The hypergeometric function has a copious amount of identities. An integral representation of the standard solution analytic at $z=0$ is given by $$_2F_1(a,b;c;z)=\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)} \int_0^\infty {\rm d}t \, \frac{t^{b-1}(t+1)^{a-c}}{(t-zt+1)^a} \, . \tag{0}$$ In total there are 4 linear transformations, that convert the integral into a version of itsself, namely $$t \to t \qquad t \to \frac{1}{(1-z)t} \tag{1}$$ $$t \to \frac{1}{t} \qquad t \to \frac{t}{1-z} \, . \tag{2}$$ (1) transforms $_2F_1$ into a version with argument $z$ and (2) into a version with argument $\frac{z}{z-1}$ (see here).
These are linear transformations, but there are also non-linear transformations such as $${_2F_1}(a,b;a-b+1,z^2)={(1+z)^{-2a}} {_2F_1}\left(a,a-b+1/2;2a-2b+1;\frac{4z}{(1+z)^2}\right) \, . \tag{3}$$
In general these quadratic transformations can be classified into 4 groups as seen in this table.
My first question is more general with regard to this classification. How do I know that my list/table is exhaustive? How do the 4 groups arise (I guess it is not just try and error)? What theory is applied here and can you give an example?
My second question is more specific about this example in (3) above. I wanted to use (0) and find an integral transformation that turns (0) into the RHS of (3), just like it is possible in the linear case, namely (1) and (2). For that I put $u=\frac{4z}{(1+z)^2}$, which gives rise to the quadratic equation in $z$ $$z^2+2z(1-2/u)+1=0$$ of which one solution behaves like $4/u$ about $u=0$ and the other $$z=\frac{2-u-2\sqrt{1-u}}{u} \tag{4}$$ has the correct behaviour of $u/4$ about $u=0$. We calculate $$1+z=\frac2u (1-\sqrt{1-u}) = \frac{2}{1+\sqrt{1-u}} \\ (1+z)^2 = 4z/u = \frac{4}{2-u+2\sqrt{1-u}} \\ (1-z)=\frac2u (u-1+\sqrt{1-u}) \\ \frac{1-z}{1+z} = \frac{u}{1-\sqrt{1-u}} - 1 = \sqrt{1-u}$$ and plug (4) into the integral in (0) (with $z^2$ instead of $z$ and $c=a-b+1$) and get $$\int_0^\infty {\rm d}t \, \frac{t^{b-1}(t+1)^{b-1}}{\left(t(1-z^2)+1\right)^a} = (1+z)^{-2a} \int_0^\infty {\rm d}t \, \frac{t^{b-1}(t+1)^{b-1}}{\left(t \, \frac{1-z}{1+z}+\frac{1}{(1+z)^2}\right)^a} \\ =(1+z)^{-2a} \int_0^\infty {\rm d}t \, \frac{t^{b-1}(t+1)^{b-1}}{\left(t\sqrt{1-u}+1/2-u/4+\frac12\sqrt{1-u}\right)^a} \, .$$ Now I'm not sure how to get rid of $\sqrt{1-u}$, because we eventually want something linear in $u$ with $a'=a$. Also I thought it maybe a non-linear transformation that does the job, because the only linear transformations that map $(0,\infty)$ onto itsself are of the form $t\mapsto ct$ and $t\mapsto c/t$ for some constant $c$ (which are the ones used in (1) and (2)) and I don't think these work.
I know it is more straight forward to show that the LHS and RHS of (3) solve the same DGL and each of them can be written in hypergeometric form by a suitable choice of variable (i.e. $z^2$ and $4z/(1+z)^2$). Then one of them is written as a general solution of the other and the analyticity at $z=0$ is inspected. However, I'm particularly interested in a solution transforming the integral.