Integrand size = 98, antiderivative size = 30 \[ \int \frac {e^{\log ^2\left (\frac {900 x^4+60 x^2 \log (x)+\log ^2(x)}{100 x^4}\right )} \left (-30 e x^3-e x \log (x)+(e (16-4 x)+e (-32+8 x) \log (x)) \log \left (\frac {900 x^4+60 x^2 \log (x)+\log ^2(x)}{100 x^4}\right )\right )}{30 x^3+x \log (x)} \, dx=e^{1+\log ^2\left (\frac {1}{4} \left (6+\frac {\log (x)}{5 x^2}\right )^2\right )} (4-x) \]
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Leaf count is larger than twice the leaf count of optimal. \(129\) vs. \(2(30)=60\).
Time = 0.08 (sec) , antiderivative size = 129, normalized size of antiderivative = 4.30, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.010, Rules used = {2326} \[ \int \frac {e^{\log ^2\left (\frac {900 x^4+60 x^2 \log (x)+\log ^2(x)}{100 x^4}\right )} \left (-30 e x^3-e x \log (x)+(e (16-4 x)+e (-32+8 x) \log (x)) \log \left (\frac {900 x^4+60 x^2 \log (x)+\log ^2(x)}{100 x^4}\right )\right )}{30 x^3+x \log (x)} \, dx=\frac {e^{\log ^2\left (\frac {900 x^4+60 x^2 \log (x)+\log ^2(x)}{100 x^4}\right )} (e (4-x)-2 e (4-x) \log (x)) \left (900 x^4+60 x^2 \log (x)+\log ^2(x)\right )}{x^4 \left (30 x^3+x \log (x)\right ) \left (\frac {1800 x^3+30 x+60 x \log (x)+\frac {\log (x)}{x}}{x^4}-\frac {2 \left (900 x^4+60 x^2 \log (x)+\log ^2(x)\right )}{x^5}\right )} \]
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Rule 2326
Rubi steps \begin{align*} \text {integral}& = \frac {e^{\log ^2\left (\frac {900 x^4+60 x^2 \log (x)+\log ^2(x)}{100 x^4}\right )} (e (4-x)-2 e (4-x) \log (x)) \left (900 x^4+60 x^2 \log (x)+\log ^2(x)\right )}{x^4 \left (30 x^3+x \log (x)\right ) \left (\frac {30 x+1800 x^3+\frac {\log (x)}{x}+60 x \log (x)}{x^4}-\frac {2 \left (900 x^4+60 x^2 \log (x)+\log ^2(x)\right )}{x^5}\right )} \\ \end{align*}
Time = 0.10 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {e^{\log ^2\left (\frac {900 x^4+60 x^2 \log (x)+\log ^2(x)}{100 x^4}\right )} \left (-30 e x^3-e x \log (x)+(e (16-4 x)+e (-32+8 x) \log (x)) \log \left (\frac {900 x^4+60 x^2 \log (x)+\log ^2(x)}{100 x^4}\right )\right )}{30 x^3+x \log (x)} \, dx=-e^{1+\log ^2\left (\frac {\left (30 x^2+\log (x)\right )^2}{100 x^4}\right )} (-4+x) \]
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Leaf count of result is larger than twice the leaf count of optimal. \(62\) vs. \(2(25)=50\).
Time = 6.43 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.10
method | result | size |
parallelrisch | \(-{\mathrm e} x \,{\mathrm e}^{\ln \left (\frac {\ln \left (x \right )^{2}+60 x^{2} \ln \left (x \right )+900 x^{4}}{100 x^{4}}\right )^{2}}+4 \,{\mathrm e} \,{\mathrm e}^{\ln \left (\frac {\ln \left (x \right )^{2}+60 x^{2} \ln \left (x \right )+900 x^{4}}{100 x^{4}}\right )^{2}}\) | \(63\) |
risch | \(\text {Expression too large to display}\) | \(7474\) |
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Time = 0.24 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \frac {e^{\log ^2\left (\frac {900 x^4+60 x^2 \log (x)+\log ^2(x)}{100 x^4}\right )} \left (-30 e x^3-e x \log (x)+(e (16-4 x)+e (-32+8 x) \log (x)) \log \left (\frac {900 x^4+60 x^2 \log (x)+\log ^2(x)}{100 x^4}\right )\right )}{30 x^3+x \log (x)} \, dx=-{\left (x - 4\right )} e^{\left (\log \left (\frac {900 \, x^{4} + 60 \, x^{2} \log \left (x\right ) + \log \left (x\right )^{2}}{100 \, x^{4}}\right )^{2} + 1\right )} \]
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Timed out. \[ \int \frac {e^{\log ^2\left (\frac {900 x^4+60 x^2 \log (x)+\log ^2(x)}{100 x^4}\right )} \left (-30 e x^3-e x \log (x)+(e (16-4 x)+e (-32+8 x) \log (x)) \log \left (\frac {900 x^4+60 x^2 \log (x)+\log ^2(x)}{100 x^4}\right )\right )}{30 x^3+x \log (x)} \, dx=\text {Timed out} \]
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Leaf count of result is larger than twice the leaf count of optimal. 123 vs. \(2 (23) = 46\).
