Finding a bound on the integral of function of exponentials I am currently trying to show that an function has a finite integral. I am trying to find the iterated integral of:
$$
\frac{e^{\alpha + 0.3\beta}}{1+e^{\alpha + 0.3\beta}}\frac{1}{1+e^{\alpha + 0.3\beta}}\frac{e^{\alpha + 0.05\beta}}{1+e^{\alpha + 0.05\beta}}\frac{1}{1+e^{\alpha + 0.05\beta}}
$$
over $-\infty < \alpha, \beta <\infty$. 
I am able to see that $\frac{e^{\alpha + 0.3\beta}}{1+e^{\alpha + 0.3\beta}}\frac{1}{1+e^{\alpha + 0.3\beta}} < e^{|\alpha + 0.3\beta|}$, and that $\frac{e^{\alpha + 0.05\beta}}{1+e^{\alpha + 0.05\beta}}\frac{1}{1+e^{\alpha + 0.05\beta}} < e^{|\alpha + 0.05\beta|}$.
Hence, I have that:
$$
\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \frac{e^{\alpha + 0.3\beta}}{1+e^{\alpha + 0.3\beta}}\frac{1}{1+e^{\alpha + 0.3\beta}}\frac{e^{\alpha + 0.05\beta}}{1+e^{\alpha + 0.05\beta}}\frac{1}{1+e^{\alpha + 0.05\beta}} d\alpha d\beta
$$
$$
< \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty}e^{|\alpha + 0.3\beta|} e^{|\alpha + 0.05\beta|}d\alpha d\beta
$$
However, I am not sure where to proceed from here. One option seems to be doing the Jacobian and using a transformation. Is there any easy way to bound this integral? Thanks!

HINT
I would separate the regions so that you know which sign the absolute values take and integrate directly. It's somewhat tediuos but will be exact instead of the upper bound.
