Closed form for $\int_{-1}^1\frac{\ln\left(2+x\,\sqrt3\right)}{\sqrt{1-x^2}\,\left(2+x\,\sqrt3\right)^n}dx$ I'm trying to find a closed form for the following integral:
$$\mathcal{J}(n)=\int_{-1}^1\frac{\ln\left(2+x\,\sqrt3\right)}{\sqrt{1-x^2}\,\left(2+x\,\sqrt3\right)^n}dx\tag1$$
I have conjectured values of $\mathcal{J}(n)$ (supported by numerical inegration) for some integer values of $n$:
$$\begin{align}&\mathcal{J}(1)\stackrel?=-\pi\ln\left(\frac32\right),\\&\mathcal{J}(2)\stackrel?=-\pi\left(1+2\ln\left(\frac32\right)\right),\\&\mathcal{J}(3)\stackrel?=-\pi\left(\frac{15}4+\frac{11}2\ln\left(\frac32\right)\right),\\&\mathcal{J}(4)\stackrel?=-\pi\left(\frac{77}6+17\ln\left(\frac32\right)\right).\end{align}\tag2$$
These values suggest that a general form for $n\in\mathbb{N}$ is
$$\mathcal{J}(n)\stackrel?=-\pi\left(a_n+b_n\ln\left(\frac32\right)\right),\tag3$$
where $a_n, b_n$ are some rational coefficients. Moreover, I conjecture that
$$b_n\stackrel?={_2F_1}\left(1-n,n;\,1;\,-\frac12\right).\tag4$$

Are my conjectures true? Can we find a formula or recurrence relation for $a_n$? Can we find a general formula for $\mathcal{J}(z)$ for non-integer values of $z$?
 A: with Maple's help, I get
$$
\mathcal{J}(n)=
\frac{\pi}{{2}^{n}}\, \left[ 
{\mbox{$_2$F$_1$}\left(\frac{n}{2},\frac{n+1}{2};\,1;\frac{3}{4}\right)\log 2}-{\frac {d}{dn}}\;
{\mbox{$_2$F$_1$}\left(\frac{n}{2},\frac{n+1}{2};\,1;\frac{3}{4}\right)} \right] 
$$
That derivative in there makes it somewhat unsatisfactory.
added 
Note
$$
{}_2F_1\left(\frac{n}{2},\frac{n+1}{2};1;\frac{3}{4}\right)
= 2^n P_{n-1}(2) =
2^n\;{}_2F_1\left(n,-n+1;1;-\frac{1}{2}\right)
$$
where $P_{n-1}$ is a Legendre function.  This perhaps (?) justifies the expression for $b_n$ in the question.  But these alternate expressions also presumably have no convenient expression for the derivative with respect to $n$.
A: $\newcommand{\+}{^{\dagger}}
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$\ds{{\cal J}\pars{n} \equiv \int_{-1}^{1}
     {\ln\pars{2 + x\root{3}} \over \root{1-x^{2}}\pars{2 + x\root{3}}^{n}}\,\dd x}$

Let's consider $\ds{{\cal K}\pars{\mu} = \int_{-1}^{1}
     {\pars{2 + x\root{3}}^{\mu} \over \root{1-x^{2}}}\,\dd x}$ such that
  $\ds{{\cal J}\pars{n} = \lim_{\mu\ \to\ -n}\totald{{\cal K}\pars{\mu}}{\mu}}$

\begin{align}
{\cal K}\pars{\mu}&=\int_{-\pi/2}^{\pi/2}\bracks{\root{3}\sin\pars{\theta} + 2}^{\mu}
\,\dd\theta
=2^{\mu}
\int_{-\infty}^{\infty}\pars{{\root{3} \over 2}\,{2t \over 1 + t^{2}} + 1}^{\mu}\,{2\,\dd t \over 1 + t^{2}}
\\[3mm]&=2^{\mu + 1}\int_{-\infty}^{\infty}
{\pars{t^{2} + \root{3}t + 1}^{\mu} \over \pars{1 + t^{2}}^{\mu + 1}}\,\dd t
=2^{\mu + 1}\int_{-\infty}^{\infty}
{\pars{t - w}^{\mu}\pars{t - w^{*}}^{\mu} \over \pars{t - \ic}^{\mu + 1}\pars{t + \ic}^{\mu + 1}}\,\dd t
\end{align}
where $\ds{w \equiv -\root{3} + \ic}$. The integral in the upper complex plane has two branch cut at $\ic$ and $w$. I guess it will be helpful.
A: Following Lucian's comment, Let's write $$
I(n)=\int^1_{-1}(2+\sqrt3x)^{-n}\frac{dx}{\sqrt{1-x^2}}
$$
so that $J(n)=-I'(n)$.
Mathematica calculates that $$
I(n)=2^{-n}\pi~{}_2F_1(\frac{1+n}{2},\frac{n}{2};1;\frac34)\\
=(2/3)^n\pi~{}_2F_1(n,n;1;\frac13)\\
=\pi~{}_2F_1(n,1-n;1;-\frac12).
$$
From DLMF 15.8.15 with $a=b=n,z=1/3$ and DLMF 15.8.1.
One should be able to get $a_n$ and $b_n$ from this.
Update 2014.03.08: From the above expression we know that $I(n)=I(1-n)$. Also, Using DLMF15.8.22, we can write $I_n$ in a form where $n$ only appears once in the paremeters. That is, we have $I(n)=\pi~{}_2F_1(n,1-n;1;-\frac12)=(2-\sqrt3)^n\pi~{}_2F_1(n,\frac12;1;4\sqrt3-6).$
Using the Contiguous Functions for 2F1 functions (DLMF 15.5.11), we know that 
$$
n(4\sqrt3-7){}_2F_1(n+1,\frac12;1;4\sqrt3-6)+(2n-1+(\frac12-n)(4\sqrt3-6)){}_2F_1(n,\frac12;1;4\sqrt3-6)+(1-n){}_2F_1(n-1,\frac12;1;4\sqrt3-6)=0.$$
That is, $$-nI(n+1)+(4n-2)I(n)-(n-1)I(n-1)=0.$$
Taking derivative with respect to $n$, we have
$$-I(n+1)+4I(n)-I(n-1)+(-nI'(n+1)+(4n-2)I'(n)-(n-1)I'(n-1))=0.$$
If we can somehow get a closed form for $-I(n+1)+4I(n)-I(n-1)$, we will have a recurrance relation for $I'(n)$. I think (and should be able to prove) that $-I(n+1)+4I(n)-I(n-1)=\pi~{}_2F_1(n,1-n;2;-\frac12)$, which will always be a rational multiple of $\pi$. That gives a possible recurrence relation concerning $b_n$: $$-nb_{n+1}+(4n-2)b_n-(n-1)b_{n-1}=0.$$
OP's original guess is that $b_n=I(n)$. We see here that $b_n$ and $I_n$ obey the same recurrence relation, so that's one step towards proving OP's claim.