Time = 0.43 (sec) , antiderivative size = 123, normalized size of antiderivative = 4.10 \[ \int \frac {e^{\log ^2\left (\frac {900 x^4+60 x^2 \log (x)+\log ^2(x)}{100 x^4}\right )} \left (-30 e x^3-e x \log (x)+(e (16-4 x)+e (-32+8 x) \log (x)) \log \left (\frac {900 x^4+60 x^2 \log (x)+\log ^2(x)}{100 x^4}\right )\right )}{30 x^3+x \log (x)} \, dx=-{\left (2^{8 \, \log \left (5\right )} x e^{\left (4 \, \log \left (5\right )^{2} + 4 \, \log \left (2\right )^{2} + 1\right )} - 2^{8 \, \log \left (5\right ) + 2} e^{\left (4 \, \log \left (5\right )^{2} + 4 \, \log \left (2\right )^{2} + 1\right )}\right )} e^{\left (-8 \, \log \left (5\right ) \log \left (30 \, x^{2} + \log \left (x\right )\right ) - 8 \, \log \left (2\right ) \log \left (30 \, x^{2} + \log \left (x\right )\right ) + 4 \, \log \left (30 \, x^{2} + \log \left (x\right )\right )^{2} + 16 \, \log \left (5\right ) \log \left (x\right ) + 16 \, \log \left (2\right ) \log \left (x\right ) - 16 \, \log \left (30 \, x^{2} + \log \left (x\right )\right ) \log \left (x\right ) + 16 \, \log \left (x\right )^{2}\right )} \]
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\[ \int \frac {e^{\log ^2\left (\frac {900 x^4+60 x^2 \log (x)+\log ^2(x)}{100 x^4}\right )} \left (-30 e x^3-e x \log (x)+(e (16-4 x)+e (-32+8 x) \log (x)) \log \left (\frac {900 x^4+60 x^2 \log (x)+\log ^2(x)}{100 x^4}\right )\right )}{30 x^3+x \log (x)} \, dx=\int { -\frac {{\left (30 \, x^{3} e + x e \log \left (x\right ) - 4 \, {\left (2 \, {\left (x - 4\right )} e \log \left (x\right ) - {\left (x - 4\right )} e\right )} \log \left (\frac {900 \, x^{4} + 60 \, x^{2} \log \left (x\right ) + \log \left (x\right )^{2}}{100 \, x^{4}}\right )\right )} e^{\left (\log \left (\frac {900 \, x^{4} + 60 \, x^{2} \log \left (x\right ) + \log \left (x\right )^{2}}{100 \, x^{4}}\right )^{2}\right )}}{30 \, x^{3} + x \log \left (x\right )} \,d x } \]
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Time = 12.67 (sec) , antiderivative size = 103, normalized size of antiderivative = 3.43 \[ \int \frac {e^{\log ^2\left (\frac {900 x^4+60 x^2 \log (x)+\log ^2(x)}{100 x^4}\right )} \left (-30 e x^3-e x \log (x)+(e (16-4 x)+e (-32+8 x) \log (x)) \log \left (\frac {900 x^4+60 x^2 \log (x)+\log ^2(x)}{100 x^4}\right )\right )}{30 x^3+x \log (x)} \, dx=\frac {{\mathrm {e}}^{{\ln \left (\frac {1}{x^4}\right )}^2+{\ln \left (900\,x^4+60\,x^2\,\ln \left (x\right )+{\ln \left (x\right )}^2\right )}^2+4\,{\ln \left (10\right )}^2}\,\left (4\,\mathrm {e}-x\,\mathrm {e}\right )\,{\left (\frac {1}{x^4}\right )}^{2\,\ln \left (900\,x^4+60\,x^2\,\ln \left (x\right )+{\ln \left (x\right )}^2\right )}}{{\left (\frac {1}{x^4}\right )}^{4\,\ln \left (10\right )}\,{\left (900\,x^4+60\,x^2\,\ln \left (x\right )+{\ln \left (x\right )}^2\right )}^{4\,\ln \left (10\right )}} \]
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