\(\int \frac {x (-2+x^3)}{\sqrt [3]{-1+x^3} (-1+x^3+x^6)} \, dx\) [6]

Optimal result

Integrand size = 26, antiderivative size = 90 \[ \int \frac {x \left (-2+x^3\right )}{\sqrt [3]{-1+x^3} \left (-1+x^3+x^6\right )} \, dx=\frac {\arctan \left (\frac {\sqrt {3} \sqrt [3]{-1+x^3}}{-2 x^2+\sqrt [3]{-1+x^3}}\right )}{\sqrt {3}}+\frac {1}{3} \log \left (x^2+\sqrt [3]{-1+x^3}\right )-\frac {1}{6} \log \left (x^4-x^2 \sqrt [3]{-1+x^3}+\left (-1+x^3\right )^{2/3}\right ) \] Output:

1/3*arctan(3^(1/2)*(x^3-1)^(1/3)/(-2*x^2+(x^3-1)^(1/3)))*3^(1/2)+1/3*ln(x^ 
2+(x^3-1)^(1/3))-1/6*ln(x^4-x^2*(x^3-1)^(1/3)+(x^3-1)^(2/3))
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00 \[ \int \frac {x \left (-2+x^3\right )}{\sqrt [3]{-1+x^3} \left (-1+x^3+x^6\right )} \, dx=\frac {\arctan \left (\frac {\sqrt {3} \sqrt [3]{-1+x^3}}{-2 x^2+\sqrt [3]{-1+x^3}}\right )}{\sqrt {3}}+\frac {1}{3} \log \left (x^2+\sqrt [3]{-1+x^3}\right )-\frac {1}{6} \log \left (x^4-x^2 \sqrt [3]{-1+x^3}+\left (-1+x^3\right )^{2/3}\right ) \] Input:

Integrate[(x*(-2 + x^3))/((-1 + x^3)^(1/3)*(-1 + x^3 + x^6)),x]
 

Output:

ArcTan[(Sqrt[3]*(-1 + x^3)^(1/3))/(-2*x^2 + (-1 + x^3)^(1/3))]/Sqrt[3] + L 
og[x^2 + (-1 + x^3)^(1/3)]/3 - Log[x^4 - x^2*(-1 + x^3)^(1/3) + (-1 + x^3) 
^(2/3)]/6
 

Rubi [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.

Time = 0.67 (sec) , antiderivative size = 261, normalized size of antiderivative = 2.90, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {7279, 2009}

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.

\(\displaystyle \int \frac {x \left (x^3-2\right )}{\sqrt [3]{x^3-1} \left (x^6+x^3-1\right )} \, dx\)

\(\Big \downarrow \) 7279

\(\displaystyle \int \left (\frac {x^4}{\sqrt [3]{x^3-1} \left (x^6+x^3-1\right )}-\frac {2 x}{\sqrt [3]{x^3-1} \left (x^6+x^3-1\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 \sqrt [3]{1-x^3} x^2 \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{3},1,\frac {5}{3},x^3,-\frac {2 x^3}{1+\sqrt {5}}\right )}{\sqrt {5} \left (1+\sqrt {5}\right ) \sqrt [3]{x^3-1}}+\frac {\sqrt [3]{1-x^3} x^2 \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{3},1,\frac {5}{3},x^3,-\frac {2 x^3}{1+\sqrt {5}}\right )}{2 \sqrt {5} \sqrt [3]{x^3-1}}-\frac {2 \sqrt [3]{1-x^3} x^2 \operatorname {AppellF1}\left (\frac {2}{3},1,\frac {1}{3},\frac {5}{3},-\frac {2 x^3}{1-\sqrt {5}},x^3\right )}{\sqrt {5} \left (1-\sqrt {5}\right ) \sqrt [3]{x^3-1}}-\frac {\sqrt [3]{1-x^3} x^2 \operatorname {AppellF1}\left (\frac {2}{3},1,\frac {1}{3},\frac {5}{3},-\frac {2 x^3}{1-\sqrt {5}},x^3\right )}{2 \sqrt {5} \sqrt [3]{x^3-1}}\)

Input:

Int[(x*(-2 + x^3))/((-1 + x^3)^(1/3)*(-1 + x^3 + x^6)),x]
 

Output:

