Integrand size = 76, antiderivative size = 34 \[ \int \frac {e^5+4 \log \left (\frac {-1+7 \log (2)}{\log (2)}\right )}{8 e^{10}-8 e^5 x+2 x^2+\left (32 e^5-16 x\right ) \log \left (\frac {-1+7 \log (2)}{\log (2)}\right )+32 \log ^2\left (\frac {-1+7 \log (2)}{\log (2)}\right )} \, dx=\frac {1}{2 \left (2-\frac {-e^5+x}{\frac {x}{4}+\log \left (7-\frac {1}{\log (2)}\right )}\right )} \]
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Time = 0.02 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.18, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {12, 2006, 27, 32} \[ \int \frac {e^5+4 \log \left (\frac {-1+7 \log (2)}{\log (2)}\right )}{8 e^{10}-8 e^5 x+2 x^2+\left (32 e^5-16 x\right ) \log \left (\frac {-1+7 \log (2)}{\log (2)}\right )+32 \log ^2\left (\frac {-1+7 \log (2)}{\log (2)}\right )} \, dx=-\frac {e^5+4 \log \left (7-\frac {1}{\log (2)}\right )}{2 \left (x-2 \left (e^5+2 \log \left (7-\frac {1}{\log (2)}\right )\right )\right )} \]
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Rule 12
Rule 27
Rule 32
Rule 2006
Rubi steps \begin{align*} \text {integral}& = \left (e^5+4 \log \left (7-\frac {1}{\log (2)}\right )\right ) \int \frac {1}{8 e^{10}-8 e^5 x+2 x^2+\left (32 e^5-16 x\right ) \log \left (\frac {-1+7 \log (2)}{\log (2)}\right )+32 \log ^2\left (\frac {-1+7 \log (2)}{\log (2)}\right )} \, dx \\ & = \left (e^5+4 \log \left (7-\frac {1}{\log (2)}\right )\right ) \int \frac {1}{2 x^2-8 x \left (e^5+2 \log \left (7-\frac {1}{\log (2)}\right )\right )+8 \left (e^5+2 \log \left (7-\frac {1}{\log (2)}\right )\right )^2} \, dx \\ & = \left (e^5+4 \log \left (7-\frac {1}{\log (2)}\right )\right ) \int \frac {1}{2 \left (2 e^5-x+4 \log \left (7-\frac {1}{\log (2)}\right )\right )^2} \, dx \\ & = \frac {1}{2} \left (e^5+4 \log \left (7-\frac {1}{\log (2)}\right )\right ) \int \frac {1}{\left (2 e^5-x+4 \log \left (7-\frac {1}{\log (2)}\right )\right )^2} \, dx \\ & = -\frac {e^5+4 \log \left (7-\frac {1}{\log (2)}\right )}{2 \left (x-2 \left (e^5+2 \log \left (7-\frac {1}{\log (2)}\right )\right )\right )} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.15 \[ \int \frac {e^5+4 \log \left (\frac {-1+7 \log (2)}{\log (2)}\right )}{8 e^{10}-8 e^5 x+2 x^2+\left (32 e^5-16 x\right ) \log \left (\frac {-1+7 \log (2)}{\log (2)}\right )+32 \log ^2\left (\frac {-1+7 \log (2)}{\log (2)}\right )} \, dx=-\frac {e^5+4 \log \left (7-\frac {1}{\log (2)}\right )}{2 \left (-2 e^5+x-4 \log \left (7-\frac {1}{\log (2)}\right )\right )} \]
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Time = 0.51 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.29
method | result | size |
gosper | \(\frac {4 \ln \left (\frac {7 \ln \left (2\right )-1}{\ln \left (2\right )}\right )+{\mathrm e}^{5}}{8 \ln \left (\frac {7 \ln \left (2\right )-1}{\ln \left (2\right )}\right )+4 \,{\mathrm e}^{5}-2 x}\) | \(44\) |
parallelrisch | \(\frac {4 \ln \left (\frac {7 \ln \left (2\right )-1}{\ln \left (2\right )}\right )+{\mathrm e}^{5}}{8 \ln \left (\frac {7 \ln \left (2\right )-1}{\ln \left (2\right )}\right )+4 \,{\mathrm e}^{5}-2 x}\) | \(44\) |
norman | \(\frac {\frac {{\mathrm e}^{5}}{2}+2 \ln \left (7 \ln \left (2\right )-1\right )-2 \ln \left (\ln \left (2\right )\right )}{4 \ln \left (\frac {7 \ln \left (2\right )-1}{\ln \left (2\right )}\right )+2 \,{\mathrm e}^{5}-x}\) | \(45\) |
risch | \(\frac {\ln \left (7 \ln \left (2\right )-1\right )}{{\mathrm e}^{5}+2 \ln \left (7 \ln \left (2\right )-1\right )-2 \ln \left (\ln \left (2\right )\right )-\frac {x}{2}}-\frac {\ln \left (\ln \left (2\right )\right )}{{\mathrm e}^{5}+2 \ln \left (7 \ln \left (2\right )-1\right )-2 \ln \left (\ln \left (2\right )\right )-\frac {x}{2}}+\frac {{\mathrm e}^{5}}{4 \,{\mathrm e}^{5}+8 \ln \left (7 \ln \left (2\right )-1\right )-8 \ln \left (\ln \left (2\right )\right )-2 x}\) | \(85\) |
