Integrand size = 83, antiderivative size = 24 \[ \int \frac {-189-198 x+e^4 \left (22 x+23 x^2\right )+\left (-9-9 x+e^4 \left (x+x^2\right )\right ) \log \left (\frac {9 x+9 x^2+e^4 \left (-x^2-x^3\right )}{e^4}\right )}{-180-180 x+e^4 \left (20 x+20 x^2\right )} \, dx=3+x+\frac {1}{20} x \log \left (\left (\frac {9}{e^4}-x\right ) \left (x+x^2\right )\right ) \]
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Time = 0.31 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.072, Rules used = {6873, 6820, 12, 6874, 907, 2603} \[ \int \frac {-189-198 x+e^4 \left (22 x+23 x^2\right )+\left (-9-9 x+e^4 \left (x+x^2\right )\right ) \log \left (\frac {9 x+9 x^2+e^4 \left (-x^2-x^3\right )}{e^4}\right )}{-180-180 x+e^4 \left (20 x+20 x^2\right )} \, dx=\frac {4 x}{5}+\frac {1}{20} x \log \left (x (x+1) \left (9-e^4 x\right )\right ) \]
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Rule 12
Rule 907
Rule 2603
Rule 6820
Rule 6873
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {189+198 x-e^4 \left (22 x+23 x^2\right )-\left (-9-9 x+e^4 \left (x+x^2\right )\right ) \log \left (\frac {9 x+9 x^2+e^4 \left (-x^2-x^3\right )}{e^4}\right )}{180+20 \left (9-e^4\right ) x-20 e^4 x^2} \, dx \\ & = \int \frac {153-18 \left (-9+e^4\right ) x-19 e^4 x^2-(1+x) \left (-9+e^4 x\right ) \log \left (-x (1+x) \left (-9+e^4 x\right )\right )}{20 (1+x) \left (9-e^4 x\right )} \, dx \\ & = \frac {1}{20} \int \frac {153-18 \left (-9+e^4\right ) x-19 e^4 x^2-(1+x) \left (-9+e^4 x\right ) \log \left (-x (1+x) \left (-9+e^4 x\right )\right )}{(1+x) \left (9-e^4 x\right )} \, dx \\ & = \frac {1}{20} \int \left (\frac {153+18 \left (9-e^4\right ) x-19 e^4 x^2}{(1+x) \left (9-e^4 x\right )}+\log \left (-x (1+x) \left (-9+e^4 x\right )\right )\right ) \, dx \\ & = \frac {1}{20} \int \frac {153+18 \left (9-e^4\right ) x-19 e^4 x^2}{(1+x) \left (9-e^4 x\right )} \, dx+\frac {1}{20} \int \log \left (-x (1+x) \left (-9+e^4 x\right )\right ) \, dx \\ & = \frac {1}{20} x \log \left (x (1+x) \left (9-e^4 x\right )\right )-\frac {1}{20} \int \frac {9+2 \left (9-e^4\right ) x-3 e^4 x^2}{(1+x) \left (9-e^4 x\right )} \, dx+\frac {1}{20} \int \left (19+\frac {1}{-1-x}+\frac {9}{-9+e^4 x}\right ) \, dx \\ & = \frac {19 x}{20}-\frac {1}{20} \log (1+x)+\frac {9 \log \left (9-e^4 x\right )}{20 e^4}+\frac {1}{20} x \log \left (x (1+x) \left (9-e^4 x\right )\right )-\frac {1}{20} \int \left (3+\frac {1}{-1-x}+\frac {9}{-9+e^4 x}\right ) \, dx \\ & = \frac {4 x}{5}+\frac {1}{20} x \log \left (x (1+x) \left (9-e^4 x\right )\right ) \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {-189-198 x+e^4 \left (22 x+23 x^2\right )+\left (-9-9 x+e^4 \left (x+x^2\right )\right ) \log \left (\frac {9 x+9 x^2+e^4 \left (-x^2-x^3\right )}{e^4}\right )}{-180-180 x+e^4 \left (20 x+20 x^2\right )} \, dx=\frac {1}{20} \left (16 x+x \log \left (-x (1+x) \left (-9+e^4 x\right )\right )\right ) \]
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Time = 4.17 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33
method | result | size |
default | \(\frac {4 x}{5}+\frac {x \ln \left (x \left (-x^{2} {\mathrm e}^{4}-x \,{\mathrm e}^{4}+9 x +9\right )\right )}{20}\) | \(32\) |
parts | \(\frac {4 x}{5}+\frac {x \ln \left (x \left (-x^{2} {\mathrm e}^{4}-x \,{\mathrm e}^{4}+9 x +9\right )\right )}{20}\) | \(32\) |
risch | \(x +\frac {x \ln \left (\left (\left (-x^{3}-x^{2}\right ) {\mathrm e}^{4}+9 x^{2}+9 x \right ) {\mathrm e}^{-4}\right )}{20}\) | \(33\) |
