Integrand size = 39, antiderivative size = 27 \[ \int \frac {-20+e^{2 e^{4 x}} \left (5+40 e^{4 x} x-40 e^{4 x} \log (\log (4))\right )}{4 e^2} \, dx=\frac {5 \left (4-e^{2 e^{4 x}}\right ) (-x+\log (\log (4)))}{4 e^2} \]
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Time = 0.03 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.59, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {12, 2326} \[ \int \frac {-20+e^{2 e^{4 x}} \left (5+40 e^{4 x} x-40 e^{4 x} \log (\log (4))\right )}{4 e^2} \, dx=\frac {5}{4} e^{-4 x+2 e^{4 x}-2} \left (e^{4 x} x-e^{4 x} \log (\log (4))\right )-\frac {5 x}{e^2} \]
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Rule 12
Rule 2326
Rubi steps \begin{align*} \text {integral}& = \frac {\int \left (-20+e^{2 e^{4 x}} \left (5+40 e^{4 x} x-40 e^{4 x} \log (\log (4))\right )\right ) \, dx}{4 e^2} \\ & = -\frac {5 x}{e^2}+\frac {\int e^{2 e^{4 x}} \left (5+40 e^{4 x} x-40 e^{4 x} \log (\log (4))\right ) \, dx}{4 e^2} \\ & = -\frac {5 x}{e^2}+\frac {5}{4} e^{-2+2 e^{4 x}-4 x} \left (e^{4 x} x-e^{4 x} \log (\log (4))\right ) \\ \end{align*}
Time = 0.15 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {-20+e^{2 e^{4 x}} \left (5+40 e^{4 x} x-40 e^{4 x} \log (\log (4))\right )}{4 e^2} \, dx=\frac {5 \left (-4 x+e^{2 e^{4 x}} (x-\log (\log (4)))\right )}{4 e^2} \]
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Time = 0.11 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15
method | result | size |
risch | \(-5 x \,{\mathrm e}^{-2}+\frac {\left (-5 \ln \left (2\right )-5 \ln \left (\ln \left (2\right )\right )+5 x \right ) {\mathrm e}^{-2+2 \,{\mathrm e}^{4 x}}}{4}\) | \(31\) |
parallelrisch | \(\frac {{\mathrm e}^{-2} \left (-5 \ln \left (2 \ln \left (2\right )\right ) {\mathrm e}^{2 \,{\mathrm e}^{4 x}}+5 x \,{\mathrm e}^{2 \,{\mathrm e}^{4 x}}-20 x \right )}{4}\) | \(35\) |
default | \(\frac {{\mathrm e}^{-2} \left (-20 x +\left (-5 \ln \left (2\right )-5 \ln \left (\ln \left (2\right )\right )\right ) {\mathrm e}^{2 \,{\mathrm e}^{4 x}}+5 x \,{\mathrm e}^{2 \,{\mathrm e}^{4 x}}\right )}{4}\) | \(39\) |
norman | \(-5 x \,{\mathrm e}^{-2}+\frac {5 x \,{\mathrm e}^{-2} {\mathrm e}^{2 \,{\mathrm e}^{4 x}}}{4}-\frac {5 \left (\ln \left (2\right )+\ln \left (\ln \left (2\right )\right )\right ) {\mathrm e}^{-2} {\mathrm e}^{2 \,{\mathrm e}^{4 x}}}{4}\) | \(42\) |
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Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {-20+e^{2 e^{4 x}} \left (5+40 e^{4 x} x-40 e^{4 x} \log (\log (4))\right )}{4 e^2} \, dx=\frac {5}{4} \, {\left ({\left (x - \log \left (2 \, \log \left (2\right )\right )\right )} e^{\left (2 \, e^{\left (4 \, x\right )}\right )} - 4 \, x\right )} e^{\left (-2\right )} \]
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Time = 0.15 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {-20+e^{2 e^{4 x}} \left (5+40 e^{4 x} x-40 e^{4 x} \log (\log (4))\right )}{4 e^2} \, dx=- \frac {5 x}{e^{2}} + \frac {\left (5 x - 5 \log {\left (2 \right )} - 5 \log {\left (\log {\left (2 \right )} \right )}\right ) e^{2 e^{4 x}}}{4 e^{2}} \]
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\[ \int \frac {-20+e^{2 e^{4 x}} \left (5+40 e^{4 x} x-40 e^{4 x} \log (\log (4))\right )}{4 e^2} \, dx=\int { \frac {5}{4} \, {\left ({\left (8 \, x e^{\left (4 \, x\right )} - 8 \, e^{\left (4 \, x\right )} \log \left (2 \, \log \left (2\right )\right ) + 1\right )} e^{\left (2 \, e^{\left (4 \, x\right )}\right )} - 4\right )} e^{\left (-2\right )} \,d x } \]
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Time = 0.25 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.48 \[ \int \frac {-20+e^{2 e^{4 x}} \left (5+40 e^{4 x} x-40 e^{4 x} \log (\log (4))\right )}{4 e^2} \, dx=\frac {5}{4} \, {\left (x e^{\left (2 \, e^{\left (4 \, x\right )}\right )} - e^{\left (2 \, e^{\left (4 \, x\right )}\right )} \log \left (2\right ) - e^{\left (2 \, e^{\left (4 \, x\right )}\right )} \log \left (\log \left (2\right )\right ) - 4 \, x\right )} e^{\left (-2\right )} \]
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Time = 12.44 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {-20+e^{2 e^{4 x}} \left (5+40 e^{4 x} x-40 e^{4 x} \log (\log (4))\right )}{4 e^2} \, dx=\frac {5\,x\,{\mathrm {e}}^{2\,{\mathrm {e}}^{4\,x}-2}}{4}-5\,x\,{\mathrm {e}}^{-2}-\frac {5\,\ln \left (2\,\ln \left (2\right )\right )\,{\mathrm {e}}^{2\,{\mathrm {e}}^{4\,x}-2}}{4} \]
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