Integrand size = 36, antiderivative size = 25 \[ \int e^{-e^4} \left (3 x^2+e^{e^4} \left (6 x-2 x \log (2)+2 x \log ^2(2)\right )\right ) \, dx=x \left (x+x \left (2+e^{-e^4} x-\log (2)+\log ^2(2)\right )\right ) \]
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Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {12} \[ \int e^{-e^4} \left (3 x^2+e^{e^4} \left (6 x-2 x \log (2)+2 x \log ^2(2)\right )\right ) \, dx=e^{-e^4} x^3+\frac {1}{2} x^2 \left (6+2 \log ^2(2)-\log (4)\right ) \]
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Rule 12
Rubi steps \begin{align*} \text {integral}& = e^{-e^4} \int \left (3 x^2+e^{e^4} \left (6 x-2 x \log (2)+2 x \log ^2(2)\right )\right ) \, dx \\ & = e^{-e^4} x^3+\frac {1}{2} x^2 \left (6+2 \log ^2(2)-\log (4)\right ) \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int e^{-e^4} \left (3 x^2+e^{e^4} \left (6 x-2 x \log (2)+2 x \log ^2(2)\right )\right ) \, dx=e^{-e^4} x^3+\frac {1}{2} x^2 \left (6+2 \log ^2(2)-\log (4)\right ) \]
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Time = 0.16 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00
method | result | size |
norman | \(\left (\ln \left (2\right )^{2}-\ln \left (2\right )+3\right ) x^{2}+{\mathrm e}^{-{\mathrm e}^{4}} x^{3}\) | \(25\) |
risch | \(x^{2} \ln \left (2\right )^{2}-x^{2} \ln \left (2\right )+3 x^{2}+{\mathrm e}^{-{\mathrm e}^{4}} x^{3}\) | \(31\) |
gosper | \(\left (\ln \left (2\right )^{2} {\mathrm e}^{{\mathrm e}^{4}}-\ln \left (2\right ) {\mathrm e}^{{\mathrm e}^{4}}+3 \,{\mathrm e}^{{\mathrm e}^{4}}+x \right ) x^{2} {\mathrm e}^{-{\mathrm e}^{4}}\) | \(32\) |
default | \({\mathrm e}^{-{\mathrm e}^{4}} \left (x^{3}+\frac {\left (2 \ln \left (2\right )^{2} {\mathrm e}^{{\mathrm e}^{4}}-2 \ln \left (2\right ) {\mathrm e}^{{\mathrm e}^{4}}+6 \,{\mathrm e}^{{\mathrm e}^{4}}\right ) x^{2}}{2}\right )\) | \(38\) |
parallelrisch | \({\mathrm e}^{-{\mathrm e}^{4}} \left (\ln \left (2\right )^{2} {\mathrm e}^{{\mathrm e}^{4}} x^{2}-\ln \left (2\right ) {\mathrm e}^{{\mathrm e}^{4}} x^{2}+3 x^{2} {\mathrm e}^{{\mathrm e}^{4}}+x^{3}\right )\) | \(40\) |
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Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.40 \[ \int e^{-e^4} \left (3 x^2+e^{e^4} \left (6 x-2 x \log (2)+2 x \log ^2(2)\right )\right ) \, dx={\left (x^{3} + {\left (x^{2} \log \left (2\right )^{2} - x^{2} \log \left (2\right ) + 3 \, x^{2}\right )} e^{\left (e^{4}\right )}\right )} e^{\left (-e^{4}\right )} \]
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Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int e^{-e^4} \left (3 x^2+e^{e^4} \left (6 x-2 x \log (2)+2 x \log ^2(2)\right )\right ) \, dx=\frac {x^{3}}{e^{e^{4}}} + x^{2} \left (- \log {\left (2 \right )} + \log {\left (2 \right )}^{2} + 3\right ) \]
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Time = 0.19 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.40 \[ \int e^{-e^4} \left (3 x^2+e^{e^4} \left (6 x-2 x \log (2)+2 x \log ^2(2)\right )\right ) \, dx={\left (x^{3} + {\left (x^{2} \log \left (2\right )^{2} - x^{2} \log \left (2\right ) + 3 \, x^{2}\right )} e^{\left (e^{4}\right )}\right )} e^{\left (-e^{4}\right )} \]
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none
Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.40 \[ \int e^{-e^4} \left (3 x^2+e^{e^4} \left (6 x-2 x \log (2)+2 x \log ^2(2)\right )\right ) \, dx={\left (x^{3} + {\left (x^{2} \log \left (2\right )^{2} - x^{2} \log \left (2\right ) + 3 \, x^{2}\right )} e^{\left (e^{4}\right )}\right )} e^{\left (-e^{4}\right )} \]
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Time = 12.62 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int e^{-e^4} \left (3 x^2+e^{e^4} \left (6 x-2 x \log (2)+2 x \log ^2(2)\right )\right ) \, dx={\mathrm {e}}^{-{\mathrm {e}}^4}\,x^3+\left ({\ln \left (2\right )}^2-\ln \left (2\right )+3\right )\,x^2 \]
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