Optimal. Leaf size=29 \[ \frac {e^{-x+\frac {5}{\log (3)}} (3+2 x-\log (2)) (5+\log (2))}{x} \]
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Rubi [A] time = 0.19, antiderivative size = 57, normalized size of antiderivative = 1.97, number of steps used = 6, number of rules used = 4, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.082, Rules used = {2199, 2194, 2177, 2178} \begin {gather*} \frac {e^{\frac {5}{\log (3)}} \left (15-\log ^2(2)-\log (4)\right ) 3^{-\frac {x}{\log (3)}}}{x}+2 e^{\frac {5}{\log (3)}} (5+\log (2)) 3^{-\frac {x}{\log (3)}} \end {gather*}
Antiderivative was successfully verified.
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Rule 2177
Rule 2178
Rule 2194
Rule 2199
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-2 e^{\frac {5-x \log (3)}{\log (3)}} (5+\log (2))+\frac {e^{\frac {5-x \log (3)}{\log (3)}} \left (-15+\log ^2(2)+\log (4)\right )}{x^2}+\frac {e^{\frac {5-x \log (3)}{\log (3)}} \left (-15+\log ^2(2)+\log (4)\right )}{x}\right ) \, dx\\ &=-\left ((2 (5+\log (2))) \int e^{\frac {5-x \log (3)}{\log (3)}} \, dx\right )+\left (-15+\log ^2(2)+\log (4)\right ) \int \frac {e^{\frac {5-x \log (3)}{\log (3)}}}{x^2} \, dx+\left (-15+\log ^2(2)+\log (4)\right ) \int \frac {e^{\frac {5-x \log (3)}{\log (3)}}}{x} \, dx\\ &=2\ 3^{-\frac {x}{\log (3)}} e^{\frac {5}{\log (3)}} (5+\log (2))+\frac {3^{-\frac {x}{\log (3)}} e^{\frac {5}{\log (3)}} \left (15-\log ^2(2)-\log (4)\right )}{x}-e^{\frac {5}{\log (3)}} \text {Ei}(-x) \left (15-\log ^2(2)-\log (4)\right )+\left (15-\log ^2(2)-\log (4)\right ) \int \frac {e^{\frac {5-x \log (3)}{\log (3)}}}{x} \, dx\\ &=2\ 3^{-\frac {x}{\log (3)}} e^{\frac {5}{\log (3)}} (5+\log (2))+\frac {3^{-\frac {x}{\log (3)}} e^{\frac {5}{\log (3)}} \left (15-\log ^2(2)-\log (4)\right )}{x}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.03, size = 29, normalized size = 1.00 \begin {gather*} \frac {e^{-x+\frac {5}{\log (3)}} (3+2 x-\log (2)) (5+\log (2))}{x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.54, size = 35, normalized size = 1.21 \begin {gather*} \frac {{\left (2 \, {\left (x - 1\right )} \log \relax (2) - \log \relax (2)^{2} + 10 \, x + 15\right )} e^{\left (-\frac {x \log \relax (3) - 5}{\log \relax (3)}\right )}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.16, size = 80, normalized size = 2.76 \begin {gather*} \frac {2 \, x e^{\left (-x + \frac {5}{\log \relax (3)}\right )} \log \relax (2) - e^{\left (-x + \frac {5}{\log \relax (3)}\right )} \log \relax (2)^{2} + 10 \, x e^{\left (-x + \frac {5}{\log \relax (3)}\right )} - 2 \, e^{\left (-x + \frac {5}{\log \relax (3)}\right )} \log \relax (2) + 15 \, e^{\left (-x + \frac {5}{\log \relax (3)}\right )}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.09, size = 30, normalized size = 1.03
method | result | size |
gosper | \(-\frac {{\mathrm e}^{-\frac {x \ln \relax (3)-5}{\ln \relax (3)}} \left (\ln \relax (2)+5\right ) \left (-3-2 x +\ln \relax (2)\right )}{x}\) | \(30\) |
risch | \(-\frac {\left (\ln \relax (2)^{2}-2 x \ln \relax (2)+2 \ln \relax (2)-10 x -15\right ) {\mathrm e}^{-\frac {x \ln \relax (3)-5}{\ln \relax (3)}}}{x}\) | \(37\) |
