Optimal. Leaf size=36 \[ 20+5 \left (x+\frac {2 (4-x)}{\log \left (\frac {i \pi +\log \left (-2 \left (4-e^2\right )\right )}{x}\right )}\right ) \]
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Rubi [A] time = 0.44, antiderivative size = 55, normalized size of antiderivative = 1.53, number of steps used = 13, number of rules used = 9, integrand size = 76, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {6741, 12, 6742, 2353, 2297, 2299, 2178, 2302, 30} \begin {gather*} 5 x-\frac {10 x}{\log \left (\frac {\log \left (-2 \left (4-e^2\right )\right )+i \pi }{x}\right )}+\frac {40}{\log \left (\frac {\log \left (-2 \left (4-e^2\right )\right )+i \pi }{x}\right )} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 30
Rule 2178
Rule 2297
Rule 2299
Rule 2302
Rule 2353
Rule 6741
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5 \left (8-2 x-2 x \log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )+x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )\right )}{x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \, dx\\ &=5 \int \frac {8-2 x-2 x \log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )+x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )}{x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \, dx\\ &=5 \int \left (1-\frac {2 (-4+x)}{x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )}-\frac {2}{\log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )}\right ) \, dx\\ &=5 x-10 \int \frac {-4+x}{x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \, dx-10 \int \frac {1}{\log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \, dx\\ &=5 x-10 \int \left (\frac {1}{\log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )}-\frac {4}{x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )}\right ) \, dx+\left (10 \left (i \pi +\log \left (-2 \left (4-e^2\right )\right )\right )\right ) \operatorname {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )\right )\\ &=5 x+10 \text {Ei}\left (-\log \left (\frac {i \pi +\log \left (-2 \left (4-e^2\right )\right )}{x}\right )\right ) \left (i \pi +\log \left (-2 \left (4-e^2\right )\right )\right )-10 \int \frac {1}{\log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \, dx+40 \int \frac {1}{x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \, dx\\ &=5 x+10 \text {Ei}\left (-\log \left (\frac {i \pi +\log \left (-2 \left (4-e^2\right )\right )}{x}\right )\right ) \left (i \pi +\log \left (-2 \left (4-e^2\right )\right )\right )-\frac {10 x}{\log \left (\frac {i \pi +\log \left (-2 \left (4-e^2\right )\right )}{x}\right )}+10 \int \frac {1}{\log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \, dx-40 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )\right )\\ &=5 x+10 \text {Ei}\left (-\log \left (\frac {i \pi +\log \left (-2 \left (4-e^2\right )\right )}{x}\right )\right ) \left (i \pi +\log \left (-2 \left (4-e^2\right )\right )\right )+\frac {40}{\log \left (\frac {i \pi +\log \left (-2 \left (4-e^2\right )\right )}{x}\right )}-\frac {10 x}{\log \left (\frac {i \pi +\log \left (-2 \left (4-e^2\right )\right )}{x}\right )}-\left (10 \left (i \pi +\log \left (-2 \left (4-e^2\right )\right )\right )\right ) \operatorname {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )\right )\\ &=5 x+\frac {40}{\log \left (\frac {i \pi +\log \left (-2 \left (4-e^2\right )\right )}{x}\right )}-\frac {10 x}{\log \left (\frac {i \pi +\log \left (-2 \left (4-e^2\right )\right )}{x}\right )}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.04, size = 51, normalized size = 1.42 \begin {gather*} 5 x+\frac {40}{\log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )}-\frac {10 x}{\log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.60, size = 35, normalized size = 0.97 \begin {gather*} \frac {5 \, {\left (x \log \left (\frac {\log \left (-2 \, e^{2} + 8\right )}{x}\right ) - 2 \, x + 8\right )}}{\log \left (\frac {\log \left (-2 \, e^{2} + 8\right )}{x}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.69, size = 74, normalized size = 2.06 \begin {gather*} \frac {5 \, {\left (\log \left (\frac {\log \left (-2 \, e^{2} + 8\right )}{x}\right ) \log \left (-2 \, e^{2} + 8\right )^{3} - 2 \, \log \left (-2 \, e^{2} + 8\right )^{3} + \frac {8 \, \log \left (-2 \, e^{2} + 8\right )^{3}}{x}\right )} x}{\log \left (\frac {\log \left (-2 \, e^{2} + 8\right )}{x}\right ) \log \left (-2 \, e^{2} + 8\right )^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.18, size = 27, normalized size = 0.75
method | result | size |
risch | \(5 x -\frac {10 \left (x -4\right )}{\ln \left (\frac {\ln \relax (2)+\ln \left (-{\mathrm e}^{2}+4\right )}{x}\right )}\) | \(27\) |
norman | \(\frac {40-10 x +5 x \ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )}{\ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )}\) | \(36\) |
derivativedivides | \(\frac {5 \ln \relax (2) x}{\ln \left (-2 \,{\mathrm e}^{2}+8\right )}+\frac {5 \ln \left (-{\mathrm e}^{2}+4\right ) x}{\ln \left (-2 \,{\mathrm e}^{2}+8\right )}-10 \ln \relax (2) \expIntegralEi \left (1, \ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )\right )-10 \ln \left (-{\mathrm e}^{2}+4\right ) \expIntegralEi \left (1, \ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )\right )+10 \ln \relax (2) \left (-\frac {x}{\ln \left (-2 \,{\mathrm e}^{2}+8\right ) \ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )}+\expIntegralEi \left (1, \ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )\right )\right )+10 \ln \left (-{\mathrm e}^{2}+4\right ) \left (-\frac {x}{\ln \left (-2 \,{\mathrm e}^{2}+8\right ) \ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )}+\expIntegralEi \left (1, \ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )\right )\right )+\frac {40}{\ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )}\) | \(187\) |
default | \(\frac {5 \ln \relax (2) x}{\ln \left (-2 \,{\mathrm e}^{2}+8\right )}+\frac {5 \ln \left (-{\mathrm e}^{2}+4\right ) x}{\ln \left (-2 \,{\mathrm e}^{2}+8\right )}-10 \ln \relax (2) \expIntegralEi \left (1, \ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )\right )-10 \ln \left (-{\mathrm e}^{2}+4\right ) \expIntegralEi \left (1, \ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )\right )+10 \ln \relax (2) \left (-\frac {x}{\ln \left (-2 \,{\mathrm e}^{2}+8\right ) \ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )}+\expIntegralEi \left (1, \ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )\right )\right )+10 \ln \left (-{\mathrm e}^{2}+4\right ) \left (-\frac {x}{\ln \left (-2 \,{\mathrm e}^{2}+8\right ) \ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )}+\expIntegralEi \left (1, \ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )\right )\right )+\frac {40}{\ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )}\) | \(187\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [C] time = 0.61, size = 33, normalized size = 0.92 \begin {gather*} 5 \, x - \frac {10 \, {\left (x - 4\right )}}{\log \left (i \, \pi + \log \relax (2) + \log \left (e + 2\right ) + \log \left (e - 2\right )\right ) - \log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.67, size = 25, normalized size = 0.69 \begin {gather*} 5\,x-\frac {10\,x-40}{\ln \left (\frac {\ln \left (8-2\,{\mathrm {e}}^2\right )}{x}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.12, size = 27, normalized size = 0.75 \begin {gather*} 5 x + \frac {40 - 10 x}{\log {\left (\frac {\log {\relax (2 )}}{x} + \frac {\log {\left (-4 + e^{2} \right )}}{x} + \frac {i \pi }{x} \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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