Integrand size = 48, antiderivative size = 25 \[ \int \frac {-100+210 x-30 x^2+e^2 \left (20 x^2-24 x^3+4 x^4\right )}{e^2 \left (5 x-6 x^2+x^3\right )} \, dx=2 \left (1+x^2+\frac {5 \log \left (\frac {-5+x}{(-1+x)^2 x^2}\right )}{e^2}\right ) \]
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Time = 0.06 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.40, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {12, 1608, 1642} \[ \int \frac {-100+210 x-30 x^2+e^2 \left (20 x^2-24 x^3+4 x^4\right )}{e^2 \left (5 x-6 x^2+x^3\right )} \, dx=2 x^2-\frac {20 \log (1-x)}{e^2}+\frac {10 \log (5-x)}{e^2}-\frac {20 \log (x)}{e^2} \]
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Rule 12
Rule 1608
Rule 1642
Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {-100+210 x-30 x^2+e^2 \left (20 x^2-24 x^3+4 x^4\right )}{5 x-6 x^2+x^3} \, dx}{e^2} \\ & = \frac {\int \frac {-100+210 x-30 x^2+e^2 \left (20 x^2-24 x^3+4 x^4\right )}{x \left (5-6 x+x^2\right )} \, dx}{e^2} \\ & = \frac {\int \left (\frac {10}{-5+x}-\frac {20}{-1+x}-\frac {20}{x}+4 e^2 x\right ) \, dx}{e^2} \\ & = 2 x^2-\frac {20 \log (1-x)}{e^2}+\frac {10 \log (5-x)}{e^2}-\frac {20 \log (x)}{e^2} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32 \[ \int \frac {-100+210 x-30 x^2+e^2 \left (20 x^2-24 x^3+4 x^4\right )}{e^2 \left (5 x-6 x^2+x^3\right )} \, dx=\frac {2 e^2 x^2-20 \log (1-x)+10 \log (5-x)-20 \log (x)}{e^2} \]
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Time = 0.06 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08
method | result | size |
risch | \(2 x^{2}+10 \,{\mathrm e}^{-2} \ln \left (-5+x \right )-20 \,{\mathrm e}^{-2} \ln \left (x^{2}-x \right )\) | \(27\) |
default | \({\mathrm e}^{-2} \left (2 x^{2} {\mathrm e}^{2}-20 \ln \left (x \right )-20 \ln \left (-1+x \right )+10 \ln \left (-5+x \right )\right )\) | \(30\) |
parallelrisch | \({\mathrm e}^{-2} \left (2 x^{2} {\mathrm e}^{2}-20 \ln \left (x \right )-20 \ln \left (-1+x \right )+10 \ln \left (-5+x \right )\right )\) | \(30\) |
norman | \(2 x^{2}-20 \ln \left (x \right ) {\mathrm e}^{-2}+10 \,{\mathrm e}^{-2} \ln \left (-5+x \right )-20 \,{\mathrm e}^{-2} \ln \left (-1+x \right )\) | \(35\) |
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Time = 0.23 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {-100+210 x-30 x^2+e^2 \left (20 x^2-24 x^3+4 x^4\right )}{e^2 \left (5 x-6 x^2+x^3\right )} \, dx=2 \, {\left (x^{2} e^{2} - 10 \, \log \left (x^{2} - x\right ) + 5 \, \log \left (x - 5\right )\right )} e^{\left (-2\right )} \]
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Time = 0.14 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {-100+210 x-30 x^2+e^2 \left (20 x^2-24 x^3+4 x^4\right )}{e^2 \left (5 x-6 x^2+x^3\right )} \, dx=2 x^{2} + \frac {10 \log {\left (x - 5 \right )}}{e^{2}} - \frac {20 \log {\left (x^{2} - x \right )}}{e^{2}} \]
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Time = 0.19 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {-100+210 x-30 x^2+e^2 \left (20 x^2-24 x^3+4 x^4\right )}{e^2 \left (5 x-6 x^2+x^3\right )} \, dx=2 \, {\left (x^{2} e^{2} - 10 \, \log \left (x - 1\right ) + 5 \, \log \left (x - 5\right ) - 10 \, \log \left (x\right )\right )} e^{\left (-2\right )} \]
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Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {-100+210 x-30 x^2+e^2 \left (20 x^2-24 x^3+4 x^4\right )}{e^2 \left (5 x-6 x^2+x^3\right )} \, dx=2 \, {\left (x^{2} e^{2} - 10 \, \log \left ({\left | x - 1 \right |}\right ) + 5 \, \log \left ({\left | x - 5 \right |}\right ) - 10 \, \log \left ({\left | x \right |}\right )\right )} e^{\left (-2\right )} \]
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Time = 0.13 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {-100+210 x-30 x^2+e^2 \left (20 x^2-24 x^3+4 x^4\right )}{e^2 \left (5 x-6 x^2+x^3\right )} \, dx=10\,\ln \left (x-5\right )\,{\mathrm {e}}^{-2}+2\,x^2-20\,\ln \left (x\,\left (x-1\right )\right )\,{\mathrm {e}}^{-2} \]
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