I would like to solve the following integral,
$\int_{0}^{\infty}\frac{\phi}{a+b\phi} \phi^{c-1}e^{-d\phi}d\phi$.
Note $\phi \sim Ga(c,d)$ is a gamma distributed random variable and the integral can be regarded as taking $\mathbb{E}[\frac{\phi}{a+b\phi}]$ apart from the normalizing constant $\frac{d^{c}}{\Gamma(c)}$.
I know that the if X,Y are gamma distributed RV's then $Z=\frac{X}{X+Y}$ is beta distributed. I could work the above into $\frac{\phi}{\phi+(a+[b-1]\phi)}$ but then Y is a non standard gamma - scaled and shifted.
The integral is a known Laplace transform : http://www.wolframalpha.com/input/?i=laplace+transform++x%5Ec%2F%28a%2Bb*x%29
I changed the typography of letter "$d$" to avoid confusion between two different "d" in the wording.
$$\begin{align*} \int_{\phi = 0}^\infty \frac{\phi}{a+b\phi} \phi^{c-1} e^{-s\phi} \, d\phi &= \int_{\phi = 0}^\infty \frac{\phi^c}{a+b\phi} e^{-s\phi} \, d\phi \\ &= \mathcal L_\phi \biggl(\phi, \frac{\phi^c}{a+b\phi}\biggr)(s) \\ &= \frac{a^c}{b^{c+1}} \exp\Bigl(\frac{a}{b}s\Bigr)\Gamma\Bigl(-c,\frac{a}{b}s\Bigr) \end{align*}$$ $$\boxed{\displaystyle \int_{\phi = 0}^\infty \frac{\phi}{a+b\phi} \phi^{c-1} e^{-d\phi} \, d\phi = \frac{a^c}{b^{c+1}}\exp\Bigl(\frac{a}{b}d\Bigr)\Gamma\Bigl(-c,\frac{a}{b}d\Bigr)}$$