Suppose that $A$ is a symmetric non-random matrix and $X\sim N(\mu,\Sigma)$ and $b \in R^n$ is a non-random vector. Then what is the distribution of 
$$X^tAX+b^tX \quad ?$$

The distribution without the linear term is solved in the answer here(https://stats.stackexchange.com/questions/41000/transformation-of-multivariate-normal-sum-of-chi-squared).

In the case of an invertible $A$ we can write $X^tAX+b^tX=(X-h)^tA(X-h)+k$ where $h=-\frac{1}{2}A^{-1}b$ and $k=-\frac{1}{4}b^tA^{-1}b$. However, the case where $A$ is not invertible is also of interest as it arises in practice. In the case where $n=1$ this corresponds to $A=0$ and thus the distribution above is simply a normal with mean $\mu_0 = b^t\mu$ and $\sigma_0 = b^t\Sigma b$. Is there a reducible solution in the more general case of $n>1$ for arbitrary non-invertible $A$? Perhaps an appropriate transformation can disentangle the quadratic form and the linear so that we have in some basis independent sum a normal and a linear combination of scaled non-central chi-squared with 1 degree of freedom.