norman | \(x +\frac {x \ln \left (\left (\left (-x^{3}-x^{2}\right ) {\mathrm e}^{4}+9 x^{2}+9 x \right ) {\mathrm e}^{-4}\right )}{20}\) | \(37\) |
parallelrisch | \(\frac {\left (\ln \left (\left (\left (-x^{3}-x^{2}\right ) {\mathrm e}^{4}+9 x^{2}+9 x \right ) {\mathrm e}^{-4}\right ) x \,{\mathrm e}^{8}+20 x \,{\mathrm e}^{8}-40 \,{\mathrm e}^{8}+360 \,{\mathrm e}^{4}\right ) {\mathrm e}^{-8}}{20}\) | \(64\) |
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Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {-189-198 x+e^4 \left (22 x+23 x^2\right )+\left (-9-9 x+e^4 \left (x+x^2\right )\right ) \log \left (\frac {9 x+9 x^2+e^4 \left (-x^2-x^3\right )}{e^4}\right )}{-180-180 x+e^4 \left (20 x+20 x^2\right )} \, dx=\frac {1}{20} \, x \log \left ({\left (9 \, x^{2} - {\left (x^{3} + x^{2}\right )} e^{4} + 9 \, x\right )} e^{\left (-4\right )}\right ) + x \]
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Time = 0.16 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {-189-198 x+e^4 \left (22 x+23 x^2\right )+\left (-9-9 x+e^4 \left (x+x^2\right )\right ) \log \left (\frac {9 x+9 x^2+e^4 \left (-x^2-x^3\right )}{e^4}\right )}{-180-180 x+e^4 \left (20 x+20 x^2\right )} \, dx=\frac {x \log {\left (\frac {9 x^{2} + 9 x + \left (- x^{3} - x^{2}\right ) e^{4}}{e^{4}} \right )}}{20} + x \]
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Leaf count of result is larger than twice the leaf count of optimal. 162 vs. \(2 (20) = 40\).
Time = 0.32 (sec) , antiderivative size = 162, normalized size of antiderivative = 6.75 \[ \int \frac {-189-198 x+e^4 \left (22 x+23 x^2\right )+\left (-9-9 x+e^4 \left (x+x^2\right )\right ) \log \left (\frac {9 x+9 x^2+e^4 \left (-x^2-x^3\right )}{e^4}\right )}{-180-180 x+e^4 \left (20 x+20 x^2\right )} \, dx=\frac {23}{20} \, {\left (x e^{\left (-4\right )} + \frac {81 \, \log \left (x e^{4} - 9\right )}{e^{12} + 9 \, e^{8}} - \frac {\log \left (x + 1\right )}{e^{4} + 9}\right )} e^{4} + \frac {11}{10} \, {\left (\frac {9 \, \log \left (x e^{4} - 9\right )}{e^{8} + 9 \, e^{4}} + \frac {\log \left (x + 1\right )}{e^{4} + 9}\right )} e^{4} + \frac {1}{20} \, {\left (x e^{4} \log \left (x\right ) - 7 \, x e^{4} + {\left (x e^{4} - 9\right )} \log \left (-x e^{4} + 9\right ) + {\left (x e^{4} + e^{4}\right )} \log \left (x + 1\right )\right )} e^{\left (-4\right )} - \frac {891 \, \log \left (x e^{4} - 9\right )}{10 \, {\left (e^{8} + 9 \, e^{4}\right )}} - \frac {189 \, \log \left (x e^{4} - 9\right )}{20 \, {\left (e^{4} + 9\right )}} - \frac {9 \, \log \left (x + 1\right )}{20 \, {\left (e^{4} + 9\right )}} \]
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Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {-189-198 x+e^4 \left (22 x+23 x^2\right )+\left (-9-9 x+e^4 \left (x+x^2\right )\right ) \log \left (\frac {9 x+9 x^2+e^4 \left (-x^2-x^3\right )}{e^4}\right )}{-180-180 x+e^4 \left (20 x+20 x^2\right )} \, dx=\frac {1}{20} \, x \log \left (-x^{3} e^{4} - x^{2} e^{4} + 9 \, x^{2} + 9 \, x\right ) + \frac {4}{5} \, x \]
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Time = 10.06 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {-189-198 x+e^4 \left (22 x+23 x^2\right )+\left (-9-9 x+e^4 \left (x+x^2\right )\right ) \log \left (\frac {9 x+9 x^2+e^4 \left (-x^2-x^3\right )}{e^4}\right )}{-180-180 x+e^4 \left (20 x+20 x^2\right )} \, dx=\frac {x\,\left (\ln \left ({\mathrm {e}}^{-4}\,\left (9\,x-{\mathrm {e}}^4\,\left (x^3+x^2\right )+9\,x^2\right )\right )+20\right )}{20} \]
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