Optimal. Leaf size=26 \[ e^3+\log (x)+\log \left (3+e+e^x+x-\frac {x}{5-e^2}\right ) \]
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Rubi [F] time = 0.86, 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 {-15-5 e-8 x+e^2 (3+e+2 x)+e^x \left (-5-5 x+e^2 (1+x)\right )}{-15 x-5 e x-4 x^2+e^x \left (-5 x+e^2 x\right )+e^2 \left (3 x+e x+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 {-15-5 e-8 x+e^2 (3+e+2 x)+e^x \left (-5-5 x+e^2 (1+x)\right )}{(-15-5 e) x-4 x^2+e^x \left (-5 x+e^2 x\right )+e^2 \left (3 x+e x+x^2\right )} \, dx\\ &=\int \frac {15 \left (1+\frac {e}{3}\right )+8 x-e^2 (3+e+2 x)-e^x \left (-5-5 x+e^2 (1+x)\right )}{-((-15-5 e) x)+4 x^2-e^x \left (-5 x+e^2 x\right )-e^2 \left (3 x+e x+x^2\right )} \, dx\\ &=\int \left (\frac {1+x}{x}+\frac {-11-5 e+2 e^2+e^3-\left (4-e^2\right ) x}{5 e^x \left (1-\frac {e^2}{5}\right )+15 \left (1-\frac {1}{15} e \left (-5+3 e+e^2\right )\right )+4 \left (1-\frac {e^2}{4}\right ) x}\right ) \, dx\\ &=\int \frac {1+x}{x} \, dx+\int \frac {-11-5 e+2 e^2+e^3-\left (4-e^2\right ) x}{5 e^x \left (1-\frac {e^2}{5}\right )+15 \left (1-\frac {1}{15} e \left (-5+3 e+e^2\right )\right )+4 \left (1-\frac {e^2}{4}\right ) x} \, dx\\ &=\int \left (1+\frac {1}{x}\right ) \, dx+\int \left (\frac {11 \left (1-\frac {1}{11} e \left (-5+2 e+e^2\right )\right )}{-5 e^x \left (1-\frac {e^2}{5}\right )-15 \left (1-\frac {1}{15} e \left (-5+3 e+e^2\right )\right )-4 \left (1-\frac {e^2}{4}\right ) x}+\frac {(-2-e) (2-e) x}{5 e^x \left (1-\frac {e^2}{5}\right )+15 \left (1-\frac {1}{15} e \left (-5+3 e+e^2\right )\right )+4 \left (1-\frac {e^2}{4}\right ) x}\right ) \, dx\\ &=x+\log (x)+((-2-e) (2-e)) \int \frac {x}{5 e^x \left (1-\frac {e^2}{5}\right )+15 \left (1-\frac {1}{15} e \left (-5+3 e+e^2\right )\right )+4 \left (1-\frac {e^2}{4}\right ) x} \, dx+\left (11+e \left (5-2 e-e^2\right )\right ) \int \frac {1}{-5 e^x \left (1-\frac {e^2}{5}\right )-15 \left (1-\frac {1}{15} e \left (-5+3 e+e^2\right )\right )-4 \left (1-\frac {e^2}{4}\right ) x} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.39, size = 40, normalized size = 1.54 \begin {gather*} \log (x)+\log \left (15+5 e-3 e^2-e^3+5 e^x-e^{2+x}+4 x-e^2 x\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.83, size = 28, normalized size = 1.08 \begin {gather*} \log \left ({\left (x + 3\right )} e^{2} + {\left (e^{2} - 5\right )} e^{x} - 4 \, x + e^{3} - 5 \, e - 15\right ) + \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.38, size = 36, normalized size = 1.38 \begin {gather*} \log \left (-x e^{2} + 4 \, x - e^{3} - 3 \, e^{2} + 5 \, e - e^{\left (x + 2\right )} + 5 \, e^{x} + 15\right ) + \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.07, size = 34, normalized size = 1.31
method | result | size |
risch | \(\ln \relax (x )+\ln \left ({\mathrm e}^{x}+\frac {{\mathrm e}^{3}+{\mathrm e}^{2} x +3 \,{\mathrm e}^{2}-5 \,{\mathrm e}-4 x -15}{{\mathrm e}^{2}-5}\right )\) | \(34\) |
norman | \(\ln \relax (x )+\ln \left ({\mathrm e} \,{\mathrm e}^{2}+{\mathrm e}^{2} {\mathrm e}^{x}+{\mathrm e}^{2} x +3 \,{\mathrm e}^{2}-5 \,{\mathrm e}-5 \,{\mathrm e}^{x}-4 x -15\right )\) | \(36\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.50, size = 36, normalized size = 1.38 \begin {gather*} \log \relax (x) + \log \left (\frac {x {\left (e^{2} - 4\right )} + {\left (e^{2} - 5\right )} e^{x} + e^{3} + 3 \, e^{2} - 5 \, e - 15}{e^{2} - 5}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.57, size = 36, normalized size = 1.38 \begin {gather*} \ln \left (4\,x-{\mathrm {e}}^{x+2}+5\,\mathrm {e}-3\,{\mathrm {e}}^2-{\mathrm {e}}^3+5\,{\mathrm {e}}^x-x\,{\mathrm {e}}^2+15\right )+\ln \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.27, size = 36, normalized size = 1.38 \begin {gather*} \log {\relax (x )} + \log {\left (\frac {- 4 x + x e^{2} - 15 - 5 e + e^{3} + 3 e^{2}}{-5 + e^{2}} + e^{x} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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