Integrand size = 102, antiderivative size = 27 \[ \int \frac {-9+6 x^2 \log (4 x)+3 x^2 \log ^2(4 x)+12 \log \left (e^{10} x\right )}{1+2 x+x^2+\left (-2 x^2-2 x^3\right ) \log ^2(4 x)+x^4 \log ^4(4 x)+\left (8+8 x-8 x^2 \log ^2(4 x)\right ) \log \left (e^{10} x\right )+16 \log ^2\left (e^{10} x\right )} \, dx=\frac {3 x}{1+x-x^2 \log ^2(4 x)+4 \log \left (e^{10} x\right )} \]
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\[ \int \frac {-9+6 x^2 \log (4 x)+3 x^2 \log ^2(4 x)+12 \log \left (e^{10} x\right )}{1+2 x+x^2+\left (-2 x^2-2 x^3\right ) \log ^2(4 x)+x^4 \log ^4(4 x)+\left (8+8 x-8 x^2 \log ^2(4 x)\right ) \log \left (e^{10} x\right )+16 \log ^2\left (e^{10} x\right )} \, dx=\int \frac {-9+6 x^2 \log (4 x)+3 x^2 \log ^2(4 x)+12 \log \left (e^{10} x\right )}{1+2 x+x^2+\left (-2 x^2-2 x^3\right ) \log ^2(4 x)+x^4 \log ^4(4 x)+\left (8+8 x-8 x^2 \log ^2(4 x)\right ) \log \left (e^{10} x\right )+16 \log ^2\left (e^{10} x\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {3 \left (37+4 \log (x)+2 x^2 \log (4 x)+x^2 \log ^2(4 x)\right )}{\left (41+x+4 \log (x)-x^2 \log ^2(4 x)\right )^2} \, dx \\ & = 3 \int \frac {37+4 \log (x)+2 x^2 \log (4 x)+x^2 \log ^2(4 x)}{\left (41+x+4 \log (x)-x^2 \log ^2(4 x)\right )^2} \, dx \\ & = 3 \int \left (\frac {78+x+8 \log (x)+2 x^2 \log (4 x)}{\left (-41-x-4 \log (x)+x^2 \log ^2(4 x)\right )^2}+\frac {1}{-41-x-4 \log (x)+x^2 \log ^2(4 x)}\right ) \, dx \\ & = 3 \int \frac {78+x+8 \log (x)+2 x^2 \log (4 x)}{\left (-41-x-4 \log (x)+x^2 \log ^2(4 x)\right )^2} \, dx+3 \int \frac {1}{-41-x-4 \log (x)+x^2 \log ^2(4 x)} \, dx \\ & = 3 \int \frac {1}{-41-x-4 \log (x)+x^2 \log ^2(4 x)} \, dx+3 \int \left (\frac {8 \log (x)}{\left (41+x+4 \log (x)-x^2 \log ^2(4 x)\right )^2}+\frac {78}{\left (-41-x-4 \log (x)+x^2 \log ^2(4 x)\right )^2}+\frac {x}{\left (-41-x-4 \log (x)+x^2 \log ^2(4 x)\right )^2}+\frac {2 x^2 \log (4 x)}{\left (-41-x-4 \log (x)+x^2 \log ^2(4 x)\right )^2}\right ) \, dx \\ & = 3 \int \frac {x}{\left (-41-x-4 \log (x)+x^2 \log ^2(4 x)\right )^2} \, dx+3 \int \frac {1}{-41-x-4 \log (x)+x^2 \log ^2(4 x)} \, dx+6 \int \frac {x^2 \log (4 x)}{\left (-41-x-4 \log (x)+x^2 \log ^2(4 x)\right )^2} \, dx+24 \int \frac {\log (x)}{\left (41+x+4 \log (x)-x^2 \log ^2(4 x)\right )^2} \, dx+234 \int \frac {1}{\left (-41-x-4 \log (x)+x^2 \log ^2(4 x)\right )^2} \, dx \\ \end{align*}
Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {-9+6 x^2 \log (4 x)+3 x^2 \log ^2(4 x)+12 \log \left (e^{10} x\right )}{1+2 x+x^2+\left (-2 x^2-2 x^3\right ) \log ^2(4 x)+x^4 \log ^4(4 x)+\left (8+8 x-8 x^2 \log ^2(4 x)\right ) \log \left (e^{10} x\right )+16 \log ^2\left (e^{10} x\right )} \, dx=\frac {3 x}{41+x+4 \log (x)-x^2 \log ^2(4 x)} \]
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Time = 2.33 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11
method | result | size |
parallelrisch | \(-\frac {3 x}{x^{2} \ln \left (4 x \right )^{2}-4 \ln \left (x \,{\mathrm e}^{10}\right )-x -1}\) | \(30\) |
default | \(-\frac {3 x}{4 x^{2} \ln \left (2\right )^{2}+4 x^{2} \ln \left (2\right ) \ln \left (x \right )+x^{2} \ln \left (x \right )^{2}-x -4 \ln \left (x \right )-41}\) | \(41\) |
