Optimal. Leaf size=32 \[ \left (-3-x-\left (4+e^3+x\right )^2+\left (-x+5 (2+x)^2\right ) \log (5)\right ) \log (x) \]
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Rubi [A] time = 0.10, antiderivative size = 47, normalized size of antiderivative = 1.47, number of steps used = 6, number of rules used = 2, integrand size = 71, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {14, 2313} \begin {gather*} -\left (\left (x^2 (1-5 \log (5))+x \left (9+2 e^3-19 \log (5)\right )\right ) \log (x)\right )-\left (19+8 e^3+e^6-20 \log (5)\right ) \log (x) \end {gather*}
Antiderivative was successfully verified.
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Rule 14
Rule 2313
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {-19-8 e^3-e^6-x \left (9+2 e^3-19 \log (5)\right )-x^2 (1-5 \log (5))+20 \log (5)}{x}+\left (-9-2 e^3-2 x (1-5 \log (5))+19 \log (5)\right ) \log (x)\right ) \, dx\\ &=\int \frac {-19-8 e^3-e^6-x \left (9+2 e^3-19 \log (5)\right )-x^2 (1-5 \log (5))+20 \log (5)}{x} \, dx+\int \left (-9-2 e^3-2 x (1-5 \log (5))+19 \log (5)\right ) \log (x) \, dx\\ &=-\left (\left (x \left (9+2 e^3-19 \log (5)\right )+x^2 (1-5 \log (5))\right ) \log (x)\right )-\int \left (-9-2 e^3+19 \log (5)+x (-1+5 \log (5))\right ) \, dx+\int \left (-9-2 e^3-x (1-5 \log (5))+19 \log (5)+\frac {-19-8 e^3-e^6+20 \log (5)}{x}\right ) \, dx\\ &=-\left (\left (x \left (9+2 e^3-19 \log (5)\right )+x^2 (1-5 \log (5))\right ) \log (x)\right )-\left (19+8 e^3+e^6-20 \log (5)\right ) \log (x)\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.03, size = 61, normalized size = 1.91 \begin {gather*} -19 \log (x)-8 e^3 \log (x)-e^6 \log (x)-9 x \log (x)-2 e^3 x \log (x)-x^2 \log (x)+20 \log (5) \log (x)+19 x \log (5) \log (x)+5 x^2 \log (5) \log (x) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.47, size = 35, normalized size = 1.09 \begin {gather*} -{\left (x^{2} + 2 \, {\left (x + 4\right )} e^{3} - {\left (5 \, x^{2} + 19 \, x + 20\right )} \log \relax (5) + 9 \, x + e^{6} + 19\right )} \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.17, size = 58, normalized size = 1.81 \begin {gather*} 5 \, x^{2} \log \relax (5) \log \relax (x) - x^{2} \log \relax (x) - 2 \, x e^{3} \log \relax (x) + 19 \, x \log \relax (5) \log \relax (x) - 9 \, x \log \relax (x) - e^{6} \log \relax (x) - 8 \, e^{3} \log \relax (x) + 20 \, \log \relax (5) \log \relax (x) - 19 \, \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.05, size = 47, normalized size = 1.47
method | result | size |
norman | \(\left (-{\mathrm e}^{6}-8 \,{\mathrm e}^{3}+20 \ln \relax (5)-19\right ) \ln \relax (x )+\left (5 \ln \relax (5)-1\right ) x^{2} \ln \relax (x )+\left (-2 \,{\mathrm e}^{3}+19 \ln \relax (5)-9\right ) x \ln \relax (x )\) | \(47\) |
risch | \(\left (5 x^{2} \ln \relax (5)-2 x \,{\mathrm e}^{3}+19 x \ln \relax (5)-x^{2}-9 x \right ) \ln \relax (x )-8 \ln \relax (x ) {\mathrm e}^{3}-\ln \relax (x ) {\mathrm e}^{6}+20 \ln \relax (5) \ln \relax (x )-19 \ln \relax (x )\) | \(53\) |
default | \(10 \ln \relax (5) \left (\frac {x^{2} \ln \relax (x )}{2}-\frac {x^{2}}{4}\right )-2 \,{\mathrm e}^{3} \left (x \ln \relax (x )-x \right )+19 \ln \relax (5) \left (x \ln \relax (x )-x \right )-x^{2} \ln \relax (x )+\frac {5 x^{2} \ln \relax (5)}{2}-9 x \ln \relax (x )-\ln \relax (x ) {\mathrm e}^{6}-2 x \,{\mathrm e}^{3}+19 x \ln \relax (5)-8 \ln \relax (x ) {\mathrm e}^{3}+20 \ln \relax (5) \ln \relax (x )-19 \ln \relax (x )\) | \(96\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.35, size = 93, normalized size = 2.91 \begin {gather*} \frac {5}{2} \, x^{2} \log \relax (5) - x^{2} \log \relax (x) - 2 \, {\left (x \log \relax (x) - x\right )} e^{3} - 2 \, x e^{3} + \frac {5}{2} \, {\left (2 \, x^{2} \log \relax (x) - x^{2}\right )} \log \relax (5) + 19 \, {\left (x \log \relax (x) - x\right )} \log \relax (5) + 19 \, x \log \relax (5) - 9 \, x \log \relax (x) - e^{6} \log \relax (x) - 8 \, e^{3} \log \relax (x) + 20 \, \log \relax (5) \log \relax (x) - 19 \, \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.46, size = 44, normalized size = 1.38 \begin {gather*} x^2\,\ln \relax (x)\,\left (5\,\ln \relax (5)-1\right )-\ln \relax (x)\,\left (8\,{\mathrm {e}}^3+{\mathrm {e}}^6-20\,\ln \relax (5)+19\right )-x\,\ln \relax (x)\,\left (2\,{\mathrm {e}}^3-19\,\ln \relax (5)+9\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.16, size = 49, normalized size = 1.53 \begin {gather*} \left (- x^{2} + 5 x^{2} \log {\relax (5 )} - 2 x e^{3} - 9 x + 19 x \log {\relax (5 )}\right ) \log {\relax (x )} + \left (- e^{6} - 8 e^{3} - 19 + 20 \log {\relax (5 )}\right ) \log {\relax (x )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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