Integrand size = 102, antiderivative size = 30 \[ \int \frac {e^2 \left (-20 x^4-5 x^5\right ) \log (4)+125 e x \log ^2(4)+\left (e^2 \left (-20 x^4-10 x^5\right ) \log (4)-250 e x \log ^2(4)\right ) \log (x)}{\left (e^2 \left (16 x^6+8 x^7+x^8\right )+e \left (-200 x^3-50 x^4\right ) \log (4)+625 \log ^2(4)\right ) \log ^2(x)} \, dx=\frac {x^2}{\left (-\frac {5}{e}+\frac {x^3 (4+x)}{5 \log (4)}\right ) \log (x)} \]
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\[ \int \frac {e^2 \left (-20 x^4-5 x^5\right ) \log (4)+125 e x \log ^2(4)+\left (e^2 \left (-20 x^4-10 x^5\right ) \log (4)-250 e x \log ^2(4)\right ) \log (x)}{\left (e^2 \left (16 x^6+8 x^7+x^8\right )+e \left (-200 x^3-50 x^4\right ) \log (4)+625 \log ^2(4)\right ) \log ^2(x)} \, dx=\int \frac {e^2 \left (-20 x^4-5 x^5\right ) \log (4)+125 e x \log ^2(4)+\left (e^2 \left (-20 x^4-10 x^5\right ) \log (4)-250 e x \log ^2(4)\right ) \log (x)}{\left (e^2 \left (16 x^6+8 x^7+x^8\right )+e \left (-200 x^3-50 x^4\right ) \log (4)+625 \log ^2(4)\right ) \log ^2(x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {5 e x \log (4) \left (-e x^3 (4+x)+25 \log (4)-2 \left (e x^3 (2+x)+25 \log (4)\right ) \log (x)\right )}{\left (4 e x^3+e x^4-25 \log (4)\right )^2 \log ^2(x)} \, dx \\ & = (5 e \log (4)) \int \frac {x \left (-e x^3 (4+x)+25 \log (4)-2 \left (e x^3 (2+x)+25 \log (4)\right ) \log (x)\right )}{\left (4 e x^3+e x^4-25 \log (4)\right )^2 \log ^2(x)} \, dx \\ & = (5 e \log (4)) \int \left (-\frac {x}{\left (4 e x^3+e x^4-25 \log (4)\right ) \log ^2(x)}-\frac {2 x \left (2 e x^3+e x^4+25 \log (4)\right )}{\left (4 e x^3+e x^4-25 \log (4)\right )^2 \log (x)}\right ) \, dx \\ & = -\left ((5 e \log (4)) \int \frac {x}{\left (4 e x^3+e x^4-25 \log (4)\right ) \log ^2(x)} \, dx\right )-(10 e \log (4)) \int \frac {x \left (2 e x^3+e x^4+25 \log (4)\right )}{\left (4 e x^3+e x^4-25 \log (4)\right )^2 \log (x)} \, dx \\ \end{align*}
Time = 0.35 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {e^2 \left (-20 x^4-5 x^5\right ) \log (4)+125 e x \log ^2(4)+\left (e^2 \left (-20 x^4-10 x^5\right ) \log (4)-250 e x \log ^2(4)\right ) \log (x)}{\left (e^2 \left (16 x^6+8 x^7+x^8\right )+e \left (-200 x^3-50 x^4\right ) \log (4)+625 \log ^2(4)\right ) \log ^2(x)} \, dx=\frac {5 e x^2 \log (4)}{\left (e x^3 (4+x)-25 \log (4)\right ) \log (x)} \]
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Time = 1.85 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.17
method | result | size |
default | \(-\frac {10 \,{\mathrm e} \ln \left (2\right ) x^{2}}{\left (-x^{4} {\mathrm e}-4 x^{3} {\mathrm e}+50 \ln \left (2\right )\right ) \ln \left (x \right )}\) | \(35\) |
risch | \(-\frac {10 \,{\mathrm e} \ln \left (2\right ) x^{2}}{\left (-x^{4} {\mathrm e}-4 x^{3} {\mathrm e}+50 \ln \left (2\right )\right ) \ln \left (x \right )}\) | \(35\) |
