Integrand size = 49, antiderivative size = 20 \[ \int \frac {76-6 x+40 \log (2)}{361 x^3-38 x^4+x^5+\left (380 x^3-20 x^4\right ) \log (2)+100 x^3 \log ^2(2)} \, dx=\frac {2}{x^2+x^3-10 x^2 (2+\log (2))} \]
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Time = 0.05 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.102, Rules used = {6, 2023, 1608, 27, 75} \[ \int \frac {76-6 x+40 \log (2)}{361 x^3-38 x^4+x^5+\left (380 x^3-20 x^4\right ) \log (2)+100 x^3 \log ^2(2)} \, dx=-\frac {2}{x^2 (-x+19+\log (1024))} \]
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Rule 6
Rule 27
Rule 75
Rule 1608
Rule 2023
Rubi steps \begin{align*} \text {integral}& = \int \frac {76-6 x+40 \log (2)}{-38 x^4+x^5+\left (380 x^3-20 x^4\right ) \log (2)+x^3 \left (361+100 \log ^2(2)\right )} \, dx \\ & = \int \frac {-6 x+4 (19+10 \log (2))}{x^5-2 x^4 (19+\log (1024))+x^3 (19+\log (1024))^2} \, dx \\ & = \int \frac {-6 x+4 (19+10 \log (2))}{x^3 \left (x^2-2 x (19+\log (1024))+(19+\log (1024))^2\right )} \, dx \\ & = \int \frac {-6 x+4 (19+10 \log (2))}{x^3 (-19+x-\log (1024))^2} \, dx \\ & = -\frac {2}{x^2 (19-x+\log (1024))} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int \frac {76-6 x+40 \log (2)}{361 x^3-38 x^4+x^5+\left (380 x^3-20 x^4\right ) \log (2)+100 x^3 \log ^2(2)} \, dx=\frac {2}{x^2 (-19+x-10 \log (2))} \]
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Time = 0.12 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85
method | result | size |
gosper | \(-\frac {2}{x^{2} \left (19+10 \ln \left (2\right )-x \right )}\) | \(17\) |
norman | \(-\frac {2}{x^{2} \left (19+10 \ln \left (2\right )-x \right )}\) | \(17\) |
risch | \(-\frac {2}{x^{2} \left (19+10 \ln \left (2\right )-x \right )}\) | \(17\) |
parallelrisch | \(-\frac {2}{x^{2} \left (19+10 \ln \left (2\right )-x \right )}\) | \(17\) |
default | \(\frac {2}{\left (19+10 \ln \left (2\right )\right )^{2} \left (-19-10 \ln \left (2\right )+x \right )}-\frac {2}{\left (19+10 \ln \left (2\right )\right )^{2} x}-\frac {2}{\left (19+10 \ln \left (2\right )\right ) x^{2}}\) | \(47\) |
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Time = 0.24 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {76-6 x+40 \log (2)}{361 x^3-38 x^4+x^5+\left (380 x^3-20 x^4\right ) \log (2)+100 x^3 \log ^2(2)} \, dx=\frac {2}{x^{3} - 10 \, x^{2} \log \left (2\right ) - 19 \, x^{2}} \]
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Time = 0.24 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {76-6 x+40 \log (2)}{361 x^3-38 x^4+x^5+\left (380 x^3-20 x^4\right ) \log (2)+100 x^3 \log ^2(2)} \, dx=\frac {2}{x^{3} + x^{2} \left (-19 - 10 \log {\left (2 \right )}\right )} \]
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Time = 0.19 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {76-6 x+40 \log (2)}{361 x^3-38 x^4+x^5+\left (380 x^3-20 x^4\right ) \log (2)+100 x^3 \log ^2(2)} \, dx=\frac {2}{x^{3} - x^{2} {\left (10 \, \log \left (2\right ) + 19\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (20) = 40\).
Time = 0.28 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.60 \[ \int \frac {76-6 x+40 \log (2)}{361 x^3-38 x^4+x^5+\left (380 x^3-20 x^4\right ) \log (2)+100 x^3 \log ^2(2)} \, dx=\frac {2}{{\left (100 \, \log \left (2\right )^{2} + 380 \, \log \left (2\right ) + 361\right )} {\left (x - 10 \, \log \left (2\right ) - 19\right )}} - \frac {2 \, {\left (x + 10 \, \log \left (2\right ) + 19\right )}}{{\left (100 \, \log \left (2\right )^{2} + 380 \, \log \left (2\right ) + 361\right )} x^{2}} \]
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Time = 0.13 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {76-6 x+40 \log (2)}{361 x^3-38 x^4+x^5+\left (380 x^3-20 x^4\right ) \log (2)+100 x^3 \log ^2(2)} \, dx=-\frac {2}{x^2\,\left (10\,\ln \left (2\right )+19\right )-x^3} \]
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