Integrand size = 76, antiderivative size = 21 \[ \int \frac {e^5 (3-12 x)+x-4 x^2+\left (-x+2 x^2\right ) \log \left (x-2 x^2\right )}{-x^3+2 x^4+e^{10} \left (-9 x+18 x^2\right )+e^5 \left (-6 x^2+12 x^3\right )} \, dx=\frac {\log (-x (-1+2 x))}{-3 e^5-x} \]
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Leaf count is larger than twice the leaf count of optimal. \(85\) vs. \(2(21)=42\).
Time = 0.19 (sec) , antiderivative size = 85, normalized size of antiderivative = 4.05, number of steps used = 12, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.092, Rules used = {6820, 6874, 153, 2581, 36, 31, 29} \[ \int \frac {e^5 (3-12 x)+x-4 x^2+\left (-x+2 x^2\right ) \log \left (x-2 x^2\right )}{-x^3+2 x^4+e^{10} \left (-9 x+18 x^2\right )+e^5 \left (-6 x^2+12 x^3\right )} \, dx=-\frac {\log ((1-2 x) x)}{x+3 e^5}+\frac {\left (1+12 e^5\right ) \log \left (x+3 e^5\right )}{3 e^5 \left (1+6 e^5\right )}-\frac {2 \log \left (x+3 e^5\right )}{1+6 e^5}-\frac {\log \left (x+3 e^5\right )}{3 e^5} \]
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Rule 29
Rule 31
Rule 36
Rule 153
Rule 2581
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {-\frac {\left (3 e^5+x\right ) (-1+4 x)}{x (-1+2 x)}+\log \left (x-2 x^2\right )}{\left (3 e^5+x\right )^2} \, dx \\ & = \int \left (\frac {1-4 x}{x \left (3 e^5+x\right ) (-1+2 x)}+\frac {\log ((1-2 x) x)}{\left (3 e^5+x\right )^2}\right ) \, dx \\ & = \int \frac {1-4 x}{x \left (3 e^5+x\right ) (-1+2 x)} \, dx+\int \frac {\log ((1-2 x) x)}{\left (3 e^5+x\right )^2} \, dx \\ & = -\frac {\log ((1-2 x) x)}{3 e^5+x}-2 \int \frac {1}{(1-2 x) \left (3 e^5+x\right )} \, dx+\int \frac {1}{x \left (3 e^5+x\right )} \, dx+\int \left (-\frac {1}{3 e^5 x}+\frac {1+12 e^5}{3 e^5 \left (1+6 e^5\right ) \left (3 e^5+x\right )}-\frac {4}{\left (1+6 e^5\right ) (-1+2 x)}\right ) \, dx \\ & = -\frac {2 \log (1-2 x)}{1+6 e^5}-\frac {\log (x)}{3 e^5}-\frac {\log ((1-2 x) x)}{3 e^5+x}+\frac {\left (1+12 e^5\right ) \log \left (3 e^5+x\right )}{3 e^5 \left (1+6 e^5\right )}+\frac {\int \frac {1}{x} \, dx}{3 e^5}-\frac {\int \frac {1}{3 e^5+x} \, dx}{3 e^5}-\frac {2 \int \frac {1}{3 e^5+x} \, dx}{1+6 e^5}-\frac {4 \int \frac {1}{1-2 x} \, dx}{1+6 e^5} \\ & = -\frac {\log ((1-2 x) x)}{3 e^5+x}-\frac {\log \left (3 e^5+x\right )}{3 e^5}-\frac {2 \log \left (3 e^5+x\right )}{1+6 e^5}+\frac {\left (1+12 e^5\right ) \log \left (3 e^5+x\right )}{3 e^5 \left (1+6 e^5\right )} \\ \end{align*}
Time = 0.10 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {e^5 (3-12 x)+x-4 x^2+\left (-x+2 x^2\right ) \log \left (x-2 x^2\right )}{-x^3+2 x^4+e^{10} \left (-9 x+18 x^2\right )+e^5 \left (-6 x^2+12 x^3\right )} \, dx=-\frac {\log ((1-2 x) x)}{3 e^5+x} \]
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Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90
method | result | size |
norman | \(-\frac {\ln \left (-2 x^{2}+x \right )}{3 \,{\mathrm e}^{5}+x}\) | \(19\) |
risch | \(-\frac {\ln \left (-2 x^{2}+x \right )}{3 \,{\mathrm e}^{5}+x}\) | \(19\) |
parallelrisch | \(-\frac {\ln \left (-2 x^{2}+x \right )}{3 \,{\mathrm e}^{5}+x}\) | \(19\) |
