Integrand size = 24, antiderivative size = 26 \[ \int \frac {5-2 x^2-\frac {x^2}{e^3}+2 x^3}{x^2} \, dx=7+\frac {-5+x}{x}-\frac {x}{e^3}+x^2-\log \left (e^{2 x}\right ) \]
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Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6, 14} \[ \int \frac {5-2 x^2-\frac {x^2}{e^3}+2 x^3}{x^2} \, dx=x^2-\left (2+\frac {1}{e^3}\right ) x-\frac {5}{x} \]
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Rule 6
Rule 14
Rubi steps \begin{align*} \text {integral}& = \int \frac {5+\left (-2-\frac {1}{e^3}\right ) x^2+2 x^3}{x^2} \, dx \\ & = \int \left (\frac {-1-2 e^3}{e^3}+\frac {5}{x^2}+2 x\right ) \, dx \\ & = -\frac {5}{x}-\left (2+\frac {1}{e^3}\right ) x+x^2 \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.69 \[ \int \frac {5-2 x^2-\frac {x^2}{e^3}+2 x^3}{x^2} \, dx=-\frac {5}{x}-2 x-\frac {x}{e^3}+x^2 \]
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Time = 0.11 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.69
method | result | size |
risch | \(-x \,{\mathrm e}^{-3}+x^{2}-2 x -\frac {5}{x}\) | \(18\) |
default | \(x^{2}-2 x -\frac {5}{x}-{\mathrm e}^{\ln \left (x \right )-3}\) | \(20\) |
parts | \(x^{2}-2 x -\frac {5}{x}-{\mathrm e}^{\ln \left (x \right )-3}\) | \(20\) |
norman | \(\frac {-5+x^{3}-{\mathrm e}^{-3} \left (2 \,{\mathrm e}^{3}+1\right ) x^{2}}{x}\) | \(25\) |
parallelrisch | \(\frac {x^{5}-2 x^{4}-{\mathrm e}^{\ln \left (x \right )-3} x^{3}-5 x^{2}}{x^{3}}\) | \(29\) |
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Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {5-2 x^2-\frac {x^2}{e^3}+2 x^3}{x^2} \, dx=-\frac {{\left (x^{2} - {\left (x^{3} - 2 \, x^{2} - 5\right )} e^{3}\right )} e^{\left (-3\right )}}{x} \]
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Time = 0.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {5-2 x^2-\frac {x^2}{e^3}+2 x^3}{x^2} \, dx=\frac {x^{2} e^{3} + x \left (- 2 e^{3} - 1\right ) - \frac {5 e^{3}}{x}}{e^{3}} \]
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Time = 0.18 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {5-2 x^2-\frac {x^2}{e^3}+2 x^3}{x^2} \, dx={\left (x^{2} e^{3} - x {\left (2 \, e^{3} + 1\right )}\right )} e^{\left (-3\right )} - \frac {5}{x} \]
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Time = 0.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {5-2 x^2-\frac {x^2}{e^3}+2 x^3}{x^2} \, dx={\left (x^{2} e^{6} - 2 \, x e^{6} - x e^{3}\right )} e^{\left (-6\right )} - \frac {5}{x} \]
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Time = 0.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {5-2 x^2-\frac {x^2}{e^3}+2 x^3}{x^2} \, dx=-\frac {-x^3+\left ({\mathrm {e}}^{-3}+2\right )\,x^2+5}{x} \]
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