Integrand size = 89, antiderivative size = 24 \[ \int \frac {2+2 e-e^{25+x}+2 x+(-2-2 e-4 x) \log (\log (2))+2 x \log ^2(\log (2))}{1+e^2-e^{25+x}+2 x+x^2+e (2+2 x)+\left (-2 x-2 e x-2 x^2\right ) \log (\log (2))+x^2 \log ^2(\log (2))} \, dx=\log \left (e^{25+x}-(-1-e-x+x \log (\log (2)))^2\right ) \]
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Time = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.92, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.022, Rules used = {6, 6816} \[ \int \frac {2+2 e-e^{25+x}+2 x+(-2-2 e-4 x) \log (\log (2))+2 x \log ^2(\log (2))}{1+e^2-e^{25+x}+2 x+x^2+e (2+2 x)+\left (-2 x-2 e x-2 x^2\right ) \log (\log (2))+x^2 \log ^2(\log (2))} \, dx=\log \left (x^2 \left (1+\log ^2(\log (2))\right )-2 \left (x^2+e x+x\right ) \log (\log (2))+2 x-e^{x+25}+2 e (x+1)+e^2+1\right ) \]
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Rule 6
Rule 6816
Rubi steps \begin{align*} \text {integral}& = \int \frac {2+2 e-e^{25+x}+(-2-2 e-4 x) \log (\log (2))+x \left (2+2 \log ^2(\log (2))\right )}{1+e^2-e^{25+x}+2 x+x^2+e (2+2 x)+\left (-2 x-2 e x-2 x^2\right ) \log (\log (2))+x^2 \log ^2(\log (2))} \, dx \\ & = \int \frac {2+2 e-e^{25+x}+(-2-2 e-4 x) \log (\log (2))+x \left (2+2 \log ^2(\log (2))\right )}{1+e^2-e^{25+x}+2 x+e (2+2 x)+\left (-2 x-2 e x-2 x^2\right ) \log (\log (2))+x^2 \left (1+\log ^2(\log (2))\right )} \, dx \\ & = \log \left (1+e^2-e^{25+x}+2 x+2 e (1+x)-2 \left (x+e x+x^2\right ) \log (\log (2))+x^2 \left (1+\log ^2(\log (2))\right )\right ) \\ \end{align*}
Time = 0.85 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {2+2 e-e^{25+x}+2 x+(-2-2 e-4 x) \log (\log (2))+2 x \log ^2(\log (2))}{1+e^2-e^{25+x}+2 x+x^2+e (2+2 x)+\left (-2 x-2 e x-2 x^2\right ) \log (\log (2))+x^2 \log ^2(\log (2))} \, dx=\log \left (-e^2+e^{25+x}+2 e (-1+x (-1+\log (\log (2))))-(-1+x (-1+\log (\log (2))))^2\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. \(54\) vs. \(2(24)=48\).
Time = 0.16 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.29
method | result | size |
derivativedivides | \(\ln \left (x^{2} \ln \left (\ln \left (2\right )\right )^{2}+\left (-2 x \,{\mathrm e}-2 x^{2}-2 x \right ) \ln \left (\ln \left (2\right )\right )-{\mathrm e}^{x +25}+{\mathrm e}^{2}+\left (2+2 x \right ) {\mathrm e}+x^{2}+2 x +1\right )\) | \(55\) |
default | \(\ln \left (x^{2} \ln \left (\ln \left (2\right )\right )^{2}+\left (-2 x \,{\mathrm e}-2 x^{2}-2 x \right ) \ln \left (\ln \left (2\right )\right )-{\mathrm e}^{x +25}+{\mathrm e}^{2}+\left (2+2 x \right ) {\mathrm e}+x^{2}+2 x +1\right )\) | \(55\) |
norman | \(\ln \left (x^{2} \ln \left (\ln \left (2\right )\right )^{2}-2 \ln \left (\ln \left (2\right )\right ) {\mathrm e} x -2 x^{2} \ln \left (\ln \left (2\right )\right )-2 x \ln \left (\ln \left (2\right )\right )+{\mathrm e}^{2}+2 x \,{\mathrm e}+x^{2}-{\mathrm e}^{x +25}+2 \,{\mathrm e}+2 x +1\right )\) | \(60\) |
risch | \(-25+\ln \left (-x^{2} \ln \left (\ln \left (2\right )\right )^{2}+2 \ln \left (\ln \left (2\right )\right ) {\mathrm e} x +2 x^{2} \ln \left (\ln \left (2\right )\right )+2 x \ln \left (\ln \left (2\right )\right )-{\mathrm e}^{2}-2 x \,{\mathrm e}-x^{2}-2 \,{\mathrm e}-2 x +{\mathrm e}^{x +25}-1\right )\) | \(63\) |
parallelrisch | \(\ln \left (\frac {x^{2} \ln \left (\ln \left (2\right )\right )^{2}-2 \ln \left (\ln \left (2\right )\right ) {\mathrm e} x -2 x^{2} \ln \left (\ln \left (2\right )\right )-2 x \ln \left (\ln \left (2\right )\right )+{\mathrm e}^{2}+2 x \,{\mathrm e}+x^{2}-{\mathrm e}^{x +25}+2 \,{\mathrm e}+2 x +1}{\ln \left (\ln \left (2\right )\right )^{2}-2 \ln \left (\ln \left (2\right )\right )+1}\right )\) | \(75\) |
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Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (24) = 48\).
