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 $z$?

Solutions Collecting From Web of "Closed form for $\int_{-1}^1\frac{\ln\left(2+x\,\sqrt3\right)}{\sqrt{1-x^2}\,\left(2+x\,\sqrt3\right)^n}dx$"

Define: $$J_{{n}} \left( x,y,z \right) =\int _{-1}^{1}\!{\frac {\ln \left( xz+y
\right) }{\sqrt {1-{x}^{2}} \left( xz+y \right) ^{n}}}{dx}\quad:
n\in \mathbb{Z},\, 0\le z\le y \in \mathbb{R} \tag{1}$$
and note that:

$$\begin{aligned}
{\frac {\partial }{\partial y}}J_{{n}} \left( x,y,z \right) &=\int _{-1
}^{1}\!{\frac {1}{\sqrt {1-{x}^{2}} \left( xz+y \right) ^{1+n}}}{dx}-
nJ_{{1+n}} \left( x,y,z \right)\\
&=\frac{\left( -1 \right) ^{n}}{n!}{\frac {\partial ^{n}}{\partial {y}^{n
}}}\int _{-1}^{1}\!{\frac {1}{\sqrt {-{x}^{2}+1} \left( xz+y \right) }
}{dx}-nJ_{{1+n}} \left( x,y,z \right)
\end{aligned} \tag{2}
$$
Now consider the integral in $(2)$:
$$
\begin{aligned}
\int _{-1}^{1}\!{\frac {1}{\sqrt {-{x}^{2}+1} \left( xz+y \right) }}{d
x}&=\frac{1}{2z}\,\int _{-\pi }^{\pi }\! \frac{1}{\left( \sin \left( \theta \right) +{
\frac {y}{z}} \right)}{d\theta} \,:\,x=\sin(\theta)\\
&={\frac {i}{2\sqrt {{y}^{2}-{z}^{2}
}}}\oint\!
\frac{1}{\left( w-\xi_{{1}} \right)}- \frac{1}{\left( w-\xi_{{2}} \right)}{dw}
\end{aligned} \tag{3}
$$
with:
$$\xi_{{1}}=i{\frac {y+\sqrt {{y}^{2}-{z}^{2}}}{z}},\quad\xi_{{2}}=i{\frac {y-\sqrt {{y}^{2}-{z}^{2}}}{z}},\quad w=e^{i\theta} \tag{4}$$
and so by the conditions in $(1)$ only $\xi_2$ is inside the unit circle contour and so, by the residue theorem:
$$
\begin{aligned}
\int _{-1}^{1}\!{\frac {1}{\sqrt {-{x}^{2}+1} \left( xz+y \right) }}{d
x}&={\frac {i}{2\sqrt {{y}^{2}-{z}^{2}
}}}(- 2i\pi)={\frac {\pi}{\sqrt {{y}^{2}-{z}^{2}
}}}
\end{aligned} \tag{5}
$$
and thus $(2)$ becomes:
$$J_{{1+n}} \left( x,y,z \right) =\frac{\pi}{n}\frac { \left( -1 \right) ^{n} \,
}{n!}\frac {\partial ^{n}}{\partial {y}^{n}} \left( {\frac {1}{\sqrt {{y}^
{2}-{z}^{2}}}} \right)-\frac{1}{n}{\frac {\partial }{\partial y}}
J_{{n}} \left( x,y,z \right) \tag{6}$$
Successive differentiation then yields:
$$J_{{n}} \left( x,y,z \right) = \frac{\left( -1 \right) ^{n-1}}{(n-1)!}{\frac {
\partial ^{n-1}}{\partial {y}^{n-1}}} \left( {\frac { \pi\left( \Psi
\left( n \right) +\gamma \right) }{\sqrt {{y}^{2}-{z}^{2}}}}+J_{{
1}} \left( x,y,z \right) \right) \tag{7}$$
where:
$$\Psi\left( n \right) +\gamma=\sum_{m=1}^{n-1}\frac{1}{m} \tag{8}$$
It remains to begin the recursion, to do so we note:
$$
\begin{aligned}
J_{{1}} \left( x,y,z \right) &=\int _{-1}^{1}\!{\frac {\ln \left( xz+y
\right) }{\sqrt {-{x}^{2}+1} \left( xz+y \right) }}{dx}\\
&=\frac{1}{2}{\frac {\partial }{\partial y}}\int _{-1}^{1}\!{\frac {
\ln^2 \left( xz+y \right) }{\sqrt {-{x}^{2}+1}}}{dx}\\
&=\frac{1}{4}{\frac {\partial }{\partial y}}\int _{-\pi }^{\pi }\! \ln^2
\left( \sin \left( \theta \right) z+y \right){d\theta}
\end{aligned} \tag{9}$$
we then use:
$$\ln \left( \sin \left( \theta \right) z+y \right) =\ln \left( \frac{q
z}{2} \right) -\sum _{l=1}^{\infty }\frac{1}{l} \left( {{\rm e}^{i\theta\,l}}+
\left( -1 \right) ^{l}{{\rm e}^{-i\theta\,l}} \right) \left( {\frac
{i}{q}} \right) ^{l}\\
q=\sqrt {{\frac {{y}^{2}}{{z}^{2}}}-1}+{\frac {y}{z}}\ge1 \tag{10}$$
together with the Kronecker integral:
$$\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{(n-m)i\theta}d\theta=\delta_{n,m} \tag{11}$$
to show that:
$$
\begin{aligned}
\frac{1}{4}\int _{-\pi }^{\pi }\! \ln^{2} \left( \sin \left( \theta
\right) z+y \right) {d\theta}=&\frac{\pi}{2}\ln^2 \left( \left( \sqrt {{\frac {{y}^{2}}{{z}^{2}}}-1}+{\frac {y}{z}} \right) \frac{z}{2} \right)\\& +\,\pi \mathrm{Li_2}\left( \left( {\frac {y}{z}}-\sqrt {{\frac {{y}^{2}}{{z}^{2}}}-1}\right)^{\!{2}} \right)
\end{aligned} \tag{12}
$$
where $\mathrm{Li_2}$ is a polylogarithm, and thus after differentiating $(12)$ w.r.t $y$ and simplifying we get:
$$J_{{1}} \left( x,y,z \right) = {\frac {\pi \, \left( \ln \left( 2
\right) -\ln \left( y+\sqrt {{y}^{2}-{z}^{2}} \right) +\ln \left( {
y}^{2}-{z}^{2} \right) \right) }{\sqrt {{y}^{2}-{z}^{2}}}} \tag{13}
$$
and finally:

