Integrand size = 24, antiderivative size = 23 \[ \int \left (8 e^4 x-12 x^2-\log \left (2 e^3\right )\right ) \log (\log (5)) \, dx=\left (e^4-x\right ) \left (4 x^2+\log \left (2 e^3\right )\right ) \log (\log (5)) \]
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Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {12} \[ \int \left (8 e^4 x-12 x^2-\log \left (2 e^3\right )\right ) \log (\log (5)) \, dx=-4 x^3 \log (\log (5))+4 e^4 x^2 \log (\log (5))-x (3+\log (2)) \log (\log (5)) \]
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Rule 12
Rubi steps \begin{align*} \text {integral}& = \log (\log (5)) \int \left (8 e^4 x-12 x^2-\log \left (2 e^3\right )\right ) \, dx \\ & = 4 e^4 x^2 \log (\log (5))-4 x^3 \log (\log (5))-x (3+\log (2)) \log (\log (5)) \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \left (8 e^4 x-12 x^2-\log \left (2 e^3\right )\right ) \log (\log (5)) \, dx=\left (-3 x+4 e^4 x^2-4 x^3-x \log (2)\right ) \log (\log (5)) \]
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Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04
method | result | size |
gosper | \(x \left (4 x \,{\mathrm e}^{4}-4 x^{2}-\ln \left (2 \,{\mathrm e}^{3}\right )\right ) \ln \left (\ln \left (5\right )\right )\) | \(24\) |
default | \(\left (-4 x^{3}+4 x^{2} {\mathrm e}^{4}-\ln \left (2 \,{\mathrm e}^{3}\right ) x \right ) \ln \left (\ln \left (5\right )\right )\) | \(26\) |
parallelrisch | \(\left (-4 x^{3}+4 x^{2} {\mathrm e}^{4}-\ln \left (2 \,{\mathrm e}^{3}\right ) x \right ) \ln \left (\ln \left (5\right )\right )\) | \(26\) |
risch | \(4 \ln \left (\ln \left (5\right )\right ) {\mathrm e}^{4} x^{2}-4 \ln \left (\ln \left (5\right )\right ) x^{3}-\ln \left (\ln \left (5\right )\right ) x \ln \left (2\right )-3 x \ln \left (\ln \left (5\right )\right )\) | \(34\) |
norman | \(\left (-3 \ln \left (\ln \left (5\right )\right )-\ln \left (\ln \left (5\right )\right ) \ln \left (2\right )\right ) x -4 \ln \left (\ln \left (5\right )\right ) x^{3}+4 \ln \left (\ln \left (5\right )\right ) {\mathrm e}^{4} x^{2}\) | \(35\) |
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Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \left (8 e^4 x-12 x^2-\log \left (2 e^3\right )\right ) \log (\log (5)) \, dx=-{\left (4 \, x^{3} - 4 \, x^{2} e^{4} + x \log \left (2\right ) + 3 \, x\right )} \log \left (\log \left (5\right )\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (20) = 40\).
Time = 0.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.78 \[ \int \left (8 e^4 x-12 x^2-\log \left (2 e^3\right )\right ) \log (\log (5)) \, dx=- 4 x^{3} \log {\left (\log {\left (5 \right )} \right )} + 4 x^{2} e^{4} \log {\left (\log {\left (5 \right )} \right )} + x \left (- 3 \log {\left (\log {\left (5 \right )} \right )} - \log {\left (2 \right )} \log {\left (\log {\left (5 \right )} \right )}\right ) \]
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Time = 0.18 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \left (8 e^4 x-12 x^2-\log \left (2 e^3\right )\right ) \log (\log (5)) \, dx=-{\left (4 \, x^{3} - 4 \, x^{2} e^{4} + x \log \left (2 \, e^{3}\right )\right )} \log \left (\log \left (5\right )\right ) \]
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Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \left (8 e^4 x-12 x^2-\log \left (2 e^3\right )\right ) \log (\log (5)) \, dx=-{\left (4 \, x^{3} - 4 \, x^{2} e^{4} + x \log \left (2 \, e^{3}\right )\right )} \log \left (\log \left (5\right )\right ) \]
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Time = 8.40 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \left (8 e^4 x-12 x^2-\log \left (2 e^3\right )\right ) \log (\log (5)) \, dx=-x\,\ln \left (\ln \left (5\right )\right )\,\left (4\,x^2-4\,{\mathrm {e}}^4\,x+\ln \left (2\right )+3\right ) \]
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