Optimal. Leaf size=18 \[ 4+x+\frac {\log (x)}{2+5 \left (-1+e^x\right ) x} \]
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Rubi [F] time = 2.02, 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 {2-x-20 x^2+25 x^3+25 e^{2 x} x^3+e^x \left (5 x+20 x^2-50 x^3\right )+\left (5 x+e^x \left (-5 x-5 x^2\right )\right ) \log (x)}{4 x-20 x^2+25 x^3+25 e^{2 x} x^3+e^x \left (20 x^2-50 x^3\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 {2+\left (-1+5 e^x\right ) x+20 \left (-1+e^x\right ) x^2+25 \left (-1+e^x\right )^2 x^3-5 x \left (-1+e^x (1+x)\right ) \log (x)}{x \left (2+5 \left (-1+e^x\right ) x\right )^2} \, dx\\ &=\int \left (1-\frac {\left (-2-2 x+5 x^2\right ) \log (x)}{x \left (2-5 x+5 e^x x\right )^2}-\frac {-1+\log (x)+x \log (x)}{x \left (2-5 x+5 e^x x\right )}\right ) \, dx\\ &=x-\int \frac {\left (-2-2 x+5 x^2\right ) \log (x)}{x \left (2-5 x+5 e^x x\right )^2} \, dx-\int \frac {-1+\log (x)+x \log (x)}{x \left (2-5 x+5 e^x x\right )} \, dx\\ &=x+(2 \log (x)) \int \frac {1}{\left (2-5 x+5 e^x x\right )^2} \, dx+(2 \log (x)) \int \frac {1}{x \left (2-5 x+5 e^x x\right )^2} \, dx-(5 \log (x)) \int \frac {x}{\left (2-5 x+5 e^x x\right )^2} \, dx-\int \left (-\frac {1}{x \left (2-5 x+5 e^x x\right )}+\frac {\log (x)}{2-5 x+5 e^x x}+\frac {\log (x)}{x \left (2-5 x+5 e^x x\right )}\right ) \, dx+\int \frac {-2 \int \frac {1}{\left (2+5 \left (-1+e^x\right ) x\right )^2} \, dx-2 \int \frac {1}{x \left (2+5 \left (-1+e^x\right ) x\right )^2} \, dx+5 \int \frac {x}{\left (2+5 \left (-1+e^x\right ) x\right )^2} \, dx}{x} \, dx\\ &=x+(2 \log (x)) \int \frac {1}{\left (2-5 x+5 e^x x\right )^2} \, dx+(2 \log (x)) \int \frac {1}{x \left (2-5 x+5 e^x x\right )^2} \, dx-(5 \log (x)) \int \frac {x}{\left (2-5 x+5 e^x x\right )^2} \, dx+\int \frac {1}{x \left (2-5 x+5 e^x x\right )} \, dx-\int \frac {\log (x)}{2-5 x+5 e^x x} \, dx-\int \frac {\log (x)}{x \left (2-5 x+5 e^x x\right )} \, dx+\int \left (-\frac {2 \left (\int \frac {1}{\left (2+5 \left (-1+e^x\right ) x\right )^2} \, dx+\int \frac {1}{x \left (2+5 \left (-1+e^x\right ) x\right )^2} \, dx\right )}{x}+\frac {5 \int \frac {x}{\left (2+5 \left (-1+e^x\right ) x\right )^2} \, dx}{x}\right ) \, dx\\ &=x-2 \int \frac {\int \frac {1}{\left (2+5 \left (-1+e^x\right ) x\right )^2} \, dx+\int \frac {1}{x \left (2+5 \left (-1+e^x\right ) x\right )^2} \, dx}{x} \, dx+5 \int \frac {\int \frac {x}{\left (2+5 \left (-1+e^x\right ) x\right )^2} \, dx}{x} \, dx-\log (x) \int \frac {1}{2-5 x+5 e^x x} \, dx-\log (x) \int \frac {1}{x \left (2-5 x+5 e^x x\right )} \, dx+(2 \log (x)) \int \frac {1}{\left (2-5 x+5 e^x x\right )^2} \, dx+(2 \log (x)) \int \frac {1}{x \left (2-5 x+5 e^x x\right )^2} \, dx-(5 \log (x)) \int \frac {x}{\left (2-5 x+5 e^x x\right )^2} \, dx+\int \frac {1}{x \left (2-5 x+5 e^x x\right )} \, dx+\int \frac {\int \frac {1}{2+5 \left (-1+e^x\right ) x} \, dx}{x} \, dx+\int \frac {\int \frac {1}{x \left (2+5 \left (-1+e^x\right ) x\right )} \, dx}{x} \, dx\\ &=x-2 \int \left (\frac {\int \frac {1}{\left (2+5 \left (-1+e^x\right ) x\right )^2} \, dx}{x}+\frac {\int \frac {1}{x \left (2+5 \left (-1+e^x\right ) x\right )^2} \, dx}{x}\right ) \, dx+5 \int \frac {\int \frac {x}{\left (2+5 \left (-1+e^x\right ) x\right )^2} \, dx}{x} \, dx-\log (x) \int \frac {1}{2-5 x+5 e^x x} \, dx-\log (x) \int \frac {1}{x \left (2-5 x+5 e^x x\right )} \, dx+(2 \log (x)) \int \frac {1}{\left (2-5 x+5 e^x x\right )^2} \, dx+(2 \log (x)) \int \frac {1}{x \left (2-5 x+5 e^x x\right )^2} \, dx-(5 \log (x)) \int \frac {x}{\left (2-5 x+5 e^x x\right )^2} \, dx+\int \frac {1}{x \left (2-5 x+5 e^x x\right )} \, dx+\int \frac {\int \frac {1}{2+5 \left (-1+e^x\right ) x} \, dx}{x} \, dx+\int \frac {\int \frac {1}{x \left (2+5 \left (-1+e^x\right ) x\right )} \, dx}{x} \, dx\\ &=x-2 \int \frac {\int \frac {1}{\left (2+5 \left (-1+e^x\right ) x\right )^2} \, dx}{x} \, dx-2 \int \frac {\int \frac {1}{x \left (2+5 \left (-1+e^x\right ) x\right )^2} \, dx}{x} \, dx+5 \int \frac {\int \frac {x}{\left (2+5 \left (-1+e^x\right ) x\right )^2} \, dx}{x} \, dx-\log (x) \int \frac {1}{2-5 x+5 e^x x} \, dx-\log (x) \int \frac {1}{x \left (2-5 x+5 e^x x\right )} \, dx+(2 \log (x)) \int \frac {1}{\left (2-5 x+5 e^x x\right )^2} \, dx+(2 \log (x)) \int \frac {1}{x \left (2-5 x+5 e^x x\right )^2} \, dx-(5 \log (x)) \int \frac {x}{\left (2-5 x+5 e^x x\right )^2} \, dx+\int \frac {1}{x \left (2-5 x+5 e^x x\right )} \, dx+\int \frac {\int \frac {1}{2+5 \left (-1+e^x\right ) x} \, dx}{x} \, dx+\int \frac {\int \frac {1}{x \left (2+5 \left (-1+e^x\right ) x\right )} \, dx}{x} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.42, size = 18, normalized size = 1.00 \begin {gather*} x+\frac {\log (x)}{2-5 x+5 e^x x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.54, size = 31, normalized size = 1.72 \begin {gather*} \frac {5 \, x^{2} e^{x} - 5 \, x^{2} + 2 \, x + \log \relax (x)}{5 \, x e^{x} - 5 \, x + 2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 3.68, size = 31, normalized size = 1.72 \begin {gather*} \frac {5 \, x^{2} e^{x} - 5 \, x^{2} + 2 \, x + \log \relax (x)}{5 \, x e^{x} - 5 \, x + 2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.03, size = 18, normalized size = 1.00
method | result | size |
risch | \(\frac {\ln \relax (x )}{5 \,{\mathrm e}^{x} x -5 x +2}+x\) | \(18\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.80, size = 31, normalized size = 1.72 \begin {gather*} \frac {5 \, x^{2} e^{x} - 5 \, x^{2} + 2 \, x + \log \relax (x)}{5 \, x e^{x} - 5 \, x + 2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.30, size = 17, normalized size = 0.94 \begin {gather*} x+\frac {\ln \relax (x)}{5\,x\,{\mathrm {e}}^x-5\,x+2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.29, size = 15, normalized size = 0.83 \begin {gather*} x + \frac {\log {\relax (x )}}{5 x e^{x} - 5 x + 2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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