Optimal. Leaf size=25 \[ \frac {-8+x+\log (5)}{4-\frac {e^x}{2}-x}+3 \log (x) \]
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Rubi [F] time = 1.03, 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 {192+3 e^{2 x}-112 x+12 x^2+4 x \log (5)+e^x \left (-48-6 x+2 x^2+2 x \log (5)\right )}{64 x+e^{2 x} x-32 x^2+4 x^3+e^x \left (-16 x+4 x^2\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {192+3 e^{2 x}+12 x^2+x (-112+4 \log (5))+e^x \left (-48-6 x+2 x^2+2 x \log (5)\right )}{64 x+e^{2 x} x-32 x^2+4 x^3+e^x \left (-16 x+4 x^2\right )} \, dx\\ &=\int \frac {3 e^{2 x}+4 \left (48+3 x^2+x (-28+\log (5))\right )+2 e^x \left (-24+x^2+x (-3+\log (5))\right )}{\left (8-e^x-2 x\right )^2 x} \, dx\\ &=\int \left (\frac {3}{x}+\frac {2 (-9+x+\log (5))}{-8+e^x+2 x}-\frac {4 (-5+x) (-8+x+\log (5))}{\left (-8+e^x+2 x\right )^2}\right ) \, dx\\ &=3 \log (x)+2 \int \frac {-9+x+\log (5)}{-8+e^x+2 x} \, dx-4 \int \frac {(-5+x) (-8+x+\log (5))}{\left (-8+e^x+2 x\right )^2} \, dx\\ &=3 \log (x)+2 \int \left (\frac {x}{-8+e^x+2 x}-\frac {9 \left (1-\frac {\log (5)}{9}\right )}{-8+e^x+2 x}\right ) \, dx-4 \int \left (\frac {x^2}{\left (-8+e^x+2 x\right )^2}+\frac {x (-13+\log (5))}{\left (-8+e^x+2 x\right )^2}-\frac {5 (-8+\log (5))}{\left (-8+e^x+2 x\right )^2}\right ) \, dx\\ &=3 \log (x)+2 \int \frac {x}{-8+e^x+2 x} \, dx-4 \int \frac {x^2}{\left (-8+e^x+2 x\right )^2} \, dx-(20 (8-\log (5))) \int \frac {1}{\left (-8+e^x+2 x\right )^2} \, dx-(2 (9-\log (5))) \int \frac {1}{-8+e^x+2 x} \, dx-(4 (-13+\log (5))) \int \frac {x}{\left (-8+e^x+2 x\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.26, size = 22, normalized size = 0.88 \begin {gather*} -\frac {2 (-8+x+\log (5))}{-8+e^x+2 x}+3 \log (x) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.71, size = 30, normalized size = 1.20 \begin {gather*} \frac {3 \, {\left (2 \, x + e^{x} - 8\right )} \log \relax (x) - 2 \, x - 2 \, \log \relax (5) + 16}{2 \, x + e^{x} - 8} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.15, size = 34, normalized size = 1.36 \begin {gather*} \frac {6 \, x \log \relax (x) + 3 \, e^{x} \log \relax (x) - 2 \, x - 2 \, \log \relax (5) - 24 \, \log \relax (x) + 16}{2 \, x + e^{x} - 8} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.09, size = 22, normalized size = 0.88
method | result | size |
risch | \(3 \ln \relax (x )-\frac {2 \left (x +\ln \relax (5)-8\right )}{-8+{\mathrm e}^{x}+2 x}\) | \(22\) |
norman | \(\frac {{\mathrm e}^{x}+8-2 \ln \relax (5)}{-8+{\mathrm e}^{x}+2 x}+3 \ln \relax (x )\) | \(24\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.51, size = 21, normalized size = 0.84 \begin {gather*} -\frac {2 \, {\left (x + \log \relax (5) - 8\right )}}{2 \, x + e^{x} - 8} + 3 \, \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.15, size = 34, normalized size = 1.36 \begin {gather*} 3\,\ln \relax (x)-\frac {x\,\left (\frac {\ln \relax (5)}{2}-2\right )+{\mathrm {e}}^x\,\left (\frac {\ln \relax (5)}{4}-2\right )}{2\,x+{\mathrm {e}}^x-8} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.13, size = 22, normalized size = 0.88 \begin {gather*} \frac {- 2 x - 2 \log {\relax (5 )} + 16}{2 x + e^{x} - 8} + 3 \log {\relax (x )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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