My best guess is that WolframAlpha did this:
$$
\left(1-t^{2}x\right)^{-\beta}=\left(1-t^{2}x\right)^{\frac{7}{4}-\frac{7}{4}-\beta}\ _{2}F_{1}\left(\frac{7}{4}-\frac{7}{4},\frac{7}{4}-\beta;\frac{7}{4};xt^{2}\right)\,.
$$
It seems to come out of nowhere, but this is true from the definition of the hypergeometric function (that you shared above)
$$
_{2}F_{1}\left(a,b,c;z\right)=1+\frac{ab}{c}\frac{z}{1!}+\frac{a\left(a+1\right)b\left(b+1\right)}{c\left(c+1\right)}\frac{z^{2}}{2!}+\ldots
$$
where $|z|<1$. After all, if $a=0$, then the series equals $1$ because of the first term and the rest of the terms vanishing. This is sort of how I came up with the $\displaystyle \frac{7}{4}-\frac{7}{4}$.
Next, we use the Hypergeometric Euler transformation
$$
\left(1-z\right)^{c-a-b} \ _{2}F_{1}\left(c-a,c-b;c;z\right)=\ _{2}F_{1}\left(a,b;c;z\right)
$$
to get
$$
\left(1-t^{2}x\right)^{\frac{7}{4}-\frac{7}{4}-\beta}\ _{2}F_{1}\left(\frac{7}{4}-\frac{7}{4},\frac{7}{4}-\beta;\frac{7}{4};xt^{2}\right) = \ _{2}F_{1}\left(\frac{7}{4},\beta;\frac{7}{4};xt^{2}\right)\,.
$$
Furthermore, we can use the identity
$$
_{2}F_{1}\left(a+1,b;c;z\right)=\frac{bz}{c}\ _{2}F_{1}\left(a+1,b+1;c+1;z\right)+\ _{2}F_{1}\left(a,b;c;z\right)
$$
and split up our expression like this:
$$
\begin{align}
_{2}F_{1}\left(\frac{7}{4},\beta,\frac{7}{4};xt^{2}\right)&=\ _{2}F_{1}\left(\frac{3}{4}+1,\beta,\frac{7}{4};xt^{2}\right) \\
&= \frac{4\beta x}{7}t^{2}\ _{2}F_{1}\left(\frac{7}{4},\beta+1,\frac{11}{4};xt^{2}\right)+\ _{2}F_{1}\left(\frac{3}{4},\beta;\frac{7}{4};xt^{2}\right)\,. \tag{1}\\
\end{align}
$$
Even though some of these steps look like I pulled them out of thin air and don't seem to initially go anywhere, the line $(1)$ will be helpful.
Next, we will prove the integral equals what you got when you plugged the integral into WolframAlpha. Denote the integral in question as $\mathcal{I}$. We get
$$
\mathcal{I} = \int_{0}^{1}\frac{\sqrt{t}}{\left(1-t^{2}x\right)^{\beta}}dt\overset{(1)}=\int_{0}^{1}\sqrt{t}\ _{2}F_{1}\left(\frac{3}{4},\beta;\frac{7}{4};xt^{2}\right)dt+\frac{4}{7}\beta x\int_{0}^{1}t^{\frac{5}{2}}\ _{2}F_{1}\left(\frac{7}{4},\beta+1,\frac{11}{4};xt^{2}\right)dt\,.
$$
With the help of the derivative of the hypergeometric function, we can integrate by parts on the first integral:
$$
u= \ _{2}F_{1}\left(\frac{3}{4},\beta;\frac{7}{4};xt^{2}\right) \implies du = \frac{\frac{3}{4}\beta}{\frac{7}{4}}\ _{2}F_{1}\left(\frac{7}{4},\beta+1;\frac{11}{4};xt^{2}\right)\cdot2xt
$$
and
$$
dv=\sqrt{t}dt \implies v = \frac{2}{3} t^{\frac{3}{2}}\,.
$$
Simplifying things nicely, we finish with
$$
\begin{align}
\mathcal{I} =& \text{ }\left[\ _{2}F_{1}\left(\frac{3}{4},\beta;\frac{7}{4};xt^{2}\right)\cdot\frac{2}{3}t^{\frac{3}{2}}\right]_{0}^{1}-\int_{0}^{1}\frac{2}{3}t^{\frac{3}{2}}\cdot\frac{6}{7}\beta xt\ _{2}F_{1}\left(\frac{7}{4},\beta+1;\frac{11}{4};xt^{2}\right)dt \\
&+\text{ } \frac{4}{7}\beta x\int_{0}^{1}t^{\frac{5}{2}}\ _{2}F_{1}\left(\frac{7}{4};\beta+1;\frac{11}{4};xt^{2}\right)dt \\
&= \ _{2}F_{1}\left(\frac{3}{4},\beta;\frac{7}{4};x\left(1\right)^{2}\right)\cdot\frac{2}{3}\left(1\right)^{\frac{3}{2}}-\ _{2}F_{1}\left(\frac{3}{4},\beta;\frac{7}{4};x\left(0\right)^{2}\right)\cdot\frac{2}{3}\left(0\right)^{\frac{3}{2}} + 0 \\
&= \frac{2}{3} \ _{2}F_{1}\left(\frac{3}{4},\beta;\frac{7}{4};x\right) \\
\end{align}
$$
and we're done!
