The fractional linear transformation of hypergeometric function can be proved by using substitution and integral expression of hypergeometric function. $$ \begin{align} {}_2F_1\left( a,b;c;x \right) &= \frac{\Gamma \left( c \right)}{\Gamma \left( c-b \right) \Gamma \left( b \right)}\int_0^1{u^{b-1}\left( 1-u \right) ^{c-b-1}\left( 1-xu \right) ^{-a}du} \quad \small\text{, let $u=1-t$} \\ &=\frac{\Gamma \left( c \right)}{\Gamma \left( c-b \right) \Gamma \left( b \right)}\int_0^1{\left( 1-t \right) ^{b-1}t^{c-b-1}\left( 1-x+xt \right) ^{-a}dt} \\ &=\left( 1-x \right) ^{-a}\frac{\Gamma \left( c \right)}{\Gamma \left( c-b \right) \Gamma \left( b \right)}\int_0^1{\left( 1-t \right) ^{b-1}t^{c-b-1}\left( 1-\frac{x}{x-1}t \right) ^{-a}dt} \\ &={\left( 1-x \right) ^{-a}} {}_2F_1\left( a,c-b;c;\frac{x}{x-1} \right) \end{align}$$
But I don't know how to prove the quadratic transformations by using the same method.
e.g. $${\sideset{_2}{_1}F}\left( a,b;a+b+\frac{1}{2};4x\left( 1-x \right) \right) = {}_2F_1 \left( 2a,2b;a+b+\frac{1}{2};x \right) $$
Is it workable to prove this formula by using substitution?