norman | \(\frac {\left (-\ln \relax (2)^{2}-2 \ln \relax (2)+15\right ) {\mathrm e}^{\frac {-x \ln \relax (3)+5}{\ln \relax (3)}}+\left (2 \ln \relax (2)+10\right ) x \,{\mathrm e}^{\frac {-x \ln \relax (3)+5}{\ln \relax (3)}}}{x}\) | \(53\) |
derivativedivides | \(-\frac {\ln \relax (2)^{2} {\mathrm e}^{\frac {5}{\ln \relax (3)}-x}}{x}+\frac {15 \,{\mathrm e}^{\frac {5}{\ln \relax (3)}-x}}{x}+10 \,{\mathrm e}^{\frac {5}{\ln \relax (3)}-x}-\frac {2 \ln \relax (2) {\mathrm e}^{\frac {5}{\ln \relax (3)}-x}}{x}+2 \ln \relax (2) {\mathrm e}^{\frac {5}{\ln \relax (3)}-x}\) | \(84\) |
default | \(-\frac {\ln \relax (2)^{2} {\mathrm e}^{\frac {5}{\ln \relax (3)}-x}}{x}+\frac {15 \,{\mathrm e}^{\frac {5}{\ln \relax (3)}-x}}{x}+10 \,{\mathrm e}^{\frac {5}{\ln \relax (3)}-x}-\frac {2 \ln \relax (2) {\mathrm e}^{\frac {5}{\ln \relax (3)}-x}}{x}+2 \ln \relax (2) {\mathrm e}^{\frac {5}{\ln \relax (3)}-x}\) | \(84\) |
meijerg | \(\left (-2 \ln \relax (2)-10\right ) {\mathrm e}^{\frac {5}{\ln \relax (3)}} \left (1-{\mathrm e}^{-x}\right )-\left (\ln \relax (2)^{2}+2 \ln \relax (2)-15\right ) {\mathrm e}^{\frac {5}{\ln \relax (3)}} \expIntegralEi \left (1, x\right )+\ln \relax (2)^{2} {\mathrm e}^{\frac {5}{\ln \relax (3)}} \left (\frac {-2 x +2}{2 x}-\frac {{\mathrm e}^{-x}}{x}+\expIntegralEi \left (1, x\right )+1-\frac {1}{x}\right )+2 \ln \relax (2) {\mathrm e}^{\frac {5}{\ln \relax (3)}} \left (\frac {-2 x +2}{2 x}-\frac {{\mathrm e}^{-x}}{x}+\expIntegralEi \left (1, x\right )+1-\frac {1}{x}\right )-15 \,{\mathrm e}^{\frac {5}{\ln \relax (3)}} \left (\frac {-2 x +2}{2 x}-\frac {{\mathrm e}^{-x}}{x}+\expIntegralEi \left (1, x\right )+1-\frac {1}{x}\right )\) | \(165\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [C] time = 0.50, size = 115, normalized size = 3.97 \begin {gather*} {\rm Ei}\left (-x\right ) e^{\frac {5}{\log \relax (3)}} \log \relax (2)^{2} - e^{\frac {5}{\log \relax (3)}} \Gamma \left (-1, x\right ) \log \relax (2)^{2} + 2 \, {\rm Ei}\left (-x\right ) e^{\frac {5}{\log \relax (3)}} \log \relax (2) - 2 \, e^{\frac {5}{\log \relax (3)}} \Gamma \left (-1, x\right ) \log \relax (2) - 15 \, {\rm Ei}\left (-x\right ) e^{\frac {5}{\log \relax (3)}} + 15 \, e^{\frac {5}{\log \relax (3)}} \Gamma \left (-1, x\right ) + 2 \, e^{\left (-x + \frac {5}{\log \relax (3)}\right )} \log \relax (2) + 10 \, e^{\left (-x + \frac {5}{\log \relax (3)}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.23, size = 34, normalized size = 1.17 \begin {gather*} \frac {{\mathrm {e}}^{\frac {5}{\ln \relax (3)}-x}\,\left (10\,x-\ln \relax (4)+x\,\ln \relax (4)-{\ln \relax (2)}^2+15\right )}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.14, size = 34, normalized size = 1.17 \begin {gather*} \frac {\left (2 x \log {\relax (2 )} + 10 x - 2 \log {\relax (2 )} - \log {\relax (2 )}^{2} + 15\right ) e^{\frac {- x \log {\relax (3 )} + 5}{\log {\relax (3 )}}}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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