Integrand size = 119, antiderivative size = 26 \[ \int \frac {5 e^2+e^4 x^2+\left (-5 e^2+e^4 \left (-x-3 x^2\right )\right ) \log (x)+\left (e^4 x-2 e^4 x \log (x)\right ) \log \left (e^5 x\right )}{25 x^2+10 e^2 x^4+e^4 x^6+\left (10 e^2 x^3+2 e^4 x^5\right ) \log \left (e^5 x\right )+e^4 x^4 \log ^2\left (e^5 x\right )} \, dx=-4+\frac {\log (x)}{x \left (\frac {5}{e^2}+x \left (x+\log \left (e^5 x\right )\right )\right )} \]
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\[ \int \frac {5 e^2+e^4 x^2+\left (-5 e^2+e^4 \left (-x-3 x^2\right )\right ) \log (x)+\left (e^4 x-2 e^4 x \log (x)\right ) \log \left (e^5 x\right )}{25 x^2+10 e^2 x^4+e^4 x^6+\left (10 e^2 x^3+2 e^4 x^5\right ) \log \left (e^5 x\right )+e^4 x^4 \log ^2\left (e^5 x\right )} \, dx=\int \frac {5 e^2+e^4 x^2+\left (-5 e^2+e^4 \left (-x-3 x^2\right )\right ) \log (x)+\left (e^4 x-2 e^4 x \log (x)\right ) \log \left (e^5 x\right )}{25 x^2+10 e^2 x^4+e^4 x^6+\left (10 e^2 x^3+2 e^4 x^5\right ) \log \left (e^5 x\right )+e^4 x^4 \log ^2\left (e^5 x\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^2 \left (5+e^2 x (5+x)-\left (5+e^2 x (10+3 x)\right ) \log (x)-2 e^2 x \log ^2(x)\right )}{x^2 \left (5+e^2 x (5+x)+e^2 x \log (x)\right )^2} \, dx \\ & = e^2 \int \frac {5+e^2 x (5+x)-\left (5+e^2 x (10+3 x)\right ) \log (x)-2 e^2 x \log ^2(x)}{x^2 \left (5+e^2 x (5+x)+e^2 x \log (x)\right )^2} \, dx \\ & = e^2 \int \left (-\frac {2}{e^2 x^3}+\frac {-25-20 e^2 x+5 e^4 x^2+6 e^4 x^3+e^4 x^4}{e^2 x^3 \left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )^2}+\frac {15+10 e^2 x+e^2 x^2}{e^2 x^3 \left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )}\right ) \, dx \\ & = \frac {1}{x^2}+\int \frac {-25-20 e^2 x+5 e^4 x^2+6 e^4 x^3+e^4 x^4}{x^3 \left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )^2} \, dx+\int \frac {15+10 e^2 x+e^2 x^2}{x^3 \left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )} \, dx \\ & = \frac {1}{x^2}+\int \left (\frac {6 e^4}{\left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )^2}-\frac {25}{x^3 \left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )^2}-\frac {20 e^2}{x^2 \left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )^2}+\frac {5 e^4}{x \left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )^2}+\frac {e^4 x}{\left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )^2}\right ) \, dx+\int \left (\frac {15}{x^3 \left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )}+\frac {10 e^2}{x^2 \left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )}+\frac {e^2}{x \left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )}\right ) \, dx \\ & = \frac {1}{x^2}+15 \int \frac {1}{x^3 \left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )} \, dx-25 \int \frac {1}{x^3 \left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )^2} \, dx+e^2 \int \frac {1}{x \left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )} \, dx+\left (10 e^2\right ) \int \frac {1}{x^2 \left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )} \, dx-\left (20 e^2\right ) \int \frac {1}{x^2 \left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )^2} \, dx+e^4 \int \frac {x}{\left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )^2} \, dx+\left (5 e^4\right ) \int \frac {1}{x \left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )^2} \, dx+\left (6 e^4\right ) \int \frac {1}{\left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )^2} \, dx \\ \end{align*}