$$\int _{-1}^{1}\!{\frac {\ln \left( xz+y \right) }{\sqrt {-{x}^{2}+1}
\left( xz+y \right) ^{n+1}}}{dx}=\pi \frac{\left( -1 \right) ^{n}}{n!}{\frac
{\partial ^{n}}{\partial {y}^{n}}} \left( {\frac {\Psi \left( n+1
\right) +\gamma+\ln \left( 2 \right) -\ln \left( y+\sqrt {{y}^{2}-{
z}^{2}} \right) +\ln \left( {y}^{2}-{z}^{2} \right) }{\sqrt {{y}^{2}-
{z}^{2}}}} \right)$$
$$ \tag{14}$$

Equation $(14)$ can be used to find closed form for arbitrary integer $n$ and arbitrary $0<z<y\in \mathbb{R}$. It confirms the ops’s conjectured cases and gives the $n=5$ case for example as:
$$\int _{-1}^{1}\!{\frac {\ln \left( 2+x\sqrt {3} \right) }{\sqrt {-{x}
^{2}+1} \left( 2+x\sqrt {3} \right) ^{5}}}{dx}=-\pi \, \left( {\frac {
4213}{96}}+{\frac {443}{8}}\,\ln \left( 3/2 \right) \right)$$
As an extension, it may be worth applying the generalized Leibniz formula to obtain a finite sum form; the $\log$ part could be pulled out and treated separately and it’s derivatives would be rational functions of $\sqrt{y^2-z^2}$. We also note that $2^2-(\sqrt{3})^2=1$ which would presumably help.

Update

The Legendre P function for $x\in\mathbb{R}\, ,s\in\mathbb{C}$ is defined as:
$$P_s\left( y \right) =\frac{1}{\pi}\int _{0}^{1}\!{\frac { \left( y+
\sqrt {{y}^{2}-1} \left( 2\,t-1 \right) \right) ^{s}}{\sqrt {t
\left( 1-t \right) }}}{dt} \tag{i}$$
make the substitution $t=x/2+1/2$ and use the symmetry property that stems from the defining differential equation:
$$P_{-s}(y)=P_{s-1}(y)\tag{ii}$$
differentiate w.r.t to the parameter $s$ and you have:
$${\frac {\partial }{\partial s}}P_s \left( y \right) =
-\frac{1}{\pi}\int _{-1}^{1}\!{\frac { \ln
\left( y+\sqrt {{y}^{2}-1}x \right) }{\left( y+\sqrt {{y}^{2}-1}x \right)^{s+1}\sqrt {1-{x}^{2}}}}{dx}\tag{iii}$$
then using the result attributed to Bromwich, but which I took from this paper by Szmytkowski (who has written a few papers on differentiation w.r.t to the parameter) we get:
$${\frac {\partial }{\partial s}}{P_s} \left( y
\right)|_{s=n} ={P_n} \left( y \right) \ln \left( \frac{y+1}{2}
\right) +\sum _{k=1}^{n}{\frac { \left( {P_k} \left( y
\right) -P_{k-1} \left( y \right) \right) {P_{n-k}
} \left( y \right) }{k}}\tag{iv}$$
equate $\mathrm{(iii)}$ and $\mathrm{(iv)}$ and evaluate at $y=2$ and we have:

$$\int _{-1}^{1}\!{\frac {\ln \left( 2+\sqrt {3}x \right) }{ \left( 2+
\sqrt {3}x \right) ^{n+1}\sqrt {1-{x}^{2}}}}{dx}=\\ -\pi\sum _{k=1}^{n}{\frac { \left( {P_k} \left( 2
\right) -P_{k-1} \left( 2 \right) \right) {P_{n-k}
} \left( 2 \right) }{k}}-\pi{P_n} \left( 2 \right) \ln \left( \frac{3}{2}
\right)\tag{v}$$
$$a_n=\sum _{k=1}^{n}{\frac { \left( {P_k} \left( 2
\right) -P_{k-1} \left( 2 \right) \right) {P_{n-k}
} \left( 2 \right) }{k}},\quad b_n={P_n} \left( 2 \right),\quad a_n,b_n\in\mathbb{Q} \tag{vi}$$
and:
$${P_{n-1}} \left(2 \right) =
{\mbox{$_2$F$_1$}(n,-n+1;\,1;\,-1/2)}\tag{vii}$$
which confirms the conjecture.

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 $n$.

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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.

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.