Recently, I commented/answered a question here on MSE. You can check it out, but my question is slightly different.
Consider
$$f(x) = \int_0^1 \frac{\sqrt{t}}{(1 -xt^2)^{\beta}}dt$$
Plugging this into WolframAlpha,
$$f(x)=\frac23 {}_2F_1\left(\frac34,\beta;\frac74;x\right)$$
I know that
$${}_2F_1(a,b;c;z)=\frac{1}{\text{B}(b,c-b)}\int_0^1 \frac{t^{b-1}(1-z)^{c-b-1}}{(1-zt)^a}dt$$
However, the methods I've tried seemed to get me nowhere.
Question: How does WolframAlpha "find" the Hypergeometric function?
Attempt 1
I tried working backwards with WolframAlpha's reply.
$$f(x)=\sum_{n=0}^{\infty}\frac{(3/4)_n(\beta)_n}{(7/4)_n}\frac{x^n}{n!}=3\sum_{n=0}^{\infty}\frac{(\beta)_nx^n}{(4n+3)n!}$$
Simplifying further seems to be a monstrous task.
Side-question: Also, if you open the MSE question I linked, can you please verify the peculiar results, which are an interpretation of the sum I derived for $f(x)$, of my comment. I would include a quotation from the question, but I feel you need a strong understanding of the other question to really get an understanding for what I'm asking for.
Please note I'm still in school, and if I miss an elementary solution, I apologize.
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!