I am trying to find a closed form for the integral $$I\equiv\int_0^t(x+c)^p(1-2x)^{N-1}\text{d}x,$$ where $N\in\mathbb{N}$, $p>0$, $c\ge 0$ and $t\in\left(0,\frac{1}{2}\right)$.
I thought to evaluate the integral proceeding by parts, in order to lower the integer power of the second factor in the integrand. \begin{equation}\begin{split} I&=\frac{1}{p+1}\left[(t+c)^{p+1}(1-2t)^{N-1}-c^{p+1}\right]+\\[5pt] &\quad+\frac{2(N-1)}{p+1}\int_0^t(x+c)^{p+1}(1-2x)^{N-2}\text{d}x=\dots\end{split} \end{equation} By iterating $N$ times the above step the starting integral can be rewritten as a sum \begin{equation}\tag{1}\label{eq1} I=\Gamma(N)\Gamma(p+1)\sum_{j=1}^N\frac{2^{j-1}}{\Gamma(N-j+1)\Gamma(p+j+1)}\left[(t+c)^{p+j}(1-2t)^{N-j}-c^{p+j}\right]. \end{equation} I wonder whether this is the best result one can hope for, or if further simplifications can be performed, maybe by evaluating the sum in a closed form or by proceeding in a different way from the beginning in the solution of the integral.
Edit: after the mention of hypergeometric functions in the comments, I managed to rewrite the above expression in their terms. To show this I will consider only the second term in the square brakets, but the procedure is the same for the first one.
$$S_1\equiv\sum_{j=1}^N\frac{(2c)^j}{\Gamma(N-j+1)\Gamma(p+j+1)}=\left(\sum_{j=1}^{\infty}-\sum_{j=N+1}^{\infty}\right)\frac{(2c)^j}{\Gamma(N-j+1)\Gamma(p+j+1)}$$
With the substitutions $j\rightarrow j-1$ and $j\rightarrow j-(N+1)$ in the first and second sum respectively I got
$$S_1=\sum_{j=0}^{\infty}\left[\frac{(2c)^{j+1}}{\Gamma(N-j)\Gamma(p+j+2)}-\frac{(2c)^{j+N+1}}{\Gamma(-j)\Gamma(p+j+N+2)}\right],$$
where the second fraction vanishes due to the fact that $1/\Gamma(-j)=0\;\;\forall j\in\mathbb{N}$. Using now the relation
$$\Gamma(\epsilon-n)=(-1)^{n-1}\frac{\Gamma(-\epsilon)\Gamma(1+\epsilon)}{\Gamma(n+1-\epsilon)}$$
with the identifications $\epsilon=N$ and $n=j$, I found
$$S_1=-\frac{2c}{\Gamma(-N)\Gamma(N+1)}\sum_{j=0}^{\infty}\frac{\Gamma(j+1-N)}{\Gamma(p+j+2)}(-2c)^j.$$
Thanks to the formula $\Gamma(z+1)=z\Gamma(z)$ I wrote $\Gamma(-N)\Gamma(N+1)=\Gamma(1-N)\Gamma(N)$, and it is quite easy using the definition of the hypergeometric function ${}_2F_1(a,b;c;z)$ to verify that
$$S_1=\frac{2c}{\Gamma(N)\Gamma(p+2)}{}_2F_1(1,1-N;p+2;-2c).$$
Exactly in the same way I found that
$$\sum_{j=1}^N\frac{2^j}{\Gamma(N-j+1)\Gamma(p+j+1)}\left(\frac{t+c}{1-2t}\right)^j=\frac{2}{\Gamma(N)\Gamma(p+2)}\frac{t+c}{1-2t}{}_2F_1\left(1,1-N;p+2;-\frac{2(t+c)}{1-2t}\right).$$
Putting together these results the integral of interest becomes
\begin{equation}\begin{split}
I&=\frac{1}{p+1}\left[(t+c)^{p+1}(1-2t)^{N-1}{}_2F_1\left(1,1-N;p+2;-\frac{2(t+c)}{1-2t}\right)\right.\\[5pt]
&\left.\quad-c^{p+1}{}_2F_1(1,1-N;p+2;-2c)\right].\end{split}
\end{equation}