meijerg | \(-\frac {\ln \left (\frac {7 \ln \left (2\right )-1}{\ln \left (2\right )}\right ) x}{\left (-2 \,{\mathrm e}^{5}-4 \ln \left (\frac {7 \ln \left (2\right )-1}{\ln \left (2\right )}\right )\right ) \left ({\mathrm e}^{5}+2 \ln \left (\frac {7 \ln \left (2\right )-1}{\ln \left (2\right )}\right )\right ) \left (1-\frac {x}{2 \left ({\mathrm e}^{5}+2 \ln \left (\frac {7 \ln \left (2\right )-1}{\ln \left (2\right )}\right )\right )}\right )}-\frac {{\mathrm e}^{5} x}{4 \left (-2 \,{\mathrm e}^{5}-4 \ln \left (\frac {7 \ln \left (2\right )-1}{\ln \left (2\right )}\right )\right ) \left ({\mathrm e}^{5}+2 \ln \left (\frac {7 \ln \left (2\right )-1}{\ln \left (2\right )}\right )\right ) \left (1-\frac {x}{2 \left ({\mathrm e}^{5}+2 \ln \left (\frac {7 \ln \left (2\right )-1}{\ln \left (2\right )}\right )\right )}\right )}\) | \(154\) |
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Time = 0.25 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.21 \[ \int \frac {e^5+4 \log \left (\frac {-1+7 \log (2)}{\log (2)}\right )}{8 e^{10}-8 e^5 x+2 x^2+\left (32 e^5-16 x\right ) \log \left (\frac {-1+7 \log (2)}{\log (2)}\right )+32 \log ^2\left (\frac {-1+7 \log (2)}{\log (2)}\right )} \, dx=-\frac {e^{5} + 4 \, \log \left (\frac {7 \, \log \left (2\right ) - 1}{\log \left (2\right )}\right )}{2 \, {\left (x - 2 \, e^{5} - 4 \, \log \left (\frac {7 \, \log \left (2\right ) - 1}{\log \left (2\right )}\right )\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (22) = 44\).
Time = 0.18 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.35 \[ \int \frac {e^5+4 \log \left (\frac {-1+7 \log (2)}{\log (2)}\right )}{8 e^{10}-8 e^5 x+2 x^2+\left (32 e^5-16 x\right ) \log \left (\frac {-1+7 \log (2)}{\log (2)}\right )+32 \log ^2\left (\frac {-1+7 \log (2)}{\log (2)}\right )} \, dx=- \frac {- 4 \log {\left (\log {\left (2 \right )} \right )} + 4 \log {\left (-1 + 7 \log {\left (2 \right )} \right )} + e^{5}}{2 x - 4 e^{5} - 8 \log {\left (-1 + 7 \log {\left (2 \right )} \right )} + 8 \log {\left (\log {\left (2 \right )} \right )}} \]
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Time = 0.18 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.21 \[ \int \frac {e^5+4 \log \left (\frac {-1+7 \log (2)}{\log (2)}\right )}{8 e^{10}-8 e^5 x+2 x^2+\left (32 e^5-16 x\right ) \log \left (\frac {-1+7 \log (2)}{\log (2)}\right )+32 \log ^2\left (\frac {-1+7 \log (2)}{\log (2)}\right )} \, dx=-\frac {e^{5} + 4 \, \log \left (\frac {7 \, \log \left (2\right ) - 1}{\log \left (2\right )}\right )}{2 \, {\left (x - 2 \, e^{5} - 4 \, \log \left (\frac {7 \, \log \left (2\right ) - 1}{\log \left (2\right )}\right )\right )}} \]
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Time = 0.27 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.21 \[ \int \frac {e^5+4 \log \left (\frac {-1+7 \log (2)}{\log (2)}\right )}{8 e^{10}-8 e^5 x+2 x^2+\left (32 e^5-16 x\right ) \log \left (\frac {-1+7 \log (2)}{\log (2)}\right )+32 \log ^2\left (\frac {-1+7 \log (2)}{\log (2)}\right )} \, dx=-\frac {e^{5} + 4 \, \log \left (\frac {7 \, \log \left (2\right ) - 1}{\log \left (2\right )}\right )}{2 \, {\left (x - 2 \, e^{5} - 4 \, \log \left (\frac {7 \, \log \left (2\right ) - 1}{\log \left (2\right )}\right )\right )}} \]
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Time = 12.79 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.15 \[ \int \frac {e^5+4 \log \left (\frac {-1+7 \log (2)}{\log (2)}\right )}{8 e^{10}-8 e^5 x+2 x^2+\left (32 e^5-16 x\right ) \log \left (\frac {-1+7 \log (2)}{\log (2)}\right )+32 \log ^2\left (\frac {-1+7 \log (2)}{\log (2)}\right )} \, dx=-\frac {2\,\ln \left (\ln \left (128\right )-1\right )+\frac {{\mathrm {e}}^5}{2}-2\,\ln \left (\ln \left (2\right )\right )}{x-4\,\ln \left (\ln \left (128\right )-1\right )-2\,{\mathrm {e}}^5+4\,\ln \left (\ln \left (2\right )\right )} \]
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