I'm interesed in the following integral, for $a,b>0$: $$ \mathcal{I}(a,b) := \int_0^1 \frac{\ln^2(x)}{\sqrt{x(a-bx)}}\,dx $$ Mathematica could evaluate it in term of hypergeometric functions, but I'm looking for a simpler closed-form.

If it is too difficult, then it would be nice to see a proof for the following two special cases: $$\begin{align} \mathcal{I}(1,1) &= \int_0^1 \frac{\ln^2(x)}{\sqrt{x(1-x)}}\,dx = \frac{\pi^3}{3}+4\pi\ln^2(2)\\ \mathcal{I}(4,1) &= \int_0^1 \frac{\ln^2(x)}{\sqrt{x(4-x)}}\,dx = \frac{7\pi^3}{27} \end{align}$$ Any other simple special case are welcome.

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    $\begingroup$ $\mathcal{I}(1,1)$ can be computed by differentiating twice (with respect to $\alpha$) $\int_{0}^{1}x^{\alpha-\frac{1}{2}}(1-x)^{-1/2}\,dx$, that is a value of the Euler beta function. $\endgroup$ – Jack D'Aurizio Sep 16 '15 at 17:57
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    $\begingroup$ And I think $\mathcal{I}(4,1)$ can be recovered from the Taylor expansion of $\frac{1}{\sqrt{4-x}}$ and the well known representation of the squared arcsine proved here: math.stackexchange.com/questions/878477/… $\endgroup$ – Jack D'Aurizio Sep 16 '15 at 18:02
  • $\begingroup$ By the way, if you want, write \mathrm dx to generate $\mathrm dx$ as opposed to $dx$. $\endgroup$ – Mr Pie Apr 25 '18 at 4:59
  • $\begingroup$ @user477343 Thank you for the advise, but I prefer the $dx$ notation. $\endgroup$ – user153012 Apr 25 '18 at 5:28
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    $\begingroup$ Ok. I used to edit lots of posts from $dx$ to $\mathrm dx$ until I realised that there are users like you who prefer the former notation. Now, I am careful :) $\endgroup$ – Mr Pie Apr 25 '18 at 5:39

Too Long for a comment. (see edit, it now contains the closed form.)

This is equivalent to a one variable function.

Substitute $x\to x^2$, and then $x=\frac1{a}\sin(t)$ to get $$ \int_0^1 \frac{\ln^2(x)}{\sqrt{x(1-a^2x)}}dx=8\int_0^1 \frac{\ln^2(x)}{\sqrt{1-a^2x^2}}dx=\frac{8}{a}\int_0^{\sin^{-1}(a)}\ln^2\left(\frac{\sin t}{a}\right)dt \\=\frac{8\ln^2(a)\sin^{-1}(a)}{a}-\frac{16\ln(a)}{a}\int_0^{\sin^{-1}(a)}\ln\sin x\,\,dx+\frac{8}{a}\int_0^{\sin^{-1}(a)}\ln^2\sin x\,\,dx$$

For certain algebraic $a$'s ,I think we can find a closed form.

For example, the case $a=\frac{\sqrt{2}}{2}$ corresponds to $\mathcal{I}(2,1)$, and using the (wonderful) results obtained by @RandomVariable, we have $$\mathcal{I}(2,1)=\frac{\pi}{2}\ln^2(2)-4\ln(2)(G+\frac{\pi}{2}\ln2)+8\left(\frac{\pi^{3}}{192} + G\frac{ \log(2)}{2} + \frac{3 \pi}{16} \log^{2}(2) - \text{Im} \ \text{Li}_{3}(1-i)\right)\,\,\left(=\frac{\pi^3}{24}-8\Im\operatorname{Li_3}(1-i)\right)$$


After some work I've been able to find the closed form. Towards the end, there's a huge cancellation which bothers me- there must be a straightforward way, but I'm blind.

I'll sketch how I found it: consider $\displaystyle \int_0^1 \frac{\ln^2(x)}{\sqrt{x(1-\sin(\pi\theta)^2x)}}dx=\frac{8}{\sin(\pi\theta)}\int_0^{\pi\theta} \ln^2\left(\frac{\sin(x)}{\sin(\pi\theta)}\right)dx\tag{1}$

Since $\displaystyle \ln(\sin x)=-\ln2-\sum_{n=1}^{\infty} \frac{\cos(2nx)}{n}$ we have $\displaystyle \int_0^{\pi\theta} \ln(\sin x)dx=-\pi\theta\ln2-\frac12\operatorname{Cl_2}(2\pi\theta)\tag{2}$.

