Integrand size = 28, antiderivative size = 73 \[ \int \frac {x \left (-3+x^2\right )}{\left (-1+x^2\right )^{2/3} \left (1-x^2+x^3\right )} \, dx=\sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{-1+x^2}}\right )+\log \left (-x+\sqrt [3]{-1+x^2}\right )-\frac {1}{2} \log \left (x^2+x \sqrt [3]{-1+x^2}+\left (-1+x^2\right )^{2/3}\right ) \]
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\[ \int \frac {x \left (-3+x^2\right )}{\left (-1+x^2\right )^{2/3} \left (1-x^2+x^3\right )} \, dx=\int \frac {x \left (-3+x^2\right )}{\left (-1+x^2\right )^{2/3} \left (1-x^2+x^3\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{\left (-1+x^2\right )^{2/3}}-\frac {1+3 x-x^2}{\left (-1+x^2\right )^{2/3} \left (1-x^2+x^3\right )}\right ) \, dx \\ & = \int \frac {1}{\left (-1+x^2\right )^{2/3}} \, dx-\int \frac {1+3 x-x^2}{\left (-1+x^2\right )^{2/3} \left (1-x^2+x^3\right )} \, dx \\ & = \frac {\left (3 \sqrt {x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+x^3}} \, dx,x,\sqrt [3]{-1+x^2}\right )}{2 x}-\int \left (\frac {1}{\left (-1+x^2\right )^{2/3} \left (1-x^2+x^3\right )}+\frac {3 x}{\left (-1+x^2\right )^{2/3} \left (1-x^2+x^3\right )}-\frac {x^2}{\left (-1+x^2\right )^{2/3} \left (1-x^2+x^3\right )}\right ) \, dx \\ & = \frac {3^{3/4} \sqrt {2+\sqrt {3}} \left (1+\sqrt [3]{-1+x^2}\right ) \sqrt {\frac {1-\sqrt [3]{-1+x^2}+\left (-1+x^2\right )^{2/3}}{\left (1+\sqrt {3}+\sqrt [3]{-1+x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1-\sqrt {3}+\sqrt [3]{-1+x^2}}{1+\sqrt {3}+\sqrt [3]{-1+x^2}}\right ),-7-4 \sqrt {3}\right )}{x \sqrt {\frac {1+\sqrt [3]{-1+x^2}}{\left (1+\sqrt {3}+\sqrt [3]{-1+x^2}\right )^2}}}-3 \int \frac {x}{\left (-1+x^2\right )^{2/3} \left (1-x^2+x^3\right )} \, dx-\int \frac {1}{\left (-1+x^2\right )^{2/3} \left (1-x^2+x^3\right )} \, dx+\int \frac {x^2}{\left (-1+x^2\right )^{2/3} \left (1-x^2+x^3\right )} \, dx \\ \end{align*}
Time = 1.19 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00 \[ \int \frac {x \left (-3+x^2\right )}{\left (-1+x^2\right )^{2/3} \left (1-x^2+x^3\right )} \, dx=\sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{-1+x^2}}\right )+\log \left (-x+\sqrt [3]{-1+x^2}\right )-\frac {1}{2} \log \left (x^2+x \sqrt [3]{-1+x^2}+\left (-1+x^2\right )^{2/3}\right ) \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.24 (sec) , antiderivative size = 230, normalized size of antiderivative = 3.15
method | result | size |
trager | \(\ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{2}-1\right )^{\frac {2}{3}} x +\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{2}-1\right )^{\frac {1}{3}} x^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{2}-x \left (x^{2}-1\right )^{\frac {2}{3}}+2 \left (x^{2}-1\right )^{\frac {1}{3}} x^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )}{x^{3}-x^{2}+1}\right )+\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{2}-1\right )^{\frac {2}{3}} x -2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{2}-1\right )^{\frac {1}{3}} x^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{2}+2 x \left (x^{2}-1\right )^{\frac {2}{3}}-\left (x^{2}-1\right )^{\frac {1}{3}} x^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )}{x^{3}-x^{2}+1}\right )\) | \(230\) |
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Time = 0.25 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.01 \[ \int \frac {x \left (-3+x^2\right )}{\left (-1+x^2\right )^{2/3} \left (1-x^2+x^3\right )} \, dx=-\sqrt {3} \arctan \left (\frac {\sqrt {3} x + 2 \, \sqrt {3} {\left (x^{2} - 1\right )}^{\frac {1}{3}}}{3 \, x}\right ) + \log \left (-\frac {x - {\left (x^{2} - 1\right )}^{\frac {1}{3}}}{x}\right ) - \frac {1}{2} \, \log \left (\frac {x^{2} + {\left (x^{2} - 1\right )}^{\frac {1}{3}} x + {\left (x^{2} - 1\right )}^{\frac {2}{3}}}{x^{2}}\right ) \]
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\[ \int \frac {x \left (-3+x^2\right )}{\left (-1+x^2\right )^{2/3} \left (1-x^2+x^3\right )} \, dx=\int \frac {x \left (x^{2} - 3\right )}{\left (\left (x - 1\right ) \left (x + 1\right )\right )^{\frac {2}{3}} \left (x^{3} - x^{2} + 1\right )}\, dx \]
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\[ \int \frac {x \left (-3+x^2\right )}{\left (-1+x^2\right )^{2/3} \left (1-x^2+x^3\right )} \, dx=\int { \frac {{\left (x^{2} - 3\right )} x}{{\left (x^{3} - x^{2} + 1\right )} {\left (x^{2} - 1\right )}^{\frac {2}{3}}} \,d x } \]
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\[ \int \frac {x \left (-3+x^2\right )}{\left (-1+x^2\right )^{2/3} \left (1-x^2+x^3\right )} \, dx=\int { \frac {{\left (x^{2} - 3\right )} x}{{\left (x^{3} - x^{2} + 1\right )} {\left (x^{2} - 1\right )}^{\frac {2}{3}}} \,d x } \]
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Timed out. \[ \int \frac {x \left (-3+x^2\right )}{\left (-1+x^2\right )^{2/3} \left (1-x^2+x^3\right )} \, dx=\int \frac {x\,\left (x^2-3\right )}{{\left (x^2-1\right )}^{2/3}\,\left (x^3-x^2+1\right )} \,d x \]
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