Integrand size = 150, antiderivative size = 26 \[ \int \frac {e^x (-220-20 x+40 \log (2))+e^x (44+4 x) \log (2) \log (11+x)+\left (e^x \left (-220-240 x-20 x^2+\left (-44 x-48 x^2-4 x^3\right ) \log (2)\right )+e^x \left (44+48 x+4 x^2\right ) \log (2) \log (11+x)\right ) \log \left (\frac {(1+x) \log (2)}{-5-x \log (2)+\log (2) \log (11+x)}\right )}{-55-60 x-5 x^2+\left (-11 x-12 x^2-x^3\right ) \log (2)+\left (11+12 x+x^2\right ) \log (2) \log (11+x)} \, dx=4 e^x \log \left (\frac {1+x}{-x-\frac {5}{\log (2)}+\log (11+x)}\right ) \]
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Time = 0.67 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {6873, 6820, 12, 2326} \[ \int \frac {e^x (-220-20 x+40 \log (2))+e^x (44+4 x) \log (2) \log (11+x)+\left (e^x \left (-220-240 x-20 x^2+\left (-44 x-48 x^2-4 x^3\right ) \log (2)\right )+e^x \left (44+48 x+4 x^2\right ) \log (2) \log (11+x)\right ) \log \left (\frac {(1+x) \log (2)}{-5-x \log (2)+\log (2) \log (11+x)}\right )}{-55-60 x-5 x^2+\left (-11 x-12 x^2-x^3\right ) \log (2)+\left (11+12 x+x^2\right ) \log (2) \log (11+x)} \, dx=4 e^x \log \left (-\frac {(x+1) \log (2)}{x \log (2)-\log (2) \log (x+11)+5}\right ) \]
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Rule 12
Rule 2326
Rule 6820
Rule 6873
Rubi steps \begin{align*} \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{align*}
Time = 0.13 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {e^x (-220-20 x+40 \log (2))+e^x (44+4 x) \log (2) \log (11+x)+\left (e^x \left (-220-240 x-20 x^2+\left (-44 x-48 x^2-4 x^3\right ) \log (2)\right )+e^x \left (44+48 x+4 x^2\right ) \log (2) \log (11+x)\right ) \log \left (\frac {(1+x) \log (2)}{-5-x \log (2)+\log (2) \log (11+x)}\right )}{-55-60 x-5 x^2+\left (-11 x-12 x^2-x^3\right ) \log (2)+\left (11+12 x+x^2\right ) \log (2) \log (11+x)} \, dx=4 e^x \log \left (\frac {(1+x) \log (2)}{-5-x \log (2)+\log (2) \log (11+x)}\right ) \]
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Time = 32.47 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08
method | result | size |
parallelrisch | \(4 \,{\mathrm e}^{x} \ln \left (\frac {\left (1+x \right ) \ln \left (2\right )}{\ln \left (2\right ) \ln \left (11+x \right )-x \ln \left (2\right )-5}\right )\) | \(28\) |
risch | \(-4 \,{\mathrm e}^{x} \ln \left (5+\left (x -\ln \left (11+x \right )\right ) \ln \left (2\right )\right )-2 i \pi \,\operatorname {csgn}\left (i \left (1+x \right )\right ) \operatorname {csgn}\left (\frac {i}{5+\left (x -\ln \left (11+x \right )\right ) \ln \left (2\right )}\right ) \operatorname {csgn}\left (\frac {i \left (1+x \right )}{5+\left (x -\ln \left (11+x \right )\right ) \ln \left (2\right )}\right ) {\mathrm e}^{x}+2 i \pi \,\operatorname {csgn}\left (i \left (1+x \right )\right ) \operatorname {csgn}\left (\frac {i \left (1+x \right )}{5+\left (x -\ln \left (11+x \right )\right ) \ln \left (2\right )}\right )^{2} {\mathrm e}^{x}-4 i \pi \operatorname {csgn}\left (\frac {i \left (1+x \right )}{5+\left (x -\ln \left (11+x \right )\right ) \ln \left (2\right )}\right )^{2} {\mathrm e}^{x}+2 i \pi \,\operatorname {csgn}\left (\frac {i}{5+\left (x -\ln \left (11+x \right )\right ) \ln \left (2\right )}\right ) \operatorname {csgn}\left (\frac {i \left (1+x \right )}{5+\left (x -\ln \left (11+x \right )\right ) \ln \left (2\right )}\right )^{2} {\mathrm e}^{x}+2 i \pi \operatorname {csgn}\left (\frac {i \left (1+x \right )}{5+\left (x -\ln \left (11+x \right )\right ) \ln \left (2\right )}\right )^{3} {\mathrm e}^{x}+4 i \pi \,{\mathrm e}^{x}+4 \ln \left (\ln \left (2\right )\right ) {\mathrm e}^{x}+4 \ln \left (1+x \right ) {\mathrm e}^{x}\) | \(241\) |