risch | \(\frac {12 x}{164+4 x -16 x^{2} \ln \left (2\right )^{2}-4 x^{2} \ln \left (x \right )^{2}-16 x^{2} \ln \left (2\right ) \ln \left (x \right )+16 \ln \left (x \right )}\) | \(42\) |
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Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {-9+6 x^2 \log (4 x)+3 x^2 \log ^2(4 x)+12 \log \left (e^{10} x\right )}{1+2 x+x^2+\left (-2 x^2-2 x^3\right ) \log ^2(4 x)+x^4 \log ^4(4 x)+\left (8+8 x-8 x^2 \log ^2(4 x)\right ) \log \left (e^{10} x\right )+16 \log ^2\left (e^{10} x\right )} \, dx=-\frac {3 \, x}{x^{2} \log \left (4 \, x\right )^{2} - x + 8 \, \log \left (2\right ) - 4 \, \log \left (4 \, x\right ) - 41} \]
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Time = 0.16 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {-9+6 x^2 \log (4 x)+3 x^2 \log ^2(4 x)+12 \log \left (e^{10} x\right )}{1+2 x+x^2+\left (-2 x^2-2 x^3\right ) \log ^2(4 x)+x^4 \log ^4(4 x)+\left (8+8 x-8 x^2 \log ^2(4 x)\right ) \log \left (e^{10} x\right )+16 \log ^2\left (e^{10} x\right )} \, dx=- \frac {3 x}{x^{2} \log {\left (4 x \right )}^{2} - x - 4 \log {\left (4 x \right )} - 41 + 8 \log {\left (2 \right )}} \]
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Time = 0.33 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.44 \[ \int \frac {-9+6 x^2 \log (4 x)+3 x^2 \log ^2(4 x)+12 \log \left (e^{10} x\right )}{1+2 x+x^2+\left (-2 x^2-2 x^3\right ) \log ^2(4 x)+x^4 \log ^4(4 x)+\left (8+8 x-8 x^2 \log ^2(4 x)\right ) \log \left (e^{10} x\right )+16 \log ^2\left (e^{10} x\right )} \, dx=-\frac {3 \, x}{4 \, x^{2} \log \left (2\right )^{2} + x^{2} \log \left (x\right )^{2} + 4 \, {\left (x^{2} \log \left (2\right ) - 1\right )} \log \left (x\right ) - x - 41} \]
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Time = 0.29 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.48 \[ \int \frac {-9+6 x^2 \log (4 x)+3 x^2 \log ^2(4 x)+12 \log \left (e^{10} x\right )}{1+2 x+x^2+\left (-2 x^2-2 x^3\right ) \log ^2(4 x)+x^4 \log ^4(4 x)+\left (8+8 x-8 x^2 \log ^2(4 x)\right ) \log \left (e^{10} x\right )+16 \log ^2\left (e^{10} x\right )} \, dx=-\frac {3 \, x}{4 \, x^{2} \log \left (2\right )^{2} + 4 \, x^{2} \log \left (2\right ) \log \left (x\right ) + x^{2} \log \left (x\right )^{2} - x - 4 \, \log \left (x\right ) - 41} \]
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Timed out. \[ \int \frac {-9+6 x^2 \log (4 x)+3 x^2 \log ^2(4 x)+12 \log \left (e^{10} x\right )}{1+2 x+x^2+\left (-2 x^2-2 x^3\right ) \log ^2(4 x)+x^4 \log ^4(4 x)+\left (8+8 x-8 x^2 \log ^2(4 x)\right ) \log \left (e^{10} x\right )+16 \log ^2\left (e^{10} x\right )} \, dx=\int \frac {12\,\ln \left (x\,{\mathrm {e}}^{10}\right )+6\,x^2\,\ln \left (4\,x\right )+3\,x^2\,{\ln \left (4\,x\right )}^2-9}{2\,x+16\,{\ln \left (x\,{\mathrm {e}}^{10}\right )}^2+\ln \left (x\,{\mathrm {e}}^{10}\right )\,\left (-8\,x^2\,{\ln \left (4\,x\right )}^2+8\,x+8\right )-{\ln \left (4\,x\right )}^2\,\left (2\,x^3+2\,x^2\right )+x^2+x^4\,{\ln \left (4\,x\right )}^4+1} \,d x \]
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