It is well known that:
$$\frac{\Gamma(c)}{\Gamma(a) \Gamma(c-a)} \int_0^1 du \, \frac{u^{a-1}(1-u)^{c-a-1}}{(1-ux)^{b_1} (1-uy)^{b_2}} = \mathfrak{F} \, (a; b_1, b_2; c \, | \, x, y) = \\ =\sum_{n = 0}^{\infty} \frac{(a)_n}{(c)_n} \sum_{n_1 + n_2 = n} (b_1)_{n_1} (b_2)_{n_2} \frac{x^{n_1}}{n_1!} \frac{y^{n_2}}{n_2!}$$
where $\mathfrak{F}$ is the (first) Appell function.
Whence one is able to express:
$$\int_0^1 du \, \frac{u^{a-1}(1-u)^{b-1}}{(u^2 A - u(1-u) B + (1-u)^2 C)^{c}}$$
through $\mathfrak{F}$'s, however, the obtained expression isn't very pleasant. On this account, I'd like to ask whether there is a "nice" formula for the above integral (in terms of hypergeometric functions)?
