Integrand size = 11, antiderivative size = 96 \[ \int x \sqrt [3]{-1+x^3} \, dx=\frac {1}{3} x^2 \sqrt [3]{-1+x^3}+\frac {\arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{-1+x^3}}\right )}{3 \sqrt {3}}+\frac {1}{9} \log \left (-x+\sqrt [3]{-1+x^3}\right )-\frac {1}{18} \log \left (x^2+x \sqrt [3]{-1+x^3}+\left (-1+x^3\right )^{2/3}\right ) \]
1/3*x^2*(x^3-1)^(1/3)+1/9*arctan(3^(1/2)*x/(x+2*(x^3-1)^(1/3)))*3^(1/2)+1/ 9*ln(-x+(x^3-1)^(1/3))-1/18*ln(x^2+x*(x^3-1)^(1/3)+(x^3-1)^(2/3))
Time = 0.16 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.96 \[ \int x \sqrt [3]{-1+x^3} \, dx=\frac {1}{18} \left (6 x^2 \sqrt [3]{-1+x^3}+2 \sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{-1+x^3}}\right )+2 \log \left (-x+\sqrt [3]{-1+x^3}\right )-\log \left (x^2+x \sqrt [3]{-1+x^3}+\left (-1+x^3\right )^{2/3}\right )\right ) \]
(6*x^2*(-1 + x^3)^(1/3) + 2*Sqrt[3]*ArcTan[(Sqrt[3]*x)/(x + 2*(-1 + x^3)^( 1/3))] + 2*Log[-x + (-1 + x^3)^(1/3)] - Log[x^2 + x*(-1 + x^3)^(1/3) + (-1 + x^3)^(2/3)])/18
Time = 0.19 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.70, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {811, 853}
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.
\(\displaystyle \int x \sqrt [3]{x^3-1} \, dx\) |
\(\Big \downarrow \) 811 |
\(\displaystyle \frac {1}{3} x^2 \sqrt [3]{x^3-1}-\frac {1}{3} \int \frac {x}{\left (x^3-1\right )^{2/3}}dx\) |
\(\Big \downarrow \) 853 |
\(\displaystyle \frac {1}{3} \left (\frac {\arctan \left (\frac {\frac {2 x}{\sqrt [3]{x^3-1}}+1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{2} \log \left (x-\sqrt [3]{x^3-1}\right )\right )+\frac {1}{3} \sqrt [3]{x^3-1} x^2\) |
(x^2*(-1 + x^3)^(1/3))/3 + (ArcTan[(1 + (2*x)/(-1 + x^3)^(1/3))/Sqrt[3]]/S qrt[3] + Log[x - (-1 + x^3)^(1/3)]/2)/3
3.14.26.3.1 Defintions of rubi rules used
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c* x)^(m + 1)*((a + b*x^n)^p/(c*(m + n*p + 1))), x] + Simp[a*n*(p/(m + n*p + 1 )) Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && I GtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m , p, x]
Int[(x_)/((a_) + (b_.)*(x_)^3)^(2/3), x_Symbol] :> With[{q = Rt[b, 3]}, Sim p[-ArcTan[(1 + 2*q*(x/(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sqrt[3]*q^2), x] - Simp [Log[q*x - (a + b*x^3)^(1/3)]/(2*q^2), x]] /; FreeQ[{a, b}, x]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 2.79 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.34
method | result | size |
meijerg | \(\frac {\operatorname {signum}\left (x^{3}-1\right )^{\frac {1}{3}} x^{2} \operatorname {hypergeom}\left (\left [-\frac {1}{3}, \frac {2}{3}\right ], \left [\frac {5}{3}\right ], x^{3}\right )}{2 {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {1}{3}}}\) | \(33\) |
risch | \(\frac {x^{2} \left (x^{3}-1\right )^{\frac {1}{3}}}{3}-\frac {{\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {2}{3}} x^{2} \operatorname {hypergeom}\left (\left [\frac {2}{3}, \frac {2}{3}\right ], \left [\frac {5}{3}\right ], x^{3}\right )}{6 \operatorname {signum}\left (x^{3}-1\right )^{\frac {2}{3}}}\) | \(46\) |
pseudoelliptic | \(\frac {-6 x^{2} \left (x^{3}-1\right )^{\frac {1}{3}}+2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x +2 \left (x^{3}-1\right )^{\frac {1}{3}}\right )}{3 x}\right )-2 \ln \left (\frac {-x +\left (x^{3}-1\right )^{\frac {1}{3}}}{x}\right )+\ln \left (\frac {x^{2}+x \left (x^{3}-1\right )^{\frac {1}{3}}+\left (x^{3}-1\right )^{\frac {2}{3}}}{x^{2}}\right )}{18 \left (-x +\left (x^{3}-1\right )^{\frac {1}{3}}\right ) \left (x^{2}+x \left (x^{3}-1\right )^{\frac {1}{3}}+\left (x^{3}-1\right )^{\frac {2}{3}}\right )}\) | \(119\) |