Time = 0.24 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.08 \[ \int \frac {2+2 e-e^{25+x}+2 x+(-2-2 e-4 x) \log (\log (2))+2 x \log ^2(\log (2))}{1+e^2-e^{25+x}+2 x+x^2+e (2+2 x)+\left (-2 x-2 e x-2 x^2\right ) \log (\log (2))+x^2 \log ^2(\log (2))} \, dx=\log \left (-x^{2} \log \left (\log \left (2\right )\right )^{2} - x^{2} - 2 \, {\left (x + 1\right )} e + 2 \, {\left (x^{2} + x e + x\right )} \log \left (\log \left (2\right )\right ) - 2 \, x - e^{2} + e^{\left (x + 25\right )} - 1\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (20) = 40\).
Time = 0.13 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.92 \[ \int \frac {2+2 e-e^{25+x}+2 x+(-2-2 e-4 x) \log (\log (2))+2 x \log ^2(\log (2))}{1+e^2-e^{25+x}+2 x+x^2+e (2+2 x)+\left (-2 x-2 e x-2 x^2\right ) \log (\log (2))+x^2 \log ^2(\log (2))} \, dx=\log {\left (- x^{2} + 2 x^{2} \log {\left (\log {\left (2 \right )} \right )} - x^{2} \log {\left (\log {\left (2 \right )} \right )}^{2} - 2 e x - 2 x + 2 e x \log {\left (\log {\left (2 \right )} \right )} + 2 x \log {\left (\log {\left (2 \right )} \right )} + e^{x + 25} - e^{2} - 2 e - 1 \right )} \]
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Time = 0.19 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.96 \[ \int \frac {2+2 e-e^{25+x}+2 x+(-2-2 e-4 x) \log (\log (2))+2 x \log ^2(\log (2))}{1+e^2-e^{25+x}+2 x+x^2+e (2+2 x)+\left (-2 x-2 e x-2 x^2\right ) \log (\log (2))+x^2 \log ^2(\log (2))} \, dx=\log \left (x^{2} \log \left (\log \left (2\right )\right )^{2} + x^{2} + 2 \, {\left (x + 1\right )} e - 2 \, {\left (x^{2} + x e + x\right )} \log \left (\log \left (2\right )\right ) + 2 \, x + e^{2} - e^{\left (x + 25\right )} + 1\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (24) = 48\).
Time = 0.28 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.50 \[ \int \frac {2+2 e-e^{25+x}+2 x+(-2-2 e-4 x) \log (\log (2))+2 x \log ^2(\log (2))}{1+e^2-e^{25+x}+2 x+x^2+e (2+2 x)+\left (-2 x-2 e x-2 x^2\right ) \log (\log (2))+x^2 \log ^2(\log (2))} \, dx=\log \left (-x^{2} \log \left (\log \left (2\right )\right )^{2} + 2 \, x^{2} \log \left (\log \left (2\right )\right ) + 2 \, x e \log \left (\log \left (2\right )\right ) - x^{2} - 2 \, x e + 2 \, x \log \left (\log \left (2\right )\right ) - 2 \, x - e^{2} - 2 \, e + e^{\left (x + 25\right )} - 1\right ) \]
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Time = 12.70 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.21 \[ \int \frac {2+2 e-e^{25+x}+2 x+(-2-2 e-4 x) \log (\log (2))+2 x \log ^2(\log (2))}{1+e^2-e^{25+x}+2 x+x^2+e (2+2 x)+\left (-2 x-2 e x-2 x^2\right ) \log (\log (2))+x^2 \log ^2(\log (2))} \, dx=\ln \left (2\,x-{\mathrm {e}}^{x+25}+2\,\mathrm {e}+{\mathrm {e}}^2-2\,x^2\,\ln \left (\ln \left (2\right )\right )+2\,x\,\mathrm {e}+x^2\,{\ln \left (\ln \left (2\right )\right )}^2+x^2-2\,x\,\ln \left (\ln \left (2\right )\right )\,\left (\mathrm {e}+1\right )+1\right ) \]
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