomial expansion will result in an infinite sum, whereas I am looking for a tractable solution. I tried looking for the solution using Mathematica, but that went in vain.

Can anybody give some helpful insights? Thanks!

Edit (after having the solution, for good): $a, g \in \mathbb Z^{+}$.

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Substitute $x = cz+d$. If $a$ is a positive integer, you can expand $z^a = (x-d)^a/c^a$ in powers of $x$, and use

$$ \int x^p \exp(-bx) dx = -b^{-1-p} \Gamma(p+1,bx) $$

where $\Gamma$ is the incomplete Gamma function. If $a$ is not a positive integer, I doubt there is a closed-form solution.