Simplifying a ratio of incomplete beta functions

Question

Can the following be simplified?

$$ \frac{\int_0^a t^{x+1} (1-t)^{y} dt}{\int_0^a t^{x} (1-t)^{y} dt} \qquad \big((x,y)\in(0,\infty)^2; a\in(0,1]\big) $$

Note: If it helps to assume that $0<x<y$ and/or that $(x,y)\in\mathbb{N}^2$ please do!

Answer 1

I don't know that I would call this a simplification, per se, but the beta function can be expressed in terms of the (Gauss) hypergeometric function. Using the notation of the NIST Handbook of Math Functions, we have

$$ B_x(a,b)=\int_0^x t^{a-1}(1-t)^{b-1} dt=\frac{x^a}{a}F(a,1-b;a+1;x) $$

So that

$$ \frac{B_x(a+1,b)}{B_x(a,b)}=\bigg(\frac{a}{a+1}\bigg)x\frac{F(a+1,1-b;a+2;x)}{F(a,1-b;a+1;x)} $$

This doesn't seem like much of an improvement, but there are no integrals. Some further simplification may accrue by noting that

$$ F(a,b;c;x)=(1-x)^{c-a-b}F(c-a,c-b;c;x) $$

especially in light of the fact that $c=a+1$ in both cases. Finally, there are all kinds of identities for specific values or relations between $a,b,c$ that may provide solutions for particular values of your parameters.