Evalaute $\int_{-\alpha}^{\alpha}{1\over x}\sqrt{\alpha+x\over \alpha-x}\ln^n\left({\alpha+x\over \alpha-x}\right)\mathrm dx=(-\pi)^{n+1}F(n)$ Proposed:
$$\int_{-\alpha}^{\alpha}{x^k}\sqrt{\alpha+x\over \alpha-x}\ln^n\left({\alpha+x\over \alpha-x}\right)\mathrm dx=F(k,n)\tag1$$
Let $k=-1$ and $F(-1,n)=F(n)$

$$\int_{-\alpha}^{\alpha}{1\over x}\sqrt{\alpha+x\over \alpha-x}\ln^n\left({\alpha+x\over \alpha-x}\right)\mathrm dx=(-\pi)^{n+1}F(n)\tag2$$
  Where $n,\alpha\ge1$

We have the following
$n=1\implies$ $F(1)=1$
$n=2\implies$ $F(2)=-1$
$n=3\implies$ $F(3)=2$
$n=4\implies$ $F(4)=-5$
$n=5\implies$ $F(5)=16$
$n=6\implies$ $F(6)=-61$
How can we find the closed form of $(2)?$
$u={\alpha+x\over \alpha-x}\implies dx={(\alpha-x)^2\over 2\alpha}$
$x=\alpha\cdot{u-1\over u+1}$
$(\alpha-x)^2={4\alpha^2\over (u+1)^2}$
$(2)\implies$
$$2\int_{0}^{\infty}{\sqrt{u}\over u^2-1}\cdot\ln^n(u)\mathrm du\tag3$$
$$....$$
 A: The comment of Simply Beautiful Art gives a big hint on how to proceed. I cannot give a general formula to (1), although it is possible albeit complicated.
Let $$I(k,t)=\int_{-1}^{+1}x^k\left(\frac{1+x}{1-x}\right)^t dx \quad\quad J(k,n)=\int_{-1}^{+1}x^k\sqrt{\frac{1+x}{1-x}}\ln^n\left(\frac{1+x}{1-x}\right)dx$$
making $u = \frac{1+x}{1-x}$ gives 
$$I(k,t) = 2\int_{0}^{\infty} \frac{u^t(u-1)^k}{(u+1)^{2+k}} dx$$
When $k$ is a positive integer, we can expand and note that $$\int_{0}^{\infty} \frac{u^a}{(u+1)^b} du = B(a+1,b-a-1)$$ with the beta function.
Since the derivative of gamma function at half-integer value can be calculated, the closed form for each $J(k,n)$ can also be found.
For instances

$$J(1,1) = 2\pi \quad J(2,1) = \frac{5\pi}{3} \quad J(2,1) = \frac{5\pi}{3} \quad \\ J(4,1) = \frac{89\pi}{60} \quad J(5,1) = \frac{89\pi}{60} \quad J(6,1) = \frac{381\pi}{280}$$

$$\begin{aligned}
J(1,2) = J(2,2) = 4\pi + \frac{\pi^3}{2} \\
J(3,2) = J(4,2) = \frac{14\pi}{3} + \frac{3\pi^3}{8} \\
J(5,2) = J(6,2) = \frac{439\pi}{90} + \frac{5\pi^3}{16} \\
\end{aligned}$$

$$\begin{aligned}
J(1,3) & = 6\pi^3 \\
J(2,3) &= J(3,3) = 8\pi + 5\pi^3 \\
J(4,3) &= J(5,3) = 12\pi + \frac{89\pi^3}{20} \\
\end{aligned}$$
