Integrand size = 108, antiderivative size = 26 \[ \int \frac {9 e^{10} x^3 \log (4)+\left (-27-36 x-18 x^2\right ) \log (5)+\left (18 e^{10} x^2 \log (4)-9 x \log (5)\right ) \log (x)+9 e^{10} x \log (4) \log ^2(x)}{9 x^3+6 x^4+x^5+\left (18 x^2+12 x^3+2 x^4\right ) \log (x)+\left (9 x+6 x^2+x^3\right ) \log ^2(x)} \, dx=e-\frac {9 \left (e^{10} \log (4)-\frac {\log (5)}{x+\log (x)}\right )}{3+x} \]
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\[ \int \frac {9 e^{10} x^3 \log (4)+\left (-27-36 x-18 x^2\right ) \log (5)+\left (18 e^{10} x^2 \log (4)-9 x \log (5)\right ) \log (x)+9 e^{10} x \log (4) \log ^2(x)}{9 x^3+6 x^4+x^5+\left (18 x^2+12 x^3+2 x^4\right ) \log (x)+\left (9 x+6 x^2+x^3\right ) \log ^2(x)} \, dx=\int \frac {9 e^{10} x^3 \log (4)+\left (-27-36 x-18 x^2\right ) \log (5)+\left (18 e^{10} x^2 \log (4)-9 x \log (5)\right ) \log (x)+9 e^{10} x \log (4) \log ^2(x)}{9 x^3+6 x^4+x^5+\left (18 x^2+12 x^3+2 x^4\right ) \log (x)+\left (9 x+6 x^2+x^3\right ) \log ^2(x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {9 \left (e^{10} x^3 \log (4)-\left (3+4 x+2 x^2\right ) \log (5)+x \left (-\log (5)+e^{10} x \log (16)\right ) \log (x)+e^{10} x \log (4) \log ^2(x)\right )}{x (3+x)^2 (x+\log (x))^2} \, dx \\ & = 9 \int \frac {e^{10} x^3 \log (4)-\left (3+4 x+2 x^2\right ) \log (5)+x \left (-\log (5)+e^{10} x \log (16)\right ) \log (x)+e^{10} x \log (4) \log ^2(x)}{x (3+x)^2 (x+\log (x))^2} \, dx \\ & = 9 \int \left (\frac {e^{10} \log (4)}{(3+x)^2}-\frac {(1+x) \log (5)}{x (3+x) (x+\log (x))^2}-\frac {\log (5)}{(3+x)^2 (x+\log (x))}\right ) \, dx \\ & = -\frac {9 e^{10} \log (4)}{3+x}-(9 \log (5)) \int \frac {1+x}{x (3+x) (x+\log (x))^2} \, dx-(9 \log (5)) \int \frac {1}{(3+x)^2 (x+\log (x))} \, dx \\ & = -\frac {9 e^{10} \log (4)}{3+x}-(9 \log (5)) \int \frac {1}{(3+x)^2 (x+\log (x))} \, dx-(9 \log (5)) \int \left (\frac {1}{3 x (x+\log (x))^2}+\frac {2}{3 (3+x) (x+\log (x))^2}\right ) \, dx \\ & = -\frac {9 e^{10} \log (4)}{3+x}-(3 \log (5)) \int \frac {1}{x (x+\log (x))^2} \, dx-(6 \log (5)) \int \frac {1}{(3+x) (x+\log (x))^2} \, dx-(9 \log (5)) \int \frac {1}{(3+x)^2 (x+\log (x))} \, dx \\ \end{align*}
Time = 5.07 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.27 \[ \int \frac {9 e^{10} x^3 \log (4)+\left (-27-36 x-18 x^2\right ) \log (5)+\left (18 e^{10} x^2 \log (4)-9 x \log (5)\right ) \log (x)+9 e^{10} x \log (4) \log ^2(x)}{9 x^3+6 x^4+x^5+\left (18 x^2+12 x^3+2 x^4\right ) \log (x)+\left (9 x+6 x^2+x^3\right ) \log ^2(x)} \, dx=\frac {9 \left (-e^{10} x \log (4)+\log (5)-e^{10} \log (4) \log (x)\right )}{(3+x) (x+\log (x))} \]
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Time = 2.35 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08
method | result | size |
risch | \(-\frac {18 \,{\mathrm e}^{10} \ln \left (2\right )}{3+x}+\frac {9 \ln \left (5\right )}{\left (3+x \right ) \left (x +\ln \left (x \right )\right )}\) | \(28\) |
default | \(-\frac {18 \,{\mathrm e}^{10} \ln \left (2\right )}{3+x}+\frac {9 \ln \left (5\right )}{\left (3+x \right ) \left (x +\ln \left (x \right )\right )}\) | \(30\) |
norman | \(\frac {-18 x \,{\mathrm e}^{10} \ln \left (2\right )-18 \,{\mathrm e}^{10} \ln \left (2\right ) \ln \left (x \right )+9 \ln \left (5\right )}{x \ln \left (x \right )+x^{2}+3 \ln \left (x \right )+3 x}\) | \(43\) |
