I am interested in analytically computing the covariance of the variable $\frac{X}{\lVert X \rVert}$, where $X \in \mathbb{R}^d \sim N(\mu, I_d)$.
I have already found an analytic expression for the second moment matrix $\mathbb{E}\left( \frac{XX^T}{\lVert X \rVert^2} \right)$ (here), and for the first moment $\mathbb{E}\left( \frac{X}{\lVert X \rVert} \right)$ (here):
\begin{equation} \mathbb{E}\left( \frac{XX^T}{\lVert X \rVert^2} \right) = \frac{1}{d}{} A I_d + \frac{1}{d+2} B \mu\mu^T \end{equation}
\begin{equation} \mathbb{E}\left( \frac{X}{\lVert X \rVert}\right) = \frac{1}{\sqrt{2}} C \mu \end{equation}
Where we have $$A = {}_{1}F_1\left(1; \frac{d}{2}+1; \frac{-||\mu||^2}{2}\right)$$
$$B = {}_{1}F_1\left(1; \frac{d}{2}+2; \frac{-||\mu||^2}{2}\right)$$
$$ C = \frac{\Gamma\left(\frac{d}{2}+1-\frac{1}{2}\right)}{\Gamma\left(\frac{d}{2}+1\right)}{}_1F_1\left(\frac{1}{2};\frac{d+2}{2};\frac{-||\mu||^2}{2}\right)$$
Thus, we have that
$$Cov \left( \frac{X}{\lVert X \rVert} \right) = \mathbb{E}\left( \frac{XX^T}{\lVert X \rVert^2} \right) - \mathbb{E}\left( \frac{X}{\lVert X \rVert} \right) \mathbb{E}\left( \frac{X}{\lVert X \rVert} \right)^T = \frac{1}{d}{} A I_n + \left( \frac{1}{d+2} B - \frac{1}{2} C^2 \right) \mu\mu^T $$
Although the formula above works well, I am wondering whether there can be some simplification in either the terms $\left( \frac{1}{d+2} B - \frac{1}{2} C^2 \right)$ or just $C^2$, which involve Kummer's confluent hypergeometric function ${}_1F_1(a;b;z)$ and the gamma function.
Thus, is there any way to simplify the following formula using some ${}_1F_1(a;b;z)$ properties?
$$ \left( \frac{1}{d+2} \cdot {}_{1}F_1\left(1; \frac{d}{2}+2; \frac{-||\mu||^2}{2}\right) - \frac{1}{2} \cdot \left(\frac{\Gamma\left(\frac{d}{2}+1-\frac{1}{2}\right)}{\Gamma\left(\frac{d}{2}+1\right)}\right)^2 \cdot \left({}_1F_1\left(\frac{1}{2};\frac{d+2}{2};\frac{-||\mu||^2}{2}\right)\right)^2 \right) $$
I'm particularly interested in knowing whether that term is different than 0 and its sign at different values of $||\mu||$ and $d$, but this seems difficult to do with that expression.