Optimal. Leaf size=27 \[ \log \left (3+e^2+x+\left (4-e^x\right ) x-\frac {4 x}{4-\log (4)}\right ) \]
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Rubi [F] time = 2.79, 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 {16-5 \log (4)+e^x (-4-4 x+(1+x) \log (4))}{12+4 e^2+16 x+\left (-3-e^2-5 x\right ) \log (4)+e^x (-4 x+x \log (4))} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {16 \left (1-\frac {5 \log (2)}{8}\right )+e^x (-4-4 x+(1+x) \log (4))}{12 \left (1+\frac {e^2}{3}\right )+16 x+\left (-3-e^2-5 x\right ) \log (4)+e^x (-4 x+x \log (4))} \, dx\\ &=\int \left (\frac {1+x}{x}+\frac {-12-x^2 (16-5 \log (4))-e^2 (4-\log (4))-x \left (12+e^2 (4-\log (4))-\log (64)\right )+\log (64)}{x \left (16 x \left (1-\frac {5 \log (2)}{8}\right )-4 e^x x \left (1-\frac {\log (2)}{2}\right )+12 \left (1+\frac {1}{12} \left (-e^2 (-4+\log (4))-3 \log (4)\right )\right )\right )}\right ) \, dx\\ &=\int \frac {1+x}{x} \, dx+\int \frac {-12-x^2 (16-5 \log (4))-e^2 (4-\log (4))-x \left (12+e^2 (4-\log (4))-\log (64)\right )+\log (64)}{x \left (16 x \left (1-\frac {5 \log (2)}{8}\right )-4 e^x x \left (1-\frac {\log (2)}{2}\right )+12 \left (1+\frac {1}{12} \left (-e^2 (-4+\log (4))-3 \log (4)\right )\right )\right )} \, dx\\ &=\int \left (1+\frac {1}{x}\right ) \, dx+\int \left (\frac {x (-16+5 \log (4))}{16 x \left (1-\frac {5 \log (2)}{8}\right )-4 e^x x \left (1-\frac {\log (2)}{2}\right )+12 \left (1+\frac {1}{12} \left (-e^2 (-4+\log (4))-3 \log (4)\right )\right )}+\frac {12 \left (1+\frac {1}{12} \left (-e^2 (-4+\log (4))-\log (64)\right )\right )}{-16 x \left (1-\frac {5 \log (2)}{8}\right )+4 e^x x \left (1-\frac {\log (2)}{2}\right )-12 \left (1+\frac {1}{12} \left (-e^2 (-4+\log (4))-\log (64)\right )\right )}+\frac {-12-e^2 (4-\log (4))+\log (64)}{x \left (16 x \left (1-\frac {5 \log (2)}{8}\right )-4 e^x x \left (1-\frac {\log (2)}{2}\right )+12 \left (1+\frac {1}{12} \left (-e^2 (-4+\log (4))-3 \log (4)\right )\right )\right )}\right ) \, dx\\ &=x+\log (x)+(-16+5 \log (4)) \int \frac {x}{16 x \left (1-\frac {5 \log (2)}{8}\right )-4 e^x x \left (1-\frac {\log (2)}{2}\right )+12 \left (1+\frac {1}{12} \left (-e^2 (-4+\log (4))-3 \log (4)\right )\right )} \, dx+\left (12+e^2 (4-\log (4))-\log (64)\right ) \int \frac {1}{-16 x \left (1-\frac {5 \log (2)}{8}\right )+4 e^x x \left (1-\frac {\log (2)}{2}\right )-12 \left (1+\frac {1}{12} \left (-e^2 (-4+\log (4))-\log (64)\right )\right )} \, dx+\left (-12-e^2 (4-\log (4))+\log (64)\right ) \int \frac {1}{x \left (16 x \left (1-\frac {5 \log (2)}{8}\right )-4 e^x x \left (1-\frac {\log (2)}{2}\right )+12 \left (1+\frac {1}{12} \left (-e^2 (-4+\log (4))-3 \log (4)\right )\right )\right )} \, dx\\ &=x+\log (x)+(-16+5 \log (4)) \int \frac {x}{-3 \left (1+\frac {e^2}{3}\right ) (-4+\log (4))+e^x x (-4+\log (4))-2 x (-8+\log (32))} \, dx+\left (12+e^2 (4-\log (4))-\log (64)\right ) \int \frac {1}{-16 x \left (1-\frac {5 \log (2)}{8}\right )+4 e^x x \left (1-\frac {\log (2)}{2}\right )-12 \left (1+\frac {1}{12} \left (-e^2 (-4+\log (4))-\log (64)\right )\right )} \, dx+\left (-12-e^2 (4-\log (4))+\log (64)\right ) \int \frac {1}{x \left (-3 \left (1+\frac {e^2}{3}\right ) (-4+\log (4))+e^x x (-4+\log (4))-2 x (-8+\log (32))\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 1.07, size = 40, normalized size = 1.48 \begin {gather*} \log \left (12+4 e^2+16 x-4 e^x x-3 \log (4)-e^2 \log (4)-5 x \log (4)+e^x x \log (4)\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.70, size = 39, normalized size = 1.44 \begin {gather*} \log \relax (x) + \log \left (\frac {{\left (x \log \relax (2) - 2 \, x\right )} e^{x} - {\left (5 \, x + e^{2} + 3\right )} \log \relax (2) + 8 \, x + 2 \, e^{2} + 6}{x}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.22, size = 36, normalized size = 1.33 \begin {gather*} \log \left (x e^{x} \log \relax (2) - 2 \, x e^{x} - 5 \, x \log \relax (2) - e^{2} \log \relax (2) + 8 \, x + 2 \, e^{2} - 3 \, \log \relax (2) + 6\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.08, size = 37, normalized size = 1.37
method | result | size |
norman | \(\ln \left (-x \ln \relax (2) {\mathrm e}^{x}+{\mathrm e}^{2} \ln \relax (2)+5 x \ln \relax (2)+2 \,{\mathrm e}^{x} x -2 \,{\mathrm e}^{2}+3 \ln \relax (2)-8 x -6\right )\) | \(37\) |
risch | \(\ln \relax (x )+\ln \left ({\mathrm e}^{x}-\frac {{\mathrm e}^{2} \ln \relax (2)+5 x \ln \relax (2)-2 \,{\mathrm e}^{2}+3 \ln \relax (2)-8 x -6}{x \left (\ln \relax (2)-2\right )}\right )\) | \(42\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.49, size = 45, normalized size = 1.67 \begin {gather*} \log \relax (x) + \log \left (\frac {x {\left (\log \relax (2) - 2\right )} e^{x} - x {\left (5 \, \log \relax (2) - 8\right )} - {\left (\log \relax (2) - 2\right )} e^{2} - 3 \, \log \relax (2) + 6}{x {\left (\log \relax (2) - 2\right )}}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 7.21, size = 34, normalized size = 1.26 \begin {gather*} \ln \left (16\,x+4\,{\mathrm {e}}^2-2\,\ln \relax (2)\,\left (5\,x+{\mathrm {e}}^2+3\right )-{\mathrm {e}}^x\,\left (4\,x-2\,x\,\ln \relax (2)\right )+12\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.58, size = 44, normalized size = 1.63 \begin {gather*} \log {\relax (x )} + \log {\left (e^{x} + \frac {- 5 x \log {\relax (2 )} + 8 x - e^{2} \log {\relax (2 )} - 3 \log {\relax (2 )} + 6 + 2 e^{2}}{- 2 x + x \log {\relax (2 )}} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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