Transformation Identities of the $_2F_1$ function From Wolfram Functions we have the following identities for the hypergeometric function $_2F_1$:
$$\begin{align}
_2F_1\left(a,c-b;c;\tfrac{z}{z-1}\right)&=(1-z)^a\,_2F_1(a,b;c;z)\tag1\\
_2F_1\left(a,a+\tfrac12;c;z(2-z)\right)&=\left(1-\tfrac{z}2\right)^{-2a}\,_2F_1\left(2a,2a-c+1;c;\tfrac{z}{2-z}\right)\tag2\\
_2F_1\left(a,b;2b;\tfrac{4z}{(z+1)^2}\right)&=(1+z)^{2a}\,_2F_1\left(a,a-b+\tfrac12;b+\tfrac12;z^2\right),\tag3
\end{align}$$
where $$_2F_1(a,b;c;z)=\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}\int_0^1t^{b-1}(1-t)^{c-b-1}(1-zt)^{-a}dt.\tag{$\star$}$$
As a recreational project, I have been trying to prove the three above identities using $(\star)$.
So far, I've been able to prove $(1)$. The proof is very simple and uses transformation $t\mapsto1-t$ in the integral. The identity follows immediately after simplification.
This prompted me to try to prove the other two identities using substitutions in the integral $(\star)$
At this point, I have been able to rewrite $(2)$ as
$$\,_2F_1\left(2a,2a-c+1;c;z\right)=(1+z)^{-2a}\,_2F_1\left(a,a+\tfrac12;c;\tfrac{4z}{(z+1)^2}\right)$$
via the substitution $\tfrac{z}{2-z}\mapsto z$.This rewritten form looks very similar to $(3)$, suggesting that they are related. Beyond this I have no leads.
I have however examined the function
$$U(\alpha,\beta,\gamma,\delta;z)=\int_0^1(1+t)^\alpha t^\beta (1-t)^\gamma (1-zt)^\delta dt,$$
considering its transformation identities given by substitutions which send $[0,1]$ to itself in the integral.
This seems related, as clearly
$$_2F_1(a,b;c;z)=\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}U(0,b-1,c-b-1,-a;z).$$
For example, the substitution $t\mapsto \tfrac{1-t}{1+t}$ gives the identity
$$U(\alpha,\beta,\gamma,\delta;z)=2^{1+\alpha+\gamma}(1-z)^{-\delta}U\left(-\alpha-\beta-\gamma-\delta-2,\gamma,\beta,\delta;\frac{z+1}{z-1}\right).$$
And while setting $\alpha=\beta+\gamma+\delta+2=0$ yields
$$\,_2F_1(-\delta,\beta+1;\beta+\gamma+2;z)=2^{\gamma+1}(1-z)^{-\delta}\,_2F_1\left(-\delta,\gamma+1;\beta+\gamma+2;\frac{z+1}{z-1}\right),$$
this is not one of the transformation identities I was trying to prove.
So, I would like some hints on how to prove $(2)$ and $(3)$, if it is possible just using the integral $(\star)$.
 A: A way to prove the second identity is to use the series representation $$\,_2F_1\left(a,b;c;z\right)=\sum_{n=0}^{\infty} \frac{(a)_{n}(b)_{n}}{(c)_{n}} \frac{z^{n}}{n!} \, , \quad |z| <1, $$ where $(a)_{n} = a(a+1) \cdots (a+n-1) = \frac{\Gamma(a+n)}{\Gamma(a)}.$
As was mentioned in the question, the second identity can be rewritten as $$\,_2F_1\left(2a,2a-c+1;c;z\right)=(1+z)^{-2a}\,_2F_1\left(a,a+\frac{1}{2};c;\frac{4z}{(1+z)^2}\right). $$
Starting on the right side of the identity, we have
$\begin{align} \frac{\,_2F_1\left(a,a+\frac{1}{2};c;\frac{4z}{(z+1)^2}\right)}{(1+z)^{2a}} &= \frac{1}{(1+z)^{2a}}\sum_{n=0}^{\infty} \frac{(a)_{n} \left(a+ \frac{1}{2} \right)_{n}}{n!(c)_{n}} \frac{(4z)^{n}}{(1+z)^{2n}} \\ &= \sum_{n=0}^{\infty} \frac{(a)_{n} \left(a+ \frac{1}{2} \right)_{n}}{n!(c)_{n}} (4z)^{n}\frac{1}{(1+z)^{2a+2n}} \\ &\overset{(1)}{=} \sum_{n=0}^{\infty} \frac{(a)_{n} \left(a+ \frac{1}{2} \right)_{n}}{n!(c)_{n}} (4z)^{n} \sum_{m=0}^{\infty} \binom{2a+2n+m-1 }{m} (-z)^{m} \\  &\overset{(2)}= \sum_{k=0}^{\infty} \sum_{j=0}^{k} \frac{(a)_{j}\left(a+ \frac{1}{2} \right)_{j}}{j!(c)_{j}} 4^{j} \binom{2a+k+j-1}{k-j} (-1)^{k-j} z^{k} \\ & = \sum_{k=0}^{\infty} \sum_{j=0}^{k} \frac{(a)_{j}\left(a+ \frac{1}{2} \right)_{j}}{j!(c)_{j}} 4^{j} \frac{\Gamma(2a+k+j)}{\Gamma(2a+2j)}\frac{1}{(k-j)!}(-1)^{k-j} z^{k} \\ & \overset{(3)}{=} \sum_{k=0}^{\infty} \sum_{j=0}^{k} \frac{1}{\Gamma(2a)j!(c)_{j}} \Gamma(2a+k+j) \frac{1}{(k-j)!} (-1)^{k-j} z^{k} \\ &= \sum_{k=0}^{\infty} \sum_{j=0}^{k} \frac{1}{\Gamma(2a)j!(c)_{j}} \Gamma(2a+k+j) \frac{(-k)_{j}}{k!}(-1)^{k} z^{k} \\ &= \sum_{k=0}^{\infty} \sum_{j=0}^{k} \frac{1}{\Gamma(2a)j!(c)_{j}} (2a+k)_{j} \Gamma(2a+k)\frac{(-k)_{j}}{k!}(-1)^{k} z^{k} \\ &= \sum_{k=0}^{\infty} \frac{(-1)^{k}(2a)_{k}}{k!}z^{k} \sum_{j=0}^{k} \frac{(2a+k)_{j}(-k)_{j}}{j!(c)_{j}} \\ & \overset{(4)}{=} \sum_{k=0}^{\infty} \frac{(-1)^{k} (2a)_{k}}{k!}z^{k} \frac{(c-2a-k)_{k}}{(c)_{k}} \\ &\overset{(5)}{=} \sum_{k=0}^{\infty} \frac{(2a)_{k}(2a-c+1)_{k}}{k!(c)_{k}} z^{k} \\ &= \,_2F_1\left(2a,2a-c+1;c;z\right).\end{align}$

$(1)$ Generalized binomial theorem
$(2)$ Cauchy product
$(3)$ Gamma duplication formula
$(4)$ Chu-Vandermonde identity
$(5)$ $(-x)_{k} = (-1)^{k}(x-k+1)_{k}$
