Closed form for $\int_0^t(x+c)^p(1-2x)^{N-1}\text{d}x$ 
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}

So just to have everything here, let me show a general approach for the following  $$\sf I=\int_e^f (a+x)^p(b-x)^qdx$$
Either one of the substitution $\sf \frac{b-x}{b+a}=y$ or $\sf \frac{a+x}{b+a}=y$ should work fine as a start.
I'll take the first one and get $\sf x=b-y(a+b)\Rightarrow dx=-(a+b)dy$.
Also for better view, denote $\sf \frac{b-f}{b+a}=k$ and $\sf \frac{b-e}{b+a}=n$ and assume $\sf 0<e<f$.
$$\sf \Rightarrow  I=(a+b)\int_{k}^{n}(a+b-y(a+b))^p(b-b+y(a+b))^qdy$$
$$\sf =(a+b)^{p+q+1}\int_k^n(1-y)^py^qdy=(a+b)^{p+q+1} \left(\int_0^n-\int_0^k\right)$$
$$\sf =(a+b)^{p+q+1}\left(B\left(\frac{b-e}{b+a};q+1,p+1\right)-B\left(\frac{b-f}{b+a};q+1,p+1\right)\right)$$
Where $\sf B(z;\alpha,\beta)$ is the Incomplete Beta function also called Chebyshev Integral.

In particular your integral is equal to:
$$\tt 2^{N-1}\int_0^t (x+c)^p(1/2-x)^{N-1}dx$$
$$\tt =2^{N-1}\left(c+\frac12\right)^{p+N}\left(B\left(\frac{1}{1+2c};N,p+1\right)-B\left(\frac{1-2t}{1+2c};N,p+1\right)\right)$$
Hopefully I haven't done any computation mistakes, but feel free to ask if anything is unclear.
