Compute $\int_{-\infty}^{+\infty}(1+\frac{1}{v^2})\exp(-\frac{u^2}{2\sigma_1^2\sigma_2^2}(\sigma_2v+\frac{\sigma_1}{v})^2)\,dv$ 
Compute $$\int_{-\infty}^{+\infty}(1+\frac{1}{v^2})e^{-\frac{u^2}{2\sigma_1^2\sigma_2^2}(\sigma_2v+\frac{\sigma_1}{v})^2}\,dv$$

Does this integral has a close form solution? What if $\sigma_1=\sigma_2=1$, $u=\sqrt2$, i.e, $$\int_{-\infty}^{+\infty}(1+\frac{1}{v^2})e^{-(v+\frac{1}{v})^2}dv$$

Let $I$ denote the integral. Then by the substitution $v \to \frac{1}{v}$ we find that
\begin{align*}
I
&= \int_{-\infty}^{+\infty} \exp \bigg\{ -\frac{u^2}{2\sigma_1^2\sigma_2^2}\left(\sigma_2 v+\frac{\sigma_1}{v} \right)^2 \bigg\} \,dv \\
&\qquad+ \int_{-\infty}^{+\infty} \exp \bigg\{ -\frac{u^2}{2\sigma_1^2\sigma_2^2}\left(\sigma_1 v+\frac{\sigma_2}{v} \right)^2 \bigg\} \,dv.
\end{align*}
Then using the relation
$$ \left( \sigma_i v+\frac{\sigma_j}{v} \right)^2 = \sigma_i^2 \left( v - \frac{\sigma_j}{\sigma_i v} \right)^2 + 4\sigma_i \sigma_j $$
for $\{i,j\} = \{1,2\}$, we can write
\begin{align*}
I
&= e^{-\frac{2u^2}{\sigma_1 \sigma_2}} \bigg( \int_{-\infty}^{+\infty} \exp \bigg\{ -\frac{u^2}{2\sigma_1^2}\left(v - \frac{\sigma_1}{\sigma_2 v} \right)^2 \bigg\} \,dv \\
&\hspace{5em} + \int_{-\infty}^{+\infty} \exp \bigg\{ -\frac{u^2}{2\sigma_2^2}\left(v - \frac{\sigma_2}{\sigma_1 v} \right)^2 \bigg\} \,dv \bigg).
\end{align*}
Applying the Glasser's Master Theorem, we find that
\begin{align*}
I
&= e^{-\frac{2u^2}{\sigma_1 \sigma_2}} \bigg( \int_{-\infty}^{+\infty} \exp \bigg\{ -\frac{u^2}{2\sigma_1^2}v^2 \bigg\} \,dv + \int_{-\infty}^{+\infty} \exp \bigg\{ -\frac{u^2}{2\sigma_2^2}v^2 \bigg\} \,dv \bigg) \\
&= \sqrt{2\pi } \left( \frac{\sigma_1 + \sigma_2}{|u|} \right) \exp\bigg\{ -\frac{2u^2}{\sigma_1 \sigma_2} \bigg\}.
\end{align*}
