Integrand size = 90, antiderivative size = 24 \[ \int \frac {-1000-1000 x-250 x^2+(-80-40 x) \log ^3(x)+(-2000-1000 x) \log (2 x)-1000 \log ^2(2 x)}{\left (4 x+4 x^2+x^3\right ) \log ^3(x)+\left (8 x+4 x^2\right ) \log ^3(x) \log (2 x)+4 x \log ^3(x) \log ^2(2 x)} \, dx=5 \left (\frac {25}{\log ^2(x)}+\frac {4}{1+\frac {x}{2}+\log (2 x)}\right ) \]
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Time = 0.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {6820, 12, 14, 2339, 30, 6818} \[ \int \frac {-1000-1000 x-250 x^2+(-80-40 x) \log ^3(x)+(-2000-1000 x) \log (2 x)-1000 \log ^2(2 x)}{\left (4 x+4 x^2+x^3\right ) \log ^3(x)+\left (8 x+4 x^2\right ) \log ^3(x) \log (2 x)+4 x \log ^3(x) \log ^2(2 x)} \, dx=\frac {125}{\log ^2(x)}+\frac {40}{x+2 \log (2 x)+2} \]
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Rule 12
Rule 14
Rule 30
Rule 2339
Rule 6818
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \int \frac {10 \left (-\frac {25}{\log ^3(x)}-\frac {4 (2+x)}{(2+x+2 \log (2 x))^2}\right )}{x} \, dx \\ & = 10 \int \frac {-\frac {25}{\log ^3(x)}-\frac {4 (2+x)}{(2+x+2 \log (2 x))^2}}{x} \, dx \\ & = 10 \int \left (-\frac {25}{x \log ^3(x)}-\frac {4 (2+x)}{x (2+x+2 \log (2 x))^2}\right ) \, dx \\ & = -\left (40 \int \frac {2+x}{x (2+x+2 \log (2 x))^2} \, dx\right )-250 \int \frac {1}{x \log ^3(x)} \, dx \\ & = \frac {40}{2+x+2 \log (2 x)}-250 \text {Subst}\left (\int \frac {1}{x^3} \, dx,x,\log (x)\right ) \\ & = \frac {125}{\log ^2(x)}+\frac {40}{2+x+2 \log (2 x)} \\ \end{align*}
Time = 5.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {-1000-1000 x-250 x^2+(-80-40 x) \log ^3(x)+(-2000-1000 x) \log (2 x)-1000 \log ^2(2 x)}{\left (4 x+4 x^2+x^3\right ) \log ^3(x)+\left (8 x+4 x^2\right ) \log ^3(x) \log (2 x)+4 x \log ^3(x) \log ^2(2 x)} \, dx=10 \left (\frac {25}{2 \log ^2(x)}+\frac {4}{2+x+\log (4)+2 \log (x)}\right ) \]
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Time = 1.89 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96
method | result | size |
parts | \(\frac {80}{4+2 x +4 \ln \left (2 x \right )}+\frac {125}{\ln \left (x \right )^{2}}\) | \(23\) |
parallelrisch | \(\frac {500+250 x +500 \ln \left (2 x \right )+80 \ln \left (x \right )^{2}}{2 \ln \left (x \right )^{2} \left (2+2 \ln \left (2 x \right )+x \right )}\) | \(35\) |
default | \(-\frac {10 \left (-4 \ln \left (x \right )^{2}-25 \ln \left (x \right )-25-\frac {25 x}{2}-25 \ln \left (2\right )\right )}{\ln \left (x \right )^{2} \left (x +2+2 \ln \left (x \right )+2 \ln \left (2\right )\right )}\) | \(39\) |
risch | \(\frac {250+250 \ln \left (2\right )+125 x +250 \ln \left (x \right )+40 \ln \left (x \right )^{2}}{\ln \left (x \right )^{2} \left (x +2+2 \ln \left (x \right )+2 \ln \left (2\right )\right )}\) | \(39\) |
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Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (20) = 40\).
Time = 0.28 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.75 \[ \int \frac {-1000-1000 x-250 x^2+(-80-40 x) \log ^3(x)+(-2000-1000 x) \log (2 x)-1000 \log ^2(2 x)}{\left (4 x+4 x^2+x^3\right ) \log ^3(x)+\left (8 x+4 x^2\right ) \log ^3(x) \log (2 x)+4 x \log ^3(x) \log ^2(2 x)} \, dx=\frac {5 \, {\left (8 \, \log \left (x\right )^{2} + 25 \, x + 50 \, \log \left (2\right ) + 50 \, \log \left (x\right ) + 50\right )}}{{\left (x + 2 \, \log \left (2\right ) + 2\right )} \log \left (x\right )^{2} + 2 \, \log \left (x\right )^{3}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (20) = 40\).
Time = 0.13 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.71 \[ \int \frac {-1000-1000 x-250 x^2+(-80-40 x) \log ^3(x)+(-2000-1000 x) \log (2 x)-1000 \log ^2(2 x)}{\left (4 x+4 x^2+x^3\right ) \log ^3(x)+\left (8 x+4 x^2\right ) \log ^3(x) \log (2 x)+4 x \log ^3(x) \log ^2(2 x)} \, dx=\frac {125 x + 40 \log {\left (x \right )}^{2} + 250 \log {\left (x \right )} + 250 \log {\left (2 \right )} + 250}{\left (x + 2 \log {\left (2 \right )} + 2\right ) \log {\left (x \right )}^{2} + 2 \log {\left (x \right )}^{3}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (20) = 40\).
Time = 0.32 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.75 \[ \int \frac {-1000-1000 x-250 x^2+(-80-40 x) \log ^3(x)+(-2000-1000 x) \log (2 x)-1000 \log ^2(2 x)}{\left (4 x+4 x^2+x^3\right ) \log ^3(x)+\left (8 x+4 x^2\right ) \log ^3(x) \log (2 x)+4 x \log ^3(x) \log ^2(2 x)} \, dx=\frac {5 \, {\left (8 \, \log \left (x\right )^{2} + 25 \, x + 50 \, \log \left (2\right ) + 50 \, \log \left (x\right ) + 50\right )}}{{\left (x + 2 \, \log \left (2\right ) + 2\right )} \log \left (x\right )^{2} + 2 \, \log \left (x\right )^{3}} \]
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none
Time = 0.30 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {-1000-1000 x-250 x^2+(-80-40 x) \log ^3(x)+(-2000-1000 x) \log (2 x)-1000 \log ^2(2 x)}{\left (4 x+4 x^2+x^3\right ) \log ^3(x)+\left (8 x+4 x^2\right ) \log ^3(x) \log (2 x)+4 x \log ^3(x) \log ^2(2 x)} \, dx=\frac {40}{x + 2 \, \log \left (2\right ) + 2 \, \log \left (x\right ) + 2} + \frac {125}{\log \left (x\right )^{2}} \]
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Time = 11.96 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {-1000-1000 x-250 x^2+(-80-40 x) \log ^3(x)+(-2000-1000 x) \log (2 x)-1000 \log ^2(2 x)}{\left (4 x+4 x^2+x^3\right ) \log ^3(x)+\left (8 x+4 x^2\right ) \log ^3(x) \log (2 x)+4 x \log ^3(x) \log ^2(2 x)} \, dx=\frac {125}{{\ln \left (x\right )}^2}+\frac {40}{x+\ln \left (4\right )+2\,\ln \left (x\right )+2} \]
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