Integrand size = 31, antiderivative size = 25 \[ \int \frac {1}{5} \left (5+5 x+e^{2 x^2} \left (-2 x-4 x^3\right )-2 x \log (x)\right ) \, dx=-2+x+\frac {1}{5} x^2 \left (3-e^{2 x^2}-\log (x)\right ) \]
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Time = 0.05 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {12, 1607, 2258, 2240, 2243, 2341} \[ \int \frac {1}{5} \left (5+5 x+e^{2 x^2} \left (-2 x-4 x^3\right )-2 x \log (x)\right ) \, dx=-\frac {1}{5} e^{2 x^2} x^2+\frac {3 x^2}{5}-\frac {1}{5} x^2 \log (x)+x \]
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Rule 12
Rule 1607
Rule 2240
Rule 2243
Rule 2258
Rule 2341
Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \left (5+5 x+e^{2 x^2} \left (-2 x-4 x^3\right )-2 x \log (x)\right ) \, dx \\ & = x+\frac {x^2}{2}+\frac {1}{5} \int e^{2 x^2} \left (-2 x-4 x^3\right ) \, dx-\frac {2}{5} \int x \log (x) \, dx \\ & = x+\frac {3 x^2}{5}-\frac {1}{5} x^2 \log (x)+\frac {1}{5} \int e^{2 x^2} x \left (-2-4 x^2\right ) \, dx \\ & = x+\frac {3 x^2}{5}-\frac {1}{5} x^2 \log (x)+\frac {1}{5} \int \left (-2 e^{2 x^2} x-4 e^{2 x^2} x^3\right ) \, dx \\ & = x+\frac {3 x^2}{5}-\frac {1}{5} x^2 \log (x)-\frac {2}{5} \int e^{2 x^2} x \, dx-\frac {4}{5} \int e^{2 x^2} x^3 \, dx \\ & = -\frac {1}{10} e^{2 x^2}+x+\frac {3 x^2}{5}-\frac {1}{5} e^{2 x^2} x^2-\frac {1}{5} x^2 \log (x)+\frac {2}{5} \int e^{2 x^2} x \, dx \\ & = x+\frac {3 x^2}{5}-\frac {1}{5} e^{2 x^2} x^2-\frac {1}{5} x^2 \log (x) \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {1}{5} \left (5+5 x+e^{2 x^2} \left (-2 x-4 x^3\right )-2 x \log (x)\right ) \, dx=x+\frac {3 x^2}{5}-\frac {1}{5} e^{2 x^2} x^2-\frac {1}{5} x^2 \log (x) \]
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Time = 0.04 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04
method | result | size |
default | \(x +\frac {3 x^{2}}{5}-\frac {x^{2} \ln \left (x \right )}{5}-\frac {{\mathrm e}^{2 x^{2}} x^{2}}{5}\) | \(26\) |
norman | \(x +\frac {3 x^{2}}{5}-\frac {x^{2} \ln \left (x \right )}{5}-\frac {{\mathrm e}^{2 x^{2}} x^{2}}{5}\) | \(26\) |
risch | \(x +\frac {3 x^{2}}{5}-\frac {x^{2} \ln \left (x \right )}{5}-\frac {{\mathrm e}^{2 x^{2}} x^{2}}{5}\) | \(26\) |
parallelrisch | \(x +\frac {3 x^{2}}{5}-\frac {x^{2} \ln \left (x \right )}{5}-\frac {{\mathrm e}^{2 x^{2}} x^{2}}{5}\) | \(26\) |
parts | \(x +\frac {3 x^{2}}{5}-\frac {x^{2} \ln \left (x \right )}{5}-\frac {{\mathrm e}^{2 x^{2}} x^{2}}{5}\) | \(26\) |
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Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {1}{5} \left (5+5 x+e^{2 x^2} \left (-2 x-4 x^3\right )-2 x \log (x)\right ) \, dx=-\frac {1}{5} \, x^{2} e^{\left (2 \, x^{2}\right )} - \frac {1}{5} \, x^{2} \log \left (x\right ) + \frac {3}{5} \, x^{2} + x \]
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Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {1}{5} \left (5+5 x+e^{2 x^2} \left (-2 x-4 x^3\right )-2 x \log (x)\right ) \, dx=- \frac {x^{2} e^{2 x^{2}}}{5} - \frac {x^{2} \log {\left (x \right )}}{5} + \frac {3 x^{2}}{5} + x \]
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Time = 0.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {1}{5} \left (5+5 x+e^{2 x^2} \left (-2 x-4 x^3\right )-2 x \log (x)\right ) \, dx=-\frac {1}{5} \, x^{2} e^{\left (2 \, x^{2}\right )} - \frac {1}{5} \, x^{2} \log \left (x\right ) + \frac {3}{5} \, x^{2} + x \]
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Leaf count of result is larger than twice the leaf count of optimal. 37 vs. \(2 (18) = 36\).
Time = 0.28 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48 \[ \int \frac {1}{5} \left (5+5 x+e^{2 x^2} \left (-2 x-4 x^3\right )-2 x \log (x)\right ) \, dx=-\frac {1}{5} \, x^{2} \log \left (x\right ) + \frac {3}{5} \, x^{2} - \frac {1}{10} \, {\left (2 \, x^{2} - 1\right )} e^{\left (2 \, x^{2}\right )} + x - \frac {1}{10} \, e^{\left (2 \, x^{2}\right )} \]
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Time = 12.39 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {1}{5} \left (5+5 x+e^{2 x^2} \left (-2 x-4 x^3\right )-2 x \log (x)\right ) \, dx=\frac {x\,\left (3\,x-x\,{\mathrm {e}}^{2\,x^2}-x\,\ln \left (x\right )+5\right )}{5} \]
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