Optimal. Leaf size=23 \[ \log \left (\frac {16 \left (2-(-5+x)^2\right )}{\left (4+e^x+x \log (x)\right )^2}\right ) \]
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Rubi [F]
time = 0.61, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps
used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {}
\begin {gather*} \int \frac {-86+28 x-2 x^2+e^x \left (-56+22 x-2 x^2\right )+(-46+10 x) \log (x)}{92-40 x+4 x^2+e^x \left (23-10 x+x^2\right )+\left (23 x-10 x^2+x^3\right ) \log (x)} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-86+28 x-2 x^2+e^x \left (-56+22 x-2 x^2\right )+(-46+10 x) \log (x)}{\left (23-10 x+x^2\right ) \left (4+e^x+x \log (x)\right )} \, dx\\ &=\int \left (-\frac {2 \left (28-11 x+x^2\right )}{23-10 x+x^2}+\frac {2 (3-\log (x)+x \log (x))}{4+e^x+x \log (x)}\right ) \, dx\\ &=-\left (2 \int \frac {28-11 x+x^2}{23-10 x+x^2} \, dx\right )+2 \int \frac {3-\log (x)+x \log (x)}{4+e^x+x \log (x)} \, dx\\ &=-\left (2 \int \left (1+\frac {5-x}{23-10 x+x^2}\right ) \, dx\right )+2 \int \left (\frac {3}{4+e^x+x \log (x)}-\frac {\log (x)}{4+e^x+x \log (x)}+\frac {x \log (x)}{4+e^x+x \log (x)}\right ) \, dx\\ &=-2 x-2 \int \frac {5-x}{23-10 x+x^2} \, dx-2 \int \frac {\log (x)}{4+e^x+x \log (x)} \, dx+2 \int \frac {x \log (x)}{4+e^x+x \log (x)} \, dx+6 \int \frac {1}{4+e^x+x \log (x)} \, dx\\ &=-2 x+\log \left (23-10 x+x^2\right )-2 \int \frac {\log (x)}{4+e^x+x \log (x)} \, dx+2 \int \frac {x \log (x)}{4+e^x+x \log (x)} \, dx+6 \int \frac {1}{4+e^x+x \log (x)} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.11, size = 26, normalized size = 1.13 \begin {gather*} -2 \left (-\frac {1}{2} \log \left (23-10 x+x^2\right )+\log \left (4+e^x+x \log (x)\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.08, size = 22, normalized size = 0.96
method | result | size |
norman | \(-2 \ln \left ({\mathrm e}^{x}+x \ln \left (x \right )+4\right )+\ln \left (x^{2}-10 x +23\right )\) | \(22\) |
risch | \(-2 \ln \left (x \right )+\ln \left (x^{2}-10 x +23\right )-2 \ln \left (\ln \left (x \right )+\frac {{\mathrm e}^{x}+4}{x}\right )\) | \(29\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.32, size = 21, normalized size = 0.91 \begin {gather*} \log \left (x^{2} - 10 \, x + 23\right ) - 2 \, \log \left (x \log \left (x\right ) + e^{x} + 4\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.33, size = 29, normalized size = 1.26 \begin {gather*} \log \left (x^{2} - 10 \, x + 23\right ) - 2 \, \log \left (x\right ) - 2 \, \log \left (\frac {x \log \left (x\right ) + e^{x} + 4}{x}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.10, size = 22, normalized size = 0.96 \begin {gather*} \log {\left (x^{2} - 10 x + 23 \right )} - 2 \log {\left (x \log {\left (x \right )} + e^{x} + 4 \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.44, size = 21, normalized size = 0.91 \begin {gather*} \log \left (x^{2} - 10 \, x + 23\right ) - 2 \, \log \left (x \log \left (x\right ) + e^{x} + 4\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 0.81, size = 29, normalized size = 1.26 \begin {gather*} \ln \left (x^2-10\,x+23\right )-2\,\ln \left (\frac {{\mathrm {e}}^x+x\,\ln \left (x\right )+4}{x}\right )-2\,\ln \left (x\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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