Asymptotic expansion for the following integral I was trying to find the asymptotic expansion for
$$\int_0^1 \sqrt{t(1-t)}(t+a)^{-x} \; \mathrm{d}t,$$ for $a>0$ as $x \rightarrow \infty$. I have already tried re-writing $$(t+a)^{-x}=\exp(- x\log(t+a)).$$ However, by using Laplace's method for asymptotic expansions I get that $f(t) =  \sqrt{t(1-t)}$ vanishes everywhere.
Any ideas are appreciated/welcome. Thank you.
 A: The minimum of the phase function occurs at the endpoint $t=0$. We have
$$
\log (t + a) = \log a + \frac{t}{a} + \mathcal{O}(t^2 )
$$
and
$$
\sqrt {t(1 - t)}  = \sqrt t  - \frac{1}{2}t^{3/2}  + \mathcal{O}(t^{5/2} )
$$
as $t\to 0+$. Thus, by Laplace's method (http://dlmf.nist.gov/2.3.iii), the leading order asymptotics is
$$
e^{ - x\log a} \Gamma \left( {\frac{3}{2}} \right)\frac{{a^{3/2} }}{{x^{3/2} }} = \frac{{\sqrt \pi  }}{2}\frac{{a^{3/2 - x} }}{{x^{3/2} }}.
$$
A: There ia an antiderivative which involves the Appell hypergeometric function of two variables.
Using the bounds, this leads to
$$\int_0^1 \sqrt{t(1-t)}(t+a)^{-x} \,dt=\frac{\pi}{8} \,  (a+1)^{-x} \,\, _2F_1\left(\frac{3}{2},x;3;\frac{1}{a+1}\right)$$ where appears the gaussian hypergeometric function. This is the exact solution.
I have been unable to find the asymptotics of the gaussian hypergeometric function when its second argument tends to infinity.
Since, in their answers, @gary and @Vajra  proposed simpler approaches, I give up.