trager | \(\frac {x^{2} \left (x^{3}-1\right )^{\frac {1}{3}}}{3}+\frac {\ln \left (\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {2}{3}} x -3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {1}{3}} x^{2}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+x^{3}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-1\right ) \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )}{9}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \ln \left (\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {2}{3}} x +3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {1}{3}} x^{2}+4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+3 x \left (x^{3}-1\right )^{\frac {2}{3}}+3 x^{2} \left (x^{3}-1\right )^{\frac {1}{3}}+4 x^{3}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-2\right )}{9}-\frac {\ln \left (\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {2}{3}} x +3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {1}{3}} x^{2}+4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+3 x \left (x^{3}-1\right )^{\frac {2}{3}}+3 x^{2} \left (x^{3}-1\right )^{\frac {1}{3}}+4 x^{3}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-2\right )}{9}\) | \(308\) |
Time = 0.30 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.92 \[ \int x \sqrt [3]{-1+x^3} \, dx=\frac {1}{3} \, {\left (x^{3} - 1\right )}^{\frac {1}{3}} x^{2} - \frac {1}{9} \, \sqrt {3} \arctan \left (\frac {\sqrt {3} x + 2 \, \sqrt {3} {\left (x^{3} - 1\right )}^{\frac {1}{3}}}{3 \, x}\right ) + \frac {1}{9} \, \log \left (-\frac {x - {\left (x^{3} - 1\right )}^{\frac {1}{3}}}{x}\right ) - \frac {1}{18} \, \log \left (\frac {x^{2} + {\left (x^{3} - 1\right )}^{\frac {1}{3}} x + {\left (x^{3} - 1\right )}^{\frac {2}{3}}}{x^{2}}\right ) \]
1/3*(x^3 - 1)^(1/3)*x^2 - 1/9*sqrt(3)*arctan(1/3*(sqrt(3)*x + 2*sqrt(3)*(x ^3 - 1)^(1/3))/x) + 1/9*log(-(x - (x^3 - 1)^(1/3))/x) - 1/18*log((x^2 + (x ^3 - 1)^(1/3)*x + (x^3 - 1)^(2/3))/x^2)
Result contains complex when optimal does not.
Time = 0.75 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.38 \[ \int x \sqrt [3]{-1+x^3} \, dx=- \frac {x^{2} e^{- \frac {2 i \pi }{3}} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {x^{3}} \right )}}{3 \Gamma \left (\frac {5}{3}\right )} \]
Time = 0.29 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.98 \[ \int x \sqrt [3]{-1+x^3} \, dx=-\frac {1}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (\frac {2 \, {\left (x^{3} - 1\right )}^{\frac {1}{3}}}{x} + 1\right )}\right ) - \frac {{\left (x^{3} - 1\right )}^{\frac {1}{3}}}{3 \, x {\left (\frac {x^{3} - 1}{x^{3}} - 1\right )}} - \frac {1}{18} \, \log \left (\frac {{\left (x^{3} - 1\right )}^{\frac {1}{3}}}{x} + \frac {{\left (x^{3} - 1\right )}^{\frac {2}{3}}}{x^{2}} + 1\right ) + \frac {1}{9} \, \log \left (\frac {{\left (x^{3} - 1\right )}^{\frac {1}{3}}}{x} - 1\right ) \]
-1/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*(x^3 - 1)^(1/3)/x + 1)) - 1/3*(x^3 - 1) ^(1/3)/(x*((x^3 - 1)/x^3 - 1)) - 1/18*log((x^3 - 1)^(1/3)/x + (x^3 - 1)^(2 /3)/x^2 + 1) + 1/9*log((x^3 - 1)^(1/3)/x - 1)
\[ \int x \sqrt [3]{-1+x^3} \, dx=\int { {\left (x^{3} - 1\right )}^{\frac {1}{3}} x \,d x } \]
Timed out. \[ \int x \sqrt [3]{-1+x^3} \, dx=\int x\,{\left (x^3-1\right )}^{1/3} \,d x \]