Integrand size = 149, antiderivative size = 24 \[ \int \frac {\left (-2916 x-1458 x^2\right ) \log \left (4+4 x+x^2\right )-108 x \log \left (\log \left (4+4 x+x^2\right )\right )+(-54-27 x) \log \left (4+4 x+x^2\right ) \log ^2\left (\log \left (4+4 x+x^2\right )\right )}{\left (1458 x^4+729 x^5\right ) \log \left (4+4 x+x^2\right )+\left (108 x^3+54 x^4\right ) \log \left (4+4 x+x^2\right ) \log ^2\left (\log \left (4+4 x+x^2\right )\right )+\left (2 x^2+x^3\right ) \log \left (4+4 x+x^2\right ) \log ^4\left (\log \left (4+4 x+x^2\right )\right )} \, dx=\frac {3}{x \left (3 x+\frac {1}{9} \log ^2\left (\log \left ((2+x)^2\right )\right )\right )} \]
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Time = 0.24 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.020, Rules used = {6820, 12, 6819} \[ \int \frac {\left (-2916 x-1458 x^2\right ) \log \left (4+4 x+x^2\right )-108 x \log \left (\log \left (4+4 x+x^2\right )\right )+(-54-27 x) \log \left (4+4 x+x^2\right ) \log ^2\left (\log \left (4+4 x+x^2\right )\right )}{\left (1458 x^4+729 x^5\right ) \log \left (4+4 x+x^2\right )+\left (108 x^3+54 x^4\right ) \log \left (4+4 x+x^2\right ) \log ^2\left (\log \left (4+4 x+x^2\right )\right )+\left (2 x^2+x^3\right ) \log \left (4+4 x+x^2\right ) \log ^4\left (\log \left (4+4 x+x^2\right )\right )} \, dx=\frac {27}{x \left (27 x+\log ^2\left (\log \left ((x+2)^2\right )\right )\right )} \]
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Rule 12
Rule 6819
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \int \frac {27 \left (-4 x \log \left (\log \left ((2+x)^2\right )\right )-(2+x) \log \left ((2+x)^2\right ) \left (54 x+\log ^2\left (\log \left ((2+x)^2\right )\right )\right )\right )}{x^2 (2+x) \log \left ((2+x)^2\right ) \left (27 x+\log ^2\left (\log \left ((2+x)^2\right )\right )\right )^2} \, dx \\ & = 27 \int \frac {-4 x \log \left (\log \left ((2+x)^2\right )\right )-(2+x) \log \left ((2+x)^2\right ) \left (54 x+\log ^2\left (\log \left ((2+x)^2\right )\right )\right )}{x^2 (2+x) \log \left ((2+x)^2\right ) \left (27 x+\log ^2\left (\log \left ((2+x)^2\right )\right )\right )^2} \, dx \\ & = \frac {27}{x \left (27 x+\log ^2\left (\log \left ((2+x)^2\right )\right )\right )} \\ \end{align*}
Time = 0.30 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {\left (-2916 x-1458 x^2\right ) \log \left (4+4 x+x^2\right )-108 x \log \left (\log \left (4+4 x+x^2\right )\right )+(-54-27 x) \log \left (4+4 x+x^2\right ) \log ^2\left (\log \left (4+4 x+x^2\right )\right )}{\left (1458 x^4+729 x^5\right ) \log \left (4+4 x+x^2\right )+\left (108 x^3+54 x^4\right ) \log \left (4+4 x+x^2\right ) \log ^2\left (\log \left (4+4 x+x^2\right )\right )+\left (2 x^2+x^3\right ) \log \left (4+4 x+x^2\right ) \log ^4\left (\log \left (4+4 x+x^2\right )\right )} \, dx=\frac {27}{x \left (27 x+\log ^2\left (\log \left ((2+x)^2\right )\right )\right )} \]
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Time = 1.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00
method | result | size |
parallelrisch | \(\frac {27}{\left ({\ln \left (\ln \left (x^{2}+4 x +4\right )\right )}^{2}+27 x \right ) x}\) | \(24\) |
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Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-2916 x-1458 x^2\right ) \log \left (4+4 x+x^2\right )-108 x \log \left (\log \left (4+4 x+x^2\right )\right )+(-54-27 x) \log \left (4+4 x+x^2\right ) \log ^2\left (\log \left (4+4 x+x^2\right )\right )}{\left (1458 x^4+729 x^5\right ) \log \left (4+4 x+x^2\right )+\left (108 x^3+54 x^4\right ) \log \left (4+4 x+x^2\right ) \log ^2\left (\log \left (4+4 x+x^2\right )\right )+\left (2 x^2+x^3\right ) \log \left (4+4 x+x^2\right ) \log ^4\left (\log \left (4+4 x+x^2\right )\right )} \, dx=\frac {27}{x \log \left (\log \left (x^{2} + 4 \, x + 4\right )\right )^{2} + 27 \, x^{2}} \]