default | \(\frac {\left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (9 \,{\mathrm e}^{10}+6 \textit {\_Z} \,{\mathrm e}^{5}+\textit {\_Z}^{2}\right )}{\sum }\left (-\frac {\underline {\hspace {1.25 ex}}\alpha }{-\left ({\mathrm e}^{5}\right )^{2}+{\mathrm e}^{10}}-\frac {3 \,{\mathrm e}^{5}}{-\left ({\mathrm e}^{5}\right )^{2}+{\mathrm e}^{10}}\right ) \left (\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (-2 x^{2}+x \right )-\left (\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right )-\ln \left (-\frac {x -\underline {\hspace {1.25 ex}}\alpha }{\underline {\hspace {1.25 ex}}\alpha }\right )\right ) \ln \left (\frac {x}{\underline {\hspace {1.25 ex}}\alpha }\right )+\operatorname {dilog}\left (-\frac {x -\underline {\hspace {1.25 ex}}\alpha }{\underline {\hspace {1.25 ex}}\alpha }\right )-\left (\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right )-\ln \left (\frac {2 x -2 \underline {\hspace {1.25 ex}}\alpha }{-2 \underline {\hspace {1.25 ex}}\alpha +1}\right )\right ) \ln \left (\frac {1-2 x}{-2 \underline {\hspace {1.25 ex}}\alpha +1}\right )+\operatorname {dilog}\left (\frac {2 x -2 \underline {\hspace {1.25 ex}}\alpha }{-2 \underline {\hspace {1.25 ex}}\alpha +1}\right )\right )\right )}{18}-\frac {2 \ln \left (-1+2 x \right )}{1+6 \,{\mathrm e}^{5}}-\frac {\left (-12 \,{\mathrm e}^{5}-1\right ) \ln \left (3 \,{\mathrm e}^{5}+x \right )}{3 \,{\mathrm e}^{5} \left (1+6 \,{\mathrm e}^{5}\right )}-\frac {\ln \left (x \right )}{3 \,{\mathrm e}^{5}}\) | \(213\) |
parts | \(\frac {\left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (9 \,{\mathrm e}^{10}+6 \textit {\_Z} \,{\mathrm e}^{5}+\textit {\_Z}^{2}\right )}{\sum }\left (-\frac {\underline {\hspace {1.25 ex}}\alpha }{-\left ({\mathrm e}^{5}\right )^{2}+{\mathrm e}^{10}}-\frac {3 \,{\mathrm e}^{5}}{-\left ({\mathrm e}^{5}\right )^{2}+{\mathrm e}^{10}}\right ) \left (\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (-2 x^{2}+x \right )-\left (\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right )-\ln \left (-\frac {x -\underline {\hspace {1.25 ex}}\alpha }{\underline {\hspace {1.25 ex}}\alpha }\right )\right ) \ln \left (\frac {x}{\underline {\hspace {1.25 ex}}\alpha }\right )+\operatorname {dilog}\left (-\frac {x -\underline {\hspace {1.25 ex}}\alpha }{\underline {\hspace {1.25 ex}}\alpha }\right )-\left (\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right )-\ln \left (\frac {2 x -2 \underline {\hspace {1.25 ex}}\alpha }{-2 \underline {\hspace {1.25 ex}}\alpha +1}\right )\right ) \ln \left (\frac {1-2 x}{-2 \underline {\hspace {1.25 ex}}\alpha +1}\right )+\operatorname {dilog}\left (\frac {2 x -2 \underline {\hspace {1.25 ex}}\alpha }{-2 \underline {\hspace {1.25 ex}}\alpha +1}\right )\right )\right )}{18}-\frac {2 \ln \left (-1+2 x \right )}{1+6 \,{\mathrm e}^{5}}-\frac {\left (-12 \,{\mathrm e}^{5}-1\right ) \ln \left (3 \,{\mathrm e}^{5}+x \right )}{3 \,{\mathrm e}^{5} \left (1+6 \,{\mathrm e}^{5}\right )}-\frac {\ln \left (x \right )}{3 \,{\mathrm e}^{5}}\) | \(213\) |
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Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {e^5 (3-12 x)+x-4 x^2+\left (-x+2 x^2\right ) \log \left (x-2 x^2\right )}{-x^3+2 x^4+e^{10} \left (-9 x+18 x^2\right )+e^5 \left (-6 x^2+12 x^3\right )} \, dx=-\frac {\log \left (-2 \, x^{2} + x\right )}{x + 3 \, e^{5}} \]
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Time = 0.11 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {e^5 (3-12 x)+x-4 x^2+\left (-x+2 x^2\right ) \log \left (x-2 x^2\right )}{-x^3+2 x^4+e^{10} \left (-9 x+18 x^2\right )+e^5 \left (-6 x^2+12 x^3\right )} \, dx=- \frac {\log {\left (- 2 x^{2} + x \right )}}{x + 3 e^{5}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 303 vs. \(2 (19) = 38\).
