Integrand size = 45, antiderivative size = 31 \[ \int \frac {-250+75 x+e \left (-25 x^2+30 x^3-x^4\right )}{e \left (400 x^3-160 x^4+16 x^5\right )} \, dx=\frac {1}{16} \left (x+\frac {-5+\frac {25}{e x^2}+x^2}{5-x}-\log (x)\right ) \]
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Time = 0.04 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.48, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {12, 1608, 27, 1634} \[ \int \frac {-250+75 x+e \left (-25 x^2+30 x^3-x^4\right )}{e \left (400 x^3-160 x^4+16 x^5\right )} \, dx=\frac {5}{16 e x^2}+\frac {1+20 e}{16 e (5-x)}+\frac {1}{16 e x}-\frac {\log (x)}{16} \]
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Rule 12
Rule 27
Rule 1608
Rule 1634
Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {-250+75 x+e \left (-25 x^2+30 x^3-x^4\right )}{400 x^3-160 x^4+16 x^5} \, dx}{e} \\ & = \frac {\int \frac {-250+75 x+e \left (-25 x^2+30 x^3-x^4\right )}{x^3 \left (400-160 x+16 x^2\right )} \, dx}{e} \\ & = \frac {\int \frac {-250+75 x+e \left (-25 x^2+30 x^3-x^4\right )}{16 (-5+x)^2 x^3} \, dx}{e} \\ & = \frac {\int \frac {-250+75 x+e \left (-25 x^2+30 x^3-x^4\right )}{(-5+x)^2 x^3} \, dx}{16 e} \\ & = \frac {\int \left (\frac {1+20 e}{(-5+x)^2}-\frac {10}{x^3}-\frac {1}{x^2}-\frac {e}{x}\right ) \, dx}{16 e} \\ & = \frac {1+20 e}{16 e (5-x)}+\frac {5}{16 e x^2}+\frac {1}{16 e x}-\frac {\log (x)}{16} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {-250+75 x+e \left (-25 x^2+30 x^3-x^4\right )}{e \left (400 x^3-160 x^4+16 x^5\right )} \, dx=-\frac {\frac {5 \left (5+4 e x^2\right )}{(-5+x) x^2}+e \log (x)}{16 e} \]
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Time = 0.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84
method | result | size |
risch | \(\frac {{\mathrm e}^{-1} \left (-\frac {5 x^{2} {\mathrm e}}{4}-\frac {25}{16}\right )}{x^{2} \left (-5+x \right )}-\frac {\ln \left (x \right )}{16}\) | \(26\) |
norman | \(\frac {-\frac {5 x^{2}}{4}-\frac {25 \,{\mathrm e}^{-1}}{16}}{x^{2} \left (-5+x \right )}-\frac {\ln \left (x \right )}{16}\) | \(27\) |
default | \(\frac {{\mathrm e}^{-1} \left (-\frac {20 \,{\mathrm e}+1}{-5+x}+\frac {5}{x^{2}}+\frac {1}{x}-{\mathrm e} \ln \left (x \right )\right )}{16}\) | \(35\) |
parallelrisch | \(-\frac {{\mathrm e}^{-1} \left (x^{3} {\mathrm e} \ln \left (x \right )-5 x^{2} {\mathrm e} \ln \left (x \right )+25+20 x^{2} {\mathrm e}\right )}{16 \left (-5+x \right ) x^{2}}\) | \(41\) |
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Time = 0.25 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.23 \[ \int \frac {-250+75 x+e \left (-25 x^2+30 x^3-x^4\right )}{e \left (400 x^3-160 x^4+16 x^5\right )} \, dx=-\frac {{\left (20 \, x^{2} e + {\left (x^{3} - 5 \, x^{2}\right )} e \log \left (x\right ) + 25\right )} e^{\left (-1\right )}}{16 \, {\left (x^{3} - 5 \, x^{2}\right )}} \]
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Time = 0.42 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {-250+75 x+e \left (-25 x^2+30 x^3-x^4\right )}{e \left (400 x^3-160 x^4+16 x^5\right )} \, dx=- \frac {20 e x^{2} + 25}{16 e x^{3} - 80 e x^{2}} - \frac {\log {\left (x \right )}}{16} \]
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Time = 0.17 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {-250+75 x+e \left (-25 x^2+30 x^3-x^4\right )}{e \left (400 x^3-160 x^4+16 x^5\right )} \, dx=-\frac {1}{16} \, {\left (e \log \left (x\right ) + \frac {5 \, {\left (4 \, x^{2} e + 5\right )}}{x^{3} - 5 \, x^{2}}\right )} e^{\left (-1\right )} \]
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Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {-250+75 x+e \left (-25 x^2+30 x^3-x^4\right )}{e \left (400 x^3-160 x^4+16 x^5\right )} \, dx=-\frac {1}{16} \, {\left (e \log \left ({\left | x \right |}\right ) + \frac {5 \, {\left (4 \, x^{2} e + 5\right )}}{{\left (x - 5\right )} x^{2}}\right )} e^{\left (-1\right )} \]
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Time = 12.17 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {-250+75 x+e \left (-25 x^2+30 x^3-x^4\right )}{e \left (400 x^3-160 x^4+16 x^5\right )} \, dx=\frac {20\,\mathrm {e}\,x^2+25}{80\,x^2\,\mathrm {e}-16\,x^3\,\mathrm {e}}-\frac {\ln \left (x\right )}{16} \]
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