Integrand size = 59, antiderivative size = 24 \[ \int \frac {-20-4 e^2+8 x}{\left (5 e^5 x^2+e^3 \left (50 x^2-10 x^3\right )+e \left (125 x^2-50 x^3+5 x^4\right )\right ) \log (6)} \, dx=\frac {4}{5 e \left (5+e^2-x\right ) x \log (6)} \]
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Time = 0.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {12, 1694, 267} \[ \int \frac {-20-4 e^2+8 x}{\left (5 e^5 x^2+e^3 \left (50 x^2-10 x^3\right )+e \left (125 x^2-50 x^3+5 x^4\right )\right ) \log (6)} \, dx=\frac {4}{5 e \left (-x+e^2+5\right ) x \log (6)} \]
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Rule 12
Rule 267
Rule 1694
Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {-20-4 e^2+8 x}{5 e^5 x^2+e^3 \left (50 x^2-10 x^3\right )+e \left (125 x^2-50 x^3+5 x^4\right )} \, dx}{\log (6)} \\ & = \frac {\text {Subst}\left (\int \frac {128 x}{5 e \left (25+10 e^2+e^4-4 x^2\right )^2} \, dx,x,\frac {-50 e-10 e^3}{20 e}+x\right )}{\log (6)} \\ & = \frac {128 \text {Subst}\left (\int \frac {x}{\left (25+10 e^2+e^4-4 x^2\right )^2} \, dx,x,\frac {-50 e-10 e^3}{20 e}+x\right )}{5 e \log (6)} \\ & = \frac {4}{5 e \left (5+e^2-x\right ) x \log (6)} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {-20-4 e^2+8 x}{\left (5 e^5 x^2+e^3 \left (50 x^2-10 x^3\right )+e \left (125 x^2-50 x^3+5 x^4\right )\right ) \log (6)} \, dx=\frac {4}{5 e \left (5+e^2-x\right ) x \log (6)} \]
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Time = 0.20 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96
method | result | size |
gosper | \(\frac {4 \,{\mathrm e}^{-1}}{5 x \left ({\mathrm e}^{2}+5-x \right ) \ln \left (6\right )}\) | \(23\) |
norman | \(\frac {4 \,{\mathrm e}^{-1}}{5 x \left ({\mathrm e}^{2}+5-x \right ) \ln \left (6\right )}\) | \(23\) |
parallelrisch | \(\frac {4 \,{\mathrm e}^{-1}}{5 x \left ({\mathrm e}^{2}+5-x \right ) \ln \left (6\right )}\) | \(23\) |
risch | \(\frac {4 \,{\mathrm e}^{-1}}{5 \left (\ln \left (2\right )+\ln \left (3\right )\right ) \left ({\mathrm e}^{2}+5-x \right ) x}\) | \(24\) |
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Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {-20-4 e^2+8 x}{\left (5 e^5 x^2+e^3 \left (50 x^2-10 x^3\right )+e \left (125 x^2-50 x^3+5 x^4\right )\right ) \log (6)} \, dx=\frac {4}{5 \, {\left (x e^{3} - {\left (x^{2} - 5 \, x\right )} e\right )} \log \left (6\right )} \]
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Time = 0.30 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {-20-4 e^2+8 x}{\left (5 e^5 x^2+e^3 \left (50 x^2-10 x^3\right )+e \left (125 x^2-50 x^3+5 x^4\right )\right ) \log (6)} \, dx=- \frac {4}{5 e x^{2} \log {\left (6 \right )} + x \left (- 5 e^{3} \log {\left (6 \right )} - 25 e \log {\left (6 \right )}\right )} \]
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Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {-20-4 e^2+8 x}{\left (5 e^5 x^2+e^3 \left (50 x^2-10 x^3\right )+e \left (125 x^2-50 x^3+5 x^4\right )\right ) \log (6)} \, dx=-\frac {4}{5 \, {\left (x^{2} e - x {\left (e^{3} + 5 \, e\right )}\right )} \log \left (6\right )} \]
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Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {-20-4 e^2+8 x}{\left (5 e^5 x^2+e^3 \left (50 x^2-10 x^3\right )+e \left (125 x^2-50 x^3+5 x^4\right )\right ) \log (6)} \, dx=-\frac {4 \, e^{\left (-1\right )}}{5 \, {\left (x^{2} - x e^{2} - 5 \, x\right )} \log \left (6\right )} \]
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Time = 0.23 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {-20-4 e^2+8 x}{\left (5 e^5 x^2+e^3 \left (50 x^2-10 x^3\right )+e \left (125 x^2-50 x^3+5 x^4\right )\right ) \log (6)} \, dx=\frac {4\,{\mathrm {e}}^{-1}}{5\,x\,\ln \left (6\right )\,\left ({\mathrm {e}}^2-x+5\right )} \]
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