Integrand size = 21, antiderivative size = 107 \[ \int \frac {(1+x) \sqrt [3]{-1+x^3}}{(-1+x)^2 x} \, dx=-\frac {3 \sqrt [3]{-1+x^3}}{-1+x}+\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{-1+x^3}}{-2+2 x+\sqrt [3]{-1+x^3}}\right )-\log \left (1-x+\sqrt [3]{-1+x^3}\right )+\frac {1}{2} \log \left (1-2 x+x^2+(-1+x) \sqrt [3]{-1+x^3}+\left (-1+x^3\right )^{2/3}\right ) \]
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Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.13 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.83, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {2180, 252, 251, 272, 53, 60, 632, 210, 31, 270, 267, 372, 371, 294, 337} \[ \int \frac {(1+x) \sqrt [3]{-1+x^3}}{(-1+x)^2 x} \, dx=-\frac {\arctan \left (\frac {\frac {2 x}{\sqrt [3]{x^3-1}}+1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\arctan \left (\frac {1-2 \sqrt [3]{x^3-1}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {3 \left (1-x^3\right )^{2/3} x \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {5}{3},\frac {4}{3},x^3\right )}{\left (x^3-1\right )^{2/3}}-\frac {3 \left (1-x^3\right )^{2/3} x^4 \operatorname {Hypergeometric2F1}\left (\frac {4}{3},\frac {5}{3},\frac {7}{3},x^3\right )}{4 \left (x^3-1\right )^{2/3}}-\frac {3}{\left (x^3-1\right )^{2/3}}-\frac {1}{2} \log \left (x-\sqrt [3]{x^3-1}\right )-\frac {1}{2} \log \left (\sqrt [3]{x^3-1}+1\right )-\frac {3 x^2}{\left (x^3-1\right )^{2/3}}+\frac {\log (x)}{2} \]
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Rule 31
Rule 53
Rule 60
Rule 210
Rule 251
Rule 252
Rule 267
Rule 270
Rule 272
Rule 294
Rule 337
Rule 371
Rule 372
Rule 632
Rule 2180
Rubi steps \begin{align*} \text {integral}= \int \left (\frac {3}{\left (-1+x^3\right )^{5/3}}+\frac {1}{x \left (-1+x^3\right )^{5/3}}+\frac {5 x}{\left (-1+x^3\right )^{5/3}}+\frac {5 x^2}{\left (-1+x^3\right )^{5/3}}+\frac {3 x^3}{\left (-1+x^3\right )^{5/3}}+\frac {x^4}{\left (-1+x^3\right )^{5/3}}\right ) \, dx \\ = 3 \int \frac {1}{\left (-1+x^3\right )^{5/3}} \, dx+3 \int \frac {x^3}{\left (-1+x^3\right )^{5/3}} \, dx+5 \int \frac {x}{\left (-1+x^3\right )^{5/3}} \, dx+5 \int \frac {x^2}{\left (-1+x^3\right )^{5/3}} \, dx+\int \frac {1}{x \left (-1+x^3\right )^{5/3}} \, dx+\int \frac {x^4}{\left (-1+x^3\right )^{5/3}} \, dx \\ = -\frac {5}{2 \left (-1+x^3\right )^{2/3}}-\frac {3 x^2}{\left (-1+x^3\right )^{2/3}}+\frac {1}{3} \text {Subst}\left (\int \frac {1}{(-1+x)^{5/3} x} \, dx,x,x^3\right )-\frac {\left (3 \left (1-x^3\right )^{2/3}\right ) \int \frac {1}{\left (1-x^3\right )^{5/3}} \, dx}{\left (-1+x^3\right )^{2/3}}-\frac {\left (3 \left (1-x^3\right )^{2/3}\right ) \int \frac {x^3}{\left (1-x^3\right )^{5/3}} \, dx}{\left (-1+x^3\right )^{2/3}}+\int \frac {x}{\left (-1+x^3\right )^{2/3}} \, dx \\ = -\frac {3}{\left (-1+x^3\right )^{2/3}}-\frac {3 x^2}{\left (-1+x^3\right )^{2/3}}-\frac {\arctan \left (\frac {1+\frac {2 x}{\sqrt [3]{-1+x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {3 x \left (1-x^3\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {5}{3},\frac {4}{3},x^3\right )}{\left (-1+x^3\right )^{2/3}}-\frac {3 x^4 \left (1-x^3\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {4}{3},\frac {5}{3},\frac {7}{3},x^3\right )}{4 \left (-1+x^3\right )^{2/3}}-\frac {1}{2} \log \left (x-\sqrt [3]{-1+x^3}\right )-\frac {1}{3} \text {Subst}\left (\int \frac {1}{(-1+x)^{2/3} x} \, dx,x,x^3\right ) \\ = -\frac {3}{\left (-1+x^3\right )^{2/3}}-\frac {3 x^2}{\left (-1+x^3\right )^{2/3}}-\frac {\arctan \left (\frac {1+\frac {2 x}{\sqrt [3]{-1+x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {3 x \left (1-x^3\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {5}{3},\frac {4}{3},x^3\right )}{\left (-1+x^3\right )^{2/3}}-\frac {3 x^4 \left (1-x^3\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {4}{3},\frac {5}{3},\frac {7}{3},x^3\right )}{4 \left (-1+x^3\right )^{2/3}}+\frac {\log (x)}{2}-\frac {1}{2} \log \left (x-\sqrt [3]{-1+x^3}\right )-\frac {1}{2} \text {Subst}\left (\int \frac {1}{1+x} \, dx,x,\sqrt [3]{-1+x^3}\right )-\frac {1}{2} \text {Subst}\left (\int \frac {1}{1-x+x^2} \, dx,x,\sqrt [3]{-1+x^3}\right ) \\ = -\frac {3}{\left (-1+x^3\right )^{2/3}}-\frac {3 x^2}{\left (-1+x^3\right )^{2/3}}-\frac {\arctan \left (\frac {1+\frac {2 x}{\sqrt [3]{-1+x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {3 x \left (1-x^3\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {5}{3},\frac {4}{3},x^3\right )}{\left (-1+x^3\right )^{2/3}}-\frac {3 x^4 \left (1-x^3\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {4}{3},\frac {5}{3},\frac {7}{3},x^3\right )}{4 \left (-1+x^3\right )^{2/3}}+\frac {\log (x)}{2}-\frac {1}{2} \log \left (x-\sqrt [3]{-1+x^3}\right )-\frac {1}{2} \log \left (1+\sqrt [3]{-1+x^3}\right )+\text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 \sqrt [3]{-1+x^3}\right ) \\ = -\frac {3}{\left (-1+x^3\right )^{2/3}}-\frac {3 x^2}{\left (-1+x^3\right )^{2/3}}-\frac {\arctan \left (\frac {1+\frac {2 x}{\sqrt [3]{-1+x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\arctan \left (\frac {-1+2 \sqrt [3]{-1+x^3}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {3 x \left (1-x^3\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {5}{3},\frac {4}{3},x^3\right )}{\left (-1+x^3\right )^{2/3}}-\frac {3 x^4 \left (1-x^3\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {4}{3},\frac {5}{3},\frac {7}{3},x^3\right )}{4 \left (-1+x^3\right )^{2/3}}+\frac {\log (x)}{2}-\frac {1}{2} \log \left (x-\sqrt [3]{-1+x^3}\right )-\frac {1}{2} \log \left (1+\sqrt [3]{-1+x^3}\right ) \\ \end{align*}
Time = 1.66 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00 \[ \int \frac {(1+x) \sqrt [3]{-1+x^3}}{(-1+x)^2 x} \, dx=-\frac {3 \sqrt [3]{-1+x^3}}{-1+x}+\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{-1+x^3}}{-2+2 x+\sqrt [3]{-1+x^3}}\right )-\log \left (1-x+\sqrt [3]{-1+x^3}\right )+\frac {1}{2} \log \left (1-2 x+x^2+(-1+x) \sqrt [3]{-1+x^3}+\left (-1+x^3\right )^{2/3}\right ) \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.79 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.53
method | result | size |
risch | \(-\frac {3 \left (x^{2}+x +1\right )}{\left (x^{3}-1\right )^{\frac {2}{3}}}+\frac {\left (\frac {\left (x^{3}-1\right )^{\frac {2}{3}} {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {2}{3}} x^{2} \operatorname {hypergeom}\left (\left [\frac {2}{3}, \frac {2}{3}\right ], \left [\frac {5}{3}\right ], x^{3}\right )}{2 {\left (\left (x^{3}-1\right )^{2}\right )}^{\frac {1}{3}} \operatorname {signum}\left (x^{3}-1\right )^{\frac {2}{3}}}-\frac {\left (x^{3}-1\right )^{\frac {2}{3}} {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {2}{3}} \left (\left (\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \left (3\right )}{2}+3 \ln \left (x \right )+i \pi \right ) \Gamma \left (\frac {2}{3}\right )+\frac {2 \Gamma \left (\frac {2}{3}\right ) x^{3} \operatorname {hypergeom}\left (\left [1, 1, \frac {5}{3}\right ], \left [2, 2\right ], x^{3}\right )}{3}\right )}{3 {\left (\left (x^{3}-1\right )^{2}\right )}^{\frac {1}{3}} \Gamma \left (\frac {2}{3}\right ) \operatorname {signum}\left (x^{3}-1\right )^{\frac {2}{3}}}\right ) {\left (\left (x^{3}-1\right )^{2}\right )}^{\frac {1}{3}}}{\left (x^{3}-1\right )^{\frac {2}{3}}}\) | \(164\) |
