Integrand size = 40, antiderivative size = 14 \[ \int \frac {e^{x+x \log (16)} (-5-5 \log (16))}{1-2 e^{x+x \log (16)}+e^{2 x+2 x \log (16)}} \, dx=\frac {5}{-1+e^{x (1+\log (16))}} \]
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Time = 0.04 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {12, 2320, 32} \[ \int \frac {e^{x+x \log (16)} (-5-5 \log (16))}{1-2 e^{x+x \log (16)}+e^{2 x+2 x \log (16)}} \, dx=-\frac {5}{1-(16 e)^x} \]
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Rule 12
Rule 32
Rule 2320
Rubi steps \begin{align*} \text {integral}& = -\left ((5 (1+\log (16))) \int \frac {e^{x+x \log (16)}}{1-2 e^{x+x \log (16)}+e^{2 x+2 x \log (16)}} \, dx\right ) \\ & = -\left (5 \text {Subst}\left (\int \frac {1}{(1-x)^2} \, dx,x,e^{x (1+\log (16))}\right )\right ) \\ & = -\frac {5}{1-(16 e)^x} \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.50 \[ \int \frac {e^{x+x \log (16)} (-5-5 \log (16))}{1-2 e^{x+x \log (16)}+e^{2 x+2 x \log (16)}} \, dx=\frac {5 (1+\log (16))}{\left (-1+(16 e)^x\right ) \log (16 e)} \]
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Time = 0.12 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.07
method | result | size |
norman | \(\frac {5}{{\mathrm e}^{4 x \ln \left (2\right )+x}-1}\) | \(15\) |
derivativedivides | \(-\frac {5 \left (-4 \ln \left (2\right )-1\right )}{\left (1+4 \ln \left (2\right )\right ) \left ({\mathrm e}^{4 x \ln \left (2\right )+x}-1\right )}\) | \(29\) |
default | \(-\frac {-20 \ln \left (2\right )-5}{\left (1+4 \ln \left (2\right )\right ) \left ({\mathrm e}^{4 x \ln \left (2\right )+x}-1\right )}\) | \(29\) |
parallelrisch | \(-\frac {-20 \ln \left (2\right )-5}{\left (1+4 \ln \left (2\right )\right ) \left ({\mathrm e}^{\left (1+4 \ln \left (2\right )\right ) x}-1\right )}\) | \(30\) |
risch | \(\frac {20 \ln \left (2\right )}{\left (1+4 \ln \left (2\right )\right ) \left (16^{x} {\mathrm e}^{x}-1\right )}+\frac {5}{\left (1+4 \ln \left (2\right )\right ) \left (16^{x} {\mathrm e}^{x}-1\right )}\) | \(44\) |
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Time = 0.23 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {e^{x+x \log (16)} (-5-5 \log (16))}{1-2 e^{x+x \log (16)}+e^{2 x+2 x \log (16)}} \, dx=\frac {5}{e^{\left (4 \, x \log \left (2\right ) + x\right )} - 1} \]
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Time = 0.05 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {e^{x+x \log (16)} (-5-5 \log (16))}{1-2 e^{x+x \log (16)}+e^{2 x+2 x \log (16)}} \, dx=\frac {5}{e^{x + 4 x \log {\left (2 \right )}} - 1} \]
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Time = 0.34 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {e^{x+x \log (16)} (-5-5 \log (16))}{1-2 e^{x+x \log (16)}+e^{2 x+2 x \log (16)}} \, dx=\frac {5}{e^{\left (4 \, x \log \left (2\right ) + x\right )} - 1} \]
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Time = 0.28 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.07 \[ \int \frac {e^{x+x \log (16)} (-5-5 \log (16))}{1-2 e^{x+x \log (16)}+e^{2 x+2 x \log (16)}} \, dx=\frac {5}{e^{\left (x {\left (4 \, \log \left (2\right ) + 1\right )}\right )} - 1} \]
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Time = 13.45 (sec) , antiderivative size = 144, normalized size of antiderivative = 10.29 \[ \int \frac {e^{x+x \log (16)} (-5-5 \log (16))}{1-2 e^{x+x \log (16)}+e^{2 x+2 x \log (16)}} \, dx=\frac {10\,\mathrm {atan}\left (\frac {\ln \left (256\right )+2}{\sqrt {4\,{\ln \left (16\right )}^2-{\ln \left (256\right )}^2}}-\frac {2^{4\,x}\,{\mathrm {e}}^x\,\left (\ln \left (256\right )+2\right )}{\sqrt {4\,{\ln \left (16\right )}^2-{\ln \left (256\right )}^2}}\right )}{\sqrt {4\,{\ln \left (16\right )}^2-{\ln \left (256\right )}^2}}+\frac {40\,\ln \left (2\right )\,\mathrm {atan}\left (\frac {\ln \left (256\right )-2\,2^{4\,x}\,{\mathrm {e}}^x-2\,2^{4\,x}\,{\mathrm {e}}^x\,\ln \left (16\right )+2}{\sqrt {8\,\ln \left (16\right )-4\,\ln \left (256\right )+4\,{\ln \left (16\right )}^2-{\ln \left (256\right )}^2}}\right )}{\sqrt {8\,\ln \left (16\right )-4\,\ln \left (256\right )+4\,{\ln \left (16\right )}^2-{\ln \left (256\right )}^2}} \]
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