Integrand size = 24, antiderivative size = 18 \[ \int \frac {\left (2+x^3\right ) \sqrt [3]{-x+x^4}}{\left (-1+x^3\right )^2} \, dx=-\frac {3 x^2}{2 \left (-x+x^4\right )^{2/3}} \]
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Time = 0.05 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.39, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {2081, 460} \[ \int \frac {\left (2+x^3\right ) \sqrt [3]{-x+x^4}}{\left (-1+x^3\right )^2} \, dx=\frac {3 x \sqrt [3]{x^4-x}}{2 \left (1-x^3\right )} \]
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Rule 460
Rule 2081
Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt [3]{-x+x^4} \int \frac {\sqrt [3]{x} \left (2+x^3\right )}{\left (-1+x^3\right )^{5/3}} \, dx}{\sqrt [3]{x} \sqrt [3]{-1+x^3}} \\ & = \frac {3 x \sqrt [3]{-x+x^4}}{2 \left (1-x^3\right )} \\ \end{align*}
Time = 6.35 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {\left (2+x^3\right ) \sqrt [3]{-x+x^4}}{\left (-1+x^3\right )^2} \, dx=-\frac {3 x^2}{2 \left (x \left (-1+x^3\right )\right )^{2/3}} \]
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Time = 1.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83
method | result | size |
pseudoelliptic | \(-\frac {3 x^{2}}{2 \left (x^{4}-x \right )^{\frac {2}{3}}}\) | \(15\) |
gosper | \(-\frac {3 x \left (x^{4}-x \right )^{\frac {1}{3}}}{2 \left (x^{3}-1\right )}\) | \(20\) |
trager | \(-\frac {3 x \left (x^{4}-x \right )^{\frac {1}{3}}}{2 \left (x^{3}-1\right )}\) | \(20\) |
risch | \(-\frac {3 {\left (x \left (x^{3}-1\right )\right )}^{\frac {1}{3}} x}{2 \left (x^{3}-1\right )}\) | \(20\) |
meijerg | \(\frac {3 \operatorname {signum}\left (x^{3}-1\right )^{\frac {1}{3}} x^{\frac {4}{3}} \operatorname {hypergeom}\left (\left [\frac {4}{9}, \frac {5}{3}\right ], \left [\frac {13}{9}\right ], x^{3}\right )}{2 {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {1}{3}}}+\frac {3 \operatorname {signum}\left (x^{3}-1\right )^{\frac {1}{3}} x^{\frac {13}{3}} \operatorname {hypergeom}\left (\left [\frac {13}{9}, \frac {5}{3}\right ], \left [\frac {22}{9}\right ], x^{3}\right )}{13 {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {1}{3}}}\) | \(66\) |
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none
Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {\left (2+x^3\right ) \sqrt [3]{-x+x^4}}{\left (-1+x^3\right )^2} \, dx=-\frac {3 \, {\left (x^{4} - x\right )}^{\frac {1}{3}} x}{2 \, {\left (x^{3} - 1\right )}} \]
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\[ \int \frac {\left (2+x^3\right ) \sqrt [3]{-x+x^4}}{\left (-1+x^3\right )^2} \, dx=\int \frac {\sqrt [3]{x \left (x - 1\right ) \left (x^{2} + x + 1\right )} \left (x^{3} + 2\right )}{\left (x - 1\right )^{2} \left (x^{2} + x + 1\right )^{2}}\, dx \]
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\[ \int \frac {\left (2+x^3\right ) \sqrt [3]{-x+x^4}}{\left (-1+x^3\right )^2} \, dx=\int { \frac {{\left (x^{4} - x\right )}^{\frac {1}{3}} {\left (x^{3} + 2\right )}}{{\left (x^{3} - 1\right )}^{2}} \,d x } \]
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\[ \int \frac {\left (2+x^3\right ) \sqrt [3]{-x+x^4}}{\left (-1+x^3\right )^2} \, dx=\int { \frac {{\left (x^{4} - x\right )}^{\frac {1}{3}} {\left (x^{3} + 2\right )}}{{\left (x^{3} - 1\right )}^{2}} \,d x } \]
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Time = 5.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.17 \[ \int \frac {\left (2+x^3\right ) \sqrt [3]{-x+x^4}}{\left (-1+x^3\right )^2} \, dx=-\frac {3\,x\,{\left (x^4-x\right )}^{1/3}}{2\,\left (x^3-1\right )} \]
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