Integrand size = 31, antiderivative size = 115 \[ \int \frac {2+x^2}{x \left (2-2 x+x^2\right ) \sqrt [3]{1-x+x^2}} \, dx=-\sqrt {3} \arctan \left (\frac {\frac {2}{\sqrt {3}}-\frac {2 x}{\sqrt {3}}+\frac {\sqrt [3]{1-x+x^2}}{\sqrt {3}}}{\sqrt [3]{1-x+x^2}}\right )+\log \left (-1+x+\sqrt [3]{1-x+x^2}\right )-\frac {1}{2} \log \left (1-2 x+x^2+(1-x) \sqrt [3]{1-x+x^2}+\left (1-x+x^2\right )^{2/3}\right ) \]
-3^(1/2)*arctan((2/3*3^(1/2)-2/3*x*3^(1/2)+1/3*(x^2-x+1)^(1/3)*3^(1/2))/(x ^2-x+1)^(1/3))+ln(-1+x+(x^2-x+1)^(1/3))-1/2*ln(1-2*x+x^2+(1-x)*(x^2-x+1)^( 1/3)+(x^2-x+1)^(2/3))
Time = 0.12 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.89 \[ \int \frac {2+x^2}{x \left (2-2 x+x^2\right ) \sqrt [3]{1-x+x^2}} \, dx=-\sqrt {3} \arctan \left (\frac {2-2 x+\sqrt [3]{1-x+x^2}}{\sqrt {3} \sqrt [3]{1-x+x^2}}\right )+\log \left (-1+x+\sqrt [3]{1-x+x^2}\right )-\frac {1}{2} \log \left (1-2 x+x^2-(-1+x) \sqrt [3]{1-x+x^2}+\left (1-x+x^2\right )^{2/3}\right ) \]
-(Sqrt[3]*ArcTan[(2 - 2*x + (1 - x + x^2)^(1/3))/(Sqrt[3]*(1 - x + x^2)^(1 /3))]) + Log[-1 + x + (1 - x + x^2)^(1/3)] - Log[1 - 2*x + x^2 - (-1 + x)* (1 - x + x^2)^(1/3) + (1 - x + x^2)^(2/3)]/2
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x^2+2}{x \left (x^2-2 x+2\right ) \sqrt [3]{x^2-x+1}} \, dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (\frac {1}{x \sqrt [3]{x^2-x+1}}+\frac {2}{\left (x^2-2 x+2\right ) \sqrt [3]{x^2-x+1}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \int \frac {1}{\left (x^2-2 x+2\right ) \sqrt [3]{x^2-x+1}}dx-\frac {3 \sqrt [3]{-\frac {-2 x-i \sqrt {3}+1}{x}} \sqrt [3]{-\frac {-2 x+i \sqrt {3}+1}{x}} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{3},\frac {1}{3},\frac {5}{3},\frac {1-i \sqrt {3}}{2 x},\frac {1+i \sqrt {3}}{2 x}\right )}{2\ 2^{2/3} \sqrt [3]{x^2-x+1}}\) |
3.18.6.3.1 Defintions of rubi rules used
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.05 (sec) , antiderivative size = 705, normalized size of antiderivative = 6.13
RootOf(_Z^2+_Z+1)*ln((RootOf(_Z^2+_Z+1)*(x^2-x+1)^(2/3)*x-RootOf(_Z^2+_Z+1 )*(x^2-x+1)^(1/3)*x^2+RootOf(_Z^2+_Z+1)*x^3-RootOf(_Z^2+_Z+1)*(x^2-x+1)^(2 /3)+2*RootOf(_Z^2+_Z+1)*(x^2-x+1)^(1/3)*x-3*RootOf(_Z^2+_Z+1)*x^2+2*x*(x^2 -x+1)^(2/3)-2*(x^2-x+1)^(1/3)*x^2+x^3-(x^2-x+1)^(1/3)*RootOf(_Z^2+_Z+1)+3* RootOf(_Z^2+_Z+1)*x-2*(x^2-x+1)^(2/3)+4*(x^2-x+1)^(1/3)*x-4*x^2-RootOf(_Z^ 2+_Z+1)-2*(x^2-x+1)^(1/3)+4*x-2)/x/(x^2-2*x+2))-ln(-(RootOf(_Z^2+_Z+1)*(x^ 2-x+1)^(2/3)*x-RootOf(_Z^2+_Z+1)*(x^2-x+1)^(1/3)*x^2+RootOf(_Z^2+_Z+1)*x^3 -RootOf(_Z^2+_Z+1)*(x^2-x+1)^(2/3)+2*RootOf(_Z^2+_Z+1)*(x^2-x+1)^(1/3)*x-3 *RootOf(_Z^2+_Z+1)*x^2-x*(x^2-x+1)^(2/3)+(x^2-x+1)^(1/3)*x^2-(x^2-x+1)^(1/ 3)*RootOf(_Z^2+_Z+1)+3*RootOf(_Z^2+_Z+1)*x+(x^2-x+1)^(2/3)-2*(x^2-x+1)^(1/ 3)*x+x^2-RootOf(_Z^2+_Z+1)+(x^2-x+1)^(1/3)-x+1)/x/(x^2-2*x+2))*RootOf(_Z^2 +_Z+1)-ln(-(RootOf(_Z^2+_Z+1)*(x^2-x+1)^(2/3)*x-RootOf(_Z^2+_Z+1)*(x^2-x+1 )^(1/3)*x^2+RootOf(_Z^2+_Z+1)*x^3-RootOf(_Z^2+_Z+1)*(x^2-x+1)^(2/3)+2*Root Of(_Z^2+_Z+1)*(x^2-x+1)^(1/3)*x-3*RootOf(_Z^2+_Z+1)*x^2-x*(x^2-x+1)^(2/3)+ (x^2-x+1)^(1/3)*x^2-(x^2-x+1)^(1/3)*RootOf(_Z^2+_Z+1)+3*RootOf(_Z^2+_Z+1)* x+(x^2-x+1)^(2/3)-2*(x^2-x+1)^(1/3)*x+x^2-RootOf(_Z^2+_Z+1)+(x^2-x+1)^(1/3 )-x+1)/x/(x^2-2*x+2))
Time = 0.70 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.28 \[ \int \frac {2+x^2}{x \left (2-2 x+x^2\right ) \sqrt [3]{1-x+x^2}} \, dx=-\sqrt {3} \arctan \left (\frac {4 \, \sqrt {3} {\left (x^{2} - x + 1\right )}^{\frac {2}{3}} {\left (x - 1\right )} + 2 \, \sqrt {3} {\left (x^{2} - x + 1\right )}^{\frac {1}{3}} {\left (x^{2} - 2 \, x + 1\right )} + \sqrt {3} {\left (x^{3} - 3 \, x^{2} + 3 \, x - 1\right )}}{x^{3} - 11 \, x^{2} + 11 \, x - 9}\right ) + \frac {1}{2} \, \log \left (\frac {x^{3} - 2 \, x^{2} + 3 \, {\left (x^{2} - x + 1\right )}^{\frac {2}{3}} {\left (x - 1\right )} + 3 \, {\left (x^{2} - x + 1\right )}^{\frac {1}{3}} {\left (x^{2} - 2 \, x + 1\right )} + 2 \, x}{x^{3} - 2 \, x^{2} + 2 \, x}\right ) \]
-sqrt(3)*arctan((4*sqrt(3)*(x^2 - x + 1)^(2/3)*(x - 1) + 2*sqrt(3)*(x^2 - x + 1)^(1/3)*(x^2 - 2*x + 1) + sqrt(3)*(x^3 - 3*x^2 + 3*x - 1))/(x^3 - 11* x^2 + 11*x - 9)) + 1/2*log((x^3 - 2*x^2 + 3*(x^2 - x + 1)^(2/3)*(x - 1) + 3*(x^2 - x + 1)^(1/3)*(x^2 - 2*x + 1) + 2*x)/(x^3 - 2*x^2 + 2*x))
\[ \int \frac {2+x^2}{x \left (2-2 x+x^2\right ) \sqrt [3]{1-x+x^2}} \, dx=\int \frac {x^{2} + 2}{x \left (x^{2} - 2 x + 2\right ) \sqrt [3]{x^{2} - x + 1}}\, dx \]
\[ \int \frac {2+x^2}{x \left (2-2 x+x^2\right ) \sqrt [3]{1-x+x^2}} \, dx=\int { \frac {x^{2} + 2}{{\left (x^{2} - x + 1\right )}^{\frac {1}{3}} {\left (x^{2} - 2 \, x + 2\right )} x} \,d x } \]
\[ \int \frac {2+x^2}{x \left (2-2 x+x^2\right ) \sqrt [3]{1-x+x^2}} \, dx=\int { \frac {x^{2} + 2}{{\left (x^{2} - x + 1\right )}^{\frac {1}{3}} {\left (x^{2} - 2 \, x + 2\right )} x} \,d x } \]
Timed out. \[ \int \frac {2+x^2}{x \left (2-2 x+x^2\right ) \sqrt [3]{1-x+x^2}} \, dx=\int \frac {x^2+2}{x\,{\left (x^2-x+1\right )}^{1/3}\,\left (x^2-2\,x+2\right )} \,d x \]