Why isn't $\frac{\beta^2}{\alpha^2}\int_{0}^p x^2 f(x, \alpha, \beta) dx = F(p, \alpha+2, \beta)$? I'm look at the following integral:
$$\int_{0}^p x^2 f(x,\alpha, \beta) dx $$
where $f(x, \alpha, \beta)$ is the pdf of the Gamma distribution is expressed as:
$$\frac{x^{\alpha-1}e^{-\beta x}\beta^\alpha}{\Gamma(\alpha)}$$
Now if I would multiply the integral with the fraction $\frac{\beta^2}{\alpha^2}$, wouldn't this be equal to the cdf of the Gamma distribution evaluated at $p$ with $\alpha+2$ ?:
$$\frac{\beta^2}{\alpha^2}\int_{0}^p x^2 f(x, \alpha, \beta) dx = F(p, \alpha+2, \beta)$$
, where $F(x)$ represents the cdf of the Gamma distribution evaluated at $x$.
I use the fact that $\Gamma(\alpha+1) = \alpha\Gamma(\alpha)$
But when I evaluate this in Python (with Scipy):
func = lambda x: x**2*gamma.pdf(x, a=alpha, scale=1/beta)
first_int = quad(func, 0, p)[0]*(beta**2/alpha**2)
cdf = gamma.cdf(p, a=alpha+2, scale=1/beta)

I can see this equation does not hold.
I'm obviously missing something, so if someone could point this out, that'd be great.
 A: It appears that the pdf and cdf of the gamma distribution are defined as:
\begin{align}
\text{pdf} &= \frac{\beta^{\alpha}}{\Gamma(\alpha)} \, e^{- \beta \, x} \, x^{\alpha - 1} \\
\text{cdf} &= \frac{1}{\Gamma(\alpha)} \, \gamma(\alpha, \beta \, x).
\end{align}
The integral in question is
$$ I = \frac{\beta^2}{\alpha^2} \, \int_{0}^{p} x^2 \, \frac{\beta^{\alpha}}{\Gamma(\alpha)} \, e^{- \beta \, x} \, x^{\alpha - 1} \, dx $$
which leads to
\begin{align}
I &= \frac{\beta^2}{\alpha^2} \, \int_{0}^{p} x^2 \, \frac{\beta^{\alpha}}{\Gamma(\alpha)} \, e^{- \beta \, x} \, x^{\alpha - 1} \, dx \\
&= \frac{\beta^{\alpha +2}}{\alpha^2 \, \Gamma(\alpha)} \, \int_{0}^{p} e^{- \beta \, x} \, x^{\alpha+1} \, dx \\
&= \frac{\beta^{\alpha +2}}{\alpha^2 \, \Gamma(\alpha)} \, \int_{0}^{p/\beta} e^{-u} \, \left(\frac{u}{\beta}\right)^{\alpha+1} \, \frac{du}{\beta} \\
&= \frac{1}{\alpha^2 \, \Gamma(\alpha)} \, \gamma\left(\alpha + 2, \frac{p}{\beta}\right) \\
&= \frac{1}{\alpha^2} \, \frac{1}{\Gamma(\alpha)} \, \gamma\left(\alpha+2, \beta \, \left(\frac{p}{\beta^2}\right)  \right)
\end{align}