Time = 0.57 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {5 e^2+e^4 x^2+\left (-5 e^2+e^4 \left (-x-3 x^2\right )\right ) \log (x)+\left (e^4 x-2 e^4 x \log (x)\right ) \log \left (e^5 x\right )}{25 x^2+10 e^2 x^4+e^4 x^6+\left (10 e^2 x^3+2 e^4 x^5\right ) \log \left (e^5 x\right )+e^4 x^4 \log ^2\left (e^5 x\right )} \, dx=\frac {e^2 \log (x)}{x \left (5+e^2 x (5+x)+e^2 x \log (x)\right )} \]
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Time = 4.52 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08
method | result | size |
parallelrisch | \(\frac {{\mathrm e}^{2} \ln \left (x \right )}{x \left ({\mathrm e}^{2} \ln \left (x \,{\mathrm e}^{5}\right ) x +x^{2} {\mathrm e}^{2}+5\right )}\) | \(28\) |
default | \(\frac {{\mathrm e}^{2} \ln \left (x \right )}{x \left (x^{2} {\mathrm e}^{2}+x \,{\mathrm e}^{2} \ln \left (x \right )+5 \,{\mathrm e}^{2} x +5\right )}\) | \(30\) |
risch | \(\frac {1}{x^{2}}-\frac {10+2 x^{2} {\mathrm e}^{2}+10 \,{\mathrm e}^{2} x}{x^{2} \left (10+2 x^{2} {\mathrm e}^{2}+2 x \,{\mathrm e}^{2} \ln \left (x \right )+10 \,{\mathrm e}^{2} x \right )}\) | \(47\) |
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Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {5 e^2+e^4 x^2+\left (-5 e^2+e^4 \left (-x-3 x^2\right )\right ) \log (x)+\left (e^4 x-2 e^4 x \log (x)\right ) \log \left (e^5 x\right )}{25 x^2+10 e^2 x^4+e^4 x^6+\left (10 e^2 x^3+2 e^4 x^5\right ) \log \left (e^5 x\right )+e^4 x^4 \log ^2\left (e^5 x\right )} \, dx=\frac {e^{2} \log \left (x\right )}{x^{2} e^{2} \log \left (x\right ) + {\left (x^{3} + 5 \, x^{2}\right )} e^{2} + 5 \, x} \]
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Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (20) = 40\).
Time = 0.11 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.96 \[ \int \frac {5 e^2+e^4 x^2+\left (-5 e^2+e^4 \left (-x-3 x^2\right )\right ) \log (x)+\left (e^4 x-2 e^4 x \log (x)\right ) \log \left (e^5 x\right )}{25 x^2+10 e^2 x^4+e^4 x^6+\left (10 e^2 x^3+2 e^4 x^5\right ) \log \left (e^5 x\right )+e^4 x^4 \log ^2\left (e^5 x\right )} \, dx=\frac {- x^{2} e^{2} - 5 x e^{2} - 5}{x^{4} e^{2} + x^{3} e^{2} \log {\left (x \right )} + 5 x^{3} e^{2} + 5 x^{2}} + \frac {1}{x^{2}} \]
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Time = 0.25 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {5 e^2+e^4 x^2+\left (-5 e^2+e^4 \left (-x-3 x^2\right )\right ) \log (x)+\left (e^4 x-2 e^4 x \log (x)\right ) \log \left (e^5 x\right )}{25 x^2+10 e^2 x^4+e^4 x^6+\left (10 e^2 x^3+2 e^4 x^5\right ) \log \left (e^5 x\right )+e^4 x^4 \log ^2\left (e^5 x\right )} \, dx=\frac {e^{2} \log \left (x\right )}{x^{3} e^{2} + x^{2} e^{2} \log \left (x\right ) + 5 \, x^{2} e^{2} + 5 \, x} \]
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Time = 0.25 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {5 e^2+e^4 x^2+\left (-5 e^2+e^4 \left (-x-3 x^2\right )\right ) \log (x)+\left (e^4 x-2 e^4 x \log (x)\right ) \log \left (e^5 x\right )}{25 x^2+10 e^2 x^4+e^4 x^6+\left (10 e^2 x^3+2 e^4 x^5\right ) \log \left (e^5 x\right )+e^4 x^4 \log ^2\left (e^5 x\right )} \, dx=\frac {e^{2} \log \left (x\right )}{x^{3} e^{2} + x^{2} e^{2} \log \left (x\right ) + 5 \, x^{2} e^{2} + 5 \, x} \]
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Time = 10.23 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {5 e^2+e^4 x^2+\left (-5 e^2+e^4 \left (-x-3 x^2\right )\right ) \log (x)+\left (e^4 x-2 e^4 x \log (x)\right ) \log \left (e^5 x\right )}{25 x^2+10 e^2 x^4+e^4 x^6+\left (10 e^2 x^3+2 e^4 x^5\right ) \log \left (e^5 x\right )+e^4 x^4 \log ^2\left (e^5 x\right )} \, dx=\frac {{\mathrm {e}}^2\,\ln \left (x\right )}{x\,\left (5\,x\,{\mathrm {e}}^2+x^2\,{\mathrm {e}}^2+x\,{\mathrm {e}}^2\,\ln \left (x\right )+5\right )} \]
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