I'm assuming that if you had a general formula for the moments that you wouldn't need the MGF.
Using Mathematica the general formula for the $k$-th moment is found in the following manner:
f[x_] := \[Alpha] L^\[Alpha] x^(-\[Alpha] - 1) /(1 - (L/H)^\[Alpha])
muk = Integrate[x^k f[x], {x, L, H}, Assumptions -> {\[Alpha] > 0, L > 0, H > L, k >= 0}]
$$\mu_k=\frac{\alpha \left(H^{\alpha } L^k-H^k L^{\alpha }\right)}{(\alpha -k) \left(H^{\alpha }-L^{\alpha }\right)}$$
This formula clearly has a problem when $k=\alpha$. So when $k=\alpha$ we can solve for that moment directly:
mualpha = Integrate[x^\[Alpha] f[x], {x, L, H}, Assumptions -> {\[Alpha] > 0, L > 0, H > L}]
$$\frac{\alpha L^{\alpha } \log \left(\frac{H}{L}\right)}{1-\left(\frac{L}{H}\right)^{\alpha }}$$
This is also what one obtains when taking the limit of the general formula as $k\rightarrow\alpha$.
* 2nd Update*
There is a relatively closed-form solution for the MGF but one needs to be a bit more careful when extracting the moments. First, the closed-form solution:
$$\text{mgf}=\sum _{k=0}^{\infty } \frac{t^k \left(\alpha \left(H^{\alpha } L^k-H^k L^{\alpha }\right)\right)}{k! \left((\alpha -k) \left(H^{\alpha }-L^{\alpha }\right)\right)}$$
$$=-\frac{\alpha \left(L^{\alpha } (-H t)^{\alpha } \Gamma (-\alpha ,0,-H t)-H^{\alpha } (-L t)^{\alpha } \Gamma (-\alpha ,0,-L t)\right)}{L^{\alpha }-H^{\alpha }}$$
where $\Gamma(a,z_0,z_1)$ is the generalized incomplete gamma function (just as suggested in a comment by @BGM).
mgf = Sum[(((H^\[Alpha] L^k - H^k L^\[Alpha]) \[Alpha])/((H^\[Alpha] -
L^\[Alpha]) (-k + \[Alpha]))) t^k/k!, {k, 0, \[Infinity]}]
To obtain the $k$-th moment one would usually differentiate $k$ times and then set $t=0$ but that won't work with this MGF. If $\alpha$ is not a positive integer, then you'll need to take the $k$-th derivative followed by taking the limit of that derivative as $t\rightarrow 0$. Using Mathematica commands that translates to the following when $k=3$:
Limit[D[mgf, {t, 3}], t -> 0]
$$\frac{\alpha \left(L^3 H^{\alpha }-H^3 L^{\alpha }\right)}{(\alpha -3) \left(H^{\alpha }-L^{\alpha }\right)}$$
If $\alpha$ is a positive integer (say $\alpha=5$, then two limits need to be taken:
$$\lim_{\alpha \to 5} \, \lim_{t\to 0} \, \frac{\partial ^3\text{mgf}}{\partial t^3}=\frac{5 \left(H^5 L^3-H^3 L^5\right)}{2 \left(H^5-L^5\right)}$$
with the associated Mathematica commands:
Limit[Limit[D[mgf, {t, 3}], t -> 0], \[Alpha] -> 5]
