Optimal. Leaf size=30 \[ e^3-\frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right ) \log (4 (4+x))}{x} \]
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Rubi [A] time = 3.52, antiderivative size = 26, normalized size of antiderivative = 0.87, number of steps used = 17, number of rules used = 11, integrand size = 104, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.106, Rules used = {6688, 6742, 14, 36, 29, 31, 2395, 2551, 30, 2557, 2554} \begin {gather*} -\frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right ) \log (4 x+16)}{x} \end {gather*}
Antiderivative was successfully verified.
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Rule 14
Rule 29
Rule 30
Rule 31
Rule 36
Rule 2395
Rule 2551
Rule 2554
Rule 2557
Rule 6688
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (4+5 x+x^2\right ) \log (4 (4+x))-\left (-1+2 e^x x\right ) \log \left (5-\frac {5 e^{-x}}{2 x}\right ) (-x+(4+x) \log (4 (4+x)))}{x^2 (4+x) \left (1-2 e^x x\right )} \, dx\\ &=\int \left (\frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right ) (-x+4 \log (4 (4+x))+x \log (4 (4+x)))}{x^2 (4+x)}+\frac {(1+x) \log (16+4 x)}{x^2 \left (1-2 e^x x\right )}\right ) \, dx\\ &=\int \frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right ) (-x+4 \log (4 (4+x))+x \log (4 (4+x)))}{x^2 (4+x)} \, dx+\int \frac {(1+x) \log (16+4 x)}{x^2 \left (1-2 e^x x\right )} \, dx\\ &=-\left (\log (16+4 x) \int \frac {1}{x^2 \left (-1+2 e^x x\right )} \, dx\right )-\log (16+4 x) \int \frac {1}{x \left (-1+2 e^x x\right )} \, dx+\int \frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right ) (-x+(4+x) \log (4 (4+x)))}{x^2 (4+x)} \, dx-\int \frac {-\int \frac {1}{x^2 \left (-1+2 e^x x\right )} \, dx-\int \frac {1}{x \left (-1+2 e^x x\right )} \, dx}{4+x} \, dx\\ &=-\left (\log (16+4 x) \int \frac {1}{x^2 \left (-1+2 e^x x\right )} \, dx\right )-\log (16+4 x) \int \frac {1}{x \left (-1+2 e^x x\right )} \, dx+\int \left (-\frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right )}{x (4+x)}+\frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right ) \log (16+4 x)}{x^2}\right ) \, dx-\int \left (-\frac {\int \frac {1}{x^2 \left (-1+2 e^x x\right )} \, dx}{4+x}-\frac {\int \frac {1}{x \left (-1+2 e^x x\right )} \, dx}{4+x}\right ) \, dx\\ &=-\left (\log (16+4 x) \int \frac {1}{x^2 \left (-1+2 e^x x\right )} \, dx\right )-\log (16+4 x) \int \frac {1}{x \left (-1+2 e^x x\right )} \, dx-\int \frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right )}{x (4+x)} \, dx+\int \frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right ) \log (16+4 x)}{x^2} \, dx+\int \frac {\int \frac {1}{x^2 \left (-1+2 e^x x\right )} \, dx}{4+x} \, dx+\int \frac {\int \frac {1}{x \left (-1+2 e^x x\right )} \, dx}{4+x} \, dx\\ &=-\frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right ) \log (16+4 x)}{x}-\log (16+4 x) \int \frac {1}{x^2 \left (-1+2 e^x x\right )} \, dx-\log (16+4 x) \int \frac {1}{x \left (-1+2 e^x x\right )} \, dx-\int \frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right )}{(-4-x) x} \, dx-\int \left (\frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right )}{4 x}-\frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right )}{4 (4+x)}\right ) \, dx-\int \frac {(1+x) \log (16+4 x)}{x^2 \left (1-2 e^x x\right )} \, dx+\int \frac {\int \frac {1}{x^2 \left (-1+2 e^x x\right )} \, dx}{4+x} \, dx+\int \frac {\int \frac {1}{x \left (-1+2 e^x x\right )} \, dx}{4+x} \, dx\\ &=-\frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right ) \log (16+4 x)}{x}-\frac {1}{4} \int \frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right )}{x} \, dx+\frac {1}{4} \int \frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right )}{4+x} \, dx-\int \left (-\frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right )}{4 x}+\frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right )}{4 (4+x)}\right ) \, dx+\int \frac {\int \frac {1}{x^2 \left (-1+2 e^x x\right )} \, dx}{4+x} \, dx+\int \frac {-\int \frac {1}{x^2 \left (-1+2 e^x x\right )} \, dx-\int \frac {1}{x \left (-1+2 e^x x\right )} \, dx}{4+x} \, dx+\int \frac {\int \frac {1}{x \left (-1+2 e^x x\right )} \, dx}{4+x} \, dx\\ &=-\frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right ) \log (16+4 x)}{x}+\int \frac {\int \frac {1}{x^2 \left (-1+2 e^x x\right )} \, dx}{4+x} \, dx+\int \frac {\int \frac {1}{x \left (-1+2 e^x x\right )} \, dx}{4+x} \, dx+\int \left (-\frac {\int \frac {1}{x^2 \left (-1+2 e^x x\right )} \, dx}{4+x}-\frac {\int \frac {1}{x \left (-1+2 e^x x\right )} \, dx}{4+x}\right ) \, dx\\ &=-\frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right ) \log (16+4 x)}{x}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.47, size = 33, normalized size = 1.10 \begin {gather*} \log (4+x)-\frac {\left (x+\log \left (5-\frac {5 e^{-x}}{2 x}\right )\right ) \log (4 (4+x))}{x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.78, size = 28, normalized size = 0.93 \begin {gather*} -\frac {\log \left (4 \, x + 16\right ) \log \left (\frac {5 \, {\left (2 \, x e^{x} - 1\right )} e^{\left (-x\right )}}{2 \, x}\right )}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.29, size = 69, normalized size = 2.30 \begin {gather*} \frac {2 \, \log \relax (2)^{2} - 2 \, \log \relax (2) \log \left (5 \, {\left (2 \, x e^{x} - 1\right )} e^{\left (-x\right )}\right ) + \log \relax (2) \log \left (x + 4\right ) - \log \left (5 \, {\left (2 \, x e^{x} - 1\right )} e^{\left (-x\right )}\right ) \log \left (x + 4\right ) + 2 \, \log \relax (2) \log \relax (x) + \log \left (x + 4\right ) \log \relax (x)}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.27, size = 290, normalized size = 9.67
method | result | size |
risch | \(\frac {\ln \left (4 x +16\right ) \ln \left ({\mathrm e}^{x}\right )}{x}+\frac {\ln \left (4 x +16\right ) \left (-i \pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{x} x -\frac {1}{2}\right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x} \left ({\mathrm e}^{x} x -\frac {1}{2}\right )\right )^{2}+i \pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{x} x -\frac {1}{2}\right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x} \left ({\mathrm e}^{x} x -\frac {1}{2}\right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x}\right )+i \pi \mathrm {csgn}\left (i {\mathrm e}^{-x} \left ({\mathrm e}^{x} x -\frac {1}{2}\right )\right )^{3}-i \pi \mathrm {csgn}\left (i {\mathrm e}^{-x} \left ({\mathrm e}^{x} x -\frac {1}{2}\right )\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{-x}\right )-i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{-x} \left ({\mathrm e}^{x} x -\frac {1}{2}\right )\right ) \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{x} x -\frac {1}{2}\right ) {\mathrm e}^{-x}}{x}\right )^{2}+i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{-x} \left ({\mathrm e}^{x} x -\frac {1}{2}\right )\right ) \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{x} x -\frac {1}{2}\right ) {\mathrm e}^{-x}}{x}\right ) \mathrm {csgn}\left (\frac {i}{x}\right )-i \pi \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{x} x -\frac {1}{2}\right ) {\mathrm e}^{-x}}{x}\right )^{2} \mathrm {csgn}\left (\frac {i}{x}\right )+i \pi \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{x} x -\frac {1}{2}\right ) {\mathrm e}^{-x}}{x}\right )^{3}-2 \ln \relax (5)+2 \ln \relax (x )-2 \ln \left ({\mathrm e}^{x} x -\frac {1}{2}\right )\right )}{2 x}\) | \(290\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.50, size = 58, normalized size = 1.93 \begin {gather*} -\frac {2 \, \log \relax (5) \log \relax (2) - 2 \, \log \relax (2)^{2} + {\left (2 \, \log \relax (2) + \log \left (x + 4\right )\right )} \log \left (2 \, x e^{x} - 1\right ) - {\left (x - \log \relax (5) + \log \relax (2) + \log \relax (x)\right )} \log \left (x + 4\right ) - 2 \, \log \relax (2) \log \relax (x)}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.10, size = 29, normalized size = 0.97 \begin {gather*} \ln \left (x+4\right )-\frac {\ln \left (4\,x+16\right )\,\ln \left (\frac {5\,\left (2\,x\,{\mathrm {e}}^x-1\right )}{2\,x}\right )}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.69, size = 26, normalized size = 0.87 \begin {gather*} - \frac {\log {\left (\frac {\left (5 x e^{x} - \frac {5}{2}\right ) e^{- x}}{x} \right )} \log {\left (4 x + 16 \right )}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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