Is $f(x) = \frac{(x+\alpha)^3}{(x+\beta)(x+\gamma)}$ quasiconvex?

Is the function

\begin{equation} f(x) = \frac{(x+\alpha)^3}{(x+\beta)(x+\gamma)} \end{equation}

where $0 \leq \beta \leq \alpha$ and $0 \leq \gamma\leq \alpha$ quasiconvex? $x$ can be assumed to always be positive.

Through extensive simulations, I have observed that it does seem to be quasiconvex, but I am wondering whether there is a proof or mathematical insight that shows it is always true? The function seems to look like a hyperbola for small values of positive $x$, then looks linear.

My intuition tells me that if $\alpha, \beta, \gamma$ are of the same order of magnitude, for example, then the function is approximately a linear function, which is of course quasiconvex. This is a very vague and weak justification though.

The function restricted to positive $x$ is continuous and strictly positive with a unique minimum.
To see this its derivative can be computed and this shows that has a unique minimum to the right of its rightmost vertical asymptote. The asymptotes are at $x=-\beta$ and $x=-\gamma$ both negative. Thus the inverse image of the restricted function on intervals of the form $(-\infty,a)$ are open finite intervals, thus they are convex sets.