Integrand size = 23, antiderivative size = 111 \[ \int \frac {-2+x}{\left (2+x^2\right ) \sqrt [3]{-1+x+2 x^2}} \, dx=\frac {1}{2} \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{-1+x+2 x^2}}{2+2 x+\sqrt [3]{-1+x+2 x^2}}\right )+\frac {1}{2} \log \left (-1-x+\sqrt [3]{-1+x+2 x^2}\right )-\frac {1}{4} \log \left (1+2 x+x^2+(1+x) \sqrt [3]{-1+x+2 x^2}+\left (-1+x+2 x^2\right )^{2/3}\right ) \]
1/2*3^(1/2)*arctan(3^(1/2)*(2*x^2+x-1)^(1/3)/(2+2*x+(2*x^2+x-1)^(1/3)))+1/ 2*ln(-1-x+(2*x^2+x-1)^(1/3))-1/4*ln(1+2*x+x^2+(1+x)*(2*x^2+x-1)^(1/3)+(2*x ^2+x-1)^(2/3))
Time = 0.10 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00 \[ \int \frac {-2+x}{\left (2+x^2\right ) \sqrt [3]{-1+x+2 x^2}} \, dx=\frac {1}{2} \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{-1+x+2 x^2}}{2+2 x+\sqrt [3]{-1+x+2 x^2}}\right )+\frac {1}{2} \log \left (-1-x+\sqrt [3]{-1+x+2 x^2}\right )-\frac {1}{4} \log \left (1+2 x+x^2+(1+x) \sqrt [3]{-1+x+2 x^2}+\left (-1+x+2 x^2\right )^{2/3}\right ) \]
(Sqrt[3]*ArcTan[(Sqrt[3]*(-1 + x + 2*x^2)^(1/3))/(2 + 2*x + (-1 + x + 2*x^ 2)^(1/3))])/2 + Log[-1 - x + (-1 + x + 2*x^2)^(1/3)]/2 - Log[1 + 2*x + x^2 + (1 + x)*(-1 + x + 2*x^2)^(1/3) + (-1 + x + 2*x^2)^(2/3)]/4
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.
\(\displaystyle \int \frac {x-2}{\left (x^2+2\right ) \sqrt [3]{2 x^2+x-1}} \, dx\) |
\(\Big \downarrow \) 1376 |
\(\displaystyle \int \frac {x-2}{\left (x^2+2\right ) \sqrt [3]{2 x^2+x-1}}dx\) |
3.17.32.3.1 Defintions of rubi rules used
Int[((g_.) + (h_.)*(x_))*((a_.) + (c_.)*(x_)^2)^(p_)*((d_.) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Unintegrable[(g + h*x)*(a + c*x^2)^p*(d + e*x + f*x^2)^q, x] /; FreeQ[{a, c, d, e, f, g, h, p, q}, x]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.18 (sec) , antiderivative size = 307, normalized size of antiderivative = 2.77
method | result | size |
trager | \(\frac {\ln \left (-\frac {4 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )^{2} x^{2}+16 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )^{2} x +6 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) \left (2 x^{2}+x -1\right )^{\frac {1}{3}} x -2 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) x^{2}+6 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) \left (2 x^{2}+x -1\right )^{\frac {1}{3}}-4 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) x +3 \left (2 x^{2}+x -1\right )^{\frac {2}{3}}-2 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )-2 x +1}{x^{2}+2}\right )}{2}+\operatorname {RootOf}\left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) \ln \left (-\frac {4 