Evaluating a certain integral which generalizes the ${_3F_2}$ hypergeometric function Euler gave the following well-known integral representations for the Gauss hypergeometric function ${_2F_1}$ and the generalized hypergeometric function ${_3F_2}$: for $0<\Re{\left(\beta\right)}<\Re{\left(\gamma\right)}$,
$$\small{{_2F_1}{\left(\alpha,\beta;\gamma;z\right)}=\frac{1}{\operatorname{B}{\left(\beta,\gamma-\beta\right)}}\int_{0}^{1}\frac{t^{\beta-1}\left(1-t\right)^{\gamma-\beta-1}}{\left(1-zt\right)^{\alpha}}\,\mathrm{d}t};\tag{1}$$
and for $0<\Re{\left(\mu\right)}<\Re{\left(\nu\right)}$,
$$\small{{_3F_2}{\left(\alpha,\beta,\mu;\gamma,\nu;z\right)}=\frac{1}{\operatorname{B}{\left(\mu,\nu-\mu\right)}}\int_{0}^{1}t^{\mu-1}\left(1-t\right)^{\nu-\mu-1}{_2F_1}{\left(\alpha,\beta;\gamma;zt\right)}\,\mathrm{d}t}.\tag{2}$$
I'm curious to learn if there is a way evaluate the following integral (possibly in terms of higher order generalized hypergeometric functions or the two-variable Appell functions?):

$$\small{\mathcal{I}{\left(\alpha,\beta,\gamma,z;\mu,\nu,\rho,w\right)}=\int_{0}^{1}\frac{t^{\mu-1}\left(1-t\right)^{\nu-\mu-1}}{\left(1-wt\right)^{\rho}}{_2F_1}{\left(\alpha,\beta;\gamma;zt\right)}\,\mathrm{d}t}.\tag{3}$$

Now, this integral $\mathcal{I}$ is a straightforward generalization of $(3)$, and it seems only natural to me that there is a paper on this integral out there somewhere. But if it exists it has eluded me, despite most furious Googling on my part.
If any of our resident master integrators have any insight to offer, I'd be very grateful. I'd also welcome any niche references that might be relevant here if someone happens to have any.
Cheers!
 A: In the special case $\mu=\gamma$ Prudnikov-Brychkov-Marychev (Vol. III, formula 2.21.1.20) gives an evaluation in terms of Appell's $F_3$ (of four Appel's functions, this is the one with the maximal number of parameters):
$$\mathcal I\left(\alpha,\beta,\gamma,z;\gamma,\nu,\rho,w\right)=
\frac{B\left(\gamma,\nu-\gamma\right)}{(1-w)^{\rho}}{}F_3\left(\rho,\alpha,\nu-\gamma,\beta,\nu;\frac{w}{w-1};z\right)$$
Since all arguments are ''free'' (there is no relation between them), your generalization is yet one more step beyond Appell. A relation to generalized hypergeometric functions for generic $w,z$ would be extremely surprising.
A: This answer is meant to connect the ones given by @Harry Peter and @Start wearing purple, clarifying a few questions emerged in the comments.
The integral of interest can be evaluated in the way pointed out by @Harry Peter, without forgetting to set some conditions on the parameters. First of all, for $\left|z\right|<1$ we can use the power series representation of the Gauss hypergeometric fucntion ${}_2F_1$
$$\begin{align*}\mathcal{I}(\alpha,\beta,\gamma,z;\mu,\nu,\rho,w)&=\int_0^1\frac{t^{\mu-1}(1-t)^{\nu-\mu-1}}{(1-wt)^{\rho}}{}_2F_1(\alpha,\beta,\gamma,zt)\,\mathrm{d}t\\[6pt]&=\int_0^1\sum_{n=0}^{\infty}\frac{t^{\mu+n-1}(1-t)^{\nu-\mu-1}}{(1-wt)^{\rho}}\frac{(\alpha)_n(\beta)_n}{(\gamma)_n}\frac{z^n}{n!}\,\mathrm{d}t,
\end{align*}$$
where $(d)_n$ is the (rising) Pochhammer symbol, defined by
$$(d)_n=\begin{cases}
1 &\;n=0\\
d(d+1)\cdots(d+n-1) &\;n>0.
