Integrand size = 83, antiderivative size = 26 \[ \int \frac {100-60 x-140 x^2-70 x^3-14 x^4-x^5+\left (-25+15 x+9 x^2+x^3\right ) \log (2)}{-100 x-120 x^2-60 x^3-13 x^4-x^5+\left (25 x+10 x^2+x^3\right ) \log (2)} \, dx=x+\log \left (\frac {x-(2+x)^2-\frac {x}{5+x}+\log (2)}{x}\right ) \]
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Time = 0.09 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.46, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {2099, 1601} \[ \int \frac {100-60 x-140 x^2-70 x^3-14 x^4-x^5+\left (-25+15 x+9 x^2+x^3\right ) \log (2)}{-100 x-120 x^2-60 x^3-13 x^4-x^5+\left (25 x+10 x^2+x^3\right ) \log (2)} \, dx=\log \left (x^3+8 x^2+x (20-\log (2))+5 (4-\log (2))\right )+x-\log (x)-\log (x+5) \]
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Rule 1601
Rule 2099
Rubi steps \begin{align*} \text {integral}& = \int \left (1+\frac {1}{-5-x}-\frac {1}{x}+\frac {20+16 x+3 x^2-\log (2)}{8 x^2+x^3+5 (4-\log (2))+x (20-\log (2))}\right ) \, dx \\ & = x-\log (x)-\log (5+x)+\int \frac {20+16 x+3 x^2-\log (2)}{8 x^2+x^3+5 (4-\log (2))+x (20-\log (2))} \, dx \\ & = x-\log (x)-\log (5+x)+\log \left (8 x^2+x^3+5 (4-\log (2))+x (20-\log (2))\right ) \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35 \[ \int \frac {100-60 x-140 x^2-70 x^3-14 x^4-x^5+\left (-25+15 x+9 x^2+x^3\right ) \log (2)}{-100 x-120 x^2-60 x^3-13 x^4-x^5+\left (25 x+10 x^2+x^3\right ) \log (2)} \, dx=x-\log (x)-\log (5+x)+\log \left (20+20 x+8 x^2+x^3-5 \log (2)-x \log (2)\right ) \]
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Time = 0.15 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38
method | result | size |
default | \(x -\ln \left (5+x \right )+\ln \left (x^{3}-x \ln \left (2\right )+8 x^{2}-5 \ln \left (2\right )+20 x +20\right )-\ln \left (x \right )\) | \(36\) |
parallelrisch | \(x -\ln \left (5+x \right )+\ln \left (x^{3}-x \ln \left (2\right )+8 x^{2}-5 \ln \left (2\right )+20 x +20\right )-\ln \left (x \right )\) | \(36\) |
norman | \(x -\ln \left (x \right )-\ln \left (5+x \right )+\ln \left (-x^{3}+x \ln \left (2\right )-8 x^{2}+5 \ln \left (2\right )-20 x -20\right )\) | \(37\) |
risch | \(x -\ln \left (-x^{2}-5 x \right )+\ln \left (x^{3}+8 x^{2}+\left (-\ln \left (2\right )+20\right ) x -5 \ln \left (2\right )+20\right )\) | \(38\) |
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none
Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.27 \[ \int \frac {100-60 x-140 x^2-70 x^3-14 x^4-x^5+\left (-25+15 x+9 x^2+x^3\right ) \log (2)}{-100 x-120 x^2-60 x^3-13 x^4-x^5+\left (25 x+10 x^2+x^3\right ) \log (2)} \, dx=x + \log \left (x^{3} + 8 \, x^{2} - {\left (x + 5\right )} \log \left (2\right ) + 20 \, x + 20\right ) - \log \left (x^{2} + 5 \, x\right ) \]
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Time = 1.20 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {100-60 x-140 x^2-70 x^3-14 x^4-x^5+\left (-25+15 x+9 x^2+x^3\right ) \log (2)}{-100 x-120 x^2-60 x^3-13 x^4-x^5+\left (25 x+10 x^2+x^3\right ) \log (2)} \, dx=x - \log {\left (x^{2} + 5 x \right )} + \log {\left (x^{3} + 8 x^{2} + x \left (20 - \log {\left (2 \right )}\right ) - 5 \log {\left (2 \right )} + 20 \right )} \]
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Time = 0.17 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31 \[ \int \frac {100-60 x-140 x^2-70 x^3-14 x^4-x^5+\left (-25+15 x+9 x^2+x^3\right ) \log (2)}{-100 x-120 x^2-60 x^3-13 x^4-x^5+\left (25 x+10 x^2+x^3\right ) \log (2)} \, dx=x + \log \left (x^{3} + 8 \, x^{2} - x {\left (\log \left (2\right ) - 20\right )} - 5 \, \log \left (2\right ) + 20\right ) - \log \left (x + 5\right ) - \log \left (x\right ) \]
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Time = 0.28 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.46 \[ \int \frac {100-60 x-140 x^2-70 x^3-14 x^4-x^5+\left (-25+15 x+9 x^2+x^3\right ) \log (2)}{-100 x-120 x^2-60 x^3-13 x^4-x^5+\left (25 x+10 x^2+x^3\right ) \log (2)} \, dx=x + \log \left ({\left | x^{3} + 8 \, x^{2} - x \log \left (2\right ) + 20 \, x - 5 \, \log \left (2\right ) + 20 \right |}\right ) - \log \left ({\left | x + 5 \right |}\right ) - \log \left ({\left | x \right |}\right ) \]
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Time = 13.84 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.27 \[ \int \frac {100-60 x-140 x^2-70 x^3-14 x^4-x^5+\left (-25+15 x+9 x^2+x^3\right ) \log (2)}{-100 x-120 x^2-60 x^3-13 x^4-x^5+\left (25 x+10 x^2+x^3\right ) \log (2)} \, dx=x-\ln \left (x\,\left (x+5\right )\right )+\ln \left (20\,x-\ln \left (32\right )-x\,\ln \left (2\right )+8\,x^2+x^3+20\right ) \]
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