In this note we show that for each fixed number of degrees of freedom $\nu$, the noncentral $t$ distribution with noncentrality parameter (= the mean of the numerator) $\mu$ is an elementary function relative to erf.

The pdf is, per Wikipedia,
$$
f(x) =\frac{\nu^{\frac{\nu}{2}} \exp\left (-\frac{\nu\mu^2}{2(x^2+\nu)} \right )}{\sqrt{\pi}\Gamma(\frac{\nu}{2})2^{\frac{\nu-1}{2}}(x^2+\nu)^{\frac{\nu+1}{2}}} \int_0^\infty y^\nu\exp\left (-\frac{1}{2}\left(y-\frac{\mu x}{\sqrt{x^2+\nu}}\right)^2\right ) dy.$$
The non-elementary part is the integral
$$
\int_0^\infty y^\nu\exp\left (-\frac{1}{2}\left(y-\frac{\mu x}{\sqrt{x^2+\nu}}\right)^2\right ) dy = g\left(\frac{\mu x}{\sqrt{x^2+\nu}}\right)
$$
where
$$
g(a)=\int_0^\infty y^\nu\exp\left (-\frac{1}{2}\left(y-a\right)^2\right ) dy.
$$
We claim that $g$ is elementary relative to the error function erf, or equivalently relative to the standard normal cdf $\Phi$.

Namely, by change of variable and the binomial theorem,
$$
g(a)=\sum_{k=0}^\nu a^{\nu-k} {\nu\choose k} \alpha_k
$$
where
$$
\alpha_k = \int_a^\infty x^k e^{-x^2/2}\,dx
$$
Next using $u=x^{k-1}$, $dv=x e^{-x^2/2}\,dx$ and integration by parts, we can show a recurrence relation for $\alpha_k$:
$$\alpha_k = a^{k-1}e^{-a^2/2} + (k-1)\alpha_{k-2}$$
Finally, $\alpha_0=\sqrt{2\pi}(1-\Phi(a))$ and $\alpha_1 = e^{-a^2/2}$, and we are done.