\end{cases}$$
The integral can now be performed using Euler representation, which in our case holds for $\Re(\nu+n)>\Re(\mu+n)>0$ and $\left|\mathrm{arg}(1-w)\right|<\pi$,
$$\begin{align*}\mathcal{I}(\alpha,\beta,\gamma,z;\mu,\nu,\rho,w)&=\sum_{n=0}^{\infty}\frac{(\alpha)_n(\beta)_n}{(\gamma)_n}\frac{z^n}{n!}B(n+\mu,\nu-\mu){}_2F_1(\rho,n+\mu;n+\nu;w)\\[6pt]&=\sum_{n,m=0}^{\infty}\frac{\Gamma(n+\mu)\Gamma(\nu-\mu)}{\Gamma(n+\nu)}\frac{(\alpha)_n(\beta)_n}{(\gamma)_n}\frac{(\rho)_m(n+\mu)_m}{(n+\nu)_m}\frac{z^n}{n!}\frac{w^m}{m!},
\end{align*}$$
valid for $\left|w\right|<1$. Considering that
$$(d)_n=\frac{\Gamma(d+n)}{\Gamma(d)}\quad\text{for}\;\;d\neq 0,-1,-2,\dots$$
when $\mu,\nu\neq 0,-1,-2,\dots$ we can write
$$\begin{align*}\mathcal{I}(\alpha,\beta,\gamma,z;\mu,\nu,\rho,w)&=\sum_{n,m=0}^{\infty}\frac{\Gamma(n+\mu+m)\Gamma(\nu-\mu)}{\Gamma(n+\nu+m)}\frac{(\alpha)_n(\beta)_n}{(\gamma)_n}(\rho)_m\frac{z^n}{n!}\frac{w^m}{m!}\\[6pt]&=\frac{\Gamma(\mu)\Gamma(\nu-\mu)}{\Gamma(\nu)}\sum_{n,m=0}^{\infty}\frac{(\mu)_{n+m}(\alpha)_n(\beta)_n(\rho)_m}{(\nu)_{n+m}(\gamma)_n}\frac{z^n}{n!}\frac{w^m}{m!}\\[6pt]&=B(\mu,\nu-\mu)\,\mathrm{F}^{1:2;1}_{1:1;0}\left(\left.\begin{matrix}\mu&:&\alpha,\beta&;&\rho&\\\nu&:&\gamma&;&-&\end{matrix}\right|z,w\right).
\end{align*}$$
$\mathrm{F}^{p:q;k}_{l:m;n}$ denotes Kampé de Fériet's double hypergeometric function in the (modified) notation of Burchnall and Chaundy [see Srivastava and Panda - "An integral representation for the product of two Jacobi polynomials", Eq. (26)]
$$\begin{align*}&\mathrm{F}^{p:q;k}_{l:m;n}\left(\left.\begin{matrix}(a_p)&:&(b_q)&;&(c_k)&\\(\alpha_l)&:&(\beta_m)&;&(\gamma_n)&\end{matrix}\right|x,y\right)\\[6pt]&\quad=\sum_{r,s=0}^{\infty}\frac{\prod_{j=1}^p(a_j)_{r+s}\prod_{j=1}^q(b_j)_r\prod_{j=1}^k(c_j)_s}{\prod_{j=1}^l(\alpha_j)_{r+s}\prod_{j=1}^m(\beta_j)_r\prod_{j=1}^n(\gamma_j)_s}\frac{x^r}{r!}\frac{y^s}{s!},
\end{align*}$$
where $(d_h)$ denotes the sequence of $h$ parameters $d_1,\dots,d_h$. In general, convergence of this double series is assured if one of the following conditions holds
i) $p+q<l+m+1$, $p+k<l+n+1$ and $\left|x\right|<\infty$, $\left|y\right|<\infty$
ii) $p+q=l+m+1$, $p+k=l+n+1$ and
$ 
\begin{align*}\;\;\begin{cases}\left|x\right|^{\frac{1}{p-l}}+\left|y\right|^{\frac{1}{p-l}}<1 &\text{if}\;\;p>l\\[5pt]\max\left\{\left|x\right|,\left|y\right|\right\}<1 &\text{if}\;\;p\le l.\end{cases}\end{align*}
$
In our case we have $p=1$, $q=2$, $k=1$, $l=1$, $m=1$ and $n=0$, so we are in (ii) and the double series converges only if $\max\left\{\left|z\right|,\left|w\right|\right\}<1$. Collecting all the constraints introduced we finally have
$$\mathcal{I}(\alpha,\beta,\gamma,z;\mu,\nu,\rho,w)=B(\mu,\nu-\mu)\,\mathrm{F}^{1:2;1}_{1:1;0}\left(\left.\begin{matrix}\mu&:&\alpha,\beta&;&\rho&\\\nu&:&\gamma&;&-&\end{matrix}\right|z,w\right),$$
if $\max\left\{\left|z\right|,\left|w\right|\right\}<1$, $\left|\text{arg}(1-w)\right|<\pi$ and $\Re(\nu)>\Re(\mu)>0$.

