Integrand size = 128, antiderivative size = 23 \[ \int \frac {-36 x-36 x^2+\left (-36 x-12 x^2\right ) \log \left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )+(30+10 x) \log ^2\left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )}{(3+x) \log ^2\left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )} \, dx=2 x \left (5-\frac {3 x}{\log \left (\frac {1}{e^2 x^2 (3+x)^4}\right )}\right ) \]
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\[ \int \frac {-36 x-36 x^2+\left (-36 x-12 x^2\right ) \log \left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )+(30+10 x) \log ^2\left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )}{(3+x) \log ^2\left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )} \, dx=\int \frac {-36 x-36 x^2+\left (-36 x-12 x^2\right ) \log \left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )+(30+10 x) \log ^2\left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )}{(3+x) \log ^2\left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {2 \left (60+38 x-6 x^2-\left (60+38 x+6 x^2\right ) \log \left (\frac {1}{x^2 (3+x)^4}\right )+5 (3+x) \log ^2\left (\frac {1}{x^2 (3+x)^4}\right )\right )}{(3+x) \left (2-\log \left (\frac {1}{x^2 (3+x)^4}\right )\right )^2} \, dx \\ & = 2 \int \frac {60+38 x-6 x^2-\left (60+38 x+6 x^2\right ) \log \left (\frac {1}{x^2 (3+x)^4}\right )+5 (3+x) \log ^2\left (\frac {1}{x^2 (3+x)^4}\right )}{(3+x) \left (2-\log \left (\frac {1}{x^2 (3+x)^4}\right )\right )^2} \, dx \\ & = 2 \int \left (5-\frac {18 x (1+x)}{(3+x) \left (-2+\log \left (\frac {1}{x^2 (3+x)^4}\right )\right )^2}-\frac {6 x}{-2+\log \left (\frac {1}{x^2 (3+x)^4}\right )}\right ) \, dx \\ & = 10 x-12 \int \frac {x}{-2+\log \left (\frac {1}{x^2 (3+x)^4}\right )} \, dx-36 \int \frac {x (1+x)}{(3+x) \left (-2+\log \left (\frac {1}{x^2 (3+x)^4}\right )\right )^2} \, dx \\ & = 10 x-12 \int \frac {x}{-2+\log \left (\frac {1}{x^2 (3+x)^4}\right )} \, dx-36 \int \left (-\frac {2}{\left (-2+\log \left (\frac {1}{x^2 (3+x)^4}\right )\right )^2}+\frac {x}{\left (-2+\log \left (\frac {1}{x^2 (3+x)^4}\right )\right )^2}+\frac {6}{(3+x) \left (-2+\log \left (\frac {1}{x^2 (3+x)^4}\right )\right )^2}\right ) \, dx \\ & = 10 x-12 \int \frac {x}{-2+\log \left (\frac {1}{x^2 (3+x)^4}\right )} \, dx-36 \int \frac {x}{\left (-2+\log \left (\frac {1}{x^2 (3+x)^4}\right )\right )^2} \, dx+72 \int \frac {1}{\left (-2+\log \left (\frac {1}{x^2 (3+x)^4}\right )\right )^2} \, dx-216 \int \frac {1}{(3+x) \left (-2+\log \left (\frac {1}{x^2 (3+x)^4}\right )\right )^2} \, dx \\ \end{align*}
Time = 0.20 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {-36 x-36 x^2+\left (-36 x-12 x^2\right ) \log \left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )+(30+10 x) \log ^2\left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )}{(3+x) \log ^2\left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )} \, dx=-2 \left (-5 x+\frac {3 x^2}{-2+\log \left (\frac {1}{x^2 (3+x)^4}\right )}\right ) \]
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Time = 0.85 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.83
method | result | size |
risch | \(10 x -\frac {6 x^{2}}{\ln \left (\frac {{\mathrm e}^{-2}}{x^{6}+12 x^{5}+54 x^{4}+108 x^{3}+81 x^{2}}\right )}\) | \(42\) |
norman | \(\frac {-6 x^{2}+10 x \ln \left (\frac {{\mathrm e}^{-2}}{x^{6}+12 x^{5}+54 x^{4}+108 x^{3}+81 x^{2}}\right )}{\ln \left (\frac {{\mathrm e}^{-2}}{x^{6}+12 x^{5}+54 x^{4}+108 x^{3}+81 x^{2}}\right )}\) | \(77\) |
parallelrisch | \(\frac {-24 x^{2}+40 \ln \left (\frac {{\mathrm e}^{-2}}{x^{2} \left (x^{4}+12 x^{3}+54 x^{2}+108 x +81\right )}\right ) x -240 \ln \left (\frac {{\mathrm e}^{-2}}{x^{2} \left (x^{4}+12 x^{3}+54 x^{2}+108 x +81\right )}\right )}{4 \ln \left (\frac {{\mathrm e}^{-2}}{x^{2} \left (x^{4}+12 x^{3}+54 x^{2}+108 x +81\right )}\right )}\) | \(103\) |
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Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (22) = 44\).
