Optimal. Leaf size=25 \[ x+(-10+x) x+\frac {2+e^x}{\frac {6}{x}-\log (5)} \]
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Rubi [B] time = 0.51, antiderivative size = 156, normalized size of antiderivative = 6.24, number of steps used = 15, number of rules used = 8, integrand size = 67, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.119, Rules used = {27, 6742, 43, 77, 2199, 2194, 2177, 2178} \begin {gather*} x^2-\frac {36 (8-\log (125)) \log (6-x \log (5))}{\log ^2(5)}+\frac {108 (2-\log (5)) \log (6-x \log (5))}{\log ^2(5)}+\frac {72 \log (6-x \log (5))}{\log ^2(5)}+\frac {432}{\log ^2(5) (6-x \log (5))}-\frac {108 (4-\log (125))}{\log ^2(5) (6-x \log (5))}+\frac {3 x (8-\log (125))}{\log (5)}-\frac {24 x}{\log (5)}+\frac {6 e^x}{\log (5) (6-x \log (5))}-\frac {312}{\log (5) (6-x \log (5))}-\frac {e^x}{\log (5)} \end {gather*}
Antiderivative was successfully verified.
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Rule 27
Rule 43
Rule 77
Rule 2177
Rule 2178
Rule 2194
Rule 2199
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-312+72 x+\left (108 x-24 x^2\right ) \log (5)+\left (-9 x^2+2 x^3\right ) \log ^2(5)+e^x \left (6+6 x-x^2 \log (5)\right )}{(-6+x \log (5))^2} \, dx\\ &=\int \left (-\frac {312}{(-6+x \log (5))^2}+\frac {72 x}{(-6+x \log (5))^2}-\frac {12 x (-9+2 x) \log (5)}{(-6+x \log (5))^2}+\frac {x^2 (-9+2 x) \log ^2(5)}{(-6+x \log (5))^2}-\frac {e^x \left (-6-6 x+x^2 \log (5)\right )}{(-6+x \log (5))^2}\right ) \, dx\\ &=-\frac {312}{\log (5) (6-x \log (5))}+72 \int \frac {x}{(-6+x \log (5))^2} \, dx-(12 \log (5)) \int \frac {x (-9+2 x)}{(-6+x \log (5))^2} \, dx+\log ^2(5) \int \frac {x^2 (-9+2 x)}{(-6+x \log (5))^2} \, dx-\int \frac {e^x \left (-6-6 x+x^2 \log (5)\right )}{(-6+x \log (5))^2} \, dx\\ &=-\frac {312}{\log (5) (6-x \log (5))}+72 \int \left (\frac {6}{\log (5) (-6+x \log (5))^2}+\frac {1}{\log (5) (-6+x \log (5))}\right ) \, dx-(12 \log (5)) \int \left (\frac {2}{\log ^2(5)}+\frac {18 (4-\log (125))}{\log ^2(5) (6-x \log (5))^2}+\frac {3 (-8+\log (125))}{\log ^2(5) (6-x \log (5))}\right ) \, dx+\log ^2(5) \int \left (\frac {2 x}{\log ^2(5)}-\frac {108 (-2+\log (5))}{\log ^3(5) (-6+x \log (5))}+\frac {3 (8-\log (125))}{\log ^3(5)}-\frac {108 (-4+\log (125))}{\log ^3(5) (-6+x \log (5))^2}\right ) \, dx-\int \left (\frac {e^x}{\log (5)}-\frac {6 e^x}{(-6+x \log (5))^2}+\frac {6 e^x}{\log (5) (-6+x \log (5))}\right ) \, dx\\ &=x^2-\frac {24 x}{\log (5)}+\frac {432}{\log ^2(5) (6-x \log (5))}-\frac {312}{\log (5) (6-x \log (5))}-\frac {108 (4-\log (125))}{\log ^2(5) (6-x \log (5))}+\frac {3 x (8-\log (125))}{\log (5)}+\frac {72 \log (6-x \log (5))}{\log ^2(5)}+\frac {108 (2-\log (5)) \log (6-x \log (5))}{\log ^2(5)}-\frac {36 (8-\log (125)) \log (6-x \log (5))}{\log ^2(5)}+6 \int \frac {e^x}{(-6+x \log (5))^2} \, dx-\frac {\int e^x \, dx}{\log (5)}-\frac {6 \int \frac {e^x}{-6+x \log (5)} \, dx}{\log (5)}\\ &=x^2-\frac {6 e^{\frac {6}{\log (5)}} \text {Ei}\left (-\frac {6-x \log (5)}{\log (5)}\right )}{\log ^2(5)}-\frac {e^x}{\log (5)}-\frac {24 x}{\log (5)}+\frac {432}{\log ^2(5) (6-x \log (5))}-\frac {312}{\log (5) (6-x \log (5))}+\frac {6 e^x}{\log (5) (6-x \log (5))}-\frac {108 (4-\log (125))}{\log ^2(5) (6-x \log (5))}+\frac {3 x (8-\log (125))}{\log (5)}+\frac {72 \log (6-x \log (5))}{\log ^2(5)}+\frac {108 (2-\log (5)) \log (6-x \log (5))}{\log ^2(5)}-\frac {36 (8-\log (125)) \log (6-x \log (5))}{\log ^2(5)}+\frac {6 \int \frac {e^x}{-6+x \log (5)} \, dx}{\log (5)}\\ &=x^2-\frac {e^x}{\log (5)}-\frac {24 x}{\log (5)}+\frac {432}{\log ^2(5) (6-x \log (5))}-\frac {312}{\log (5) (6-x \log (5))}+\frac {6 e^x}{\log (5) (6-x \log (5))}-\frac {108 (4-\log (125))}{\log ^2(5) (6-x \log (5))}+\frac {3 x (8-\log (125))}{\log (5)}+\frac {72 \log (6-x \log (5))}{\log ^2(5)}+\frac {108 (2-\log (5)) \log (6-x \log (5))}{\log ^2(5)}-\frac {36 (8-\log (125)) \log (6-x \log (5))}{\log ^2(5)}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.21, size = 47, normalized size = 1.88 \begin {gather*} \frac {-12-e^x x \log (5)+x^3 \log ^2(5)+18 x \log (125)-3 x^2 \log (5) (2+\log (125))}{\log (5) (-6+x \log (5))} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.60, size = 49, normalized size = 1.96 \begin {gather*} -\frac {x e^{x} \log \relax (5) - {\left (x^{3} - 9 \, x^{2}\right )} \log \relax (5)^{2} + 6 \, {\left (x^{2} - 9 \, x\right )} \log \relax (5) + 12}{x \log \relax (5)^{2} - 6 \, \log \relax (5)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.39, size = 52, normalized size = 2.08 \begin {gather*} \frac {x^{3} \log \relax (5)^{2} - 9 \, x^{2} \log \relax (5)^{2} - 6 \, x^{2} \log \relax (5) - x e^{x} \log \relax (5) + 54 \, x \log \relax (5) - 12}{x \log \relax (5)^{2} - 6 \, \log \relax (5)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.23, size = 35, normalized size = 1.40
method | result | size |
norman | \(\frac {x^{3} \ln \relax (5)+52 x +\left (-9 \ln \relax (5)-6\right ) x^{2}-{\mathrm e}^{x} x}{x \ln \relax (5)-6}\) | \(35\) |
risch | \(x^{2}-9 x -\frac {12}{\ln \relax (5) \left (x \ln \relax (5)-6\right )}-\frac {x \,{\mathrm e}^{x}}{x \ln \relax (5)-6}\) | \(35\) |
default | \(-\frac {12}{\ln \relax (5) \left (x \ln \relax (5)-6\right )}-\frac {6 \,{\mathrm e}^{x}}{\ln \relax (5)^{2} \left (x -\frac {6}{\ln \relax (5)}\right )}-9 x +x^{2}-\frac {{\mathrm e}^{x}}{\ln \relax (5)}\) | \(48\) |
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*} -{\left (\frac {432}{x \log \relax (5)^{5} - 6 \, \log \relax (5)^{4}} - \frac {x^{2} \log \relax (5) + 24 \, x}{\log \relax (5)^{3}} - \frac {216 \, \log \left (x \log \relax (5) - 6\right )}{\log \relax (5)^{4}}\right )} \log \relax (5)^{2} + 9 \, {\left (\frac {36}{x \log \relax (5)^{4} - 6 \, \log \relax (5)^{3}} - \frac {x}{\log \relax (5)^{2}} - \frac {12 \, \log \left (x \log \relax (5) - 6\right )}{\log \relax (5)^{3}}\right )} \log \relax (5)^{2} + 24 \, {\left (\frac {36}{x \log \relax (5)^{4} - 6 \, \log \relax (5)^{3}} - \frac {x}{\log \relax (5)^{2}} - \frac {12 \, \log \left (x \log \relax (5) - 6\right )}{\log \relax (5)^{3}}\right )} \log \relax (5) - 108 \, {\left (\frac {6}{x \log \relax (5)^{3} - 6 \, \log \relax (5)^{2}} - \frac {\log \left (x \log \relax (5) - 6\right )}{\log \relax (5)^{2}}\right )} \log \relax (5) - \frac {x e^{x}}{x \log \relax (5) - 6} - \frac {6 \, e^{\frac {6}{\log \relax (5)}} E_{2}\left (-\frac {x \log \relax (5) - 6}{\log \relax (5)}\right )}{{\left (x \log \relax (5) - 6\right )} \log \relax (5)} - \frac {432}{x \log \relax (5)^{3} - 6 \, \log \relax (5)^{2}} + \frac {312}{x \log \relax (5)^{2} - 6 \, \log \relax (5)} + \frac {72 \, \log \left (x \log \relax (5) - 6\right )}{\log \relax (5)^{2}} - 6 \, \int \frac {e^{x}}{x^{2} \log \relax (5)^{2} - 12 \, x \log \relax (5) + 36}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.22, size = 30, normalized size = 1.20 \begin {gather*} -\frac {x\,\left (6\,x+{\mathrm {e}}^x+9\,x\,\ln \relax (5)-x^2\,\ln \relax (5)-52\right )}{x\,\ln \relax (5)-6} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.24, size = 31, normalized size = 1.24 \begin {gather*} x^{2} - 9 x - \frac {x e^{x}}{x \log {\relax (5 )} - 6} - \frac {12}{x \log {\relax (5 )}^{2} - 6 \log {\relax (5 )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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