Let $n$ be a large positive integer and $c \ge 0$ and define $I(n,c) := \int_0^1(u-u^2)^{n/2}e^{-cu^2}du$.
Question. How to rewrite (or upper-bound) $I(n,c)$ in terms of special functions (gamma, beta, ratios thereof, etc.) ?
For example, one always has
$$ \begin{split} I(n,c) &\le I(n,0) = \int_0^1(u-u^2)^{n/2}du = \frac{\Gamma(n/2+1)^2}{\Gamma(n + 2)}. \end{split} $$
The issue with this bound is that it doesn't take advantage of $c$ (i.e I'd expect to get a better bound for large $c$).
On
There is an exact solution for $$I_n= \int_0^1(u-u^2)^{n/2}\,e^{-cu^2}\,du$$ $$I_n=\sqrt{\pi }\,\frac{ \Gamma \left(\frac{n+2}{2}\right)}{2^{n+1}\,\Gamma \left(\frac{n+3}{2}\right)}\, _2F_2\left(\frac{n+2}{4},\frac{n+4}{4};\frac{n+2}{2},\frac{n+3}{2};-\frac{c}{2}\right)$$ If $n$ is even, the formulae are not too bad. To simplify them, let
$$t=\sqrt{\frac{\pi }{2}} \text{erf}\left(\frac{\sqrt{c}}{\sqrt{2}}\right)$$ $$I_2=\frac{1}{c}-\frac{t}{c^{3/2}}$$ $$I_4=\frac{e^{-c/2}-4}{c^2}+\frac{(c+3) t}{c^{5/2}}$$ $$I_6=\frac{2 c+24-9 e^{-c/2}}{c^3}-\frac{3 (3 c+5) t}{c^{7/2}}$$ $$I_8=\frac{3 (c+29) e^{-c/2}-32 (c+6)}{c^4}+\frac{3 (c^2+30c+35) t}{c^{9/2}}$$
If we consider the case of small values of $c$, we could write $$I_n=\sum_{k=0}^\infty (-1)^k\frac{c^k }{2^k \,k!} \int_0^1 {u^{2k}} (u-u^2)^{\frac n 2}\,du$$ $$\int_0^1 {u^{2k}} (u-u^2)^{\frac n 2}\,du=\frac{\Gamma \left(\frac{n}{2}+1\right) \Gamma \left(2 k+\frac{n}{2}+1\right)}{\Gamma (2 k+n+2)}$$ This would give $$I_n \sim\frac{\Gamma \left(\frac{n}{2}+1\right)^2}{\Gamma (n+2)}\Big[1+\sum_{n=1}^\infty a_n\, c^n\Big]$$ where the first coefficients are $$\left\{-\frac{(n+4)}{8 (n+3)},\frac{(n+6) (n+8)}{128 (n+3) (n+5)},-\frac{(n+8) (n+10) (n+12)}{3072 (n+3) (n+5) (n+7)},\cdots\right\}$$ where we can see clear patterns. Notice that the numbers in denominators are the octo-factorial numbers. Since the series is alternating, it seems quite easy to find upper and lower bounds (as long as $c$ is small).
I might be a little off at the end with the constants.
Substitute $ u = w - \tfrac12 $: \begin{align} I &= \int_{-1/2}^{1/2} (\tfrac14-w^2)^{n/2} e^{-c w^2 + c w - c/4} \, dw \\&= \frac{e^{-c/4}}{2^n}\int_{-1/2}^{1/2} (1-4w^2)^{n/2} e^{-c w^2 + cw } \, dw .\end{align} Substitute $ w = \tfrac v{\sqrt n} $, and use $1-x \le e^{-x}$: \begin{align} I &= \frac{e^{-c/4}\sqrt n}{2^n}\int_{-\sqrt n/2}^{\sqrt n/2} (1-4\tfrac{v^2}{n})^{n/2} e^{-c v^2/n + c v/\sqrt n} \, dv \\&\le \frac{e^{-c/4}\sqrt n }{2^n}\int_{-\infty}^{\infty} e^{-2v^2-c v^2/n + c v/\sqrt n} \, dv \end{align} Now completing the square, we get \begin{align} -2v^2-\frac{c v^2}n + \frac{c v}{\sqrt n} &= -(2+\tfrac cn) v^2 + \tfrac c{\sqrt n} v \\&= -(2+\tfrac cn) (v - \tfrac c{2\sqrt n}(2+\tfrac cn)^{-1}) ^2 + \tfrac{c^2} {4n} (2+\tfrac cn)^{-1} \end{align} So $$ I \le J := \frac{e^{-c/4}2\sqrt{\pi n}}{2^n} \exp\left(\tfrac{c^2} {4n} (2+\tfrac cn)^{-1}\right) \frac1{\sqrt{2+\tfrac cn}}$$ Notice that as $n \to \infty$, and using Stirling's forumla $$ J \sim e^{-c/4} \frac{\Gamma(\frac n2+1)^2}{\Gamma(n+2)}, $$ so this asymptotically gives an extra $e^{-c/4}$. This is to be expected, because when you compute the integral, almost all of the mass of the integrand will be concentrated around $u = \frac12$.