Integrand size = 45, antiderivative size = 22 \[ \int \frac {-26-10 x+13 x^2+5 x^3+e^4 \left (13 x+10 x^2\right )}{e^4 \left (13 x^2+5 x^3\right )} \, dx=3+\frac {4+\frac {2}{x}+x}{e^4}+\log (x (13+5 x)) \]
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Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {12, 1607, 1634} \[ \int \frac {-26-10 x+13 x^2+5 x^3+e^4 \left (13 x+10 x^2\right )}{e^4 \left (13 x^2+5 x^3\right )} \, dx=\frac {x}{e^4}+\frac {2}{e^4 x}+\log (x)+\log (5 x+13) \]
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Rule 12
Rule 1607
Rule 1634
Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {-26-10 x+13 x^2+5 x^3+e^4 \left (13 x+10 x^2\right )}{13 x^2+5 x^3} \, dx}{e^4} \\ & = \frac {\int \frac {-26-10 x+13 x^2+5 x^3+e^4 \left (13 x+10 x^2\right )}{x^2 (13+5 x)} \, dx}{e^4} \\ & = \frac {\int \left (1-\frac {2}{x^2}+\frac {e^4}{x}+\frac {5 e^4}{13+5 x}\right ) \, dx}{e^4} \\ & = \frac {2}{e^4 x}+\frac {x}{e^4}+\log (x)+\log (13+5 x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {-26-10 x+13 x^2+5 x^3+e^4 \left (13 x+10 x^2\right )}{e^4 \left (13 x^2+5 x^3\right )} \, dx=\frac {\frac {2}{x}+x+e^4 \log (x)+e^4 \log (13+5 x)}{e^4} \]
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Time = 1.85 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05
method | result | size |
risch | \({\mathrm e}^{-4} x +\frac {2 \,{\mathrm e}^{-4}}{x}+\ln \left (5 x^{2}+13 x \right )\) | \(23\) |
default | \({\mathrm e}^{-4} \left (x +{\mathrm e}^{4} \ln \left (x \right )+\frac {2}{x}+{\mathrm e}^{4} \ln \left (13+5 x \right )\right )\) | \(27\) |
parallelrisch | \(\frac {{\mathrm e}^{-4} \left (x \,{\mathrm e}^{4} \ln \left (x \right )+{\mathrm e}^{4} \ln \left (\frac {13}{5}+x \right ) x +x^{2}+2\right )}{x}\) | \(28\) |
norman | \(\frac {{\mathrm e}^{-4} x^{2}+2 \,{\mathrm e}^{-4}}{x}+\ln \left (x \right )+\ln \left (13+5 x \right )\) | \(29\) |
meijerg | \(-\frac {10 \,{\mathrm e}^{-4} \left (-\frac {13}{5 x}-\ln \left (x \right )-\ln \left (5\right )+\ln \left (13\right )+\ln \left (1+\frac {5 x}{13}\right )\right )}{13}+\left ({\mathrm e}^{4}-\frac {10}{13}\right ) {\mathrm e}^{-4} \left (\ln \left (x \right )+\ln \left (5\right )-\ln \left (13\right )-\ln \left (1+\frac {5 x}{13}\right )\right )+\frac {13 \left (\frac {10 \,{\mathrm e}^{4}}{13}+1\right ) {\mathrm e}^{-4} \ln \left (1+\frac {5 x}{13}\right )}{5}+\frac {13 \,{\mathrm e}^{-4} \left (\frac {5 x}{13}-\ln \left (1+\frac {5 x}{13}\right )\right )}{5}\) | \(84\) |
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Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {-26-10 x+13 x^2+5 x^3+e^4 \left (13 x+10 x^2\right )}{e^4 \left (13 x^2+5 x^3\right )} \, dx=\frac {{\left (x e^{4} \log \left (5 \, x^{2} + 13 \, x\right ) + x^{2} + 2\right )} e^{\left (-4\right )}}{x} \]
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Time = 0.17 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {-26-10 x+13 x^2+5 x^3+e^4 \left (13 x+10 x^2\right )}{e^4 \left (13 x^2+5 x^3\right )} \, dx=\frac {x}{e^{4}} + \log {\left (5 x^{2} + 13 x \right )} + \frac {2}{x e^{4}} \]
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Time = 0.17 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {-26-10 x+13 x^2+5 x^3+e^4 \left (13 x+10 x^2\right )}{e^4 \left (13 x^2+5 x^3\right )} \, dx={\left (e^{4} \log \left (5 \, x + 13\right ) + e^{4} \log \left (x\right ) + x + \frac {2}{x}\right )} e^{\left (-4\right )} \]
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Time = 0.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {-26-10 x+13 x^2+5 x^3+e^4 \left (13 x+10 x^2\right )}{e^4 \left (13 x^2+5 x^3\right )} \, dx={\left (e^{4} \log \left ({\left | 5 \, x + 13 \right |}\right ) + e^{4} \log \left ({\left | x \right |}\right ) + x + \frac {2}{x}\right )} e^{\left (-4\right )} \]
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Time = 0.11 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {-26-10 x+13 x^2+5 x^3+e^4 \left (13 x+10 x^2\right )}{e^4 \left (13 x^2+5 x^3\right )} \, dx=\ln \left (x\,\left (5\,x+13\right )\right )+x\,{\mathrm {e}}^{-4}+\frac {2\,{\mathrm {e}}^{-4}}{x} \]
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