Compute $E[ \sqrt{Q} e^{-tQ} ]$ where $Q$ is non-central chi-square random variable. Let  $Q$  be non-central chi-square random variable of degree $k$ and centrality parameter $\mu$. 
How to compute  the following expectation 
\begin{align}
E\left[ \sqrt{Q} e^{-tQ} \right], t>0 
\end{align} 
I was thinking this can be computed by using the fact that $Q= \sqrt{ \sum_{i=1}^k Z_i^2 }$ where $Z_i \sim \mathcal{N}(\mu_i,1)$ and $\mu= \sqrt{\sum_{i=1}^k \mu_i^2}$, and $Z_i$'s are independent. 
Using this observation we get that
\begin{align}
E\left[ \sqrt{\sum_{i=1}^k Z_i^2 } e^{-t\sum_{i=1}^k Z_i^2 } \right]= E\left[ \sqrt{\sum_{i=1}^k Z_i^2 }  \prod_{i=1}^k  e^{-tZ_i^2 } \right].
\end{align} 
However, it is not clear to me how such exressions can be  manipulated or expanded. 
I also tried direct integration but expression seem very long, so I stopped.  

Mathematica 11.3.0 produces
Mean[TransformedDistribution[Sqrt[x]*Exp[-t*x],x \[Distributed] 
 NoncentralChiSquareDistribution[k, \[Mu]],  Assumptions -> t > 0]]


$$\sqrt{2} e^{-\frac{\mu }{2}} (2 t+1)^{-\frac{k}{2}-\frac{1}{2}} \Gamma \left(\frac{k+1}{2}\right) \, _1\tilde{F}_1\left(\frac{k+1}{2};\frac{k}{2};\frac{\mu }{4 t+2}\right) $$

Edit. Thanks to @owen88 who paid my attention to the typo in my code:x*Exp[-t*x] instead of Sqrt[x]*Exp[-t*x]. Fixed.
