Integrand size = 150, antiderivative size = 27 \[ \int \frac {-\log (4)+\left (x \log (4)+e^{e^8+2 e^4 x+x^2} \left (2 e^4 x+2 x^2\right ) \log (4)\right ) \log (x)}{\left (16 x+e^{2 e^8+4 e^4 x+2 x^2} x-8 x^2+x^3+e^{e^8+2 e^4 x+x^2} \left (-8 x+2 x^2\right )\right ) \log (x)+\left (8 x-2 e^{e^8+2 e^4 x+x^2} x-2 x^2\right ) \log (x) \log (\log (x))+x \log (x) \log ^2(\log (x))} \, dx=\frac {1}{e^2}-\frac {\log (4)}{-4+e^{\left (e^4+x\right )^2}+x-\log (\log (x))} \]
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Time = 0.64 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.020, Rules used = {6820, 12, 6818} \[ \int \frac {-\log (4)+\left (x \log (4)+e^{e^8+2 e^4 x+x^2} \left (2 e^4 x+2 x^2\right ) \log (4)\right ) \log (x)}{\left (16 x+e^{2 e^8+4 e^4 x+2 x^2} x-8 x^2+x^3+e^{e^8+2 e^4 x+x^2} \left (-8 x+2 x^2\right )\right ) \log (x)+\left (8 x-2 e^{e^8+2 e^4 x+x^2} x-2 x^2\right ) \log (x) \log (\log (x))+x \log (x) \log ^2(\log (x))} \, dx=\frac {\log (4)}{-x-e^{\left (x+e^4\right )^2}+\log (\log (x))+4} \]
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Rule 12
Rule 6818
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \int \frac {\log (4) \left (-1+x \left (1+2 e^{4+e^8+2 e^4 x+x^2}+2 e^{\left (e^4+x\right )^2} x\right ) \log (x)\right )}{x \log (x) \left (4-e^{\left (e^4+x\right )^2}-x+\log (\log (x))\right )^2} \, dx \\ & = \log (4) \int \frac {-1+x \left (1+2 e^{4+e^8+2 e^4 x+x^2}+2 e^{\left (e^4+x\right )^2} x\right ) \log (x)}{x \log (x) \left (4-e^{\left (e^4+x\right )^2}-x+\log (\log (x))\right )^2} \, dx \\ & = \frac {\log (4)}{4-e^{\left (e^4+x\right )^2}-x+\log (\log (x))} \\ \end{align*}
Time = 0.17 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {-\log (4)+\left (x \log (4)+e^{e^8+2 e^4 x+x^2} \left (2 e^4 x+2 x^2\right ) \log (4)\right ) \log (x)}{\left (16 x+e^{2 e^8+4 e^4 x+2 x^2} x-8 x^2+x^3+e^{e^8+2 e^4 x+x^2} \left (-8 x+2 x^2\right )\right ) \log (x)+\left (8 x-2 e^{e^8+2 e^4 x+x^2} x-2 x^2\right ) \log (x) \log (\log (x))+x \log (x) \log ^2(\log (x))} \, dx=-\frac {\log (4)}{-4+e^{\left (e^4+x\right )^2}+x-\log (\log (x))} \]
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Time = 3.49 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00
method | result | size |
risch | \(-\frac {2 \ln \left (2\right )}{x +{\mathrm e}^{{\mathrm e}^{8}+2 x \,{\mathrm e}^{4}+x^{2}}-\ln \left (\ln \left (x \right )\right )-4}\) | \(27\) |
parallelrisch | \(-\frac {2 \ln \left (2\right )}{x +{\mathrm e}^{{\mathrm e}^{8}+2 x \,{\mathrm e}^{4}+x^{2}}-\ln \left (\ln \left (x \right )\right )-4}\) | \(29\) |
