Integrand size = 42, antiderivative size = 23 \[ \int \frac {-4-4 x+\left (-2 x^2+e^3 (1+x)\right ) \log (400)}{-4 x+\left (e^3 x-2 x^2\right ) \log (400)} \, dx=-3+x+\log \left (\frac {x}{-e^3+2 x+\frac {4}{\log (400)}}\right ) \]
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Time = 0.04 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {2009, 1607, 907} \[ \int \frac {-4-4 x+\left (-2 x^2+e^3 (1+x)\right ) \log (400)}{-4 x+\left (e^3 x-2 x^2\right ) \log (400)} \, dx=x+\log (x)-\log \left (2 x \log (400)+4-e^3 \log (400)\right ) \]
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Rule 907
Rule 1607
Rule 2009
Rubi steps \begin{align*} \text {integral}& = \int \frac {-4+e^3 \log (400)-2 x^2 \log (400)-x \left (4-e^3 \log (400)\right )}{-2 x^2 \log (400)-x \left (4-e^3 \log (400)\right )} \, dx \\ & = \int \frac {-4+e^3 \log (400)-2 x^2 \log (400)-x \left (4-e^3 \log (400)\right )}{x \left (-4+e^3 \log (400)-2 x \log (400)\right )} \, dx \\ & = \int \left (1+\frac {1}{x}+\frac {2 \log (400)}{-4+e^3 \log (400)-2 x \log (400)}\right ) \, dx \\ & = x+\log (x)-\log \left (4-e^3 \log (400)+2 x \log (400)\right ) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {-4-4 x+\left (-2 x^2+e^3 (1+x)\right ) \log (400)}{-4 x+\left (e^3 x-2 x^2\right ) \log (400)} \, dx=x+\log (x)-\log \left (4-e^3 \log (400)+2 x \log (400)\right ) \]
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Time = 0.14 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87
method | result | size |
norman | \(x -\ln \left (\ln \left (20\right ) {\mathrm e}^{3}-2 x \ln \left (20\right )-2\right )+\ln \left (x \right )\) | \(20\) |
default | \(x +\ln \left (x \right )-\ln \left (-\ln \left (20\right ) {\mathrm e}^{3}+2 x \ln \left (20\right )+2\right )\) | \(21\) |
parallelrisch | \(x +\ln \left (x \right )-\ln \left (-\frac {\ln \left (20\right ) {\mathrm e}^{3}-2 x \ln \left (20\right )-2}{2 \ln \left (20\right )}\right )\) | \(26\) |
risch | \(x -\ln \left (\left (4 \ln \left (2\right )+2 \ln \left (5\right )\right ) x -{\mathrm e}^{3} \ln \left (5\right )-2 \,{\mathrm e}^{3} \ln \left (2\right )+2\right )+\ln \left (-x \right )\) | \(35\) |
meijerg | \(-\frac {\left (\ln \left (20\right ) {\mathrm e}^{3}-2\right )^{2} \left (-\frac {2 x \ln \left (20\right )}{\ln \left (20\right ) {\mathrm e}^{3}-2}-\ln \left (1-\frac {2 x \ln \left (20\right )}{\ln \left (20\right ) {\mathrm e}^{3}-2}\right )\right )}{\left (2 \ln \left (20\right ) {\mathrm e}^{3}-4\right ) \ln \left (20\right )}-\frac {\left (\ln \left (20\right ) {\mathrm e}^{3}-2\right ) \ln \left (1-\frac {2 x \ln \left (20\right )}{\ln \left (20\right ) {\mathrm e}^{3}-2}\right )}{2 \ln \left (20\right )}+\frac {2 \ln \left (20\right ) {\mathrm e}^{3} \left (\ln \left (x \right )+\ln \left (2\right )+\ln \left (\ln \left (20\right )\right )-\ln \left (\ln \left (20\right ) {\mathrm e}^{3}-2\right )+i \pi -\ln \left (1-\frac {2 x \ln \left (20\right )}{\ln \left (20\right ) {\mathrm e}^{3}-2}\right )\right )}{2 \ln \left (20\right ) {\mathrm e}^{3}-4}-\frac {4 \left (\ln \left (x \right )+\ln \left (2\right )+\ln \left (\ln \left (20\right )\right )-\ln \left (\ln \left (20\right ) {\mathrm e}^{3}-2\right )+i \pi -\ln \left (1-\frac {2 x \ln \left (20\right )}{\ln \left (20\right ) {\mathrm e}^{3}-2}\right )\right )}{2 \ln \left (20\right ) {\mathrm e}^{3}-4}\) | \(201\) |
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Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {-4-4 x+\left (-2 x^2+e^3 (1+x)\right ) \log (400)}{-4 x+\left (e^3 x-2 x^2\right ) \log (400)} \, dx=x - \log \left ({\left (2 \, x - e^{3}\right )} \log \left (20\right ) + 2\right ) + \log \left (x\right ) \]
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Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {-4-4 x+\left (-2 x^2+e^3 (1+x)\right ) \log (400)}{-4 x+\left (e^3 x-2 x^2\right ) \log (400)} \, dx=x + \log {\left (x \right )} - \log {\left (x + \frac {- 2 e^{3} \log {\left (20 \right )} + 4}{4 \log {\left (20 \right )}} \right )} \]
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Time = 0.21 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {-4-4 x+\left (-2 x^2+e^3 (1+x)\right ) \log (400)}{-4 x+\left (e^3 x-2 x^2\right ) \log (400)} \, dx=x - \log \left (2 \, x \log \left (20\right ) - e^{3} \log \left (20\right ) + 2\right ) + \log \left (x\right ) \]
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Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {-4-4 x+\left (-2 x^2+e^3 (1+x)\right ) \log (400)}{-4 x+\left (e^3 x-2 x^2\right ) \log (400)} \, dx=x - \log \left ({\left | 2 \, x \log \left (20\right ) - e^{3} \log \left (20\right ) + 2 \right |}\right ) + \log \left ({\left | x \right |}\right ) \]
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Time = 0.21 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.43 \[ \int \frac {-4-4 x+\left (-2 x^2+e^3 (1+x)\right ) \log (400)}{-4 x+\left (e^3 x-2 x^2\right ) \log (400)} \, dx=x-\mathrm {atan}\left (\frac {-{\mathrm {e}}^3\,\ln \left (20\right )\,2{}\mathrm {i}+x\,\ln \left (20\right )\,8{}\mathrm {i}+4{}\mathrm {i}}{2\,{\mathrm {e}}^3\,\ln \left (20\right )-4}\right )\,2{}\mathrm {i} \]
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