Optimal. Leaf size=24 \[ \frac {1}{2} \left (\frac {4 (-1+x)}{15 \left (-16+e^3\right )}+2 x\right ) \log (x) \]
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Rubi [A] time = 0.03, antiderivative size = 39, normalized size of antiderivative = 1.62, number of steps used = 8, number of rules used = 5, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {6, 12, 14, 43, 2295} \begin {gather*} \frac {\left (238-15 e^3\right ) x \log (x)}{15 \left (16-e^3\right )}+\frac {2 \log (x)}{15 \left (16-e^3\right )} \end {gather*}
Antiderivative was successfully verified.
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Rule 6
Rule 12
Rule 14
Rule 43
Rule 2295
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-2-238 x+15 e^3 x+\left (-238 x+15 e^3 x\right ) \log (x)}{\left (-240+15 e^3\right ) x} \, dx\\ &=\int \frac {-2+\left (-238+15 e^3\right ) x+\left (-238 x+15 e^3 x\right ) \log (x)}{\left (-240+15 e^3\right ) x} \, dx\\ &=-\frac {\int \frac {-2+\left (-238+15 e^3\right ) x+\left (-238 x+15 e^3 x\right ) \log (x)}{x} \, dx}{15 \left (16-e^3\right )}\\ &=-\frac {\int \left (\frac {-2-\left (238-15 e^3\right ) x}{x}+\left (-238+15 e^3\right ) \log (x)\right ) \, dx}{15 \left (16-e^3\right )}\\ &=-\frac {\int \frac {-2-\left (238-15 e^3\right ) x}{x} \, dx}{15 \left (16-e^3\right )}+\frac {\left (238-15 e^3\right ) \int \log (x) \, dx}{15 \left (16-e^3\right )}\\ &=-\frac {\left (238-15 e^3\right ) x}{15 \left (16-e^3\right )}+\frac {\left (238-15 e^3\right ) x \log (x)}{15 \left (16-e^3\right )}-\frac {\int \left (-238+15 e^3-\frac {2}{x}\right ) \, dx}{15 \left (16-e^3\right )}\\ &=\frac {2 \log (x)}{15 \left (16-e^3\right )}+\frac {\left (238-15 e^3\right ) x \log (x)}{15 \left (16-e^3\right )}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.02, size = 26, normalized size = 1.08 \begin {gather*} \frac {2 \log (x)+\left (238-15 e^3\right ) x \log (x)}{240-15 e^3} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.50, size = 20, normalized size = 0.83 \begin {gather*} \frac {{\left (15 \, x e^{3} - 238 \, x - 2\right )} \log \relax (x)}{15 \, {\left (e^{3} - 16\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.20, size = 25, normalized size = 1.04 \begin {gather*} \frac {15 \, x e^{3} \log \relax (x) - 238 \, x \log \relax (x) - 2 \, \log \relax (x)}{15 \, {\left (e^{3} - 16\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 29, normalized size = 1.21
method | result | size |
norman | \(-\frac {2 \ln \relax (x )}{15 \left ({\mathrm e}^{3}-16\right )}+\frac {\left (15 \,{\mathrm e}^{3}-238\right ) x \ln \relax (x )}{15 \,{\mathrm e}^{3}-240}\) | \(29\) |
risch | \(-\frac {2 \ln \relax (x )}{15 \left ({\mathrm e}^{3}-16\right )}+\frac {\left (15 \,{\mathrm e}^{3}-238\right ) x \ln \relax (x )}{15 \,{\mathrm e}^{3}-240}\) | \(29\) |
default | \(\frac {{\mathrm e}^{3} \left (x \ln \relax (x )-x \right )}{{\mathrm e}^{3}-16}-\frac {238 \left (x \ln \relax (x )-x \right )}{15 \left ({\mathrm e}^{3}-16\right )}+\frac {{\mathrm e}^{3} x}{{\mathrm e}^{3}-16}-\frac {238 x}{15 \left ({\mathrm e}^{3}-16\right )}-\frac {2 \ln \relax (x )}{15 \left ({\mathrm e}^{3}-16\right )}\) | \(64\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.42, size = 62, normalized size = 2.58 \begin {gather*} {\left (\frac {x \log \relax (x)}{e^{3} - 16} - \frac {x}{e^{3} - 16}\right )} e^{3} + \frac {x e^{3}}{e^{3} - 16} - \frac {238 \, x \log \relax (x)}{15 \, {\left (e^{3} - 16\right )}} - \frac {2 \, \log \left (x e^{3} - 16 \, x\right )}{15 \, {\left (e^{3} - 16\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.32, size = 22, normalized size = 0.92 \begin {gather*} -\frac {\ln \relax (x)\,\left (238\,x-15\,x\,{\mathrm {e}}^3+2\right )}{15\,\left ({\mathrm {e}}^3-16\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.14, size = 31, normalized size = 1.29 \begin {gather*} \frac {\left (- 238 x + 15 x e^{3}\right ) \log {\relax (x )}}{-240 + 15 e^{3}} - \frac {2 \log {\relax (x )}}{-240 + 15 e^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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