Fix $t>0$, let $B$ be a Brownian motion and let $\sigma$ be a previsible process such that $$\mathbb{E}\left[\text{exp}\left(\frac{1}{2}\int_0^t\sigma_s^2ds\right)\right]<\infty.$$
Then is $$\mathbb{E}\left[\text{exp}\left(\int_0^t\sigma_sdB_s\right)\right]<\infty\qquad?$$
This holds when $\sigma$ and $B$ are independent, but I'm not sure whether it holds in general.
Thoughts
By Novikov's condition we know that, when the above condition holds, the Doleans-Dade exponential is a martingale: $$M_t=\text{exp}\left(\int_0^t\sigma_sdB_s-\frac{1}{2}\int_0^t\sigma_s^2ds\right),$$ but it's not clear to me whether the result follows from this.
[I saw the question Moment generating function of a stochastic integral, but this question appears to be somewhat different.]