Integrand size = 65, antiderivative size = 28 \[ \int \frac {2 x^2+\log ^{e^{2 x} x}(3) \left (-4+8 x-4 x^2+e^{2 x} \left (4 x-12 x^3+8 x^4\right ) \log (\log (3))\right )}{x^2-2 x^3+x^4} \, dx=\frac {2 x}{x-x^2}+\frac {4 \log ^{e^{2 x} x}(3)}{x} \]
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Time = 0.45 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.82, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {1608, 27, 6820, 2326} \[ \int \frac {2 x^2+\log ^{e^{2 x} x}(3) \left (-4+8 x-4 x^2+e^{2 x} \left (4 x-12 x^3+8 x^4\right ) \log (\log (3))\right )}{x^2-2 x^3+x^4} \, dx=\frac {2}{1-x}+\frac {4 e^{2 x} (2 x+1) \log ^{e^{2 x} x}(3)}{x \left (2 e^{2 x} x+e^{2 x}\right )} \]
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Rule 27
Rule 1608
Rule 2326
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \int \frac {2 x^2+\log ^{e^{2 x} x}(3) \left (-4+8 x-4 x^2+e^{2 x} \left (4 x-12 x^3+8 x^4\right ) \log (\log (3))\right )}{x^2 \left (1-2 x+x^2\right )} \, dx \\ & = \int \frac {2 x^2+\log ^{e^{2 x} x}(3) \left (-4+8 x-4 x^2+e^{2 x} \left (4 x-12 x^3+8 x^4\right ) \log (\log (3))\right )}{(-1+x)^2 x^2} \, dx \\ & = \int \left (\frac {2}{(-1+x)^2}+\frac {4 \log ^{e^{2 x} x}(3) \left (-1+e^{2 x} x (1+2 x) \log (\log (3))\right )}{x^2}\right ) \, dx \\ & = \frac {2}{1-x}+4 \int \frac {\log ^{e^{2 x} x}(3) \left (-1+e^{2 x} x (1+2 x) \log (\log (3))\right )}{x^2} \, dx \\ & = \frac {2}{1-x}+\frac {4 e^{2 x} (1+2 x) \log ^{e^{2 x} x}(3)}{x \left (e^{2 x}+2 e^{2 x} x\right )} \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.75 \[ \int \frac {2 x^2+\log ^{e^{2 x} x}(3) \left (-4+8 x-4 x^2+e^{2 x} \left (4 x-12 x^3+8 x^4\right ) \log (\log (3))\right )}{x^2-2 x^3+x^4} \, dx=-\frac {2}{-1+x}+\frac {4 e^{2 x} (1+2 x) \log ^{e^{2 x} x}(3)}{x \left (e^{2 x}+2 e^{2 x} x\right )} \]
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Time = 0.68 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82
method | result | size |
risch | \(-\frac {2}{-1+x}+\frac {4 \ln \left (3\right )^{x \,{\mathrm e}^{2 x}}}{x}\) | \(23\) |
norman | \(\frac {-2 x +4 \,{\mathrm e}^{x \,{\mathrm e}^{2 x} \ln \left (\ln \left (3\right )\right )} x -4 \,{\mathrm e}^{x \,{\mathrm e}^{2 x} \ln \left (\ln \left (3\right )\right )}}{x \left (-1+x \right )}\) | \(39\) |
parallelrisch | \(\frac {-2 x +4 \,{\mathrm e}^{x \,{\mathrm e}^{2 x} \ln \left (\ln \left (3\right )\right )} x -4 \,{\mathrm e}^{x \,{\mathrm e}^{2 x} \ln \left (\ln \left (3\right )\right )}}{x \left (-1+x \right )}\) | \(39\) |
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Time = 0.23 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {2 x^2+\log ^{e^{2 x} x}(3) \left (-4+8 x-4 x^2+e^{2 x} \left (4 x-12 x^3+8 x^4\right ) \log (\log (3))\right )}{x^2-2 x^3+x^4} \, dx=\frac {2 \, {\left (2 \, {\left (x - 1\right )} \log \left (3\right )^{x e^{\left (2 \, x\right )}} - x\right )}}{x^{2} - x} \]
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Time = 0.11 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {2 x^2+\log ^{e^{2 x} x}(3) \left (-4+8 x-4 x^2+e^{2 x} \left (4 x-12 x^3+8 x^4\right ) \log (\log (3))\right )}{x^2-2 x^3+x^4} \, dx=- \frac {2}{x - 1} + \frac {4 e^{x e^{2 x} \log {\left (\log {\left (3 \right )} \right )}}}{x} \]
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Time = 0.32 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {2 x^2+\log ^{e^{2 x} x}(3) \left (-4+8 x-4 x^2+e^{2 x} \left (4 x-12 x^3+8 x^4\right ) \log (\log (3))\right )}{x^2-2 x^3+x^4} \, dx=\frac {2 \, {\left (2 \, {\left (x - 1\right )} \log \left (3\right )^{x e^{\left (2 \, x\right )}} - x\right )}}{x^{2} - x} \]
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\[ \int \frac {2 x^2+\log ^{e^{2 x} x}(3) \left (-4+8 x-4 x^2+e^{2 x} \left (4 x-12 x^3+8 x^4\right ) \log (\log (3))\right )}{x^2-2 x^3+x^4} \, dx=\int { \frac {2 \, {\left (x^{2} + 2 \, {\left ({\left (2 \, x^{4} - 3 \, x^{3} + x\right )} e^{\left (2 \, x\right )} \log \left (\log \left (3\right )\right ) - x^{2} + 2 \, x - 1\right )} \log \left (3\right )^{x e^{\left (2 \, x\right )}}\right )}}{x^{4} - 2 \, x^{3} + x^{2}} \,d x } \]
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Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {2 x^2+\log ^{e^{2 x} x}(3) \left (-4+8 x-4 x^2+e^{2 x} \left (4 x-12 x^3+8 x^4\right ) \log (\log (3))\right )}{x^2-2 x^3+x^4} \, dx=\frac {4\,{\ln \left (3\right )}^{x\,{\mathrm {e}}^{2\,x}}}{x}-\frac {2}{x-1} \]
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