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Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {e^x (-220-20 x+40 \log (2))+e^x (44+4 x) \log (2) \log (11+x)+\left (e^x \left (-220-240 x-20 x^2+\left (-44 x-48 x^2-4 x^3\right ) \log (2)\right )+e^x \left (44+48 x+4 x^2\right ) \log (2) \log (11+x)\right ) \log \left (\frac {(1+x) \log (2)}{-5-x \log (2)+\log (2) \log (11+x)}\right )}{-55-60 x-5 x^2+\left (-11 x-12 x^2-x^3\right ) \log (2)+\left (11+12 x+x^2\right ) \log (2) \log (11+x)} \, dx=4 \, e^{x} \log \left (-\frac {{\left (x + 1\right )} \log \left (2\right )}{x \log \left (2\right ) - \log \left (2\right ) \log \left (x + 11\right ) + 5}\right ) \]
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Time = 14.42 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {e^x (-220-20 x+40 \log (2))+e^x (44+4 x) \log (2) \log (11+x)+\left (e^x \left (-220-240 x-20 x^2+\left (-44 x-48 x^2-4 x^3\right ) \log (2)\right )+e^x \left (44+48 x+4 x^2\right ) \log (2) \log (11+x)\right ) \log \left (\frac {(1+x) \log (2)}{-5-x \log (2)+\log (2) \log (11+x)}\right )}{-55-60 x-5 x^2+\left (-11 x-12 x^2-x^3\right ) \log (2)+\left (11+12 x+x^2\right ) \log (2) \log (11+x)} \, dx=4 e^{x} \log {\left (\frac {\left (x + 1\right ) \log {\left (2 \right )}}{- x \log {\left (2 \right )} + \log {\left (2 \right )} \log {\left (x + 11 \right )} - 5} \right )} \]
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Time = 0.33 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35 \[ \int \frac {e^x (-220-20 x+40 \log (2))+e^x (44+4 x) \log (2) \log (11+x)+\left (e^x \left (-220-240 x-20 x^2+\left (-44 x-48 x^2-4 x^3\right ) \log (2)\right )+e^x \left (44+48 x+4 x^2\right ) \log (2) \log (11+x)\right ) \log \left (\frac {(1+x) \log (2)}{-5-x \log (2)+\log (2) \log (11+x)}\right )}{-55-60 x-5 x^2+\left (-11 x-12 x^2-x^3\right ) \log (2)+\left (11+12 x+x^2\right ) \log (2) \log (11+x)} \, dx=-4 \, e^{x} \log \left (-x \log \left (2\right ) + \log \left (2\right ) \log \left (x + 11\right ) - 5\right ) + 4 \, e^{x} \log \left (x + 1\right ) + 4 \, e^{x} \log \left (\log \left (2\right )\right ) \]
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Time = 0.59 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35 \[ \int \frac {e^x (-220-20 x+40 \log (2))+e^x (44+4 x) \log (2) \log (11+x)+\left (e^x \left (-220-240 x-20 x^2+\left (-44 x-48 x^2-4 x^3\right ) \log (2)\right )+e^x \left (44+48 x+4 x^2\right ) \log (2) \log (11+x)\right ) \log \left (\frac {(1+x) \log (2)}{-5-x \log (2)+\log (2) \log (11+x)}\right )}{-55-60 x-5 x^2+\left (-11 x-12 x^2-x^3\right ) \log (2)+\left (11+12 x+x^2\right ) \log (2) \log (11+x)} \, dx=-4 \, e^{x} \log \left (x \log \left (2\right ) - \log \left (2\right ) \log \left (x + 11\right ) + 5\right ) + 4 \, e^{x} \log \left (-x \log \left (2\right ) - \log \left (2\right )\right ) \]
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Time = 0.88 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {e^x (-220-20 x+40 \log (2))+e^x (44+4 x) \log (2) \log (11+x)+\left (e^x \left (-220-240 x-20 x^2+\left (-44 x-48 x^2-4 x^3\right ) \log (2)\right )+e^x \left (44+48 x+4 x^2\right ) \log (2) \log (11+x)\right ) \log \left (\frac {(1+x) \log (2)}{-5-x \log (2)+\log (2) \log (11+x)}\right )}{-55-60 x-5 x^2+\left (-11 x-12 x^2-x^3\right ) \log (2)+\left (11+12 x+x^2\right ) \log (2) \log (11+x)} \, dx=4\,\ln \left (-\frac {\ln \left (2\right )\,\left (x+1\right )}{x\,\ln \left (2\right )-\ln \left (x+11\right )\,\ln \left (2\right )+5}\right )\,{\mathrm {e}}^x \]
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