Energy and momentum coefficients, power-law profile

Let $d$ be the height above the bed, $d_0$ the full flow depth and $u_0$ the velocity at $d = d_0$ (the surface). Take a unit width, so $dA = dd$ and $A = d_0$.

Mean velocity

$$V = \frac{1}{d_0}\int_0^{d_0} u_0\left(\frac{d}{d_0}\right)^n dd = \frac{u_0}{d_0}\cdot\frac{d_0}{n+1} = \frac{u_0}{n+1}$$

Energy coefficient (kinetic energy correction factor)

$$\alpha = \frac{\int u^3\,dA}{V^3 A}$$ $$\int_0^{d_0} u_0^3\left(\frac{d}{d_0}\right)^{3n} dd = \frac{u_0^3\,d_0}{3n+1}$$ $$\alpha = \frac{u_0^3\,d_0/(3n+1)}{\left[u_0/(n+1)\right]^3 d_0} = \frac{(n+1)^3}{3n+1}$$

Momentum coefficient (Boussinesq coefficient)

$$\beta = \frac{\int u^2\,dA}{V^2 A}$$ $$\int_0^{d_0} u_0^2\left(\frac{d}{d_0}\right)^{2n} dd = \frac{u_0^2\,d_0}{2n+1}$$ $$\beta = \frac{u_0^2\,d_0/(2n+1)}{\left[u_0/(n+1)\right]^2 d_0} = \frac{(n+1)^2}{2n+1}$$

Answer:

$$\boxed{\alpha = \frac{(n+1)^3}{3n+1},\qquad \beta = \frac{(n+1)^2}{2n+1}}$$

Check: for uniform velocity ($n = 0$), $\alpha = \beta = 1$. For $n = 1/7$, $\alpha = \dfrac{(8/7)^3}{10/7} = 1.045$ and $\beta = \dfrac{(8/7)^2}{9/7} = 1.016$, typical of real channels.