Optimal. Leaf size=26 \[ 64 x^2 \left (5+x+\log ^2\left (\frac {5}{\log \left (\log \left (-3+e^{4/x}\right )\right )}\right )\right ) \]
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Rubi [F] time = 1.32, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {\left (-1920 x-576 x^2+e^{4/x} \left (640 x+192 x^2\right )\right ) \log \left (-3+e^{4/x}\right ) \log \left (\log \left (-3+e^{4/x}\right )\right )+512 e^{4/x} \log \left (\frac {5}{\log \left (\log \left (-3+e^{4/x}\right )\right )}\right )+\left (-384 x+128 e^{4/x} x\right ) \log \left (-3+e^{4/x}\right ) \log \left (\log \left (-3+e^{4/x}\right )\right ) \log ^2\left (\frac {5}{\log \left (\log \left (-3+e^{4/x}\right )\right )}\right )}{\left (-3+e^{4/x}\right ) \log \left (-3+e^{4/x}\right ) \log \left (\log \left (-3+e^{4/x}\right )\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int 64 \left (\frac {8 e^{4/x} \log \left (\frac {5}{\log \left (\log \left (-3+e^{4/x}\right )\right )}\right )}{\left (-3+e^{4/x}\right ) \log \left (-3+e^{4/x}\right ) \log \left (\log \left (-3+e^{4/x}\right )\right )}+x \left (10+3 x+2 \log ^2\left (\frac {5}{\log \left (\log \left (-3+e^{4/x}\right )\right )}\right )\right )\right ) \, dx\\ &=64 \int \left (\frac {8 e^{4/x} \log \left (\frac {5}{\log \left (\log \left (-3+e^{4/x}\right )\right )}\right )}{\left (-3+e^{4/x}\right ) \log \left (-3+e^{4/x}\right ) \log \left (\log \left (-3+e^{4/x}\right )\right )}+x \left (10+3 x+2 \log ^2\left (\frac {5}{\log \left (\log \left (-3+e^{4/x}\right )\right )}\right )\right )\right ) \, dx\\ &=64 \int x \left (10+3 x+2 \log ^2\left (\frac {5}{\log \left (\log \left (-3+e^{4/x}\right )\right )}\right )\right ) \, dx+512 \int \frac {e^{4/x} \log \left (\frac {5}{\log \left (\log \left (-3+e^{4/x}\right )\right )}\right )}{\left (-3+e^{4/x}\right ) \log \left (-3+e^{4/x}\right ) \log \left (\log \left (-3+e^{4/x}\right )\right )} \, dx\\ &=64 \int \left (x (10+3 x)+2 x \log ^2\left (\frac {5}{\log \left (\log \left (-3+e^{4/x}\right )\right )}\right )\right ) \, dx+512 \int \frac {e^{4/x} \log \left (\frac {5}{\log \left (\log \left (-3+e^{4/x}\right )\right )}\right )}{\left (-3+e^{4/x}\right ) \log \left (-3+e^{4/x}\right ) \log \left (\log \left (-3+e^{4/x}\right )\right )} \, dx\\ &=64 \int x (10+3 x) \, dx+128 \int x \log ^2\left (\frac {5}{\log \left (\log \left (-3+e^{4/x}\right )\right )}\right ) \, dx+512 \int \frac {e^{4/x} \log \left (\frac {5}{\log \left (\log \left (-3+e^{4/x}\right )\right )}\right )}{\left (-3+e^{4/x}\right ) \log \left (-3+e^{4/x}\right ) \log \left (\log \left (-3+e^{4/x}\right )\right )} \, dx\\ &=64 \int \left (10 x+3 x^2\right ) \, dx+128 \int x \log ^2\left (\frac {5}{\log \left (\log \left (-3+e^{4/x}\right )\right )}\right ) \, dx+512 \int \frac {e^{4/x} \log \left (\frac {5}{\log \left (\log \left (-3+e^{4/x}\right )\right )}\right )}{\left (-3+e^{4/x}\right ) \log \left (-3+e^{4/x}\right ) \log \left (\log \left (-3+e^{4/x}\right )\right )} \, dx\\ &=320 x^2+64 x^3+128 \int x \log ^2\left (\frac {5}{\log \left (\log \left (-3+e^{4/x}\right )\right )}\right ) \, dx+512 \int \frac {e^{4/x} \log \left (\frac {5}{\log \left (\log \left (-3+e^{4/x}\right )\right )}\right )}{\left (-3+e^{4/x}\right ) \log \left (-3+e^{4/x}\right ) \log \left (\log \left (-3+e^{4/x}\right )\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.21, size = 33, normalized size = 1.27 \begin {gather*} 64 \left (5 x^2+x^3+x^2 \log ^2\left (\frac {5}{\log \left (\log \left (-3+e^{4/x}\right )\right )}\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.64, size = 33, normalized size = 1.27 \begin {gather*} 64 \, x^{2} \log \left (\frac {5}{\log \left (\log \left (e^{\frac {4}{x}} - 3\right )\right )}\right )^{2} + 64 \, x^{3} + 320 \, x^{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {64 \, {\left (2 \, {\left (x e^{\frac {4}{x}} - 3 \, x\right )} \log \left (\frac {5}{\log \left (\log \left (e^{\frac {4}{x}} - 3\right )\right )}\right )^{2} \log \left (e^{\frac {4}{x}} - 3\right ) \log \left (\log \left (e^{\frac {4}{x}} - 3\right )\right ) - {\left (9 \, x^{2} - {\left (3 \, x^{2} + 10 \, x\right )} e^{\frac {4}{x}} + 30 \, x\right )} \log \left (e^{\frac {4}{x}} - 3\right ) \log \left (\log \left (e^{\frac {4}{x}} - 3\right )\right ) + 8 \, e^{\frac {4}{x}} \log \left (\frac {5}{\log \left (\log \left (e^{\frac {4}{x}} - 3\right )\right )}\right )\right )}}{{\left (e^{\frac {4}{x}} - 3\right )} \log \left (e^{\frac {4}{x}} - 3\right ) \log \left (\log \left (e^{\frac {4}{x}} - 3\right )\right )}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.54, size = 57, normalized size = 2.19
method | result | size |
risch | \(64 x^{2} \ln \left (\ln \left (\ln \left ({\mathrm e}^{\frac {4}{x}}-3\right )\right )\right )^{2}-128 \ln \relax (5) x^{2} \ln \left (\ln \left (\ln \left ({\mathrm e}^{\frac {4}{x}}-3\right )\right )\right )+64 x^{2} \ln \relax (5)^{2}+64 x^{3}+320 x^{2}\) | \(57\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.55, size = 53, normalized size = 2.04 \begin {gather*} -128 \, x^{2} \log \relax (5) \log \left (\log \left (\log \left (e^{\frac {4}{x}} - 3\right )\right )\right ) + 64 \, x^{2} \log \left (\log \left (\log \left (e^{\frac {4}{x}} - 3\right )\right )\right )^{2} + 64 \, {\left (\log \relax (5)^{2} + 5\right )} x^{2} + 64 \, x^{3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 8.55, size = 25, normalized size = 0.96 \begin {gather*} 64\,x^2\,\left ({\ln \left (\frac {5}{\ln \left (\ln \left ({\mathrm {e}}^{4/x}-3\right )\right )}\right )}^2+x+5\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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