How do I solve $\int\frac{x}{(x^2+x+1)^{\frac{1}{12}}}\,dx$? I'm trying to solve  $$\int_0^1\frac{x}{(x^2+x+1)^{\frac{1}{12}}}\mathrm dx$$ To calculate it I first tried to calculate the primitive function. So let $$\int\frac{x}{(x^2+x+1)^{\frac{1}{12}}}\mathrm dx$$ first I added $+1-1$ to numerator then split it and tried to obtain the derivative of $x^2+x+1$ on numerator after this passages I obtain $$=\frac{6}{11}(x^2+x+1)^{\frac{11}{12}}-\frac12\int\frac{1}{(x^2+x+1)^{\frac{1}{12}}}\mathrm dx$$ My problem starts here: how can I deal with the last integral $$\int\frac{1}{(x^2+x+1)^{\frac{1}{12}}}\mathrm dx$$ I tried to substitute $(x^2+x+1)^{\frac{1}{2}}=x+t$ but it makes it more complex..Then wolphram says that we can't express it by elementary function. How can I proceed? Note that the book where I found the integral asks to calculate it from $0$ to $1$ and it reports a solution which is $$\frac12(\sqrt{3}-\frac72+\frac{3}{2(1+2\sqrt{3})}+\ln(1+2\sqrt{3}))$$.Thank you in advance

Mathematica 11.1 gives that this integral is equal to 
$$
\frac{2}{11} \left(i 3^{17/12} \, _2F_1\left(1,\frac{11}{6};\frac{23}{12};\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)-i \sqrt{3} \, _2F_1\left(1,\frac{11}{6};\frac{23}{12};\frac{1}{6} \left(3+i
   \sqrt{3}\right)\right)-3+ 3^{23/12}\right)
$$
the function of interest is then 
$$
_2F_1\left(1,\frac{11}{6};\frac{23}{12};x\right) = \sum_{n=0}^\infty \frac{\Gamma(\frac{23}{12})\Gamma(\frac{11}{6}+n)}{\Gamma(\frac{11}{6})\Gamma(\frac{23}{12}+n)}x^n
$$
Removing the complex terms we can also the write this as 
$$
\frac{6}{11} \left( 3^{11/12} -1 + 3^{5/12}\frac{\Gamma(\frac{23}{12})}{\Gamma(\frac{11}{6})}\underbrace{\sum_{n=0}^\infty \frac{\Gamma(\frac{11}{6}+n)}{\Gamma(\frac{23}{12}+n)}\left(3^{-\frac{11}{12}-\frac{n}{2}}\sin\left(\frac{n \pi}{6}\right)-\sin\left(\frac{n\pi}{3}\right)\right)}_{Q}\right)
$$
so if you want a closed form you need to work out the value of $Q$, which doesn't look super promising.
Edit: After playing with this we can write the sum as 
$$
\int_0^1 \frac{x}{(x^2+x+1)^(1/12)}\;dx = \frac{1}{\Gamma(\frac{1}{12})}\sum_{k=0}^\infty \frac{(-1)^k}{k!}B\left(-1,2+2k,\frac{11}{12}-k\right)\Gamma\left(\frac{1}{12}-k\right)= 0.47017004410654...
$$
where $B$ is the incomplete beta function.
