Integrand size = 39, antiderivative size = 31 \[ \int \frac {-200+40 x+3 x^3+2 x^4+e^{-2+x^2} \left (-x+2 x^3\right )}{2 x^3} \, dx=\frac {1}{2} \left (\left (-2+\frac {10}{x}-x\right )^2+\frac {e^{-2+x^2}}{x}-x\right ) \]
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Time = 0.04 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.19, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {12, 14, 2326} \[ \int \frac {-200+40 x+3 x^3+2 x^4+e^{-2+x^2} \left (-x+2 x^3\right )}{2 x^3} \, dx=\frac {x^2}{2}+\frac {e^{x^2-2}}{2 x}+\frac {50}{x^2}+\frac {3 x}{2}-\frac {20}{x} \]
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Rule 12
Rule 14
Rule 2326
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \frac {-200+40 x+3 x^3+2 x^4+e^{-2+x^2} \left (-x+2 x^3\right )}{x^3} \, dx \\ & = \frac {1}{2} \int \left (\frac {e^{-2+x^2} \left (-1+2 x^2\right )}{x^2}+\frac {-200+40 x+3 x^3+2 x^4}{x^3}\right ) \, dx \\ & = \frac {1}{2} \int \frac {e^{-2+x^2} \left (-1+2 x^2\right )}{x^2} \, dx+\frac {1}{2} \int \frac {-200+40 x+3 x^3+2 x^4}{x^3} \, dx \\ & = \frac {e^{-2+x^2}}{2 x}+\frac {1}{2} \int \left (3-\frac {200}{x^3}+\frac {40}{x^2}+2 x\right ) \, dx \\ & = \frac {50}{x^2}-\frac {20}{x}+\frac {e^{-2+x^2}}{2 x}+\frac {3 x}{2}+\frac {x^2}{2} \\ \end{align*}
Time = 0.73 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {-200+40 x+3 x^3+2 x^4+e^{-2+x^2} \left (-x+2 x^3\right )}{2 x^3} \, dx=\frac {100+\left (-40+e^{-2+x^2}\right ) x+3 x^3+x^4}{2 x^2} \]
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Time = 0.07 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94
method | result | size |
parallelrisch | \(\frac {100-40 x +3 x^{3}+x^{4}+{\mathrm e}^{2} x \,{\mathrm e}^{x^{2}-4}}{2 x^{2}}\) | \(29\) |
norman | \(\frac {50-20 x +\frac {3 x^{3}}{2}+\frac {x^{4}}{2}+\frac {{\mathrm e}^{2} x \,{\mathrm e}^{x^{2}-4}}{2}}{x^{2}}\) | \(31\) |
risch | \(\frac {x^{2}}{2}+\frac {3 x}{2}+\frac {-40 x +100}{2 x^{2}}+\frac {{\mathrm e}^{x^{2}-2}}{2 x}\) | \(31\) |
default | \(\frac {x^{2}}{2}+\frac {3 x}{2}+\frac {50}{x^{2}}-\frac {20}{x}+\frac {{\mathrm e}^{2} {\mathrm e}^{-4} \sqrt {\pi }\, \operatorname {erfi}\left (x \right )}{2}-\frac {{\mathrm e}^{2} {\mathrm e}^{-4} \left (-\frac {{\mathrm e}^{x^{2}}}{x}+\sqrt {\pi }\, \operatorname {erfi}\left (x \right )\right )}{2}\) | \(53\) |
parts | \(\frac {3 x}{2}+\frac {x^{2}}{2}-\frac {20}{x}+\frac {50}{x^{2}}+\frac {{\mathrm e}^{2} \left ({\mathrm e}^{-4} \sqrt {\pi }\, \operatorname {erfi}\left (x \right )-{\mathrm e}^{-4} \left (-\frac {{\mathrm e}^{x^{2}}}{x}+\sqrt {\pi }\, \operatorname {erfi}\left (x \right )\right )\right )}{2}\) | \(53\) |
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Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {-200+40 x+3 x^3+2 x^4+e^{-2+x^2} \left (-x+2 x^3\right )}{2 x^3} \, dx=\frac {x^{4} + 3 \, x^{3} + x e^{\left (x^{2} - 2\right )} - 40 \, x + 100}{2 \, x^{2}} \]
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Time = 0.07 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {-200+40 x+3 x^3+2 x^4+e^{-2+x^2} \left (-x+2 x^3\right )}{2 x^3} \, dx=\frac {x^{2}}{2} + \frac {3 x}{2} + \frac {e^{2} e^{x^{2} - 4}}{2 x} + \frac {100 - 40 x}{2 x^{2}} \]
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Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.22 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.65 \[ \int \frac {-200+40 x+3 x^3+2 x^4+e^{-2+x^2} \left (-x+2 x^3\right )}{2 x^3} \, dx=-\frac {1}{2} i \, \sqrt {\pi } \operatorname {erf}\left (i \, x\right ) e^{\left (-2\right )} + \frac {1}{2} \, x^{2} + \frac {\sqrt {-x^{2}} e^{\left (-2\right )} \Gamma \left (-\frac {1}{2}, -x^{2}\right )}{4 \, x} + \frac {3}{2} \, x - \frac {20}{x} + \frac {50}{x^{2}} \]
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Time = 0.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16 \[ \int \frac {-200+40 x+3 x^3+2 x^4+e^{-2+x^2} \left (-x+2 x^3\right )}{2 x^3} \, dx=\frac {{\left (x^{4} e^{2} + 3 \, x^{3} e^{2} - 40 \, x e^{2} + x e^{\left (x^{2}\right )} + 100 \, e^{2}\right )} e^{\left (-2\right )}}{2 \, x^{2}} \]
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Time = 10.90 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {-200+40 x+3 x^3+2 x^4+e^{-2+x^2} \left (-x+2 x^3\right )}{2 x^3} \, dx=\frac {3\,x}{2}+\frac {x\,\left (\frac {{\mathrm {e}}^{x^2-2}}{2}-20\right )+50}{x^2}+\frac {x^2}{2} \]
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