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Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {-\log (4)+\left (x \log (4)+e^{e^8+2 e^4 x+x^2} \left (2 e^4 x+2 x^2\right ) \log (4)\right ) \log (x)}{\left (16 x+e^{2 e^8+4 e^4 x+2 x^2} x-8 x^2+x^3+e^{e^8+2 e^4 x+x^2} \left (-8 x+2 x^2\right )\right ) \log (x)+\left (8 x-2 e^{e^8+2 e^4 x+x^2} x-2 x^2\right ) \log (x) \log (\log (x))+x \log (x) \log ^2(\log (x))} \, dx=-\frac {2 \, \log \left (2\right )}{x + e^{\left (x^{2} + 2 \, x e^{4} + e^{8}\right )} - \log \left (\log \left (x\right )\right ) - 4} \]
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Time = 0.14 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {-\log (4)+\left (x \log (4)+e^{e^8+2 e^4 x+x^2} \left (2 e^4 x+2 x^2\right ) \log (4)\right ) \log (x)}{\left (16 x+e^{2 e^8+4 e^4 x+2 x^2} x-8 x^2+x^3+e^{e^8+2 e^4 x+x^2} \left (-8 x+2 x^2\right )\right ) \log (x)+\left (8 x-2 e^{e^8+2 e^4 x+x^2} x-2 x^2\right ) \log (x) \log (\log (x))+x \log (x) \log ^2(\log (x))} \, dx=- \frac {2 \log {\left (2 \right )}}{x + e^{x^{2} + 2 x e^{4} + e^{8}} - \log {\left (\log {\left (x \right )} \right )} - 4} \]
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Time = 0.34 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {-\log (4)+\left (x \log (4)+e^{e^8+2 e^4 x+x^2} \left (2 e^4 x+2 x^2\right ) \log (4)\right ) \log (x)}{\left (16 x+e^{2 e^8+4 e^4 x+2 x^2} x-8 x^2+x^3+e^{e^8+2 e^4 x+x^2} \left (-8 x+2 x^2\right )\right ) \log (x)+\left (8 x-2 e^{e^8+2 e^4 x+x^2} x-2 x^2\right ) \log (x) \log (\log (x))+x \log (x) \log ^2(\log (x))} \, dx=-\frac {2 \, \log \left (2\right )}{x + e^{\left (x^{2} + 2 \, x e^{4} + e^{8}\right )} - \log \left (\log \left (x\right )\right ) - 4} \]
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Time = 0.47 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {-\log (4)+\left (x \log (4)+e^{e^8+2 e^4 x+x^2} \left (2 e^4 x+2 x^2\right ) \log (4)\right ) \log (x)}{\left (16 x+e^{2 e^8+4 e^4 x+2 x^2} x-8 x^2+x^3+e^{e^8+2 e^4 x+x^2} \left (-8 x+2 x^2\right )\right ) \log (x)+\left (8 x-2 e^{e^8+2 e^4 x+x^2} x-2 x^2\right ) \log (x) \log (\log (x))+x \log (x) \log ^2(\log (x))} \, dx=-\frac {2 \, \log \left (2\right )}{x + e^{\left (x^{2} + 2 \, x e^{4} + e^{8}\right )} - \log \left (\log \left (x\right )\right ) - 4} \]
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Time = 9.19 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {-\log (4)+\left (x \log (4)+e^{e^8+2 e^4 x+x^2} \left (2 e^4 x+2 x^2\right ) \log (4)\right ) \log (x)}{\left (16 x+e^{2 e^8+4 e^4 x+2 x^2} x-8 x^2+x^3+e^{e^8+2 e^4 x+x^2} \left (-8 x+2 x^2\right )\right ) \log (x)+\left (8 x-2 e^{e^8+2 e^4 x+x^2} x-2 x^2\right ) \log (x) \log (\log (x))+x \log (x) \log ^2(\log (x))} \, dx=-\frac {2\,\ln \left (2\right )}{x-\ln \left (\ln \left (x\right )\right )+{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{2\,x\,{\mathrm {e}}^4}\,{\mathrm {e}}^{{\mathrm {e}}^8}-4} \]
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