Optimal. Leaf size=30 \[ -4+\log \left (-3 \left (4+e^x\right )+\frac {\left (3+e^3+\frac {-3+x}{x}\right ) x}{4+x}\right ) \]
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Rubi [F] time = 1.27, 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 {-19-4 e^3+e^x \left (48+24 x+3 x^2\right )}{204+83 x+8 x^2+e^3 \left (-4 x-x^2\right )+e^x \left (48+24 x+3 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 {-19 \left (1+\frac {4 e^3}{19}\right )+e^x \left (48+24 x+3 x^2\right )}{(4+x) \left (51+12 e^x+3 e^x x+8 \left (1-\frac {e^3}{8}\right ) x\right )} \, dx\\ &=\int \left (1+\frac {-223-4 e^3-\left (83-4 e^3\right ) x-\left (8-e^3\right ) x^2}{(4+x) \left (51+12 e^x+3 e^x x+8 \left (1-\frac {e^3}{8}\right ) x\right )}\right ) \, dx\\ &=x+\int \frac {-223-4 e^3-\left (83-4 e^3\right ) x-\left (8-e^3\right ) x^2}{(4+x) \left (51+12 e^x+3 e^x x+8 \left (1-\frac {e^3}{8}\right ) x\right )} \, dx\\ &=x+\int \left (\frac {51}{-51-12 e^x-3 e^x x-8 \left (1-\frac {e^3}{8}\right ) x}+\frac {\left (-8+e^3\right ) x}{51+12 e^x+3 e^x x+8 \left (1-\frac {e^3}{8}\right ) x}+\frac {-19-4 e^3}{(4+x) \left (51+12 e^x+3 e^x x+8 \left (1-\frac {e^3}{8}\right ) x\right )}\right ) \, dx\\ &=x+51 \int \frac {1}{-51-12 e^x-3 e^x x-8 \left (1-\frac {e^3}{8}\right ) x} \, dx+\left (-19-4 e^3\right ) \int \frac {1}{(4+x) \left (51+12 e^x+3 e^x x+8 \left (1-\frac {e^3}{8}\right ) x\right )} \, dx+\left (-8+e^3\right ) \int \frac {x}{51+12 e^x+3 e^x x+8 \left (1-\frac {e^3}{8}\right ) x} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.32, size = 30, normalized size = 1.00 \begin {gather*} -\log (4+x)+\log \left (51+12 e^x+8 x-e^3 x+3 e^x x\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.03, size = 24, normalized size = 0.80 \begin {gather*} \log \left (-\frac {x e^{3} - 3 \, {\left (x + 4\right )} e^{x} - 8 \, x - 51}{x + 4}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.22, size = 26, normalized size = 0.87 \begin {gather*} \log \left (x e^{3} - 3 \, x e^{x} - 8 \, x - 12 \, e^{x} - 51\right ) - \log \left (x + 4\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.10, size = 21, normalized size = 0.70
method | result | size |
risch | \(\ln \left ({\mathrm e}^{x}-\frac {x \,{\mathrm e}^{3}-8 x -51}{3 \left (4+x \right )}\right )\) | \(21\) |
norman | \(-\ln \left (4+x \right )+\ln \left (x \,{\mathrm e}^{3}-3 \,{\mathrm e}^{x} x -8 x -12 \,{\mathrm e}^{x}-51\right )\) | \(27\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.61, size = 23, normalized size = 0.77 \begin {gather*} \log \left (-\frac {x {\left (e^{3} - 8\right )} - 3 \, {\left (x + 4\right )} e^{x} - 51}{3 \, {\left (x + 4\right )}}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.44, size = 27, normalized size = 0.90 \begin {gather*} \ln \left (8\,x+12\,{\mathrm {e}}^x-x\,{\mathrm {e}}^3+3\,x\,{\mathrm {e}}^x+51\right )-\ln \left (x+4\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.37, size = 19, normalized size = 0.63 \begin {gather*} \log {\left (e^{x} + \frac {- x e^{3} + 8 x + 51}{3 x + 12} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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