(x^2*(1 - x^3)^(1/3)*AppellF1[2/3, 1/3, 1, 5/3, x^3, (-2*x^3)/(1 + Sqrt[5] 
)])/(2*Sqrt[5]*(-1 + x^3)^(1/3)) + (2*x^2*(1 - x^3)^(1/3)*AppellF1[2/3, 1/ 
3, 1, 5/3, x^3, (-2*x^3)/(1 + Sqrt[5])])/(Sqrt[5]*(1 + Sqrt[5])*(-1 + x^3) 
^(1/3)) - (x^2*(1 - x^3)^(1/3)*AppellF1[2/3, 1, 1/3, 5/3, (-2*x^3)/(1 - Sq 
rt[5]), x^3])/(2*Sqrt[5]*(-1 + x^3)^(1/3)) - (2*x^2*(1 - x^3)^(1/3)*Appell 
F1[2/3, 1, 1/3, 5/3, (-2*x^3)/(1 - Sqrt[5]), x^3])/(Sqrt[5]*(1 - Sqrt[5])* 
(-1 + x^3)^(1/3))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7279

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]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.51 (sec) , antiderivative size = 342, normalized size of antiderivative = 3.80

method	result	size
trager	\(\operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \ln \left (-\frac {3 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{6}-3 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {1}{3}} x^{4}+x^{6}+3 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {2}{3}} x^{2}-2 \left (x^{3}-1\right )^{\frac {1}{3}} x^{4}+2 \left (x^{3}-1\right )^{\frac {2}{3}} x^{2}-x^{3}+1}{x^{6}+x^{3}-1}\right )-\frac {\ln \left (\frac {3 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{6}-3 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {1}{3}} x^{4}+3 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {2}{3}} x^{2}+\left (x^{3}-1\right )^{\frac {1}{3}} x^{4}-\left (x^{3}-1\right )^{\frac {2}{3}} x^{2}+x^{3}-1}{x^{6}+x^{3}-1}\right )}{3}-\ln \left (\frac {3 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{6}-3 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {1}{3}} x^{4}+3 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {2}{3}} x^{2}+\left (x^{3}-1\right )^{\frac {1}{3}} x^{4}-\left (x^{3}-1\right )^{\frac {2}{3}} x^{2}+x^{3}-1}{x^{6}+x^{3}-1}\right ) \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )\)	\(342\)

Input:

int(x*(x^3-2)/(x^3-1)^(1/3)/(x^6+x^3-1),x,method=_RETURNVERBOSE)
 

Output:

RootOf(9*_Z^2+3*_Z+1)*ln(-(3*RootOf(9*_Z^2+3*_Z+1)*x^6-3*RootOf(9*_Z^2+3*_ 
Z+1)*(x^3-1)^(1/3)*x^4+x^6+3*RootOf(9*_Z^2+3*_Z+1)*(x^3-1)^(2/3)*x^2-2*(x^ 
3-1)^(1/3)*x^4+2*(x^3-1)^(2/3)*x^2-x^3+1)/(x^6+x^3-1))-1/3*ln((3*RootOf(9* 
_Z^2+3*_Z+1)*x^6-3*RootOf(9*_Z^2+3*_Z+1)*(x^3-1)^(1/3)*x^4+3*RootOf(9*_Z^2 
+3*_Z+1)*(x^3-1)^(2/3)*x^2+(x^3-1)^(1/3)*x^4-(x^3-1)^(2/3)*x^2+x^3-1)/(x^6 
+x^3-1))-ln((3*RootOf(9*_Z^2+3*_Z+1)*x^6-3*RootOf(9*_Z^2+3*_Z+1)*(x^3-1)^( 
1/3)*x^4+3*RootOf(9*_Z^2+3*_Z+1)*(x^3-1)^(2/3)*x^2+(x^3-1)^(1/3)*x^4-(x^3- 
1)^(2/3)*x^2+x^3-1)/(x^6+x^3-1))*RootOf(9*_Z^2+3*_Z+1)
 

Fricas [A] (verification not implemented)

