TITLE: Closed form $\int_{-1}^{1} \frac{\ln (\sqrt{3} x +2)}{\sqrt{1-x^{2}} (\sqrt{3} x + 2)^{n}}\ dx$
QUESTION [13 upvotes]: Does the following integral 
$$\int_{-1}^{1} \frac{\ln (\sqrt{3} x +2)}{\sqrt{1-x^{2}} (\sqrt{3} x + 2)^{n}}\ dx, \; \; n \in \mathbb{N}$$
have a nice closed form? Basically I cannot tackle it in any direction. Symmetry is useless . Applyig parts , well , gets things worse than they actually are. Could complex analysis help us here? That is, integrating around a dog bone contour ... I highly doubt it but it is just a thought.
Any help on this one?

REPLY [11 votes]: Let's integrate by $n$ first:
$$
K(n) = \int I(n) dn = \int \int_{-1}^1 \frac{\log(\sqrt{3}x+2)}{\sqrt{1-x^2}(\sqrt{3}x+2)^n} dx dn = -\int_{-1}^1 \frac{dx}{\sqrt{1-x^2}(\sqrt{3}x+2)^n} + C
$$
Long path. (original)
According to Wolfram Mathematica
$$
K(n) = -\int_{-1}^1 \frac{dx}{\sqrt{1-x^2}(\sqrt{3}x+2)^n} + C = -\frac{\pi}{2^n} \,{_2F_1}\left(\frac{n}{2},\frac{n+1}{2};1;\frac{3}{4}\right) + C.
$$
So $$
I(n) = K'(n) = \frac{\pi \log 2}{2^n} {_2F_1}\left(\frac{n}{2},\frac{n+1}{2};1;\frac{3}{4}\right) -\\
-\frac{\pi}{2^{n+1}}\left(
\frac{\partial _2F_1}{\partial a}\left(\frac{n}{2},\frac{n+1}{2};1;\frac{3}{4}\right) +
\frac{\partial _2F_1}{\partial b}\left(\frac{n}{2},\frac{n+1}{2};1;\frac{3}{4}\right)
\right)
$$
Let
$$G(n,z) = \frac{1}{2^{n}}{_2F_1}\left(\frac{n}{2},\frac{n+1}{2};1;z\right)
$$
Or rewriting in terms of Legendre functions (A&S 15.4.10)
$$
G(n,z) = \left(\frac{1}{4-4z}\right)^{n/2} P_{n-1}\left(\frac{1}{\sqrt{1-z}}\right)\\
G\left(n,\frac{3}{4}\right) = P_{n-1}\left(2\right)
$$
Short path. Note that
$$
P_\lambda(x) = \frac{1}{2\pi}\int_{-\pi}^\pi (x + \sqrt{x^2 - 1}\cos \theta)^\lambda dx = \frac{1}{\pi}\int_0^\pi (x + \sqrt{x^2 - 1}\cos \theta)^\lambda dx.
$$
And
$$
\int_{-1}^1 \frac{dx}{\sqrt{1-x^2}(\sqrt{3}x+2)^n} = \int_{0}^\pi \frac{d\theta}{(2+\sqrt{3}\cos \theta)^n} = \pi P_{-n}(2) = \pi P_{n-1}(2)
$$
Next, following this paper and this paper
$$
\frac{\partial P_{\nu}(z)}{\partial \nu}\Big|_{\nu = n} = -P_n(z) \log\frac{z+1}{2} + \frac{1}{2^{n-1}n!}\frac{d^n}{dz^n} \left[
(z^2-1)\log \frac{z+1}{2}
\right]
$$
so
$$
\frac{\partial}{\partial n} G\left(n, \frac{3}{4}\right) = 
-P_{n-1}(2) \log\frac{3}{2} + \frac{1}{2^{n-2}(n-1)!}\frac{d^{n-1}}{dz^{n-1}} \left[
(z^2-1)^{n-1}\log \frac{z+1}{2}
\right]_{z=2}
$$
And finally
$$
I(n) = \pi P_{n-1}(2) \log\frac{3}{2} - \frac{\pi}{2^{n-2}(n-1)!}\frac{d^{n-1}}{dz^{n-1}} \left[
(z^2-1)^{n-1}\log \frac{z+1}{2}
\right]_{z=2} = -\pi A_n \log\frac{3}{2} - \pi B_n
$$
Here's a table for some $n$:
$$
\begin{array}{ccc}
n & A_n & B_n \\\hline
0 & -1 & 0\\
1 & 1 & 0 \\
2 & 2 & 1 \\
3 & \frac{11}{2} & \frac{15}{4} \\
4 & 17 & \frac{77}{6} \\
5 & \frac{443}{8} & \frac{4213}{96} \\
6 & \frac{743}{4} & \frac{36353}{240} \\
7 & \frac{10159}{16} & \frac{168833}{320} \\
8 & \frac{17593}{8} & \frac{2074197}{1120} \\
9 & \frac{984467}{128} & \frac{234465461}{35840} \\
10 & \frac{1734443}{64} & \frac{3746664781}{161280} \\
\end{array}
$$