Integrand size = 30, antiderivative size = 29 \[ \int \frac {-12-4 x^2-4 x^3+e^4 \left (-3-x^2\right )}{30 x^2} \, dx=\frac {\left (1-\frac {x^2}{3}\right ) \left (2+\frac {1}{2} \left (e^4+2 x\right )\right )}{5 x} \]
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Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {12, 14} \[ \int \frac {-12-4 x^2-4 x^3+e^4 \left (-3-x^2\right )}{30 x^2} \, dx=-\frac {x^2}{15}-\frac {1}{30} \left (4+e^4\right ) x+\frac {4+e^4}{10 x} \]
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Rule 12
Rule 14
Rubi steps \begin{align*} \text {integral}& = \frac {1}{30} \int \frac {-12-4 x^2-4 x^3+e^4 \left (-3-x^2\right )}{x^2} \, dx \\ & = \frac {1}{30} \int \left (-4 \left (1+\frac {e^4}{4}\right )-\frac {3 \left (4+e^4\right )}{x^2}-4 x\right ) \, dx \\ & = \frac {4+e^4}{10 x}-\frac {1}{30} \left (4+e^4\right ) x-\frac {x^2}{15} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {-12-4 x^2-4 x^3+e^4 \left (-3-x^2\right )}{30 x^2} \, dx=\frac {1}{30} \left (\frac {3 \left (4+e^4\right )}{x}-\left (4+e^4\right ) x-2 x^2\right ) \]
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Time = 0.12 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90
method | result | size |
default | \(-\frac {x^{2}}{15}-\frac {x \,{\mathrm e}^{4}}{30}-\frac {2 x}{15}-\frac {-3 \,{\mathrm e}^{4}-12}{30 x}\) | \(26\) |
risch | \(-\frac {x \,{\mathrm e}^{4}}{30}-\frac {x^{2}}{15}-\frac {2 x}{15}+\frac {{\mathrm e}^{4}}{10 x}+\frac {2}{5 x}\) | \(27\) |
norman | \(\frac {\left (-\frac {2}{15}-\frac {{\mathrm e}^{4}}{30}\right ) x^{2}-\frac {x^{3}}{15}+\frac {{\mathrm e}^{4}}{10}+\frac {2}{5}}{x}\) | \(30\) |
gosper | \(-\frac {x^{2} {\mathrm e}^{4}+2 x^{3}+4 x^{2}-3 \,{\mathrm e}^{4}-12}{30 x}\) | \(32\) |
parallelrisch | \(-\frac {x^{2} {\mathrm e}^{4}+2 x^{3}+4 x^{2}-3 \,{\mathrm e}^{4}-12}{30 x}\) | \(32\) |
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Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {-12-4 x^2-4 x^3+e^4 \left (-3-x^2\right )}{30 x^2} \, dx=-\frac {2 \, x^{3} + 4 \, x^{2} + {\left (x^{2} - 3\right )} e^{4} - 12}{30 \, x} \]
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Time = 0.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {-12-4 x^2-4 x^3+e^4 \left (-3-x^2\right )}{30 x^2} \, dx=- \frac {x^{2}}{15} - \frac {x \left (4 + e^{4}\right )}{30} - \frac {- 3 e^{4} - 12}{30 x} \]
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Time = 0.19 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {-12-4 x^2-4 x^3+e^4 \left (-3-x^2\right )}{30 x^2} \, dx=-\frac {1}{15} \, x^{2} - \frac {1}{30} \, x {\left (e^{4} + 4\right )} + \frac {e^{4} + 4}{10 \, x} \]
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Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79 \[ \int \frac {-12-4 x^2-4 x^3+e^4 \left (-3-x^2\right )}{30 x^2} \, dx=-\frac {1}{15} \, x^{2} - \frac {1}{30} \, x e^{4} - \frac {2}{15} \, x + \frac {e^{4} + 4}{10 \, x} \]
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Time = 0.07 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {-12-4 x^2-4 x^3+e^4 \left (-3-x^2\right )}{30 x^2} \, dx=\frac {\frac {{\mathrm {e}}^4}{10}+\frac {2}{5}}{x}-\frac {x^2}{15}-x\,\left (\frac {{\mathrm {e}}^4}{30}+\frac {2}{15}\right ) \]
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