Integrand size = 38, antiderivative size = 23 \[ \int \frac {1}{15} \left (-243 x^2+5 x^4+e^x \left (162 x+81 x^2-4 x^3-x^4\right )\right ) \, dx=5+\frac {27}{5} \left (e^x-x\right ) x \left (x-\frac {x^3}{81}\right ) \]
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Time = 0.07 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.52, number of steps used = 18, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {12, 2227, 2207, 2225} \[ \int \frac {1}{15} \left (-243 x^2+5 x^4+e^x \left (162 x+81 x^2-4 x^3-x^4\right )\right ) \, dx=\frac {x^5}{15}-\frac {e^x x^4}{15}-\frac {27 x^3}{5}+\frac {27 e^x x^2}{5} \]
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Rule 12
Rule 2207
Rule 2225
Rule 2227
Rubi steps \begin{align*} \text {integral}& = \frac {1}{15} \int \left (-243 x^2+5 x^4+e^x \left (162 x+81 x^2-4 x^3-x^4\right )\right ) \, dx \\ & = -\frac {27 x^3}{5}+\frac {x^5}{15}+\frac {1}{15} \int e^x \left (162 x+81 x^2-4 x^3-x^4\right ) \, dx \\ & = -\frac {27 x^3}{5}+\frac {x^5}{15}+\frac {1}{15} \int \left (162 e^x x+81 e^x x^2-4 e^x x^3-e^x x^4\right ) \, dx \\ & = -\frac {27 x^3}{5}+\frac {x^5}{15}-\frac {1}{15} \int e^x x^4 \, dx-\frac {4}{15} \int e^x x^3 \, dx+\frac {27}{5} \int e^x x^2 \, dx+\frac {54}{5} \int e^x x \, dx \\ & = \frac {54 e^x x}{5}+\frac {27 e^x x^2}{5}-\frac {27 x^3}{5}-\frac {4 e^x x^3}{15}-\frac {e^x x^4}{15}+\frac {x^5}{15}+\frac {4}{15} \int e^x x^3 \, dx+\frac {4}{5} \int e^x x^2 \, dx-\frac {54 \int e^x \, dx}{5}-\frac {54}{5} \int e^x x \, dx \\ & = -\frac {54 e^x}{5}+\frac {31 e^x x^2}{5}-\frac {27 x^3}{5}-\frac {e^x x^4}{15}+\frac {x^5}{15}-\frac {4}{5} \int e^x x^2 \, dx-\frac {8}{5} \int e^x x \, dx+\frac {54 \int e^x \, dx}{5} \\ & = -\frac {8 e^x x}{5}+\frac {27 e^x x^2}{5}-\frac {27 x^3}{5}-\frac {e^x x^4}{15}+\frac {x^5}{15}+\frac {8 \int e^x \, dx}{5}+\frac {8}{5} \int e^x x \, dx \\ & = \frac {8 e^x}{5}+\frac {27 e^x x^2}{5}-\frac {27 x^3}{5}-\frac {e^x x^4}{15}+\frac {x^5}{15}-\frac {8 \int e^x \, dx}{5} \\ & = \frac {27 e^x x^2}{5}-\frac {27 x^3}{5}-\frac {e^x x^4}{15}+\frac {x^5}{15} \\ \end{align*}
Time = 0.18 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {1}{15} \left (-243 x^2+5 x^4+e^x \left (162 x+81 x^2-4 x^3-x^4\right )\right ) \, dx=-\frac {1}{15} \left (e^x-x\right ) x^2 \left (-81+x^2\right ) \]
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Time = 0.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13
method | result | size |
default | \(\frac {27 \,{\mathrm e}^{x} x^{2}}{5}-\frac {{\mathrm e}^{x} x^{4}}{15}-\frac {27 x^{3}}{5}+\frac {x^{5}}{15}\) | \(26\) |
norman | \(\frac {27 \,{\mathrm e}^{x} x^{2}}{5}-\frac {{\mathrm e}^{x} x^{4}}{15}-\frac {27 x^{3}}{5}+\frac {x^{5}}{15}\) | \(26\) |
parallelrisch | \(\frac {27 \,{\mathrm e}^{x} x^{2}}{5}-\frac {{\mathrm e}^{x} x^{4}}{15}-\frac {27 x^{3}}{5}+\frac {x^{5}}{15}\) | \(26\) |
parts | \(\frac {27 \,{\mathrm e}^{x} x^{2}}{5}-\frac {{\mathrm e}^{x} x^{4}}{15}-\frac {27 x^{3}}{5}+\frac {x^{5}}{15}\) | \(26\) |
risch | \(\frac {\left (-x^{4}+81 x^{2}\right ) {\mathrm e}^{x}}{15}+\frac {x^{5}}{15}-\frac {27 x^{3}}{5}\) | \(27\) |
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Time = 0.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {1}{15} \left (-243 x^2+5 x^4+e^x \left (162 x+81 x^2-4 x^3-x^4\right )\right ) \, dx=\frac {1}{15} \, x^{5} - \frac {27}{5} \, x^{3} - \frac {1}{15} \, {\left (x^{4} - 81 \, x^{2}\right )} e^{x} \]
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Time = 0.06 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {1}{15} \left (-243 x^2+5 x^4+e^x \left (162 x+81 x^2-4 x^3-x^4\right )\right ) \, dx=\frac {x^{5}}{15} - \frac {27 x^{3}}{5} + \frac {\left (- x^{4} + 81 x^{2}\right ) e^{x}}{15} \]
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Time = 0.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {1}{15} \left (-243 x^2+5 x^4+e^x \left (162 x+81 x^2-4 x^3-x^4\right )\right ) \, dx=\frac {1}{15} \, x^{5} - \frac {27}{5} \, x^{3} - \frac {1}{15} \, {\left (x^{4} - 81 \, x^{2}\right )} e^{x} \]
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Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {1}{15} \left (-243 x^2+5 x^4+e^x \left (162 x+81 x^2-4 x^3-x^4\right )\right ) \, dx=\frac {1}{15} \, x^{5} - \frac {27}{5} \, x^{3} - \frac {1}{15} \, {\left (x^{4} - 81 \, x^{2}\right )} e^{x} \]
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Time = 13.34 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.70 \[ \int \frac {1}{15} \left (-243 x^2+5 x^4+e^x \left (162 x+81 x^2-4 x^3-x^4\right )\right ) \, dx=\frac {x^2\,\left (x-{\mathrm {e}}^x\right )\,\left (x^2-81\right )}{15} \]
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