Your integral admits a closed form.
Theorem. Let $a$ and $b$ be any real numbers. Then
$$\int_0^{\infty }\!\! e^{\large-\left(\frac{a-\log x}{b}\right)^2}\!\! \left(\text{erf}\left(\frac{a-\log x}{b}\right)+1\right) \mathrm{d}x=\sqrt{\pi}\:|b|\:e^{{\large {a+\frac{b^2}{4}}}}\left(1-\text{erf}\!\left(\! \frac{\sqrt{2}\:b}{4}\!\right)\! \right) \tag1$$
where $\displaystyle \text{erf}(z)=\frac{2}{\sqrt{\pi }}\int _0^ze^{\large-t^2}\mathrm{d}t.$
Proof. Let's denote the left hand side of $(1)$ by $I(a,b)$.
By the change of variable $$\displaystyle U:=\frac{a-\log x}{b}, \quad dx=-b\:e^a \:e^{-b U}dU,$$ we have
$$
\begin{align}
I(a,b)&=|b|\:e^a \int_{-\infty }^{+\infty } e^{\large-U^2-bU} \left(\text{erf}\left(U\right)+1\right) \mathrm{d}U\\\\
&=|b|\:e^a \int_{-\infty }^{+\infty } e^{\large-U^2-bU} \text{erf}(U)\: \mathrm{d}U+|b|\:e^a \int_{-\infty }^{+\infty } e^{\large-U^2-bU} \mathrm{d}U \tag2
\end{align}
$$ Clearly, by the gaussian integral:
$$
\int_{-\infty }^{+\infty } e^{\large-U^2-bU} \mathrm{d}U =e^{\large b^2/4}\int_{-\infty }^{+\infty } e^{\large-(U+b/2)^2} \mathrm{d}U =\sqrt{\pi}\:e^{ b^2/4} .\tag3
$$ Set
$$
f(b):=\int_{-\infty }^{+\infty } e^{\large-U^2-bU} \text{erf}(U)\: \mathrm{d}U \tag4
$$ and observe that $$ f(0)=\frac{\sqrt{\pi}}{4}\left[(\text{erf}(U))^2\right]_{-\infty }^{+\infty }=0. \tag5$$ Differentiating $(4)$ with respect to $b$ and performing an integration by parts gives
$$
\begin{align}
f'(b)&=-\int_{-\infty }^{+\infty } U\:e^{\large-U^2-bU} \text{erf}(U)\:\mathrm{d}U\\\\
f'(b)&=\left[\frac 12 e^{\large-U^2}\left(e^{\large-bU} \text{erf}(U)\right)\right]_{-\infty }^{+\infty }-\frac 12\int_{-\infty }^{+\infty } e^{\large-U^2} \left(-b\:e^{\large-bU} \text{erf}(U)+\frac{2}{\sqrt{\pi }}e^{\large-bU}e^{\large-U^2}\right)\mathrm{d}U\\\\
f'(b)&=\frac{b}{2}f(b)-\frac{1}{\sqrt{\pi }}\int_{-\infty }^{+\infty } e^{\large-2U^2-bU}\mathrm{d}U
\end{align}
$$ or
$$
f'(b)=\frac{b}{2}f(b)-\frac{\sqrt{2}}{2}\:e^{\large b^2/8}. \tag6
$$ We classically solve the ODE $(6)$, using $(5)$, to obtain
$$
f(b)=-\sqrt{\pi }\:e^{\large b^2/4}\text{erf}\!\left(\! \frac{\sqrt{2}\:b}{4}\!\right) \tag7
$$ then plugging $(7)$, $(4)$ and $(3)$ into $(2)$ gives $(1)$ as desired.
Without loss of generality we can assume $c>0$ , since $c<0$ only changes the sign of the integral.
