Is there a way to calculate the definite integral for any $x\in\mathbb{R}$ and $\sigma\in\mathbb{R}^+$: $$f_\sigma(x)=\int_{-x/\sigma}^{\infty} \frac{dz}{\sqrt{2\pi}} e^{-z^2/2} \sqrt{x+\sigma z} =\int_{0}^{\infty} \frac{dz}{\sqrt{2\pi\sigma^2}} e^{-(z-x)^2/\left(2\sigma^2\right)} \sqrt{z} $$
The regular approaches (i.e. Gamma integrals) work only for $x=0$. Any result in terms of well-known functions is, of course, acceptable.
Performing the change of variables $z = \sigma t$ in $$ f_\sigma (x) = \frac{1}{{\sqrt {2\pi } \sigma }}\int_0^{ + \infty } {\sqrt z \exp \left( { - \frac{{(z - x)^2 }}{{2\sigma ^2 }}} \right){\rm d}z} $$ yields $$ f_\sigma (x) = \sqrt {\frac{\sigma }{{2\pi }}} \exp \left( { - \frac{{x^2 }}{{2\sigma ^2 }}} \right)\int_0^{ + \infty } {\sqrt t \exp \left( { - \frac{{t^2 }}{2} + \frac{x}{\sigma }t} \right){\rm d}t} . $$ By $(12.5.1)$ $$ f_\sigma (x) = \frac{1}{2}\sqrt {\frac{\sigma }{2}} \exp \left( { - \frac{{x^2 }}{{4\sigma ^2 }}} \right)U\!\left( {1, - \frac{x}{\sigma }} \right), $$ where $U(a,z)$ is the parabolic cylinder function.