Integrand size = 52, antiderivative size = 22 \[ \int \frac {-32 e^9+32 e^{9+x} x+160 e^9 x \log (x)+80 e^{9+x} x^2 \log ^2(x)}{2 x+5 x^2 \log ^2(x)} \, dx=16 e^9 \left (e^x+\log \left (\frac {2}{5 x}+\log ^2(x)\right )\right ) \]
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\[ \int \frac {-32 e^9+32 e^{9+x} x+160 e^9 x \log (x)+80 e^{9+x} x^2 \log ^2(x)}{2 x+5 x^2 \log ^2(x)} \, dx=\int \frac {-32 e^9+32 e^{9+x} x+160 e^9 x \log (x)+80 e^{9+x} x^2 \log ^2(x)}{2 x+5 x^2 \log ^2(x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {-32 e^9+32 e^{9+x} x+160 e^9 x \log (x)+80 e^{9+x} x^2 \log ^2(x)}{x^2 \left (\frac {2}{x}+5 \log ^2(x)\right )} \, dx \\ & = \int \frac {-32 e^9+32 e^{9+x} x+160 e^9 x \log (x)+80 e^{9+x} x^2 \log ^2(x)}{x \left (2+5 x \log ^2(x)\right )} \, dx \\ & = \int \frac {16 e^9 \left (-2+2 e^x x+10 x \log (x)+5 e^x x^2 \log ^2(x)\right )}{x \left (2+5 x \log ^2(x)\right )} \, dx \\ & = \left (16 e^9\right ) \int \frac {-2+2 e^x x+10 x \log (x)+5 e^x x^2 \log ^2(x)}{x \left (2+5 x \log ^2(x)\right )} \, dx \\ & = \left (16 e^9\right ) \int \left (e^x+\frac {2 (-1+5 x \log (x))}{x \left (2+5 x \log ^2(x)\right )}\right ) \, dx \\ & = \left (16 e^9\right ) \int e^x \, dx+\left (32 e^9\right ) \int \frac {-1+5 x \log (x)}{x \left (2+5 x \log ^2(x)\right )} \, dx \\ & = 16 e^{9+x}+\left (32 e^9\right ) \int \left (-\frac {1}{x \left (2+5 x \log ^2(x)\right )}+\frac {5 \log (x)}{2+5 x \log ^2(x)}\right ) \, dx \\ & = 16 e^{9+x}-\left (32 e^9\right ) \int \frac {1}{x \left (2+5 x \log ^2(x)\right )} \, dx+\left (160 e^9\right ) \int \frac {\log (x)}{2+5 x \log ^2(x)} \, dx \\ \end{align*}
Time = 0.16 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {-32 e^9+32 e^{9+x} x+160 e^9 x \log (x)+80 e^{9+x} x^2 \log ^2(x)}{2 x+5 x^2 \log ^2(x)} \, dx=16 e^9 \left (e^x-\log (x)+\log \left (2+5 x \log ^2(x)\right )\right ) \]
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Time = 0.75 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05
method | result | size |
risch | \(16 \,{\mathrm e}^{x +9}+16 \,{\mathrm e}^{9} \ln \left (\ln \left (x \right )^{2}+\frac {2}{5 x}\right )\) | \(23\) |
parallelrisch | \(16 \,{\mathrm e}^{9} \ln \left (x \ln \left (x \right )^{2}+\frac {2}{5}\right )-16 \,{\mathrm e}^{9} \ln \left (x \right )+16 \,{\mathrm e}^{9} {\mathrm e}^{x}\) | \(27\) |
norman | \(-16 \,{\mathrm e}^{9} \ln \left (x \right )+16 \,{\mathrm e}^{9} {\mathrm e}^{x}+16 \,{\mathrm e}^{9} \ln \left (5 x \ln \left (x \right )^{2}+2\right )\) | \(28\) |
default | \(32 \,{\mathrm e}^{9} \left (-\frac {\ln \left (x \right )}{2}+\frac {\ln \left (5 x \ln \left (x \right )^{2}+2\right )}{2}\right )+16 \,{\mathrm e}^{9} {\mathrm e}^{x}\) | \(29\) |
parts | \(32 \,{\mathrm e}^{9} \left (-\frac {\ln \left (x \right )}{2}+\frac {\ln \left (5 x \ln \left (x \right )^{2}+2\right )}{2}\right )+16 \,{\mathrm e}^{9} {\mathrm e}^{x}\) | \(29\) |
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Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {-32 e^9+32 e^{9+x} x+160 e^9 x \log (x)+80 e^{9+x} x^2 \log ^2(x)}{2 x+5 x^2 \log ^2(x)} \, dx=16 \, e^{9} \log \left (\frac {5 \, x \log \left (x\right )^{2} + 2}{x}\right ) + 16 \, e^{\left (x + 9\right )} \]
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Time = 0.15 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {-32 e^9+32 e^{9+x} x+160 e^9 x \log (x)+80 e^{9+x} x^2 \log ^2(x)}{2 x+5 x^2 \log ^2(x)} \, dx=16 e^{9} e^{x} + 16 e^{9} \log {\left (\log {\left (x \right )}^{2} + \frac {2}{5 x} \right )} \]
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Time = 0.22 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {-32 e^9+32 e^{9+x} x+160 e^9 x \log (x)+80 e^{9+x} x^2 \log ^2(x)}{2 x+5 x^2 \log ^2(x)} \, dx=16 \, e^{9} \log \left (\frac {5 \, x \log \left (x\right )^{2} + 2}{5 \, x}\right ) + 16 \, e^{\left (x + 9\right )} \]
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Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {-32 e^9+32 e^{9+x} x+160 e^9 x \log (x)+80 e^{9+x} x^2 \log ^2(x)}{2 x+5 x^2 \log ^2(x)} \, dx=16 \, e^{9} \log \left (5 \, x \log \left (x\right )^{2} + 2\right ) - 16 \, e^{9} \log \left (x\right ) + 16 \, e^{\left (x + 9\right )} \]
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Time = 11.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {-32 e^9+32 e^{9+x} x+160 e^9 x \log (x)+80 e^{9+x} x^2 \log ^2(x)}{2 x+5 x^2 \log ^2(x)} \, dx=16\,{\mathrm {e}}^9\,{\mathrm {e}}^x+16\,\ln \left (\frac {5\,x\,{\ln \left (x\right )}^2+2}{x}\right )\,{\mathrm {e}}^9 \]
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