Optimal. Leaf size=19 \[ x+\frac {(-3+x)^2}{\left (5+e^x-\log (x)\right )^2} \]
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Rubi [F] time = 3.11, 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 {-18-83 x-15 e^{2 x} x-e^{3 x} x-12 x^2+e^x \left (-51 x-14 x^2+2 x^3\right )+\left (69 x+30 e^x x+3 e^{2 x} x+2 x^2\right ) \log (x)+\left (-15 x-3 e^x x\right ) \log ^2(x)+x \log ^3(x)}{-125 x-75 e^x x-15 e^{2 x} x-e^{3 x} x+\left (75 x+30 e^x x+3 e^{2 x} x\right ) \log (x)+\left (-15 x-3 e^x x\right ) \log ^2(x)+x \log ^3(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 {18+\left (83+51 e^x+15 e^{2 x}+e^{3 x}\right ) x+2 \left (6+7 e^x\right ) x^2-2 e^x x^3-x \left (69+30 e^x+3 e^{2 x}+2 x\right ) \log (x)+3 \left (5+e^x\right ) x \log ^2(x)-x \log ^3(x)}{x \left (5+e^x-\log (x)\right )^3} \, dx\\ &=\int \left (1-\frac {2 \left (12-7 x+x^2\right )}{\left (5+e^x-\log (x)\right )^2}-\frac {2 (-3+x)^2 (-1-5 x+x \log (x))}{x \left (5+e^x-\log (x)\right )^3}\right ) \, dx\\ &=x-2 \int \frac {12-7 x+x^2}{\left (5+e^x-\log (x)\right )^2} \, dx-2 \int \frac {(-3+x)^2 (-1-5 x+x \log (x))}{x \left (5+e^x-\log (x)\right )^3} \, dx\\ &=x-2 \int \left (\frac {12}{\left (5+e^x-\log (x)\right )^2}-\frac {7 x}{\left (5+e^x-\log (x)\right )^2}+\frac {x^2}{\left (5+e^x-\log (x)\right )^2}\right ) \, dx-2 \int \left (-\frac {6 (-1-5 x+x \log (x))}{\left (5+e^x-\log (x)\right )^3}+\frac {9 (-1-5 x+x \log (x))}{x \left (5+e^x-\log (x)\right )^3}+\frac {x (-1-5 x+x \log (x))}{\left (5+e^x-\log (x)\right )^3}\right ) \, dx\\ &=x-2 \int \frac {x^2}{\left (5+e^x-\log (x)\right )^2} \, dx-2 \int \frac {x (-1-5 x+x \log (x))}{\left (5+e^x-\log (x)\right )^3} \, dx+12 \int \frac {-1-5 x+x \log (x)}{\left (5+e^x-\log (x)\right )^3} \, dx+14 \int \frac {x}{\left (5+e^x-\log (x)\right )^2} \, dx-18 \int \frac {-1-5 x+x \log (x)}{x \left (5+e^x-\log (x)\right )^3} \, dx-24 \int \frac {1}{\left (5+e^x-\log (x)\right )^2} \, dx\\ &=x-2 \int \frac {x^2}{\left (5+e^x-\log (x)\right )^2} \, dx-2 \int \left (-\frac {x}{\left (5+e^x-\log (x)\right )^3}-\frac {5 x^2}{\left (5+e^x-\log (x)\right )^3}+\frac {x^2 \log (x)}{\left (5+e^x-\log (x)\right )^3}\right ) \, dx+12 \int \left (-\frac {1}{\left (5+e^x-\log (x)\right )^3}-\frac {5 x}{\left (5+e^x-\log (x)\right )^3}+\frac {x \log (x)}{\left (5+e^x-\log (x)\right )^3}\right ) \, dx+14 \int \frac {x}{\left (5+e^x-\log (x)\right )^2} \, dx-18 \int \left (-\frac {5}{\left (5+e^x-\log (x)\right )^3}-\frac {1}{x \left (5+e^x-\log (x)\right )^3}+\frac {\log (x)}{\left (5+e^x-\log (x)\right )^3}\right ) \, dx-24 \int \frac {1}{\left (5+e^x-\log (x)\right )^2} \, dx\\ &=x+2 \int \frac {x}{\left (5+e^x-\log (x)\right )^3} \, dx-2 \int \frac {x^2}{\left (5+e^x-\log (x)\right )^2} \, dx-2 \int \frac {x^2 \log (x)}{\left (5+e^x-\log (x)\right )^3} \, dx+10 \int \frac {x^2}{\left (5+e^x-\log (x)\right )^3} \, dx-12 \int \frac {1}{\left (5+e^x-\log (x)\right )^3} \, dx+12 \int \frac {x \log (x)}{\left (5+e^x-\log (x)\right )^3} \, dx+14 \int \frac {x}{\left (5+e^x-\log (x)\right )^2} \, dx+18 \int \frac {1}{x \left (5+e^x-\log (x)\right )^3} \, dx-18 \int \frac {\log (x)}{\left (5+e^x-\log (x)\right )^3} \, dx-24 \int \frac {1}{\left (5+e^x-\log (x)\right )^2} \, dx-60 \int \frac {x}{\left (5+e^x-\log (x)\right )^3} \, dx+90 \int \frac {1}{\left (5+e^x-\log (x)\right )^3} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.07, size = 19, normalized size = 1.00 \begin {gather*} x+\frac {(-3+x)^2}{\left (-5-e^x+\log (x)\right )^2} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.63, size = 67, normalized size = 3.53 \begin {gather*} -\frac {x \log \relax (x)^{2} + x^{2} + x e^{\left (2 \, x\right )} + 10 \, x e^{x} - 2 \, {\left (x e^{x} + 5 \, x\right )} \log \relax (x) + 19 \, x + 9}{2 \, {\left (e^{x} + 5\right )} \log \relax (x) - \log \relax (x)^{2} - e^{\left (2 \, x\right )} - 10 \, e^{x} - 25} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.42, size = 99, normalized size = 5.21 \begin {gather*} \frac {4 \, x e^{x} \log \relax (x) - 2 \, x \log \relax (x)^{2} - 2 \, x^{2} - 2 \, x e^{\left (2 \, x\right )} - 20 \, x e^{x} + 20 \, x \log \relax (x) + 6 \, e^{x} \log \relax (x) - 3 \, \log \relax (x)^{2} - 38 \, x - 3 \, e^{\left (2 \, x\right )} - 30 \, e^{x} + 30 \, \log \relax (x) - 93}{2 \, {\left (2 \, e^{x} \log \relax (x) - \log \relax (x)^{2} - e^{\left (2 \, x\right )} - 10 \, e^{x} + 10 \, \log \relax (x) - 25\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.03, size = 22, normalized size = 1.16
method | result | size |
risch | \(x +\frac {x^{2}-6 x +9}{\left (5+{\mathrm e}^{x}-\ln \relax (x )\right )^{2}}\) | \(22\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.77, size = 67, normalized size = 3.53 \begin {gather*} -\frac {x \log \relax (x)^{2} + x^{2} + x e^{\left (2 \, x\right )} - 2 \, {\left (x \log \relax (x) - 5 \, x\right )} e^{x} - 10 \, x \log \relax (x) + 19 \, x + 9}{2 \, {\left (\log \relax (x) - 5\right )} e^{x} - \log \relax (x)^{2} - e^{\left (2 \, x\right )} + 10 \, \log \relax (x) - 25} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int \frac {83\,x+15\,x\,{\mathrm {e}}^{2\,x}+x\,{\mathrm {e}}^{3\,x}-x\,{\ln \relax (x)}^3-\ln \relax (x)\,\left (69\,x+3\,x\,{\mathrm {e}}^{2\,x}+30\,x\,{\mathrm {e}}^x+2\,x^2\right )+{\ln \relax (x)}^2\,\left (15\,x+3\,x\,{\mathrm {e}}^x\right )+12\,x^2+{\mathrm {e}}^x\,\left (-2\,x^3+14\,x^2+51\,x\right )+18}{-x\,{\ln \relax (x)}^3+\left (15\,x+3\,x\,{\mathrm {e}}^x\right )\,{\ln \relax (x)}^2+\left (-75\,x-3\,x\,{\mathrm {e}}^{2\,x}-30\,x\,{\mathrm {e}}^x\right )\,\ln \relax (x)+125\,x+15\,x\,{\mathrm {e}}^{2\,x}+x\,{\mathrm {e}}^{3\,x}+75\,x\,{\mathrm {e}}^x} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.34, size = 36, normalized size = 1.89 \begin {gather*} x + \frac {x^{2} - 6 x + 9}{\left (10 - 2 \log {\relax (x )}\right ) e^{x} + e^{2 x} + \log {\relax (x )}^{2} - 10 \log {\relax (x )} + 25} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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