Expectation of the product of polynomial & exponential transformations of normal r.v Let $X \sim \mathcal{N}(\mu, \sigma^2)$. Are there any (1) general formula and (2) references to the general formula for
$$ \mathbb{E} (X^n e^{tX}),\; n \in \mathbb{N}, t \in \mathbb{R}$$
in particular for $n = 1$ and $n = 2$?

I know $\mathbb{E}(X^n)$ are the normal moments ($\mu$, $(\sigma^2 + \mu^2)$, ... for increasing $n$). I also know $\mathbb{E}(e^{tX})$ is the moment generating function for a normal, which evaluates to
$$ \mathbb{E}(e^{tX}) = \exp\left(\mu t + \frac{1}{2}\sigma^2 t^2\right).$$
Clearly, directly multiplying both will not work as the two parts are dependent. I also considered Delta method and making $X^n e^{tX}$ a derivative of something and using the exchangeability of expectation and derivatives, but few pages of calculations in and it does not look promising.
I also looked at the table of normal integrals by Owen (1980), but am unable to find anything of the form (ignoring the constants)
$$ \int x^n \exp(tx) \exp\left(-\frac{1}{2}\frac{(x-\mu)^2}{\sigma^2}\right) \textrm{d}x .$$
 A: As hinted by @whuber (with thanks), the key is to transform the expectation to the (pure) moment of another normal distribution.
We first recognise the expectation in question is, by definition
$$ 
\mathbb{E}_{X \sim \mathcal{N}(\mu, \sigma^2)}(X^n e^{tX}) = 
\int x^n \exp(tx)\frac{1}{\sqrt{2\pi \sigma^2}}\exp\left(-\frac{1}{2}\frac{(x-\mu)^2}{\sigma^2}\right) \,\textrm{d}x .$$
We combine the exponential terms on the RHS to obtain
$$ 
\int x^n \frac{1}{\sqrt{2\pi \sigma^2}} \exp\left(-\frac{1}{2}\left[\frac{(x-\mu)^2}{\sigma^2} - 2tx\right]\right) \,\textrm{d}x
$$
Completing the square within the square brackets we have
$$ 
\int x^n \frac{1}{\sqrt{2\pi \sigma^2}} \exp\left(-\frac{1}{2}\left[\frac{(x-(\mu+t\sigma^2))^2}{\sigma^2}\right]\right) \exp\left(\mu t + \frac{1}{2}\sigma^2 t^2\right) \,\textrm{d}x
$$
The rightmost exponential term can be moved out of the integral. What is left inside is the moment of a different normal $\mathcal{N}(\mu + t\sigma^2, \sigma^2)$ by definition. Thus
$$ \mathbb{E}_{X \sim \mathcal{N}(\mu, \sigma^2)}(X^n e^{tX}) = 
\mathbb{E}_{X \sim \mathcal{N}(\mu + t\sigma^2, \sigma^2)}(X^n) \cdot \exp\left(\mu t + \frac{1}{2}\sigma^2 t^2\right)
$$
(For completeness) for $n = 1, 2$,
$$\mathbb{E}_{X \sim \mathcal{N}(\mu + t\sigma^2, \sigma^2)}(X) = \mu + t\sigma^2$$
$$\mathbb{E}_{X \sim \mathcal{N}(\mu + t\sigma^2, \sigma^2)}(X^2) = Var(X) + \mathbb{E}^2_{X \sim \mathcal{N}(\mu + t\sigma^2, \sigma^2)}(X) = \sigma^2 + (\mu+t\sigma^2)^2.$$
A: Let $X$ be a random variable with moment generating function (mgf) $m(t),$ where $t$ is a real number. Then,
$$E[X^n exp(tX)] = E[\frac{d^n}{dt^n} exp(tX)] = \frac{d^n}{dt^n} m(t),$$ which is the n-th derivative with respect to $t$ of $m(t)$ and $E$ denotes expectation.
In this specific example, $m(t)$ is the mgf of a normal random variable.
