Computing $ \int_a^b \frac{x^p}{\sqrt{(x-a)(b-x)}} \mathrm dx$ I would like to compute the integral:
$$ \int_a^b \frac{x^p}{\sqrt{(x-a)(b-x)}} \mathrm dx$$
where $$ a<b$$ and $$ p\in\mathbb{N}$$
$$ \int_a^b \frac{x^p}{\sqrt{(x-a)(b-x)}} \mathrm dx=\frac{2}{b-a} \int_a^b \frac{x^p \mathrm dx}{\sqrt{1-\left(\frac{2x-(a+b)}{b-a}\right)^2}} $$
The substitution $x\rightarrow \frac{2x-(a+b)}{b-a}$ gives:
$$ \frac{1}{2^p} \int_{-1}^{1} \frac{((b-a)x+a+b)^p}{\sqrt{1-x^2}} \mathrm dx$$
I tried to make an integration by parts:
$$ \frac{1}{2^p}\left(2^{p-1}\pi(a^p+b^p)-p(b-a) \int_{-1}^1 \arcsin(x)((b-a)x+a+b)^{p-1} \mathrm dx \right)$$
What about the integral $$ \int_{-1}^1 \arcsin(x)((b-a)x+a+b)^{p-1} \mathrm dx $$?
 A: I would proceed by complex integration. I will give a sketchy proof as the detailed discussion would take pages, but involves many standard moves. Consider the contour consisting of the upper edge of $[a,b]$ ($\arg z=0$) a small circle around $z=a$, then the lower edge of $[a,b]$ ($\arg z$ decreases by $\pi i$ due to factor $\left(z-a\right)^{-1/2}$, the other factors make no contribution) and a small circle around $z=b$. Integrating along this contour amounts to integrating  along a large circle surrounding the infinity.

It should be relatively easy to show that the integrals along the small circles tend to 0 as the radii tend to 0 hence, by the residue theorem:
$$-I+e^{-\pi i}I=2\pi i R_{\infty}$$
$$-Ie^{-\frac{\pi i}{2}}\left(e^{\frac{\pi i}{2}}-e^{-\frac{\pi i}{2}}\right)=2\pi i R_{\infty}$$
$$I=-\frac{e^{\frac{\pi i}{2}}}{\sin\frac{\pi}{2}}R_\infty=-\pi e^{\frac{\pi i}{2}}R_{\infty}$$
Now it remains to find $R_{\infty}$ which is the coefficient at $z^{-1}$ taken with the opposite sign. Expanding:
$$\frac{z^{p}}{\sqrt{\left(z-a\right)\left(b-z\right)}}=e^{-\frac{\pi i}{2}}z^{p-1}\left(1-\frac{a}{z}\right)^{-\frac{1}{2}}\left(1-\frac{b}{z}\right)^{-\frac{1}{2}}=e^{-\frac{\pi i}{2}}z^{p-1}\sum_{k=0}^{\infty}\left(\begin{array}{c}
k-\frac{1}{2}\\
k
\end{array}\right)\left(\frac{a}{z}\right)^{k}\times \sum_{k=0}^{\infty}\left(\begin{array}{c}
k-\frac{1}{2}\\
k
\end{array}\right)\left(\frac{b}{z}\right)^{k}$$
So we need to take the coefficient at $z^{-p}$ in the expansion. Finally, as the exponents cancel out we obtain:
$$I=\pi \sum_{k=0}^{p}\left(\begin{array}{c}
k-\frac{1}{2}\\
k
\end{array}\right)\left(\begin{array}{c}
p-k+\frac{1}{2}\\
p-k
\end{array}\right)a^{k}b^{p-k}$$
which should be equivalent to the form stated in the previous answer.
A: I would probably find it easiest to reduce the given integral to Euler's integral for Gauss's hypergeometric function:
$$
  \int_a^b \frac{ x^p \mathrm{d} x}{\sqrt{(b-x)(x-a)}} \stackrel{x = a+(b-a) u}{=} \int_0^1 \frac{(a+(b-a)u)^p }{\sqrt{u(1-u)}} \mathrm{d} u = \\ a^p \int_0^1 u^{-\frac{1}{2}} \left(1-u\right)^{-\frac{1}{2}} \left( 1- \left(1-\frac{b}{a}\right) u \right)^p \mathrm{d} u 
$$
This is the case of the Euler integral representation of the Gauss's hypergeometric function with parameters $\alpha = \frac{1}{2}$, $\gamma=1$ and $\beta = -p$  and $z=1-\frac{b}{a}$, hence
$$
  \int_a^b \frac{ x^p \mathrm{d} x}{\sqrt{(b-x)(x-a)}} = a^p \cdot \underbrace{\mathrm{B}\left(\frac{1}{2}, \frac{1}{2}\right)}_{\pi} \cdot {}_2 F_1\left(\frac{1}{2}, -p ; 1; 1 - \frac{b}{a}\right)
$$
The hypergeometric series terminates for $p\in \mathbb{N}$, and the result coincides with that of @Marvis, but the above result extends to all the complex values of $p$:
In[29]:= With[{a = 2.4, b = 3.6, p = 3.4}, 
 a^p \[Pi] Hypergeometric2F1[1/2, -p, 1, 1 - b/a]]

