Given $a, b, c \in \mathbb R$, what is an analytic formula for the expectation $\mathbb E_{z \sim \mathcal N(0,1)} [e^{az^2+2bz + c}]$ ?
Observation
In the special case when $a=0$, $2b=\sigma$, and $c=\mu$ , it is known that $\mathbb E_{z \sim \mathcal N(0,1)} [e^{\sigma z + \mu}]=e^{\sigma^2/2+\mu}$ (mean of log-normal distribution).
Detailed Solution (based on hints given in the other answers)
Let $\alpha := \frac{1}{2}-a$. It's clear that the sought-for integral diverges unless $a < 1/2$. So let $a < 1/2$, so that $\alpha > 0$. Then direct computation yields $$ \begin{split} \mathbb E_{z \sim \mathcal N(0,1)} [e^{az^2+2bz + c}] &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{az^2+2bz+c}e^{-z^2/2}dz=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\alpha z^2 + 2bz + c}dz\\ &=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-(\sqrt{\alpha}z-\frac{b}{\sqrt{\alpha} })^2 + c-b^2/\alpha}dz = \frac{e^{c+b^2/\alpha}}{\sqrt{2\alpha\pi}}\int_{-\infty}^\infty e^{-x^2}dx=\frac{e^{c+b^2/\alpha}}{\sqrt{2\alpha\pi}}\sqrt{\pi}\\ &=\frac{e^{c+b^2/\alpha}}{\sqrt{2\alpha}}, \end{split} $$ where we've used the change of variable $x=\sqrt{\alpha}z-b/\sqrt{\alpha}$ and that the fact $\int_{-\infty}^\infty e^{-x^2}dx=\sqrt{\pi}$.
Reasoning along similar lines, one may prove an analogous result for higher-order moments. Viz,
If $x = e^{az^2+2bz + c}$, with $z \sim N(0, 1)$, then the expectation of $x^k$ is given by $$ \mathbb E[x^k] = \begin{cases}+\infty,&\mbox{ if }a > 0,\text{ and }ka \ge \dfrac{1}{2},\\ \dfrac{e^{kc+k^2b^2/\alpha_k}}{\sqrt{2\alpha_k}},&\mbox{ else,} \end{cases} $$ where $\alpha_k := 1/2 - ka > 0$ for $ka < 1/2$.
Hint
If $X\sim \mathcal N(0,1)$,
$$\mathbb E[f(X)]=\int_{\mathbb R}f(x)e^{-\frac{x^2}{2}}\,\mathrm d x.$$
Moreover, $$\int_{\mathbb R}e^{-x^2}\,\mathrm d x=\pi.$$
All these formulas will allow you to conclude.