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )^{2} x^{2}+16 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )^{2} x +6 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) \left (2 x^{2}+x -1\right )^{\frac {1}{3}} x +6 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) \left (2 x^{2}+x -1\right )^{\frac {1}{3}}-4 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) x +3 \left (2 x^{2}+x -1\right )^{\frac {2}{3}}-x^{2}+2 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )-2 x -1}{x^{2}+2}\right )\) | \(307\) |
1/2*ln(-(4*RootOf(4*_Z^2+2*_Z+1)^2*x^2+16*RootOf(4*_Z^2+2*_Z+1)^2*x+6*Root Of(4*_Z^2+2*_Z+1)*(2*x^2+x-1)^(1/3)*x-2*RootOf(4*_Z^2+2*_Z+1)*x^2+6*RootOf (4*_Z^2+2*_Z+1)*(2*x^2+x-1)^(1/3)-4*RootOf(4*_Z^2+2*_Z+1)*x+3*(2*x^2+x-1)^ (2/3)-2*RootOf(4*_Z^2+2*_Z+1)-2*x+1)/(x^2+2))+RootOf(4*_Z^2+2*_Z+1)*ln(-(4 *RootOf(4*_Z^2+2*_Z+1)^2*x^2+16*RootOf(4*_Z^2+2*_Z+1)^2*x+6*RootOf(4*_Z^2+ 2*_Z+1)*(2*x^2+x-1)^(1/3)*x+6*RootOf(4*_Z^2+2*_Z+1)*(2*x^2+x-1)^(1/3)-4*Ro otOf(4*_Z^2+2*_Z+1)*x+3*(2*x^2+x-1)^(2/3)-x^2+2*RootOf(4*_Z^2+2*_Z+1)-2*x- 1)/(x^2+2))
Time = 0.68 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.96 \[ \int \frac {-2+x}{\left (2+x^2\right ) \sqrt [3]{-1+x+2 x^2}} \, dx=-\frac {1}{2} \, \sqrt {3} \arctan \left (-\frac {4 \, \sqrt {3} {\left (2 \, x^{2} + x - 1\right )}^{\frac {1}{3}} {\left (x + 1\right )} + \sqrt {3} {\left (2 \, x - 1\right )} - 2 \, \sqrt {3} {\left (2 \, x^{2} + x - 1\right )}^{\frac {2}{3}}}{8 \, x^{2} + 18 \, x + 7}\right ) + \frac {1}{4} \, \log \left (\frac {x^{2} - 3 \, {\left (2 \, x^{2} + x - 1\right )}^{\frac {1}{3}} {\left (x + 1\right )} + 3 \, {\left (2 \, x^{2} + x - 1\right )}^{\frac {2}{3}} + 2}{x^{2} + 2}\right ) \]
-1/2*sqrt(3)*arctan(-(4*sqrt(3)*(2*x^2 + x - 1)^(1/3)*(x + 1) + sqrt(3)*(2 *x - 1) - 2*sqrt(3)*(2*x^2 + x - 1)^(2/3))/(8*x^2 + 18*x + 7)) + 1/4*log(( x^2 - 3*(2*x^2 + x - 1)^(1/3)*(x + 1) + 3*(2*x^2 + x - 1)^(2/3) + 2)/(x^2 + 2))
\[ \int \frac {-2+x}{\left (2+x^2\right ) \sqrt [3]{-1+x+2 x^2}} \, dx=\int \frac {x - 2}{\sqrt [3]{\left (x + 1\right ) \left (2 x - 1\right )} \left (x^{2} + 2\right )}\, dx \]
\[ \int \frac {-2+x}{\left (2+x^2\right ) \sqrt [3]{-1+x+2 x^2}} \, dx=\int { \frac {x - 2}{{\left (2 \, x^{2} + x - 1\right )}^{\frac {1}{3}} {\left (x^{2} + 2\right )}} \,d x } \]
\[ \int \frac {-2+x}{\left (2+x^2\right ) \sqrt [3]{-1+x+2 x^2}} \, dx=\int { \frac {x - 2}{{\left (2 \, x^{2} + x - 1\right )}^{\frac {1}{3}} {\left (x^{2} + 2\right )}} \,d x } \]
Timed out. \[ \int \frac {-2+x}{\left (2+x^2\right ) \sqrt [3]{-1+x+2 x^2}} \, dx=\int \frac {x-2}{\left (x^2+2\right )\,{\left (2\,x^2+x-1\right )}^{1/3}} \,d x \]