Out[29]= 142.388

In[30]:= With[{a = 2.4, b = 3.6, p = 3.4}, 
 NIntegrate[x^p/Sqrt[(b - x) (x - a)], {x, a, b}]]

Out[30]= 142.388

A: Let $x = \sin(\theta)$ in your second to last integral. Then your integral reduces to
$${1 \over 2^p}\int_{-{\pi \over 2}}^{\pi \over 2} ((b - a)\sin(\theta) + (a + b))^p\,d\theta$$
If you expand the integrand you get terms of the form $Const \times \int_{-{\pi \over 2}}^{\pi \over 2}\sin^k(\theta)\,d\theta$ which there are well known formulas for.
A: Let $x = a + (b-a)t$. Then $$I(p) = \int_0^1 \dfrac{(a+(b-a)t)^p}{\sqrt{t(1-t)}} dt$$
We have that $$\int_0^1 \dfrac{t^k dt}{\sqrt{t(1-t)}} = \sqrt{\pi} \dfrac{\Gamma(k+1/2)}{\Gamma(k+1)}$$
The above simplification is possible due to the following reason. Recall that the $\beta$ function is defined as $$\beta(x,y) = \int_0^1 t^{x-1} (1-t)^{y-1} dt$$
The $\beta$-function is closely related to the $\Gamma$ function through the relation $$\beta(x,y) = \dfrac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}.$$ The proof for the above claim can be seen here.
Hence, we get that $$\int_0^1 \dfrac{t^k dt}{\sqrt{t(1-t)}} = \int_0^1 t^{k-1/2} (1-t)^{-1/2} dt = \int_0^1 t^{(k+1/2)-1} (1-t)^{1/2-1} dt\\ = \beta(k+1/2,1/2) = \dfrac{\Gamma(k+1/2) \Gamma(1/2)}{\Gamma(k+1)} = \sqrt{\pi} \dfrac{\Gamma(k+1/2)}{\Gamma(k+1)}$$
Further,
$$\Gamma(1/2) = \sqrt{\pi}$$
You can look up here for some particular values of the $\Gamma$ function.
$$\Gamma(k+1/2) = \dfrac{(2k-1)(2k-3)\cdots 1}{2^k} \sqrt{\pi} \text{ where } k \in \mathbb{Z}^+$$
The above is so since $\Gamma(z+1) = z \Gamma(z)$ and $\Gamma(1/2) = \sqrt{\pi}$.
$$\Gamma(k+1) = k! \text{ where } k \in \mathbb{Z}^+$$
$$\dfrac{\Gamma(k+1/2)}{\Gamma(k+1)} = \dfrac{(2k-1)(2k-3)\cdots 1}{2^k k!} \sqrt{\pi} = \dfrac1{4^k} \binom{2k}{k} \sqrt{\pi}$$
Hence, $$I(p) = \int_0^1 \sum_{k=0}^{p} \binom{p}{k} a^{p-k}(b-a)^{k} \dfrac{t^{k}}{\sqrt{t(1-t)}} dt\\ = \displaystyle \sum_{k=0}^{p} \left(\binom{p}{k} a^{p-k}(b-a)^{k} \int_0^1  \dfrac{t^{k}}{\sqrt{t(1-t)}} dt \right)\\
= \pi \times \left( \sum_{k=0}^{p} \left( \binom{p}{k}\binom{2k}{k} \dfrac{a^{p-k}(b-a)^{k}}{4^k} \right) \right)$$
We have that
$$I(0) = \pi$$
$$I(1) = \dfrac{a+b}{2} \pi$$
$$I(2) = \dfrac{3a^2+2ab+3b^2}{8} \pi$$ 
$$I(3) = \dfrac{5a^3+3a^2b+3ab^2+5b^3}{16} \pi$$
I am not sure if you can simplify this further in terms of elementary functions.
