Integrand size = 32, antiderivative size = 28 \[ \int \frac {-2-x+2 x^2}{(-1+x) x \sqrt [4]{-x^2+x^3}} \, dx=\frac {4 (-1+2 x) \left (-x^2+x^3\right )^{3/4}}{(-1+x) x^2} \]
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Time = 0.09 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {2081, 963, 75} \[ \int \frac {-2-x+2 x^2}{(-1+x) x \sqrt [4]{-x^2+x^3}} \, dx=\frac {4}{\sqrt [4]{x^3-x^2}}-\frac {8 (1-x)}{\sqrt [4]{x^3-x^2}} \]
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Rule 75
Rule 963
Rule 2081
Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt [4]{-1+x} \sqrt {x}\right ) \int \frac {-2-x+2 x^2}{(-1+x)^{5/4} x^{3/2}} \, dx}{\sqrt [4]{-x^2+x^3}} \\ & = \frac {4}{\sqrt [4]{-x^2+x^3}}-\frac {\left (4 \sqrt [4]{-1+x} \sqrt {x}\right ) \int \frac {-1-\frac {x}{2}}{\sqrt [4]{-1+x} x^{3/2}} \, dx}{\sqrt [4]{-x^2+x^3}} \\ & = \frac {4}{\sqrt [4]{-x^2+x^3}}-\frac {8 (1-x)}{\sqrt [4]{-x^2+x^3}} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 2 in optimal.
Time = 10.03 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.50 \[ \int \frac {-2-x+2 x^2}{(-1+x) x \sqrt [4]{-x^2+x^3}} \, dx=-\frac {4 \sqrt {x} \left (2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {1}{4},\frac {3}{4},1-x\right )-\operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},1-x\right )-2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {3}{2},\frac {3}{4},1-x\right )\right )}{\sqrt [4]{(-1+x) x^2}} \]
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Time = 0.95 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.61
method | result | size |
risch | \(\frac {-4+8 x}{\left (\left (x -1\right ) x^{2}\right )^{\frac {1}{4}}}\) | \(17\) |
gosper | \(\frac {-4+8 x}{\left (x^{3}-x^{2}\right )^{\frac {1}{4}}}\) | \(19\) |
trager | \(\frac {4 \left (-1+2 x \right ) \left (x^{3}-x^{2}\right )^{\frac {3}{4}}}{\left (x -1\right ) x^{2}}\) | \(27\) |
meijerg | \(-\frac {4 \left (-\operatorname {signum}\left (x -1\right )\right )^{\frac {1}{4}} \operatorname {hypergeom}\left (\left [-\frac {1}{2}, \frac {5}{4}\right ], \left [\frac {1}{2}\right ], x\right )}{\operatorname {signum}\left (x -1\right )^{\frac {1}{4}} \sqrt {x}}-\frac {4 \left (-\operatorname {signum}\left (x -1\right )\right )^{\frac {1}{4}} x^{\frac {3}{2}} \operatorname {hypergeom}\left (\left [\frac {5}{4}, \frac {3}{2}\right ], \left [\frac {5}{2}\right ], x\right )}{3 \operatorname {signum}\left (x -1\right )^{\frac {1}{4}}}+\frac {2 \left (-\operatorname {signum}\left (x -1\right )\right )^{\frac {1}{4}} \sqrt {x}\, \operatorname {hypergeom}\left (\left [\frac {1}{2}, \frac {5}{4}\right ], \left [\frac {3}{2}\right ], x\right )}{\operatorname {signum}\left (x -1\right )^{\frac {1}{4}}}\) | \(80\) |
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none
Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.64 \[ \int \frac {-2-x+2 x^2}{(-1+x) x \sqrt [4]{-x^2+x^3}} \, dx=\frac {4 \, {\left (2 \, x - 1\right )}}{{\left (x^{3} - x^{2}\right )}^{\frac {1}{4}}} \]
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\[ \int \frac {-2-x+2 x^2}{(-1+x) x \sqrt [4]{-x^2+x^3}} \, dx=\int \frac {2 x^{2} - x - 2}{x \sqrt [4]{x^{2} \left (x - 1\right )} \left (x - 1\right )}\, dx \]
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\[ \int \frac {-2-x+2 x^2}{(-1+x) x \sqrt [4]{-x^2+x^3}} \, dx=\int { \frac {2 \, x^{2} - x - 2}{{\left (x^{3} - x^{2}\right )}^{\frac {1}{4}} {\left (x - 1\right )} x} \,d x } \]
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\[ \int \frac {-2-x+2 x^2}{(-1+x) x \sqrt [4]{-x^2+x^3}} \, dx=\int { \frac {2 \, x^{2} - x - 2}{{\left (x^{3} - x^{2}\right )}^{\frac {1}{4}} {\left (x - 1\right )} x} \,d x } \]
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Time = 5.64 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.61 \[ \int \frac {-2-x+2 x^2}{(-1+x) x \sqrt [4]{-x^2+x^3}} \, dx=\frac {8\,x-4}{{\left (x^3-x^2\right )}^{1/4}} \]
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