Integral $\int\limits_0^a x^b (1 - c x)^d\ \cosh x\,\mathrm dx$ How does one calculate
$$\frac{2\pi^{\frac{m-1}{2}}}{\Gamma \left(\frac{m-1}{2} \right)} \left(\frac{\alpha}{\kappa} \right)^{\frac{1}{\alpha}-1}\int\limits_0^{\kappa/\alpha}x^{m+\frac{1}{\alpha}-2}\cosh(x)\left(1-\left(\frac{\alpha}{\kappa}x\right)\right)^{\frac{2}{\alpha}}dx ?$$
 A: Since there is no context, there are no bounds on the parameters.  Consequently, I have invented bounds and evluated a closely related integral, which evaluation is dependent on those bounds.  If $c \in \mathbb{R}$ and $c > 0$, $\Re(b) > -1$, and $\Re(d) > -1$, 
$$  \int_0^{1/c} \; x^b (1 - c x)^d \cosh(x) \,\mathrm{d}x = 2^{-1-b-d}c^{-1-b}\pi \Gamma(b+1)\Gamma(d+1) \cdot {}_2\overline{F}_3\left( \frac{b+1}{2}, \frac{b+2}{2}; \frac{1}{2}, \frac{1}{2}(b+d+2), \frac{1}{2}(b+d+3); \frac{1}{4 c^2} \right)  \text{,}$$
where ${}_p \overline{F}_q(\vec{a}; \vec{b}; z)$ is the regularized generalized hypergeometric function.
A: Doing basically the same as @Eric Towers, for the integral 
$$I=\frac{2\pi^{\frac{m-1}{2}}}{\Gamma \left(\frac{m-1}{2} \right)} \left(\frac{\alpha}{\kappa} \right)^{\frac{1}{\alpha}-1}\int\limits_0^{\kappa/\alpha}x^{m+\frac{1}{\alpha}-2}\cosh(x)\left(1-\frac{\alpha}{\kappa}x\right)^{\frac{2}{\alpha}}\,dx $$
we have
$$I=K\, _2F_3\left(\frac{1}{2 \alpha }+\frac{m-1}{2},\frac{1}{2 \alpha
   }+\frac{m}{2};\frac{1}{2},\frac{3}{2 \alpha }+\frac{m}{2},\frac{3}{2 \alpha
   }+\frac{m+1}{2};\frac{\kappa ^2}{4 \alpha ^2}\right)$$
where
$$K=\frac 2 {\sqrt \pi}\left(\frac { \kappa\sqrt \pi}{\alpha}\right)^m \,\,\frac{ \Gamma \left(1+\frac{2}{\alpha }\right)\,
   \Gamma \left(m+\frac{1}{\alpha
   }-1\right)}{\Gamma \left(\frac{m-1}{2}\right)\, \Gamma
   \left(m+\frac{3}{\alpha }\right)}$$
provided $\qquad \alpha  \kappa >0\land \Re\left(m+\frac{1}{\alpha }\right)>1\land
   \Re\left(\frac{1}{\alpha }\right)>-\frac{1}{2}$