parallelrisch | \(-\frac {10 \,{\mathrm e} \ln \left (2\right ) x^{2}}{\left (-x^{4} {\mathrm e}-4 x^{3} {\mathrm e}+50 \ln \left (2\right )\right ) \ln \left (x \right )}\) | \(35\) |
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Time = 0.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {e^2 \left (-20 x^4-5 x^5\right ) \log (4)+125 e x \log ^2(4)+\left (e^2 \left (-20 x^4-10 x^5\right ) \log (4)-250 e x \log ^2(4)\right ) \log (x)}{\left (e^2 \left (16 x^6+8 x^7+x^8\right )+e \left (-200 x^3-50 x^4\right ) \log (4)+625 \log ^2(4)\right ) \log ^2(x)} \, dx=\frac {10 \, x^{2} e \log \left (2\right )}{{\left ({\left (x^{4} + 4 \, x^{3}\right )} e - 50 \, \log \left (2\right )\right )} \log \left (x\right )} \]
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Time = 0.08 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {e^2 \left (-20 x^4-5 x^5\right ) \log (4)+125 e x \log ^2(4)+\left (e^2 \left (-20 x^4-10 x^5\right ) \log (4)-250 e x \log ^2(4)\right ) \log (x)}{\left (e^2 \left (16 x^6+8 x^7+x^8\right )+e \left (-200 x^3-50 x^4\right ) \log (4)+625 \log ^2(4)\right ) \log ^2(x)} \, dx=\frac {10 e x^{2} \log {\left (2 \right )}}{\left (e x^{4} + 4 e x^{3} - 50 \log {\left (2 \right )}\right ) \log {\left (x \right )}} \]
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Time = 0.30 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \frac {e^2 \left (-20 x^4-5 x^5\right ) \log (4)+125 e x \log ^2(4)+\left (e^2 \left (-20 x^4-10 x^5\right ) \log (4)-250 e x \log ^2(4)\right ) \log (x)}{\left (e^2 \left (16 x^6+8 x^7+x^8\right )+e \left (-200 x^3-50 x^4\right ) \log (4)+625 \log ^2(4)\right ) \log ^2(x)} \, dx=\frac {10 \, x^{2} e \log \left (2\right )}{{\left (x^{4} e + 4 \, x^{3} e - 50 \, \log \left (2\right )\right )} \log \left (x\right )} \]
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Timed out. \[ \int \frac {e^2 \left (-20 x^4-5 x^5\right ) \log (4)+125 e x \log ^2(4)+\left (e^2 \left (-20 x^4-10 x^5\right ) \log (4)-250 e x \log ^2(4)\right ) \log (x)}{\left (e^2 \left (16 x^6+8 x^7+x^8\right )+e \left (-200 x^3-50 x^4\right ) \log (4)+625 \log ^2(4)\right ) \log ^2(x)} \, dx=\text {Timed out} \]
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Time = 14.95 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \frac {e^2 \left (-20 x^4-5 x^5\right ) \log (4)+125 e x \log ^2(4)+\left (e^2 \left (-20 x^4-10 x^5\right ) \log (4)-250 e x \log ^2(4)\right ) \log (x)}{\left (e^2 \left (16 x^6+8 x^7+x^8\right )+e \left (-200 x^3-50 x^4\right ) \log (4)+625 \log ^2(4)\right ) \log ^2(x)} \, dx=\frac {10\,x^2\,\mathrm {e}\,\ln \left (2\right )}{\ln \left (x\right )\,\left (\mathrm {e}\,x^4+4\,\mathrm {e}\,x^3-50\,\ln \left (2\right )\right )} \]
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