Time = 0.23 (sec) , antiderivative size = 73, normalized size of antiderivative = 3.17 \[ \int \frac {-36 x-36 x^2+\left (-36 x-12 x^2\right ) \log \left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )+(30+10 x) \log ^2\left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )}{(3+x) \log ^2\left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )} \, dx=-\frac {2 \, {\left (3 \, x^{2} - 5 \, x \log \left (\frac {e^{\left (-2\right )}}{x^{6} + 12 \, x^{5} + 54 \, x^{4} + 108 \, x^{3} + 81 \, x^{2}}\right )\right )}}{\log \left (\frac {e^{\left (-2\right )}}{x^{6} + 12 \, x^{5} + 54 \, x^{4} + 108 \, x^{3} + 81 \, x^{2}}\right )} \]
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Time = 0.09 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.61 \[ \int \frac {-36 x-36 x^2+\left (-36 x-12 x^2\right ) \log \left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )+(30+10 x) \log ^2\left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )}{(3+x) \log ^2\left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )} \, dx=- \frac {6 x^{2}}{\log {\left (\frac {1}{\left (x^{6} + 12 x^{5} + 54 x^{4} + 108 x^{3} + 81 x^{2}\right ) e^{2}} \right )}} + 10 x \]
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Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48 \[ \int \frac {-36 x-36 x^2+\left (-36 x-12 x^2\right ) \log \left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )+(30+10 x) \log ^2\left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )}{(3+x) \log ^2\left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )} \, dx=\frac {3 \, x^{2} + 20 \, x \log \left (x + 3\right ) + 10 \, x \log \left (x\right ) + 10 \, x}{2 \, \log \left (x + 3\right ) + \log \left (x\right ) + 1} \]
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Time = 0.34 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.65 \[ \int \frac {-36 x-36 x^2+\left (-36 x-12 x^2\right ) \log \left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )+(30+10 x) \log ^2\left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )}{(3+x) \log ^2\left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )} \, dx=10 \, x + \frac {6 \, x^{2}}{\log \left (x^{6} + 12 \, x^{5} + 54 \, x^{4} + 108 \, x^{3} + 81 \, x^{2}\right ) + 2} \]
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Time = 8.31 (sec) , antiderivative size = 99, normalized size of antiderivative = 4.30 \[ \int \frac {-36 x-36 x^2+\left (-36 x-12 x^2\right ) \log \left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )+(30+10 x) \log ^2\left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )}{(3+x) \log ^2\left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )} \, dx=14\,x+\frac {4}{x+1}+2\,x^2-\frac {6\,x^2+\frac {2\,x^2\,\ln \left (\frac {{\mathrm {e}}^{-2}}{x^6+12\,x^5+54\,x^4+108\,x^3+81\,x^2}\right )\,\left (x+3\right )}{x+1}}{\ln \left (\frac {{\mathrm {e}}^{-2}}{x^6+12\,x^5+54\,x^4+108\,x^3+81\,x^2}\right )} \]
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