Optimal. Leaf size=27 \[ \frac {5 \left (4-e^{2 e^{4 x}}\right ) (-x+\log (\log (4)))}{4 e^2} \]
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Rubi [A] time = 0.05, antiderivative size = 43, normalized size of antiderivative = 1.59, number of steps used = 3, number of rules used = 2, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {12, 2288} \begin {gather*} \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} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 2288
Rubi steps
\begin {gather*} \begin {aligned} \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 {aligned} \end {gather*}
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Mathematica [A] time = 0.11, size = 28, normalized size = 1.04 \begin {gather*} \frac {5 \left (-4 x+e^{2 e^{4 x}} (x-\log (\log (4)))\right )}{4 e^2} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.84, size = 25, normalized size = 0.93 \begin {gather*} \frac {5}{4} \, {\left ({\left (x - \log \left (2 \, \log \relax (2)\right )\right )} e^{\left (2 \, e^{\left (4 \, x\right )}\right )} - 4 \, x\right )} e^{\left (-2\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.25, size = 40, normalized size = 1.48 \begin {gather*} \frac {5}{4} \, {\left (x e^{\left (2 \, e^{\left (4 \, x\right )}\right )} - e^{\left (2 \, e^{\left (4 \, x\right )}\right )} \log \relax (2) - e^{\left (2 \, e^{\left (4 \, x\right )}\right )} \log \left (\log \relax (2)\right ) - 4 \, x\right )} e^{\left (-2\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.07, size = 31, normalized size = 1.15
method | result | size |
risch | \(-5 x \,{\mathrm e}^{-2}+\frac {\left (-5 \ln \relax (2)-5 \ln \left (\ln \relax (2)\right )+5 x \right ) {\mathrm e}^{-2+2 \,{\mathrm e}^{4 x}}}{4}\) | \(31\) |
norman | \(-5 x \,{\mathrm e}^{-2}+\frac {5 x \,{\mathrm e}^{-2} {\mathrm e}^{2 \,{\mathrm e}^{4 x}}}{4}-\frac {5 \left (\ln \relax (2)+\ln \left (\ln \relax (2)\right )\right ) {\mathrm e}^{-2} {\mathrm e}^{2 \,{\mathrm e}^{4 x}}}{4}\) | \(42\) |
default | \(\frac {{\mathrm e}^{-2} \left (-20 x -5 \,{\mathrm e}^{2 \,{\mathrm e}^{4 x}} \ln \relax (2)-5 \,{\mathrm e}^{2 \,{\mathrm e}^{4 x}} \ln \left (\ln \relax (2)\right )+5 x \,{\mathrm e}^{2 \,{\mathrm e}^{4 x}}\right )}{4}\) | \(44\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {5}{16} \, {\left (4 \, x e^{\left (2 \, e^{\left (4 \, x\right )}\right )} - 4 \, e^{\left (2 \, e^{\left (4 \, x\right )}\right )} \log \left (2 \, \log \relax (2)\right ) - 16 \, x + {\rm Ei}\left (2 \, e^{\left (4 \, x\right )}\right ) - 4 \, \int e^{\left (2 \, e^{\left (4 \, x\right )}\right )}\,{d x}\right )} e^{\left (-2\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.56, size = 34, normalized size = 1.26 \begin {gather*} \frac {5\,x\,{\mathrm {e}}^{2\,{\mathrm {e}}^{4\,x}-2}}{4}-5\,x\,{\mathrm {e}}^{-2}-\frac {5\,\ln \left (2\,\ln \relax (2)\right )\,{\mathrm {e}}^{2\,{\mathrm {e}}^{4\,x}-2}}{4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.26, size = 34, normalized size = 1.26 \begin {gather*} - \frac {5 x}{e^{2}} + \frac {\left (5 x - 5 \log {\relax (2 )} - 5 \log {\left (\log {\relax (2 )} \right )}\right ) e^{2 e^{4 x}}}{4 e^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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