Optimal. Leaf size=30 \[ \frac {2}{-2+\frac {3+x}{2 \log \left (3+e^{1-x}+x-\log (2)\right )}} \]
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Rubi [F] time = 6.28, 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 {12+e^{1-x} (-12-4 x)+4 x+\left (-12-4 e^{1-x}-4 x+4 \log (2)\right ) \log \left (3+e^{1-x}+x-\log (2)\right )}{27+27 x+9 x^2+x^3+e^{1-x} \left (9+6 x+x^2\right )+\left (-9-6 x-x^2\right ) \log (2)+\left (-72+e^{1-x} (-24-8 x)-48 x-8 x^2+(24+8 x) \log (2)\right ) \log \left (3+e^{1-x}+x-\log (2)\right )+\left (48+16 e^{1-x}+16 x-16 \log (2)\right ) \log ^2\left (3+e^{1-x}+x-\log (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 {4 \left (-e+e^x\right ) (3+x)-4 \left (e+e^x (3+x-\log (2))\right ) \log \left (3+e^{1-x}+x-\log (2)\right )}{\left (e+e^x (3+x-\log (2))\right ) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2} \, dx\\ &=\int \left (\frac {4 e (3+x) (-4-x+\log (2))}{\left (e+e^x x+3 e^x \left (1-\frac {\log (2)}{3}\right )\right ) (3+x-\log (2)) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2}+\frac {4 \left (3+x-x \log \left (3+e^{1-x}+x-\log (2)\right )-3 \left (1-\frac {\log (2)}{3}\right ) \log \left (3+e^{1-x}+x-\log (2)\right )\right )}{(3+x-\log (2)) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2}\right ) \, dx\\ &=4 \int \frac {3+x-x \log \left (3+e^{1-x}+x-\log (2)\right )-3 \left (1-\frac {\log (2)}{3}\right ) \log \left (3+e^{1-x}+x-\log (2)\right )}{(3+x-\log (2)) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2} \, dx+(4 e) \int \frac {(3+x) (-4-x+\log (2))}{\left (e+e^x x+3 e^x \left (1-\frac {\log (2)}{3}\right )\right ) (3+x-\log (2)) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2} \, dx\\ &=4 \int \frac {3+x+(-3-x+\log (2)) \log \left (3+e^{1-x}+x-\log (2)\right )}{(3+x-\log (2)) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2} \, dx+(4 e) \int \frac {(3+x) (-4-x+\log (2))}{\left (e+e^x (3+x-\log (2))\right ) (3+x-\log (2)) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2} \, dx\\ &=4 \int \left (-\frac {(3+x) (-1+x-\log (2))}{4 (3+x-\log (2)) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2}+\frac {1}{4 \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )}\right ) \, dx+(4 e) \int \left (\frac {4}{\left (-e-e^x x-3 e^x \left (1-\frac {\log (2)}{3}\right )\right ) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2}+\frac {x}{\left (-e-e^x x-3 e^x \left (1-\frac {\log (2)}{3}\right )\right ) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2}+\frac {\log (2)}{\left (-e-e^x x-3 e^x \left (1-\frac {\log (2)}{3}\right )\right ) (3+x-\log (2)) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2}\right ) \, dx\\ &=(4 e) \int \frac {x}{\left (-e-e^x x-3 e^x \left (1-\frac {\log (2)}{3}\right )\right ) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2} \, dx+(16 e) \int \frac {1}{\left (-e-e^x x-3 e^x \left (1-\frac {\log (2)}{3}\right )\right ) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2} \, dx+(4 e \log (2)) \int \frac {1}{\left (-e-e^x x-3 e^x \left (1-\frac {\log (2)}{3}\right )\right ) (3+x-\log (2)) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2} \, dx-\int \frac {(3+x) (-1+x-\log (2))}{(3+x-\log (2)) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2} \, dx+\int \frac {1}{3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )} \, dx\\ &=(4 e) \int \frac {x}{\left (-e-e^x (3+x-\log (2))\right ) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2} \, dx+(16 e) \int \frac {1}{\left (-e-e^x (3+x-\log (2))\right ) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2} \, dx+(4 e \log (2)) \int \frac {1}{\left (-e-e^x (3+x-\log (2))\right ) (3+x-\log (2)) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2} \, dx-\int \left (-\frac {1}{\left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2}+\frac {x}{\left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2}-\frac {4 \log (2)}{(3+x-\log (2)) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2}\right ) \, dx+\int \frac {1}{3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )} \, dx\\ &=(4 e) \int \frac {x}{\left (-e-e^x (3+x-\log (2))\right ) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2} \, dx+(16 e) \int \frac {1}{\left (-e-e^x (3+x-\log (2))\right ) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2} \, dx+(4 \log (2)) \int \frac {1}{(3+x-\log (2)) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2} \, dx+(4 e \log (2)) \int \frac {1}{\left (-e-e^x (3+x-\log (2))\right ) (3+x-\log (2)) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2} \, dx+\int \frac {1}{\left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2} \, dx-\int \frac {x}{\left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2} \, dx+\int \frac {1}{3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.11, size = 26, normalized size = 0.87 \begin {gather*} \frac {3+x}{3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.08, size = 25, normalized size = 0.83 \begin {gather*} \frac {x + 3}{x - 4 \, \log \left (x + e^{\left (-x + 1\right )} - \log \relax (2) + 3\right ) + 3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.72, size = 25, normalized size = 0.83 \begin {gather*} \frac {x + 3}{x - 4 \, \log \left (x + e^{\left (-x + 1\right )} - \log \relax (2) + 3\right ) + 3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.04, size = 26, normalized size = 0.87
method | result | size |
risch | \(\frac {3+x}{x -4 \ln \left ({\mathrm e}^{1-x}-\ln \relax (2)+3+x \right )+3}\) | \(26\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.61, size = 27, normalized size = 0.90 \begin {gather*} \frac {x + 3}{5 \, x - 4 \, \log \left ({\left (x - \log \relax (2) + 3\right )} e^{x} + e\right ) + 3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F(-1)] time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \text {Hanged} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.34, size = 22, normalized size = 0.73 \begin {gather*} \frac {- x - 3}{- x + 4 \log {\left (x + e^{1 - x} - \log {\relax (2 )} + 3 \right )} - 3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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