Integrand size = 30, antiderivative size = 29 \[ \int \frac {x^2 \left (-4+x^3\right )}{\left (-1+x^3\right )^{3/4} \left (1-x^3+x^4\right )} \, dx=2 \arctan \left (\frac {x}{\sqrt [4]{-1+x^3}}\right )-2 \text {arctanh}\left (\frac {x}{\sqrt [4]{-1+x^3}}\right ) \]
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\[ \int \frac {x^2 \left (-4+x^3\right )}{\left (-1+x^3\right )^{3/4} \left (1-x^3+x^4\right )} \, dx=\int \frac {x^2 \left (-4+x^3\right )}{\left (-1+x^3\right )^{3/4} \left (1-x^3+x^4\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{\left (-1+x^3\right )^{3/4}}+\frac {x}{\left (-1+x^3\right )^{3/4}}-\frac {1+x+4 x^2-x^3}{\left (-1+x^3\right )^{3/4} \left (1-x^3+x^4\right )}\right ) \, dx \\ & = \int \frac {1}{\left (-1+x^3\right )^{3/4}} \, dx+\int \frac {x}{\left (-1+x^3\right )^{3/4}} \, dx-\int \frac {1+x+4 x^2-x^3}{\left (-1+x^3\right )^{3/4} \left (1-x^3+x^4\right )} \, dx \\ & = \frac {\left (1-x^3\right )^{3/4} \int \frac {1}{\left (1-x^3\right )^{3/4}} \, dx}{\left (-1+x^3\right )^{3/4}}+\frac {\left (1-x^3\right )^{3/4} \int \frac {x}{\left (1-x^3\right )^{3/4}} \, dx}{\left (-1+x^3\right )^{3/4}}-\int \left (\frac {1}{\left (-1+x^3\right )^{3/4} \left (1-x^3+x^4\right )}+\frac {x}{\left (-1+x^3\right )^{3/4} \left (1-x^3+x^4\right )}+\frac {4 x^2}{\left (-1+x^3\right )^{3/4} \left (1-x^3+x^4\right )}-\frac {x^3}{\left (-1+x^3\right )^{3/4} \left (1-x^3+x^4\right )}\right ) \, dx \\ & = \frac {x \left (1-x^3\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {3}{4},\frac {4}{3},x^3\right )}{\left (-1+x^3\right )^{3/4}}+\frac {x^2 \left (1-x^3\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {3}{4},\frac {5}{3},x^3\right )}{2 \left (-1+x^3\right )^{3/4}}-4 \int \frac {x^2}{\left (-1+x^3\right )^{3/4} \left (1-x^3+x^4\right )} \, dx-\int \frac {1}{\left (-1+x^3\right )^{3/4} \left (1-x^3+x^4\right )} \, dx-\int \frac {x}{\left (-1+x^3\right )^{3/4} \left (1-x^3+x^4\right )} \, dx+\int \frac {x^3}{\left (-1+x^3\right )^{3/4} \left (1-x^3+x^4\right )} \, dx \\ \end{align*}
Time = 2.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {x^2 \left (-4+x^3\right )}{\left (-1+x^3\right )^{3/4} \left (1-x^3+x^4\right )} \, dx=2 \arctan \left (\frac {x}{\sqrt [4]{-1+x^3}}\right )-2 \text {arctanh}\left (\frac {x}{\sqrt [4]{-1+x^3}}\right ) \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.97 (sec) , antiderivative size = 153, normalized size of antiderivative = 5.28
method | result | size |
trager | \(\ln \left (\frac {2 \left (x^{3}-1\right )^{\frac {3}{4}} x -2 x^{2} \sqrt {x^{3}-1}+2 \left (x^{3}-1\right )^{\frac {1}{4}} x^{3}-x^{4}-x^{3}+1}{x^{4}-x^{3}+1}\right )+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{3}-1}\, x^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{4}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{3}-2 \left (x^{3}-1\right )^{\frac {3}{4}} x +2 \left (x^{3}-1\right )^{\frac {1}{4}} x^{3}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )}{x^{4}-x^{3}+1}\right )\) | \(153\) |
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Time = 0.25 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.66 \[ \int \frac {x^2 \left (-4+x^3\right )}{\left (-1+x^3\right )^{3/4} \left (1-x^3+x^4\right )} \, dx=-2 \, \arctan \left (\frac {{\left (x^{3} - 1\right )}^{\frac {1}{4}}}{x}\right ) - \log \left (\frac {x + {\left (x^{3} - 1\right )}^{\frac {1}{4}}}{x}\right ) + \log \left (-\frac {x - {\left (x^{3} - 1\right )}^{\frac {1}{4}}}{x}\right ) \]
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Timed out. \[ \int \frac {x^2 \left (-4+x^3\right )}{\left (-1+x^3\right )^{3/4} \left (1-x^3+x^4\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {x^2 \left (-4+x^3\right )}{\left (-1+x^3\right )^{3/4} \left (1-x^3+x^4\right )} \, dx=\int { \frac {{\left (x^{3} - 4\right )} x^{2}}{{\left (x^{4} - x^{3} + 1\right )} {\left (x^{3} - 1\right )}^{\frac {3}{4}}} \,d x } \]
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\[ \int \frac {x^2 \left (-4+x^3\right )}{\left (-1+x^3\right )^{3/4} \left (1-x^3+x^4\right )} \, dx=\int { \frac {{\left (x^{3} - 4\right )} x^{2}}{{\left (x^{4} - x^{3} + 1\right )} {\left (x^{3} - 1\right )}^{\frac {3}{4}}} \,d x } \]
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Timed out. \[ \int \frac {x^2 \left (-4+x^3\right )}{\left (-1+x^3\right )^{3/4} \left (1-x^3+x^4\right )} \, dx=\int \frac {x^2\,\left (x^3-4\right )}{{\left (x^3-1\right )}^{3/4}\,\left (x^4-x^3+1\right )} \,d x \]
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