Time = 1.42 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.16 \[ \int \frac {x \left (-2+x^3\right )}{\sqrt [3]{-1+x^3} \left (-1+x^3+x^6\right )} \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {\sqrt {3} x^{6} + 2 \, \sqrt {3} {\left (x^{3} - 1\right )}^{\frac {1}{3}} x^{4} + 4 \, \sqrt {3} {\left (x^{3} - 1\right )}^{\frac {2}{3}} x^{2}}{x^{6} - 8 \, x^{3} + 8}\right ) + \frac {1}{6} \, \log \left (\frac {x^{6} + 3 \, {\left (x^{3} - 1\right )}^{\frac {1}{3}} x^{4} + x^{3} + 3 \, {\left (x^{3} - 1\right )}^{\frac {2}{3}} x^{2} - 1}{x^{6} + x^{3} - 1}\right ) \] Input:

integrate(x*(x^3-2)/(x^3-1)^(1/3)/(x^6+x^3-1),x, algorithm="fricas")
 

Output:

-1/3*sqrt(3)*arctan((sqrt(3)*x^6 + 2*sqrt(3)*(x^3 - 1)^(1/3)*x^4 + 4*sqrt( 
3)*(x^3 - 1)^(2/3)*x^2)/(x^6 - 8*x^3 + 8)) + 1/6*log((x^6 + 3*(x^3 - 1)^(1 
/3)*x^4 + x^3 + 3*(x^3 - 1)^(2/3)*x^2 - 1)/(x^6 + x^3 - 1))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x \left (-2+x^3\right )}{\sqrt [3]{-1+x^3} \left (-1+x^3+x^6\right )} \, dx=\text {Timed out} \] Input:

integrate(x*(x**3-2)/(x**3-1)**(1/3)/(x**6+x**3-1),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {x \left (-2+x^3\right )}{\sqrt [3]{-1+x^3} \left (-1+x^3+x^6\right )} \, dx=\int { \frac {{\left (x^{3} - 2\right )} x}{{\left (x^{6} + x^{3} - 1\right )} {\left (x^{3} - 1\right )}^{\frac {1}{3}}} \,d x } \] Input:

integrate(x*(x^3-2)/(x^3-1)^(1/3)/(x^6+x^3-1),x, algorithm="maxima")
 

Output:

integrate((x^3 - 2)*x/((x^6 + x^3 - 1)*(x^3 - 1)^(1/3)), x)
 

Giac [F]

\[ \int \frac {x \left (-2+x^3\right )}{\sqrt [3]{-1+x^3} \left (-1+x^3+x^6\right )} \, dx=\int { \frac {{\left (x^{3} - 2\right )} x}{{\left (x^{6} + x^{3} - 1\right )} {\left (x^{3} - 1\right )}^{\frac {1}{3}}} \,d x } \] Input:

integrate(x*(x^3-2)/(x^3-1)^(1/3)/(x^6+x^3-1),x, algorithm="giac")
 

Output:

integrate((x^3 - 2)*x/((x^6 + x^3 - 1)*(x^3 - 1)^(1/3)), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x \left (-2+x^3\right )}{\sqrt [3]{-1+x^3} \left (-1+x^3+x^6\right )} \, dx=\int \frac {x\,\left (x^3-2\right )}{{\left (x^3-1\right )}^{1/3}\,\left (x^6+x^3-1\right )} \,d x \] Input:

int((x*(x^3 - 2))/((x^3 - 1)^(1/3)*(x^3 + x^6 - 1)),x)
 

Output:

int((x*(x^3 - 2))/((x^3 - 1)^(1/3)*(x^3 + x^6 - 1)), x)
 

Reduce [F]

\[ \int \frac {x \left (-2+x^3\right )}{\sqrt [3]{-1+x^3} \left (-1+x^3+x^6\right )} \, dx=\int \frac {x^{4}}{\left (x^{3}-1\right )^{\frac {1}{3}} x^{6}+\left (x^{3}-1\right )^{\frac {1}{3}} x^{3}-\left (x^{3}-1\right )^{\frac {1}{3}}}d x -2 \left (\int \frac {x}{\left (x^{3}-1\right )^{\frac {1}{3}} x^{6}+\left (x^{3}-1\right )^{\frac {1}{3}} x^{3}-\left (x^{3}-1\right )^{\frac {1}{3}}}d x \right ) \] Input:

int(x*(x^3-2)/(x^3-1)^(1/3)/(x^6+x^3-1),x)
 

Output:

int(x**4/((x**3 - 1)**(1/3)*x**6 + (x**3 - 1)**(1/3)*x**3 - (x**3 - 1)**(1 
/3)),x) - 2*int(x/((x**3 - 1)**(1/3)*x**6 + (x**3 - 1)**(1/3)*x**3 - (x**3 
 - 1)**(1/3)),x)