The student-t CDf has a hypergeometric function as a component
$$_2F_1\left(\frac{1}{2}, \frac{\nu + 1}{2}; \frac{3}{2}; -\frac{x^2}{\nu}\right)$$
where $\nu$ is the distributions degree of freedom.
Abramowitz and Stegun (10th edition, p.556, 15.1.7) have a simplification that is extremely close to this
$$_2F_1\left(\frac{1}{2}, \frac{1}{2}; \frac{3}{2}; -z^2\right) = z^{-1}\ln[z + (1+z^2)^\frac{1}{2}]$$
The only problem is that the second component of $\frac{1}{2}$ in the Abramowitz simplification is constant, whereas we have $\frac{\nu+1}{2}$.
Is there any way to apply the Abramowitz simplification to our student-t CDF component? How can we simplify our student-t CDF component into some closed form?
There are closed forms for some values of $\nu$. Here is a partial list (some of them are not found) $$\left( \begin{array}{cc} \color{blue} {\nu} &\color{blue} { \, _2F_1\left(\frac{1}{2},\frac{\nu +1}{2};\frac{3}{2};-\frac{x^2}{\nu }\right)}\\ 1 & \frac{\tan ^{-1}(x)}{x} \\ 2 & \frac{1}{\sqrt{\frac{x^2}{2}+1}} \\ 3 & \frac{1}{2} \left(\frac{\sqrt{3} \tan ^{-1}\left(\frac{x}{\sqrt{3}}\right)}{x}+\frac{3}{x^2+3}\right) \\ 4 & \frac{4 \left(x^2+6\right)}{3 \left(x^2+4\right)^{3/2}} \\ 6 & \frac{2 x^4+30 x^2+135}{135 \left(\frac{x^2}{6}+1\right)^{5/2}} \\ 8 & \frac{32 \sqrt{2} \left(x^6+28 x^4+280 x^2+1120\right)}{35 \left(x^2+8\right)^{7/2}} \\ 10 & \frac{16 \sqrt{\frac{2}{5}} \left(8 x^8+360 x^6+6300 x^4+52500 x^2+196875\right)}{63 \left(x^2+10\right)^{9/2}} \\ 12 & \frac{x^{10}+66 x^8+1782 x^6+24948 x^4+187110 x^2+673596}{673596 \left(\frac{x^2}{12}+1\right)^{11/2}} \end{array} \right)$$