TITLE: Closed form for ${\large\int}_0^1\frac{\ln^3x}{\sqrt{x^2-x+1}}dx$
QUESTION [21 upvotes]: This is a follow-up to my earlier question Closed form for ${\large\int}_0^1\frac{\ln^2x}{\sqrt{1-x+x^2}}dx$.
Is there a closed form for this integral?
$$I=\int_0^1\frac{\ln^3x}{\sqrt{x^2-x+1}}dx\tag1$$
Mathematica and Maple cannot evaluate it directly. A numeric approximation for it is
$$I\approx-6.1665252325192513801994672415450909679747097867356795481...\tag2$$
(click here to see more digits).
As I mentioned in the earlier question, Mathematica is able to find a closed form for a parameterized integral in terms of the Appell hypergeometric function:
$$I(a)=\int_0^1\frac{x^a}{\sqrt{x^2-x+1}}dx\\=\frac1{a+1}F_1\left(a+1;\frac{1}{2},\frac{1}{2};a+2;(-1)^{\small1/3},-(-1)^{\small2/3}\right),\tag3$$
but taking a derivative from this looks a hard problem.

REPLY [14 votes]: I followed the same approach as I used in an answer to another
question, and
expanded your integral in multiple polylogarithms of weight 4, then
used some patterns in their values of weight 3 to guess terms that
might appear in the integral. Then I used an integer relation
algorithm to express your integral in terms of logs, zeta functions
and polylogarithms of small rational arguments with a tolerance of
about $10^{-200}$, and I found this value, which is correct to $3000$
digits.
There are $27$ polylogarithm terms there in total, and while I managed
to simplify them somewhat, I never quite managed to evaluate some of them
except by an integer relation algorithm.
Here it is:
$$\textstyle\def\Li{\mathrm{Li}}
-6 \Li_2(\frac{1}{3}) \zeta (2)+27 \Li_4(\frac{3}{4})+36 \Li_4(\frac{2}{3})-4 \Li_4(\frac{1}{2})-18 \Li_4(\frac{1}{3})-\frac{9}{2} \Li_4(\frac{1}{4})+6 \Li_2(\frac{1}{3}){}^2+6 \Li_2(\frac{1}{3}) \log ^2 3+24 \Li_2(\frac{1}{3}) \log ^2 2-48 \Li_3(\frac{2}{3}) \log3+96 \Li_3(\frac{2}{3}) \log2-48 \Li_3(\frac{1}{3}) \log3+72 \Li_3(\frac{1}{3}) \log2-24 \Li_2(\frac{1}{3}) \log2 \log3+78 \zeta (3) \log3-142 \zeta (3) \log2-\frac{151}{4} \zeta (4)-69 \zeta (2) \log ^2 3-122 \zeta (2) \log ^2 2+192 \zeta (2) \log2 \log3+\frac{73}{4} \log ^4 3+\frac{89}{6} \log ^4 2-70 \log2 \log ^3 3-56 \log ^3 2 \log3+93 \log ^2 2 \log ^2 3
$$
Here is the equivalent Mathematica expression to save people typing:
(-151*Pi^4)/360 - (61*Pi^2*Log[2]^2)/3 + (89*Log[2]^4)/6 + 32*Pi^2*Log[2]*Log[3] -  56*Log[2]^3*Log[3] - (23*Pi^2*Log[3]^2)/2 + 93*Log[2]^2*Log[3]^2 - 70*Log[2]*Log[3]^3 +  (73*Log[3]^4)/4 - Pi^2*PolyLog[2, 1/3] + 24*Log[2]^2*PolyLog[2, 1/3] -  24*Log[2]*Log[3]*PolyLog[2, 1/3] + 6*Log[3]^2*PolyLog[2, 1/3] + 6*PolyLog[2, 1/3]^2 +  72*Log[2]*PolyLog[3, 1/3] - 48*Log[3]*PolyLog[3, 1/3] + 96*Log[2]*PolyLog[3, 2/3] -  48*Log[3]*PolyLog[3, 2/3] - (9*PolyLog[4, 1/4])/2 - 18*PolyLog[4, 1/3] - 4*PolyLog[4, 1/2] +  36*PolyLog[4, 2/3] + 27*PolyLog[4, 3/4] - 142*Log[2]*Zeta[3] + 78*Log[3]*Zeta[3]