Expected ratio of x'Ax and x'AAx on a unit sphere? Suppose $A$ is symmetric positive definite matrix. Is there a nice expression for the first moment of the following quantity?
$$\frac{x^TAx}{x^TA^2 x}$$
Where $x$ is distributed as $\text{Normal}(0,I_n)$. This is the ratio of two quadratic forms evaluated on the surface of the sphere.
When $A$ has eigenvalues $\langle 1, \frac{1}{2}\rangle$, this expectation is equal to $\frac{4}{3}$, visualized below (notebook)

Edit Aug 18
To expand on hyperplane's answer, we can take $A$ to be diagonal without loss of generality and write solution as
$$E\left[\frac{x^TAx}{x^TA^2 x}\right]=\langle a, z\rangle$$
Where
$$\begin{align}
a_i=&\frac{1}{A_{ii}}\\
z_i=&E_{y\sim\mathcal{N}(0,A)} \frac{y_i^2}{\|y\|^2}
\end{align}
$$

$A\in\mathbb{R}^{n\times n}$ is symmetric positive, so there exists an orthonormal base $U=u_1,...,u_n$ and scalars $\lambda_1,...,\lambda_n$ s.t. $A=UDU^T$, with $D=\begin{pmatrix}
\lambda_1 &  & 0 \\
 & \ddots &  \\
0 &  & \lambda_n \\
\end{pmatrix}$. This is the spectral decomposition.
With these, we can decompose $$x^TAx=x^TUDU^Tx$$ and more importantly
$$x^TA^2x=x^TUDU^TUDU^Tx$$. As U is orthonormal, $U^TU=I$ and thus we get the denominator as $x^TA^2x=x^TUD^2U^Tx$.
Now we denote $w=U^Tx$. As a transformation, we get $w\sim N(U^T\mu_x, U^T\Sigma_xU)$. Plug in $x\sim N(0,I_n)$ and we get $w\sim N(0,I_n)$, again using $U^TU=I$.
The numerator is $w^TDw=\sum_{i=1}^{n}{\lambda_iw_i^Tw_i}$. Denote $g=\sum_{i=1}^{n}{\lambda_i}$, this is simply a $\chi^2_{ng}$ variable ($w^Tw\sim\chi^2_n$, we sum $g$ of those). Similarly, the denominator is a $\chi^2_{nk}$ variable, where $k=\sum_{i=1}^{n}{\lambda_i^2}$.
Overall, $\frac{x^TAx}{x^TA^2x}=\frac{w^TDw}{w^TD^2w}$ is a ratio of $\chi^2_{ng}$ variable and a $\chi^2_{nk}$ variable, so it has a beta prime distribution with parameters $\left(\alpha=\frac{ng}{2},\beta=\frac{nk}{2}\right)$. Assuming $nk>2$, the mean is
$$ \frac{\alpha}{\beta-1} = \frac{\frac{ng}{2}}{\frac{nk}{2}-1}=\frac{ng}{nk-2} $$
That's the first moment of $\frac{x^TAx}{x^TA^2x}$. Hope this helps.
