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

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.

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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.