trager | \(-\frac {3 \left (x^{3}-1\right )^{\frac {1}{3}}}{-1+x}-\ln \left (\frac {12 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{2}-30 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x -99 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {2}{3}}+127 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {1}{3}} x -40 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{2}+12 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2}-127 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {1}{3}}+157 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x +127 \left (x^{3}-1\right )^{\frac {2}{3}}-28 \left (x^{3}-1\right )^{\frac {1}{3}} x -87 x^{2}-40 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )+28 \left (x^{3}-1\right )^{\frac {1}{3}}-58 x -87}{x}\right )+\operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \ln \left (\frac {46 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{2}-115 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x +99 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {2}{3}}+28 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {1}{3}} x -173 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{2}+46 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2}-28 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {1}{3}}+214 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x +28 \left (x^{3}-1\right )^{\frac {2}{3}}-127 \left (x^{3}-1\right )^{\frac {1}{3}} x +145 x^{2}-173 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )+127 \left (x^{3}-1\right )^{\frac {1}{3}}-87 x +145}{x}\right )\) | \(384\) |
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Time = 0.40 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.02 \[ \int \frac {(1+x) \sqrt [3]{-1+x^3}}{(-1+x)^2 x} \, dx=-\frac {2 \, \sqrt {3} {\left (x - 1\right )} \arctan \left (-\frac {4 \, \sqrt {3} {\left (x^{3} - 1\right )}^{\frac {1}{3}} {\left (x - 1\right )} + \sqrt {3} {\left (x^{2} + x + 1\right )} - 2 \, \sqrt {3} {\left (x^{3} - 1\right )}^{\frac {2}{3}}}{3 \, {\left (3 \, x^{2} - 5 \, x + 3\right )}}\right ) + {\left (x - 1\right )} \log \left (\frac {{\left (x^{3} - 1\right )}^{\frac {1}{3}} {\left (x - 1\right )} + x - {\left (x^{3} - 1\right )}^{\frac {2}{3}}}{x}\right ) + 6 \, {\left (x^{3} - 1\right )}^{\frac {1}{3}}}{2 \, {\left (x - 1\right )}} \]
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\[ \int \frac {(1+x) \sqrt [3]{-1+x^3}}{(-1+x)^2 x} \, dx=\int \frac {\sqrt [3]{\left (x - 1\right ) \left (x^{2} + x + 1\right )} \left (x + 1\right )}{x \left (x - 1\right )^{2}}\, dx \]
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\[ \int \frac {(1+x) \sqrt [3]{-1+x^3}}{(-1+x)^2 x} \, dx=\int { \frac {{\left (x^{3} - 1\right )}^{\frac {1}{3}} {\left (x + 1\right )}}{{\left (x - 1\right )}^{2} x} \,d x } \]
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\[ \int \frac {(1+x) \sqrt [3]{-1+x^3}}{(-1+x)^2 x} \, dx=\int { \frac {{\left (x^{3} - 1\right )}^{\frac {1}{3}} {\left (x + 1\right )}}{{\left (x - 1\right )}^{2} x} \,d x } \]
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Timed out. \[ \int \frac {(1+x) \sqrt [3]{-1+x^3}}{(-1+x)^2 x} \, dx=\int \frac {{\left (x^3-1\right )}^{1/3}\,\left (x+1\right )}{x\,{\left (x-1\right )}^2} \,d x \]
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