Optimal. Leaf size=26 \[ -2+\frac {\log (5) \left (-2-x \left (4 \left (5+e^x\right )+2 x\right ) \log (x)\right )}{x} \]
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Rubi [A] time = 0.07, antiderivative size = 30, normalized size of antiderivative = 1.15, number of steps used = 8, number of rules used = 3, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.064, Rules used = {14, 2288, 2295} \begin {gather*} -4 e^x \log (5) \log (x)-2 x \log (5) \log (x)-20 \log (5) \log (x)-\frac {2 \log (5)}{x} \end {gather*}
Antiderivative was successfully verified.
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Rule 14
Rule 2288
Rule 2295
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {4 e^x \log (5) (1+x \log (x))}{x}-\frac {2 \log (5) \left (-1+10 x+x^2+x^2 \log (x)\right )}{x^2}\right ) \, dx\\ &=-\left ((2 \log (5)) \int \frac {-1+10 x+x^2+x^2 \log (x)}{x^2} \, dx\right )-(4 \log (5)) \int \frac {e^x (1+x \log (x))}{x} \, dx\\ &=-4 e^x \log (5) \log (x)-(2 \log (5)) \int \left (\frac {-1+10 x+x^2}{x^2}+\log (x)\right ) \, dx\\ &=-4 e^x \log (5) \log (x)-(2 \log (5)) \int \frac {-1+10 x+x^2}{x^2} \, dx-(2 \log (5)) \int \log (x) \, dx\\ &=2 x \log (5)-4 e^x \log (5) \log (x)-2 x \log (5) \log (x)-(2 \log (5)) \int \left (1-\frac {1}{x^2}+\frac {10}{x}\right ) \, dx\\ &=-\frac {2 \log (5)}{x}-20 \log (5) \log (x)-4 e^x \log (5) \log (x)-2 x \log (5) \log (x)\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.05, size = 19, normalized size = 0.73 \begin {gather*} -2 \log (5) \left (\frac {1}{x}+\left (10+2 e^x+x\right ) \log (x)\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.03, size = 29, normalized size = 1.12 \begin {gather*} -\frac {2 \, {\left ({\left (2 \, x e^{x} \log \relax (5) + {\left (x^{2} + 10 \, x\right )} \log \relax (5)\right )} \log \relax (x) + \log \relax (5)\right )}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.13, size = 32, normalized size = 1.23 \begin {gather*} -\frac {2 \, {\left (x^{2} \log \relax (5) \log \relax (x) + 2 \, x e^{x} \log \relax (5) \log \relax (x) + 10 \, x \log \relax (5) \log \relax (x) + \log \relax (5)\right )}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 30, normalized size = 1.15
method | result | size |
default | \(-4 \ln \relax (5) {\mathrm e}^{x} \ln \relax (x )-2 x \ln \relax (5) \ln \relax (x )-20 \ln \relax (5) \ln \relax (x )-\frac {2 \ln \relax (5)}{x}\) | \(30\) |
risch | \(\left (-2 x \ln \relax (5)-4 \,{\mathrm e}^{x} \ln \relax (5)\right ) \ln \relax (x )-\frac {2 \ln \relax (5) \left (10 x \ln \relax (x )+1\right )}{x}\) | \(31\) |
norman | \(\frac {-20 x \ln \relax (5) \ln \relax (x )-2 x^{2} \ln \relax (5) \ln \relax (x )-4 x \ln \relax (5) {\mathrm e}^{x} \ln \relax (x )-2 \ln \relax (5)}{x}\) | \(35\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.39, size = 39, normalized size = 1.50 \begin {gather*} -4 \, e^{x} \log \relax (5) \log \relax (x) - 2 \, {\left (x \log \relax (x) - x\right )} \log \relax (5) - 2 \, x \log \relax (5) - 20 \, \log \relax (5) \log \relax (x) - \frac {2 \, \log \relax (5)}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.27, size = 30, normalized size = 1.15 \begin {gather*} -2\,\ln \relax (5)\,\left (10\,\ln \relax (x)+2\,{\mathrm {e}}^x\,\ln \relax (x)\right )-\frac {2\,\ln \relax (5)}{x}-2\,x\,\ln \relax (5)\,\ln \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.45, size = 37, normalized size = 1.42 \begin {gather*} - 2 x \log {\relax (5 )} \log {\relax (x )} - 4 e^{x} \log {\relax (5 )} \log {\relax (x )} - 20 \log {\relax (5 )} \log {\relax (x )} - \frac {2 \log {\relax (5 )}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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