Time = 0.27 (sec) , antiderivative size = 303, normalized size of antiderivative = 14.43 \[ \int \frac {e^5 (3-12 x)+x-4 x^2+\left (-x+2 x^2\right ) \log \left (x-2 x^2\right )}{-x^3+2 x^4+e^{10} \left (-9 x+18 x^2\right )+e^5 \left (-6 x^2+12 x^3\right )} \, dx=-\frac {1}{3} \, {\left (e^{\left (-10\right )} \log \left (x\right ) - \frac {{\left (12 \, e^{5} + 1\right )} \log \left (x + 3 \, e^{5}\right )}{36 \, e^{20} + 12 \, e^{15} + e^{10}} - \frac {36 \, \log \left (2 \, x - 1\right )}{36 \, e^{10} + 12 \, e^{5} + 1} + \frac {3}{x {\left (6 \, e^{10} + e^{5}\right )} + 18 \, e^{15} + 3 \, e^{10}}\right )} e^{5} - 12 \, {\left (\frac {2 \, \log \left (2 \, x - 1\right )}{36 \, e^{10} + 12 \, e^{5} + 1} - \frac {2 \, \log \left (x + 3 \, e^{5}\right )}{36 \, e^{10} + 12 \, e^{5} + 1} + \frac {1}{x {\left (6 \, e^{5} + 1\right )} + 18 \, e^{10} + 3 \, e^{5}}\right )} e^{5} + \frac {1}{3} \, e^{\left (-5\right )} \log \left (x\right ) - \frac {{\left (12 \, e^{5} + 1\right )} \log \left (x + 3 \, e^{5}\right )}{3 \, {\left (6 \, e^{10} + e^{5}\right )}} - \frac {{\left (6 \, e^{5} + 1\right )} \log \left (x\right ) - {\left (2 \, x - 1\right )} \log \left (-2 \, x + 1\right )}{x {\left (6 \, e^{5} + 1\right )} + 18 \, e^{10} + 3 \, e^{5}} + \frac {12 \, e^{5}}{x {\left (6 \, e^{5} + 1\right )} + 18 \, e^{10} + 3 \, e^{5}} - \frac {2 \, \log \left (2 \, x - 1\right )}{36 \, e^{10} + 12 \, e^{5} + 1} + \frac {2 \, \log \left (x + 3 \, e^{5}\right )}{36 \, e^{10} + 12 \, e^{5} + 1} + \frac {1}{x {\left (6 \, e^{5} + 1\right )} + 18 \, e^{10} + 3 \, e^{5}} \]
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Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {e^5 (3-12 x)+x-4 x^2+\left (-x+2 x^2\right ) \log \left (x-2 x^2\right )}{-x^3+2 x^4+e^{10} \left (-9 x+18 x^2\right )+e^5 \left (-6 x^2+12 x^3\right )} \, dx=-\frac {\log \left (-2 \, x^{2} + x\right )}{x + 3 \, e^{5}} \]
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Time = 13.92 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {e^5 (3-12 x)+x-4 x^2+\left (-x+2 x^2\right ) \log \left (x-2 x^2\right )}{-x^3+2 x^4+e^{10} \left (-9 x+18 x^2\right )+e^5 \left (-6 x^2+12 x^3\right )} \, dx=-\frac {\ln \left (x-2\,x^2\right )}{x+3\,{\mathrm {e}}^5} \]
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