Optimal. Leaf size=26 \[ 4 e^x \log \left (\frac {1+x}{-x-\frac {5}{\log (2)}+\log (11+x)}\right ) \]
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Rubi [A] time = 1.10, antiderivative size = 29, normalized size of antiderivative = 1.12, number of steps used = 4, number of rules used = 4, integrand size = 150, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {6741, 6688, 12, 2288} \begin {gather*} 4 e^x \log \left (-\frac {(x+1) \log (2)}{x \log (2)-\log (2) \log (x+11)+5}\right ) \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 2288
Rule 6688
Rule 6741
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-e^x (-220-20 x+40 \log (2))-e^x (44+4 x) \log (2) \log (11+x)-\left (e^x \left (-220-240 x-20 x^2+\left (-44 x-48 x^2-4 x^3\right ) \log (2)\right )+e^x \left (44+48 x+4 x^2\right ) \log (2) \log (11+x)\right ) \log \left (\frac {(1+x) \log (2)}{-5-x \log (2)+\log (2) \log (11+x)}\right )}{\left (11+12 x+x^2\right ) (5+x \log (2)-\log (2) \log (11+x))} \, dx\\ &=\int \frac {4 e^x \left (5 (11+x-\log (4))-(11+x) \log (2) \log (11+x)+\left (11+12 x+x^2\right ) (5+x \log (2)-\log (2) \log (11+x)) \log \left (\frac {(1+x) \log (2)}{-5-x \log (2)+\log (2) \log (11+x)}\right )\right )}{\left (11+12 x+x^2\right ) (5+x \log (2)-\log (2) \log (11+x))} \, dx\\ &=4 \int \frac {e^x \left (5 (11+x-\log (4))-(11+x) \log (2) \log (11+x)+\left (11+12 x+x^2\right ) (5+x \log (2)-\log (2) \log (11+x)) \log \left (\frac {(1+x) \log (2)}{-5-x \log (2)+\log (2) \log (11+x)}\right )\right )}{\left (11+12 x+x^2\right ) (5+x \log (2)-\log (2) \log (11+x))} \, dx\\ &=4 e^x \log \left (-\frac {(1+x) \log (2)}{5+x \log (2)-\log (2) \log (11+x)}\right )\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.13, size = 28, normalized size = 1.08 \begin {gather*} 4 e^x \log \left (\frac {(1+x) \log (2)}{-5-x \log (2)+\log (2) \log (11+x)}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.62, size = 28, normalized size = 1.08 \begin {gather*} 4 \, e^{x} \log \left (-\frac {{\left (x + 1\right )} \log \relax (2)}{x \log \relax (2) - \log \relax (2) \log \left (x + 11\right ) + 5}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.75, size = 35, normalized size = 1.35 \begin {gather*} -4 \, e^{x} \log \left (x \log \relax (2) - \log \relax (2) \log \left (x + 11\right ) + 5\right ) + 4 \, e^{x} \log \left (-x \log \relax (2) - \log \relax (2)\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.18, size = 241, normalized size = 9.27
method | result | size |
risch | \(-4 \,{\mathrm e}^{x} \ln \left (5+\left (x -\ln \left (11+x \right )\right ) \ln \relax (2)\right )-2 i \pi \,\mathrm {csgn}\left (i \left (x +1\right )\right ) \mathrm {csgn}\left (\frac {i}{5+\left (x -\ln \left (11+x \right )\right ) \ln \relax (2)}\right ) \mathrm {csgn}\left (\frac {i \left (x +1\right )}{5+\left (x -\ln \left (11+x \right )\right ) \ln \relax (2)}\right ) {\mathrm e}^{x}+2 i \pi \,\mathrm {csgn}\left (i \left (x +1\right )\right ) \mathrm {csgn}\left (\frac {i \left (x +1\right )}{5+\left (x -\ln \left (11+x \right )\right ) \ln \relax (2)}\right )^{2} {\mathrm e}^{x}-4 i \pi \mathrm {csgn}\left (\frac {i \left (x +1\right )}{5+\left (x -\ln \left (11+x \right )\right ) \ln \relax (2)}\right )^{2} {\mathrm e}^{x}+2 i \pi \,\mathrm {csgn}\left (\frac {i}{5+\left (x -\ln \left (11+x \right )\right ) \ln \relax (2)}\right ) \mathrm {csgn}\left (\frac {i \left (x +1\right )}{5+\left (x -\ln \left (11+x \right )\right ) \ln \relax (2)}\right )^{2} {\mathrm e}^{x}+2 i \pi \mathrm {csgn}\left (\frac {i \left (x +1\right )}{5+\left (x -\ln \left (11+x \right )\right ) \ln \relax (2)}\right )^{3} {\mathrm e}^{x}+4 i \pi \,{\mathrm e}^{x}+4 \ln \left (\ln \relax (2)\right ) {\mathrm e}^{x}+4 \ln \left (x +1\right ) {\mathrm e}^{x}\) | \(241\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.76, size = 35, normalized size = 1.35 \begin {gather*} -4 \, e^{x} \log \left (-x \log \relax (2) + \log \relax (2) \log \left (x + 11\right ) - 5\right ) + 4 \, e^{x} \log \left (x + 1\right ) + 4 \, e^{x} \log \left (\log \relax (2)\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.87, size = 28, normalized size = 1.08 \begin {gather*} 4\,\ln \left (-\frac {\ln \relax (2)\,\left (x+1\right )}{x\,\ln \relax (2)-\ln \left (x+11\right )\,\ln \relax (2)+5}\right )\,{\mathrm {e}}^x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 27.47, size = 27, normalized size = 1.04 \begin {gather*} 4 e^{x} \log {\left (\frac {\left (x + 1\right ) \log {\relax (2 )}}{- x \log {\relax (2 )} + \log {\relax (2 )} \log {\left (x + 11 \right )} - 5} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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