Integrand size = 29, antiderivative size = 30 \[ \int \frac {1}{5} \left (-30+10 x-2 e^3 x+15 x^2+10 x \log \left (2 x^2\right )\right ) \, dx=(-3+x)^2-x^2+x^2 \left (-\frac {e^3}{5}+x+\log \left (2 x^2\right )\right ) \]
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Time = 0.01 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {6, 12, 2341} \[ \int \frac {1}{5} \left (-30+10 x-2 e^3 x+15 x^2+10 x \log \left (2 x^2\right )\right ) \, dx=x^3+\frac {1}{5} \left (5-e^3\right ) x^2-x^2+x^2 \log \left (2 x^2\right )-6 x \]
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Rule 6
Rule 12
Rule 2341
Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{5} \left (-30+\left (10-2 e^3\right ) x+15 x^2+10 x \log \left (2 x^2\right )\right ) \, dx \\ & = \frac {1}{5} \int \left (-30+\left (10-2 e^3\right ) x+15 x^2+10 x \log \left (2 x^2\right )\right ) \, dx \\ & = -6 x+\frac {1}{5} \left (5-e^3\right ) x^2+x^3+2 \int x \log \left (2 x^2\right ) \, dx \\ & = -6 x-x^2+\frac {1}{5} \left (5-e^3\right ) x^2+x^3+x^2 \log \left (2 x^2\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {1}{5} \left (-30+10 x-2 e^3 x+15 x^2+10 x \log \left (2 x^2\right )\right ) \, dx=-6 x-\frac {e^3 x^2}{5}+x^3+x^2 \log \left (2 x^2\right ) \]
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Time = 0.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83
method | result | size |
default | \(x^{3}-6 x -\frac {x^{2} {\mathrm e}^{3}}{5}+x^{2} \ln \left (2 x^{2}\right )\) | \(25\) |
norman | \(x^{3}-6 x -\frac {x^{2} {\mathrm e}^{3}}{5}+x^{2} \ln \left (2 x^{2}\right )\) | \(25\) |
risch | \(x^{3}-6 x -\frac {x^{2} {\mathrm e}^{3}}{5}+x^{2} \ln \left (2 x^{2}\right )\) | \(25\) |
parallelrisch | \(x^{3}-6 x -\frac {x^{2} {\mathrm e}^{3}}{5}+x^{2} \ln \left (2 x^{2}\right )\) | \(25\) |
parts | \(x^{3}-6 x -\frac {x^{2} {\mathrm e}^{3}}{5}+x^{2} \ln \left (2 x^{2}\right )\) | \(25\) |
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Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {1}{5} \left (-30+10 x-2 e^3 x+15 x^2+10 x \log \left (2 x^2\right )\right ) \, dx=x^{3} - \frac {1}{5} \, x^{2} e^{3} + x^{2} \log \left (2 \, x^{2}\right ) - 6 \, x \]
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Time = 0.06 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {1}{5} \left (-30+10 x-2 e^3 x+15 x^2+10 x \log \left (2 x^2\right )\right ) \, dx=x^{3} + x^{2} \log {\left (2 x^{2} \right )} - \frac {x^{2} e^{3}}{5} - 6 x \]
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Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {1}{5} \left (-30+10 x-2 e^3 x+15 x^2+10 x \log \left (2 x^2\right )\right ) \, dx=x^{3} - \frac {1}{5} \, x^{2} e^{3} + x^{2} \log \left (2 \, x^{2}\right ) - 6 \, x \]
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Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {1}{5} \left (-30+10 x-2 e^3 x+15 x^2+10 x \log \left (2 x^2\right )\right ) \, dx=x^{3} - \frac {1}{5} \, x^{2} e^{3} + x^{2} \log \left (2 \, x^{2}\right ) - 6 \, x \]
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Time = 8.24 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {1}{5} \left (-30+10 x-2 e^3 x+15 x^2+10 x \log \left (2 x^2\right )\right ) \, dx=x^2\,\ln \left (x^2\right )-6\,x-\frac {x^2\,{\mathrm {e}}^3}{5}+x^2\,\ln \left (2\right )+x^3 \]
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