How to find $\mathbb{P}(\textbf{Y}^T\mathbf{A}\textbf{Y}>0)$ where $\mathbf{Y}$ is a vector of independent normal distributions?

Question

Say $\mathbf{Y} \sim \mathcal{N}(\mathbf{\mu}, \mathbf{D})$ where $\mathbf{D}$ is diagonal and $A$ is any real matrix, how could you calculate $\mathbb{P}(\mathbf{Y}^T\mathbf{A}\mathbf{Y}>0)$?

In the case $\mathbf{A}$ is diagonal I can see that it would be possible to write $\mathbf{Y}^T\mathbf{A}\mathbf{Y}$ as a non-central chi-squared subtract a non-central chi-squared, but I can't see how to do this for a general $\mathbf{A}$.

Edit: I've just realised as well by replacing $\mathbf{A}$ with its symmetric part, and then diagonalising $\mathbf{A}$ you can frame an alternate question which is to calculate $\mathbb{P}(\textbf{Y}^T\mathbf{D}\textbf{Y}>0)$ where $\mathbf{Y} \sim \mathcal{N}(\mathbf{\mu}, \mathbf{A})$ for diagonal $\mathbf{D}$ and general covariance matrix $\mathbf{A}$.