Working on a problem related with Apple function, I arrive to an expresion with the following integral representation
$$\int_0^1 (u-1/2)^n(1-x\, u)^a(1-y\,u)^b\,du$$
with $a,b,x,y\in\mathbb{R}$ and $n\in\mathbb{N}$. It can be expresed as a Apple function but I wonder if is posible get a closed form in terms of elementary or special functions.
Thanks in advance.
For the lovers of finite sum expressions, Mathematica generously gives us that
$$\int_{0}^1(1-xu)^a(1-yu)^bdu=\frac{1}{(b+1)(x-y)}[(1-x)^{a+1}(1-y)^{b+1} {}_2F_1(1, 2+a+b, 2+b, x\frac{1-y}{x-y})-{}_2F_1(1, 2+a+b, 2+b, \frac{x}{x-y})]$$
Now we can show that:
$$\small{I(n,a,b;x,y)=\int_{0}^1u^n(1-xu)^a(1-yu)^bdu=\frac{1}{(x-y)^n}\int_{0}^1((1-yu)-(1-xu))^n(1-xu)^a(1-yu)^bdu\\=\frac{1}{(x-y)^n}\sum_{k=0}^{n}{n\choose k}(-1)^n\int_{0}^1(1-xu)^{a+k}(1-yu)^{b+n-k}du\\=\frac{1}{(x-y)^{n+1}}\sum_{k=0}^{n}{n\choose k}\frac{(-1)^n}{b+n-k+1}[(1-x)^{a+k+1}(1-y)^{b+n-k+1}{}_2F_1(1,2+a+b+n, 2+b+n-k, x\frac{1-y}{x-y})-{}_2F_1(1, 2+a+b+n, 2+b+n-k, \frac{x}{x-y})]}$$
A similar result with a double finite sum may be obtain by expanding $(u-1/2)^n$ using the binomial theorem again
for the deceptively neat looking result:
$$\small{\int_{0}^1(u-\frac{1}{2})^n(1-xu)^a(1-yu)^bdu=\sum_{l=0}^n{n\choose l}(-\frac{1}{2})^{n-l}I(l,a,b;x,y)}$$