As already discussed in the comments, Glasser's master theorem does not apply here. We can instead write the integral as
$$ I = \mathrm{e}^{2ab} \int \limits_0^c \mathrm{e}^{-(a x + b/x)^2} \, \mathrm{d} x \, .$$
Now we would like to let $a x + b/x = t$ . Since this map is not bijective, we need to be careful here! For $0<x<\sqrt{b/a}$ the correct inverse is
$$x = \frac{1}{2a} \left(t-\sqrt{t^2-4ab}\right) \, ,$$
while for $x > \sqrt{b/a}$ we have to use
$$x = \frac{1}{2a} \left(t+\sqrt{t^2-4ab}\right) \, .$$
For $c \leq \sqrt{b/a} \, \Leftrightarrow \, a c - b/c \leq 0$ we only need the first version. The substitution yields
$$ I = \frac{\mathrm{e}^{2ab}}{2a} \int \limits_{ac+b/c}^\infty \left[\frac{t}{\sqrt{t^2-4ab}} - 1\right]\mathrm{e}^{-t^2} \, \mathrm{d} t \, . $$
Changing variables once more for the first part ($u = \sqrt{t^2 - 4ab}$) and using the complementary error function $\operatorname{erfc} = 1-\operatorname{erf}$ we obtain
$$ I = \frac{\sqrt{\pi}}{4a} \left[\mathrm{e}^{-2ab} \operatorname{erfc}(|ac - b/c|) - \mathrm{e}^{2ab} \operatorname{erfc}(ac + b/c)\right] \, .$$
For $c > \sqrt{b/a} \, \Leftrightarrow \, a c - b/c > 0$ we have to split the integral:
\begin{align}
I &= \mathrm{e}^{2ab} \left\{\int \limits_0^\sqrt{b/a} \mathrm{e}^{-(a x + b/x)^2} \, \mathrm{d} x + \int \limits_\sqrt{b/a}^c \mathrm{e}^{-(a x + b/x)^2} \, \mathrm{d} x \right\} \\
&= \frac{\mathrm{e}^{2ab}}{2a} \left\{\int \limits_{2\sqrt{ab}}^\infty \left[\frac{t}{\sqrt{t^2-4ab}} - 1\right]\mathrm{e}^{-t^2} \, \mathrm{d} t + \int \limits_{2\sqrt{ab}}^{ac+b/c} \left[1 + \frac{t}{\sqrt{t^2-4ab}}\right]\mathrm{e}^{-t^2} \, \mathrm{d} t \right\} \, .
\end{align}
Evaluating the remaining integrals as in the other case we find
$$ I = \frac{\sqrt{\pi}}{4a} \left[\mathrm{e}^{-2ab} [1+\operatorname{erf}(ac - b/c)] - \mathrm{e}^{2ab} \operatorname{erfc}(ac + b/c)\right] \, .$$
Both results can be simplified and the final result for arbitrary $c>0$ is
\begin{align}
I &= \frac{\sqrt{\pi}}{4a} \left[\mathrm{e}^{-2ab} \operatorname{erfc}(b/c - ac) - \mathrm{e}^{2ab} \operatorname{erfc}(b/c + ac)\right] \\
&= \frac{\sqrt{\pi}}{4a} \left[\mathrm{e}^{2ab} \operatorname{erf}(ac+b/c) + \mathrm{e}^{-2ab} \operatorname{erf}(ac-b/c) - 2 \sinh(2ab)\right] \, ,
\end{align}
which agrees with the result from the table (except for a constant, of course). Letting $c \to \infty$ we obtain the integral
$$ \int \limits_{-\infty}^\infty \mathrm{e}^{-a^2 x^2 - b^2 /x^2} \, \mathrm{d} x = \frac{\sqrt{\pi}}{a} \mathrm{e}^{-2ab} \, , $$
which can also be computed using the master theorem.
Best Answer
You can not compute it directly. But you can use normal distribution in other to have a very accurate estimate. In fact by supposing that $b\leq0$, the region $a\leq -\frac{x^2}{2}\leq b$ is the same as $\{\sqrt{-2b}\leq x\leq \sqrt{-2a}\}\cup\{-\sqrt{-2a}\leq x\leq -\sqrt{-2b}\}$.
So, $\begin{align*} \int_{a\leq -\frac{x^2}{2}\leq b} \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx &= \int_{\sqrt{-2b}}^{\sqrt{-2a}} \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx + \int_{-\sqrt{-2a}}^{-\sqrt{-2b}} \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx\\ &= \left[\Phi(\sqrt{-2a}) - \Phi(\sqrt{-2b})\right] + \left[\Phi(-\sqrt{-2b}) - \Phi(-\sqrt{-2a})\right]\\ &= 2\left[1+\Phi(\sqrt{-2a}) - \Phi(\sqrt{-2b})\right] \end{align*}$
where $\Phi(.)$ is the normal cumulating distribution function.