(And, as was done in @RandomVariable's answer linked above,) since $\displaystyle \Re\ln^2(1-e^{i2\pi x})=\ln^2(2\sin(\pi x))-(\frac{\pi}{2}-x)^2$, we have $$\int_0^{\pi\theta} \ln^2(\sin x)dx\\=\int_0^{\pi\theta}(\frac{\pi}{2}-x)^2dx+\Re\int_0^{\pi\theta}\ln^2(1-e^{i2\pi x})dx-\pi\theta\ln^2(2)-2\ln(2)(-\pi\theta\ln2-\frac12\operatorname{Cl_2}(2\pi\theta))$$
Since $\displaystyle \ln^2(1-x)=2\sum_{n=1}^{\infty} \frac{H_{n-1}x^n}{n}$,

and using a closed form for $\sum_{n=1}^{\infty} \frac{H_n x^n}{n^2}$ obtained by @RaymondManzoni
We have $$=\frac{\pi^3}{12}\theta(4\theta^2-6\theta+3)+\pi\theta\ln^2(2)+\operatorname{Cl_2}(2\pi\theta)\ln2+2\sum_{n=1}^{\infty} \frac{H_{n-1}}{n}\int_0^{\pi\theta} \cos(2nx)dx \\=\frac{\pi^3}{12}\theta(4\theta^2-6\theta+3)+\pi\theta\ln^2(2)+\operatorname{Cl_2}(2\pi\theta)\ln2+\Im[i\pi\theta\ln^2(1-e^{i2\pi\theta})]+\Im[\ln(1-e^{i2\pi x})\operatorname{Li_2}(1-e^{i2\pi\theta})]-\Im\operatorname{Li_3}(1-e^{i2\pi\theta}) $$ $$=\frac{\pi^3\theta^3}{3}-\pi\theta\ln^2(2\sin(\pi\theta))-\ln(\sin(\pi\theta)\operatorname{Cl_2}(2\pi\theta)+\pi\theta\ln^2(2)-\Im\operatorname{Li_3}(1-e^{i2\pi\theta}) \tag{3}$$

Putting $(2)$ and $(3)$ in $(1)$, a true miracle happens and the verbosity reduces down to $$\int_0^{\pi\theta} \ln^2\left(\frac{\sin(x)}{\sin(\pi\theta)}\right)dx=\frac{\pi^3\theta^3}{3}-\Im\operatorname{Li_3}(1-e^{i2\pi\theta})$$,

Or in terms of your original function, $$\mathcal{I}(a,b)=\frac{8}{\sqrt{b}}\left(\frac13\operatorname{arcsin}^3\sqrt{\frac{b}{a}}-\Im\operatorname{Li_3}\left(1-e^{2i\operatorname{arcsin}\sqrt{\tfrac{b}{a}}}\right)\right)$$

As a cute example, since $\sin(\pi/10)=\frac{\sqrt{5}-1}{4}$ we have

$$ \mathcal{I}(8,3-\sqrt{5})=\frac{\pi^3}{375\sqrt{3-\sqrt{5}}}-\frac{8}{\sqrt{3-\sqrt{5}}}\Im\operatorname{Li_3}(1-e^{i\pi/5})$$

Another one:
$$ \mathcal{I}(4,2-\sqrt{3})=\frac{\pi^3}{648\sqrt{2-\sqrt{3}}}-\frac{8}{\sqrt{2-\sqrt{3}}}\Im\operatorname{Li_3}\left(\frac{2-\sqrt{3}}{2}-\frac{i}{2}\right)$$

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    $\begingroup$ Thank you for your answer! +1 Your last expression has a more compact form: $$\mathcal{I}(2,1) = \frac{\pi^3}{24}-8\Im\operatorname{Li}_3\left(1-i\right).$$ $\endgroup$ – user153012 Sep 16 '15 at 20:25
  • $\begingroup$ @user153012 wow! didn't expect that much cancellation, it's possible that something else is going on here.. $\endgroup$ – nospoon Sep 16 '15 at 20:32
  • $\begingroup$ Great answer! It establishes a closed form for hypergeometric function that I was looking for for a long time. $\endgroup$ – Vladimir Reshetnikov Sep 22 '15 at 19:40
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    $\begingroup$ $${_4F_3}\!\left(\begin{array}c \tfrac12,\tfrac12,\tfrac12,\tfrac12\\\tfrac32, \tfrac32,\tfrac32\end{array}\middle|\,z\right)=\frac{\arcsin^3\left(\sqrt z\right)}{6\sqrt z}+\\ \frac 1{4\sqrt z}\begin{cases}{i \operatorname{Li}_3 \left(2z+2\sqrt{z^2-z}\right) - i \operatorname{Li}_3 \left(2z-2\sqrt{z^2-z}\right)+\pi\ln^2\left(2z-2\sqrt{z^2-z}\right)}& \color{gray}{z\geq1}\\{i \operatorname{Li}_3\left(2z-2i\sqrt{z-z^2}\right)-i \operatorname{Li}_3\left(2z+2i\sqrt{z-z^2}\right)}\phantom{\Huge|}& \color{gray} {\text{all other }z\in\mathbb C} \end{cases}$$ $\endgroup$ – Vladimir Reshetnikov Sep 22 '15 at 19:44
  • $\begingroup$ I wonder if it is possible to construct a single expression (without cases) that gives correct values for all $z\in\mathbb C$ with standard branch cuts. $\endgroup$ – Vladimir Reshetnikov Sep 22 '15 at 19:47

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