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Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {\left (-2916 x-1458 x^2\right ) \log \left (4+4 x+x^2\right )-108 x \log \left (\log \left (4+4 x+x^2\right )\right )+(-54-27 x) \log \left (4+4 x+x^2\right ) \log ^2\left (\log \left (4+4 x+x^2\right )\right )}{\left (1458 x^4+729 x^5\right ) \log \left (4+4 x+x^2\right )+\left (108 x^3+54 x^4\right ) \log \left (4+4 x+x^2\right ) \log ^2\left (\log \left (4+4 x+x^2\right )\right )+\left (2 x^2+x^3\right ) \log \left (4+4 x+x^2\right ) \log ^4\left (\log \left (4+4 x+x^2\right )\right )} \, dx=\frac {27}{27 x^{2} + x \log {\left (\log {\left (x^{2} + 4 x + 4 \right )} \right )}^{2}} \]
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Time = 0.33 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {\left (-2916 x-1458 x^2\right ) \log \left (4+4 x+x^2\right )-108 x \log \left (\log \left (4+4 x+x^2\right )\right )+(-54-27 x) \log \left (4+4 x+x^2\right ) \log ^2\left (\log \left (4+4 x+x^2\right )\right )}{\left (1458 x^4+729 x^5\right ) \log \left (4+4 x+x^2\right )+\left (108 x^3+54 x^4\right ) \log \left (4+4 x+x^2\right ) \log ^2\left (\log \left (4+4 x+x^2\right )\right )+\left (2 x^2+x^3\right ) \log \left (4+4 x+x^2\right ) \log ^4\left (\log \left (4+4 x+x^2\right )\right )} \, dx=\frac {27}{x \log \left (2\right )^{2} + 2 \, x \log \left (2\right ) \log \left (\log \left (x + 2\right )\right ) + x \log \left (\log \left (x + 2\right )\right )^{2} + 27 \, x^{2}} \]
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Time = 1.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-2916 x-1458 x^2\right ) \log \left (4+4 x+x^2\right )-108 x \log \left (\log \left (4+4 x+x^2\right )\right )+(-54-27 x) \log \left (4+4 x+x^2\right ) \log ^2\left (\log \left (4+4 x+x^2\right )\right )}{\left (1458 x^4+729 x^5\right ) \log \left (4+4 x+x^2\right )+\left (108 x^3+54 x^4\right ) \log \left (4+4 x+x^2\right ) \log ^2\left (\log \left (4+4 x+x^2\right )\right )+\left (2 x^2+x^3\right ) \log \left (4+4 x+x^2\right ) \log ^4\left (\log \left (4+4 x+x^2\right )\right )} \, dx=\frac {27}{x \log \left (\log \left (x^{2} + 4 \, x + 4\right )\right )^{2} + 27 \, x^{2}} \]
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Time = 12.62 (sec) , antiderivative size = 199, normalized size of antiderivative = 8.29 \[ \int \frac {\left (-2916 x-1458 x^2\right ) \log \left (4+4 x+x^2\right )-108 x \log \left (\log \left (4+4 x+x^2\right )\right )+(-54-27 x) \log \left (4+4 x+x^2\right ) \log ^2\left (\log \left (4+4 x+x^2\right )\right )}{\left (1458 x^4+729 x^5\right ) \log \left (4+4 x+x^2\right )+\left (108 x^3+54 x^4\right ) \log \left (4+4 x+x^2\right ) \log ^2\left (\log \left (4+4 x+x^2\right )\right )+\left (2 x^2+x^3\right ) \log \left (4+4 x+x^2\right ) \log ^4\left (\log \left (4+4 x+x^2\right )\right )} \, dx=\frac {27\,{\left (2\,\ln \left (x^2+4\,x+4\right )+x\,\ln \left (x^2+4\,x+4\right )\right )}^2\,\left (27\,x^2\,{\ln \left (x^2+4\,x+4\right )}^2+108\,x\,{\ln \left (x^2+4\,x+4\right )}^2+16\,x+108\,{\ln \left (x^2+4\,x+4\right )}^2\right )}{x\,\ln \left (x^2+4\,x+4\right )\,\left ({\ln \left (\ln \left (x^2+4\,x+4\right )\right )}^2+27\,x\right )\,\left (x+2\right )\,\left (27\,x^3\,{\ln \left (x^2+4\,x+4\right )}^3+162\,x^2\,{\ln \left (x^2+4\,x+4\right )}^3+16\,x^2\,\ln \left (x^2+4\,x+4\right )+324\,x\,{\ln \left (x^2+4\,x+4\right )}^3+32\,x\,\ln \left (x^2+4\,x+4\right )+216\,{\ln \left (x^2+4\,x+4\right )}^3\right )} \]
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