Integrand size = 130, antiderivative size = 24 \[ \int \frac {10 e^4+e^8 \left (-15 x+18 x^4\right )+12 e^8 x^3 \log (4)+2 e^8 x^2 \log ^2(4)+\left (-24 e^4 x^3-8 e^4 x^2 \log (4)\right ) \log (x)+8 x^2 \log ^2(x)}{9 e^8 x^3+6 e^8 x^2 \log (4)+e^8 x \log ^2(4)+\left (-12 e^4 x^2-4 e^4 x \log (4)\right ) \log (x)+4 x \log ^2(x)} \, dx=30+x^2+\frac {5}{x+\log (4)+2 \left (x-\frac {\log (x)}{e^4}\right )} \]
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Time = 0.44 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {6820, 6874, 6818} \[ \int \frac {10 e^4+e^8 \left (-15 x+18 x^4\right )+12 e^8 x^3 \log (4)+2 e^8 x^2 \log ^2(4)+\left (-24 e^4 x^3-8 e^4 x^2 \log (4)\right ) \log (x)+8 x^2 \log ^2(x)}{9 e^8 x^3+6 e^8 x^2 \log (4)+e^8 x \log ^2(4)+\left (-12 e^4 x^2-4 e^4 x \log (4)\right ) \log (x)+4 x \log ^2(x)} \, dx=x^2+\frac {5 e^4}{3 e^4 x-2 \log (x)+e^4 \log (4)} \]
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Rule 6818
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {10 e^4+e^8 x \left (-15+18 x^3+12 x^2 \log (4)+2 x \log ^2(4)\right )-8 e^4 x^2 (3 x+\log (4)) \log (x)+8 x^2 \log ^2(x)}{x \left (e^4 (3 x+\log (4))-2 \log (x)\right )^2} \, dx \\ & = \int \left (2 x-\frac {5 e^4 \left (-2+3 e^4 x\right )}{x \left (3 e^4 x+e^4 \log (4)-2 \log (x)\right )^2}\right ) \, dx \\ & = x^2-\left (5 e^4\right ) \int \frac {-2+3 e^4 x}{x \left (3 e^4 x+e^4 \log (4)-2 \log (x)\right )^2} \, dx \\ & = x^2+\frac {5 e^4}{3 e^4 x+e^4 \log (4)-2 \log (x)} \\ \end{align*}
Time = 1.14 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {10 e^4+e^8 \left (-15 x+18 x^4\right )+12 e^8 x^3 \log (4)+2 e^8 x^2 \log ^2(4)+\left (-24 e^4 x^3-8 e^4 x^2 \log (4)\right ) \log (x)+8 x^2 \log ^2(x)}{9 e^8 x^3+6 e^8 x^2 \log (4)+e^8 x \log ^2(4)+\left (-12 e^4 x^2-4 e^4 x \log (4)\right ) \log (x)+4 x \log ^2(x)} \, dx=x^2+\frac {5 e^4}{e^4 (3 x+\log (4))-2 \log (x)} \]
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Time = 1.04 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12
method | result | size |
risch | \(x^{2}+\frac {5 \,{\mathrm e}^{4}}{2 \,{\mathrm e}^{4} \ln \left (2\right )+3 x \,{\mathrm e}^{4}-2 \ln \left (x \right )}\) | \(27\) |
default | \(\frac {3 x^{3} {\mathrm e}^{4}-2 x^{2} \ln \left (x \right )+2 x^{2} {\mathrm e}^{4} \ln \left (2\right )+5 \,{\mathrm e}^{4}}{2 \,{\mathrm e}^{4} \ln \left (2\right )+3 x \,{\mathrm e}^{4}-2 \ln \left (x \right )}\) | \(48\) |
norman | \(\frac {3 x^{3} {\mathrm e}^{4}-2 x^{2} \ln \left (x \right )+2 x^{2} {\mathrm e}^{4} \ln \left (2\right )+5 \,{\mathrm e}^{4}}{2 \,{\mathrm e}^{4} \ln \left (2\right )+3 x \,{\mathrm e}^{4}-2 \ln \left (x \right )}\) | \(48\) |
parallelrisch | \(-\frac {-4 x^{2} {\mathrm e}^{4} \ln \left (2\right )-6 x^{3} {\mathrm e}^{4}+4 x^{2} \ln \left (x \right )-10 \,{\mathrm e}^{4}}{2 \left (2 \,{\mathrm e}^{4} \ln \left (2\right )+3 x \,{\mathrm e}^{4}-2 \ln \left (x \right )\right )}\) | \(49\) |
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none
Time = 0.26 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.92 \[ \int \frac {10 e^4+e^8 \left (-15 x+18 x^4\right )+12 e^8 x^3 \log (4)+2 e^8 x^2 \log ^2(4)+\left (-24 e^4 x^3-8 e^4 x^2 \log (4)\right ) \log (x)+8 x^2 \log ^2(x)}{9 e^8 x^3+6 e^8 x^2 \log (4)+e^8 x \log ^2(4)+\left (-12 e^4 x^2-4 e^4 x \log (4)\right ) \log (x)+4 x \log ^2(x)} \, dx=\frac {2 \, x^{2} e^{4} \log \left (2\right ) - 2 \, x^{2} \log \left (x\right ) + {\left (3 \, x^{3} + 5\right )} e^{4}}{3 \, x e^{4} + 2 \, e^{4} \log \left (2\right ) - 2 \, \log \left (x\right )} \]
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Time = 0.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {10 e^4+e^8 \left (-15 x+18 x^4\right )+12 e^8 x^3 \log (4)+2 e^8 x^2 \log ^2(4)+\left (-24 e^4 x^3-8 e^4 x^2 \log (4)\right ) \log (x)+8 x^2 \log ^2(x)}{9 e^8 x^3+6 e^8 x^2 \log (4)+e^8 x \log ^2(4)+\left (-12 e^4 x^2-4 e^4 x \log (4)\right ) \log (x)+4 x \log ^2(x)} \, dx=x^{2} - \frac {5 e^{4}}{- 3 x e^{4} + 2 \log {\left (x \right )} - 2 e^{4} \log {\left (2 \right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (23) = 46\).