In the case $\mu=\gamma$ the general result reduces to the one given by @Start wearing purple
$$\begin{align*}\mathcal{I}(\alpha,\beta,\gamma,z;\gamma,\nu,\rho,&w)=B(\gamma,\nu-\gamma)\,\mathrm{F}^{1:2;1}_{1:1;0}\left(\left.\begin{matrix}\gamma&:&\alpha,\beta&;&\rho&\\\nu&:&\gamma&;&-&\end{matrix}\right|z,w\right)\\[6pt]&=\int_0^1\frac{t^{\gamma-1}(1-t)^{\nu-\gamma-1}}{(1-wt)^{\rho}}{}_2F_1(\alpha,\beta,\gamma;zt)\,\mathrm{d}t\\[6pt]&=\frac{1}{(1-w)^{\rho}}\int_0^1t^{\gamma-1}(1-t)^{\nu-\gamma-1}\left(1-\frac{w}{w-1}+\frac{wt}{w-1}\right)^{-\rho}{}_2F_1(\alpha,\beta,\gamma;zt)\,\mathrm{d}t,
\end{align*}$$
where we have multiplied and divided by $(1-w)^{\rho}$. According to the single integral representation of Appell series $F_3$, the last expression is
$$\mathcal{I}(\alpha,\beta,\gamma,z;\gamma,\nu,\rho,w)=\frac{B(\gamma,\nu-\gamma)}{(1-w)^{\rho}}F_3\left(\rho,\alpha,\nu-\gamma,\beta;\nu;\frac{w}{w-1};z\right).$$ 
A: $\int_0^1\dfrac{t^{\mu-1}(1-t)^{\nu-\mu-1}}{(1-wt)^\rho}{_2F_1}(\alpha,\beta;\gamma;zt)~dt$
$=\int_0^1\sum\limits_{n=0}^\infty\dfrac{(\alpha)_n(\beta)_nz^nt^{n+\mu-1}(1-t)^{\nu-\mu-1}}{(\gamma)_nn!(1-wt)^\rho}dt$
$=\sum\limits_{n=0}^\infty\dfrac{(\alpha)_n(\beta)_nz^nB(n+\mu,\nu-\mu){_2F_1}(\rho,n+\mu;n+\nu;w)}{(\gamma)_nn!}$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{\Gamma(n+\mu)\Gamma(\nu-\mu)(\alpha)_n(\beta)_n(\rho)_k(n+\mu)_kz^nw^k}{\Gamma(n+\nu)(\gamma)_n(n+\nu)_kn!k!}$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{\Gamma(n+k+\mu)\Gamma(\nu-\mu)(\alpha)_n(\beta)_n(\rho)_kz^nw^k}{\Gamma(n+k+\nu)(\gamma)_nn!k!}$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{\Gamma(\mu)\Gamma(\nu-\mu)(\mu)_{n+k}(\alpha)_n(\beta)_n(\rho)_kz^nw^k}{\Gamma(\nu)(\nu)_{n+k}(\gamma)_nn!k!}$
$=B(\mu,\nu-\mu)\mathrm{F}^{1:2;1}_{1:1;0}\Bigg(\begin{matrix}\mu&:&\alpha,\beta&;&\rho&\\\nu&:&\gamma&;&-&\end{matrix}\Bigg|z,w\Bigg)$