parallelrisch | \(-\frac {18 x \,{\mathrm e}^{10} \ln \left (2\right )+18 \,{\mathrm e}^{10} \ln \left (2\right ) \ln \left (x \right )-9 \ln \left (5\right )}{x \ln \left (x \right )+x^{2}+3 \ln \left (x \right )+3 x}\) | \(44\) |
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Time = 0.28 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42 \[ \int \frac {9 e^{10} x^3 \log (4)+\left (-27-36 x-18 x^2\right ) \log (5)+\left (18 e^{10} x^2 \log (4)-9 x \log (5)\right ) \log (x)+9 e^{10} x \log (4) \log ^2(x)}{9 x^3+6 x^4+x^5+\left (18 x^2+12 x^3+2 x^4\right ) \log (x)+\left (9 x+6 x^2+x^3\right ) \log ^2(x)} \, dx=-\frac {9 \, {\left (2 \, x e^{10} \log \left (2\right ) + 2 \, e^{10} \log \left (2\right ) \log \left (x\right ) - \log \left (5\right )\right )}}{x^{2} + {\left (x + 3\right )} \log \left (x\right ) + 3 \, x} \]
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Time = 0.10 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {9 e^{10} x^3 \log (4)+\left (-27-36 x-18 x^2\right ) \log (5)+\left (18 e^{10} x^2 \log (4)-9 x \log (5)\right ) \log (x)+9 e^{10} x \log (4) \log ^2(x)}{9 x^3+6 x^4+x^5+\left (18 x^2+12 x^3+2 x^4\right ) \log (x)+\left (9 x+6 x^2+x^3\right ) \log ^2(x)} \, dx=\frac {9 \log {\left (5 \right )}}{x^{2} + 3 x + \left (x + 3\right ) \log {\left (x \right )}} - \frac {18 e^{10} \log {\left (2 \right )}}{x + 3} \]
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Time = 0.31 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42 \[ \int \frac {9 e^{10} x^3 \log (4)+\left (-27-36 x-18 x^2\right ) \log (5)+\left (18 e^{10} x^2 \log (4)-9 x \log (5)\right ) \log (x)+9 e^{10} x \log (4) \log ^2(x)}{9 x^3+6 x^4+x^5+\left (18 x^2+12 x^3+2 x^4\right ) \log (x)+\left (9 x+6 x^2+x^3\right ) \log ^2(x)} \, dx=-\frac {9 \, {\left (2 \, x e^{10} \log \left (2\right ) + 2 \, e^{10} \log \left (2\right ) \log \left (x\right ) - \log \left (5\right )\right )}}{x^{2} + {\left (x + 3\right )} \log \left (x\right ) + 3 \, x} \]
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Time = 0.29 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50 \[ \int \frac {9 e^{10} x^3 \log (4)+\left (-27-36 x-18 x^2\right ) \log (5)+\left (18 e^{10} x^2 \log (4)-9 x \log (5)\right ) \log (x)+9 e^{10} x \log (4) \log ^2(x)}{9 x^3+6 x^4+x^5+\left (18 x^2+12 x^3+2 x^4\right ) \log (x)+\left (9 x+6 x^2+x^3\right ) \log ^2(x)} \, dx=-\frac {9 \, {\left (2 \, x e^{10} \log \left (2\right ) + 2 \, e^{10} \log \left (2\right ) \log \left (x\right ) - \log \left (5\right )\right )}}{x^{2} + x \log \left (x\right ) + 3 \, x + 3 \, \log \left (x\right )} \]
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Time = 15.34 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {9 e^{10} x^3 \log (4)+\left (-27-36 x-18 x^2\right ) \log (5)+\left (18 e^{10} x^2 \log (4)-9 x \log (5)\right ) \log (x)+9 e^{10} x \log (4) \log ^2(x)}{9 x^3+6 x^4+x^5+\left (18 x^2+12 x^3+2 x^4\right ) \log (x)+\left (9 x+6 x^2+x^3\right ) \log ^2(x)} \, dx=\frac {9\,\ln \left (5\right )}{\left (x+\ln \left (x\right )\right )\,\left (x+3\right )}-\frac {18\,{\mathrm {e}}^{10}\,\ln \left (2\right )}{x+3} \]
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