Time = 0.31 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.96 \[ \int \frac {10 e^4+e^8 \left (-15 x+18 x^4\right )+12 e^8 x^3 \log (4)+2 e^8 x^2 \log ^2(4)+\left (-24 e^4 x^3-8 e^4 x^2 \log (4)\right ) \log (x)+8 x^2 \log ^2(x)}{9 e^8 x^3+6 e^8 x^2 \log (4)+e^8 x \log ^2(4)+\left (-12 e^4 x^2-4 e^4 x \log (4)\right ) \log (x)+4 x \log ^2(x)} \, dx=\frac {3 \, x^{3} e^{4} + 2 \, x^{2} e^{4} \log \left (2\right ) - 2 \, x^{2} \log \left (x\right ) + 5 \, e^{4}}{3 \, x e^{4} + 2 \, e^{4} \log \left (2\right ) - 2 \, \log \left (x\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (23) = 46\).
Time = 0.26 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.96 \[ \int \frac {10 e^4+e^8 \left (-15 x+18 x^4\right )+12 e^8 x^3 \log (4)+2 e^8 x^2 \log ^2(4)+\left (-24 e^4 x^3-8 e^4 x^2 \log (4)\right ) \log (x)+8 x^2 \log ^2(x)}{9 e^8 x^3+6 e^8 x^2 \log (4)+e^8 x \log ^2(4)+\left (-12 e^4 x^2-4 e^4 x \log (4)\right ) \log (x)+4 x \log ^2(x)} \, dx=\frac {3 \, x^{3} e^{4} + 2 \, x^{2} e^{4} \log \left (2\right ) - 2 \, x^{2} \log \left (x\right ) + 5 \, e^{4}}{3 \, x e^{4} + 2 \, e^{4} \log \left (2\right ) - 2 \, \log \left (x\right )} \]
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Timed out. \[ \int \frac {10 e^4+e^8 \left (-15 x+18 x^4\right )+12 e^8 x^3 \log (4)+2 e^8 x^2 \log ^2(4)+\left (-24 e^4 x^3-8 e^4 x^2 \log (4)\right ) \log (x)+8 x^2 \log ^2(x)}{9 e^8 x^3+6 e^8 x^2 \log (4)+e^8 x \log ^2(4)+\left (-12 e^4 x^2-4 e^4 x \log (4)\right ) \log (x)+4 x \log ^2(x)} \, dx=\int \frac {10\,{\mathrm {e}}^4-{\mathrm {e}}^8\,\left (15\,x-18\,x^4\right )-\ln \left (x\right )\,\left (24\,{\mathrm {e}}^4\,x^3+16\,{\mathrm {e}}^4\,\ln \left (2\right )\,x^2\right )+8\,x^2\,{\ln \left (x\right )}^2+24\,x^3\,{\mathrm {e}}^8\,\ln \left (2\right )+8\,x^2\,{\mathrm {e}}^8\,{\ln \left (2\right )}^2}{4\,x\,{\ln \left (x\right )}^2+9\,x^3\,{\mathrm {e}}^8-\ln \left (x\right )\,\left (12\,{\mathrm {e}}^4\,x^2+8\,{\mathrm {e}}^4\,\ln \left (2\right )\,x\right )+4\,x\,{\mathrm {e}}^8\,{\ln \left (2\right )}^2+12\,x^2\,{\mathrm {e}}^8\,\ln \left (2\right )} \,d x \]
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