Is there any closed form for $\int_{0}^{1} \frac{\ln ^nx}{\sqrt{1-x^{2}}}\,d x$? In my post, I found that $$\int_{0}^{1} \frac{\ln x}{\sqrt{1-x^{2}}} d x = -\frac{\pi}{2} \ln 2 .$$
Then I try to generalize the result to the integral by the same technique. $$
J_{n}:=\int_{0}^{1} \frac{\ln ^nx}{\sqrt{1-x^{2}}} d x
$$
Letting $x=\cos \theta $ converting $I_n$ into
\begin{aligned}
J_{n} &=\int_{0}^{\frac{\pi}{2} } \frac{\ln^n (\cos \theta)}{\sin \theta} \sin \theta d \theta \\
&=\int_{0}^{\frac{\pi}{2}} \ln ^{n}(\cos \theta) d \theta \end{aligned}
By my post,
$$\begin{aligned}
(-2)^{n}J_n&= 2 \ln 2(-2)^{n-1} J_{n-1} + (n-1) !\sum_{k=0}^{n-2} \frac{2^{n-k}-2}{k_ !} \zeta(n-k) (-2)^{k} J_k \\J_n&= -\ln 2 J_{n-1} + (n-1) !\sum_{k=0}^{n-2} \frac{(-1)^{n-k}}{k !} \left(1-\frac{1}{2^{n-k-1}} \right)\zeta(n-k)  J_k
\end{aligned}
$$
which is a reduction formula for $J_n.$
For example, $$
\begin{aligned}
\int_{0}^{1} \frac{\ln ^2x}{\sqrt{1-x^{2}}} d x&=-\ln 2 J_{1}+\frac{1}{2} \zeta(2) J_{0} \\
&=-\ln 2\left(-\frac{\pi}{2} \ln 2\right)+\frac{1}{2} \zeta(2) J_{0} \\
&=\frac{\pi \ln ^{2} 2}{2}+\frac{\pi^{3}}{24},
\end{aligned}
$$
which is checked by Wolframalpha.
My Question
Is there any closed form for the integral?  Your comments and methods are warmly welcome.
 A: Under the substitution $x^2 \to x$ we get that
$$I_{n,m} =\int_{0}^{1} \frac{\ln^n(x)}{\left(1-x^2\right)^m}\mathrm{d}x = \frac{1}{2^{n+1}}\int_{0}^{1}\ln^n(x) x^{-\frac{1}{2}}(1-x)^{-m}\mathrm{d}x $$ Recalling the definition of the Beta function we see that
\begin{align*}
B\left(\frac{1}{2}+t, 1-m\right) =& \int_0^{1} x^{-\frac{1}{2}+t}(1-x)^{-m}\mathrm{d}x \\
\mathbin{\color{blue}{\implies}}\frac{\mathrm{d}^n}{\mathrm{d}t^n}B\left(\frac{1}{2}+t, 1-m\right)\Bigg\vert_{t=0} =&\int_{0}^{1}\ln^n(x) x^{-\frac{1}{2}}(1-x)^{-m}\mathrm{d}x =2^{n+1}I_{n,m}
\end{align*}
So we get the generalization

$$\int_{0}^{1} \frac{\ln^n(x)}{\left(1-x^2\right)^m}\mathrm{d}x =\frac{1}{2^{n+1}}\frac{\mathrm{d}^n}{\mathrm{d}t^n}B\left(\frac{1}{2}+t, 1-m\right)\Bigg\vert_{t=0} \qquad \text{for }\ \  m<1,\ n \in \mathbb{N}$$

This result could technically be expanded in terms of polygamma functions using the General Leibniz rule, but I expect this will just result in a way messier expression.
A: This is not an answer
$$I_n=\int_{0}^{1} \frac{\log^n(x)}{\sqrt{1-x^{2}}}\,dx$$ Using the binomial expansion of the denominator
$$I_n=\sum_{k=0}^\infty\frac{\Gamma \left(k+\frac{1}{2}\right)}{\sqrt{\pi }\, k!}\int_{0}^{1} x^{2k}\,\log^n(x)\,dx$$
$$\int_{0}^{1} x^{2k}\,\log^n(x)\,dx=(-1)^n \frac{ \Gamma (n+1)}{(2 k+1)^{n+1} }$$
$$I_n=(-1)^n\frac{ n!}{\sqrt{\pi }}\sum_{k=0}^\infty \frac{ \Gamma \left(k+\frac{1}{2}\right)}{(2 k+1)^{n+1}\,k!}$$ which I am unable to simplify even using special function. But, for a given $n$, this leads to the results.
