Integrand size = 166, antiderivative size = 25 \[ \int \frac {4 x^2+4 x^2 \log (5)+x^2 \log ^2(5)+e^{2 x} \left (729+486 \log (5)+81 \log ^2(5)\right )+e^x \left (108 x+90 x \log (5)+18 x \log ^2(5)\right )+e^x \left (e \left (-27 x+27 x^2\right )+e \left (-9 x+9 x^2\right ) \log (5)\right ) \log (x)}{\left (4 x^3+4 x^3 \log (5)+x^3 \log ^2(5)+e^{2 x} \left (729 x+486 x \log (5)+81 x \log ^2(5)\right )+e^x \left (108 x^2+90 x^2 \log (5)+18 x^2 \log ^2(5)\right )\right ) \log (x)} \, dx=-\frac {e}{2+\log (5)+\frac {9 e^x (3+\log (5))}{x}}+\log (\log (x)) \]
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Time = 0.55 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48, number of steps used = 10, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {6, 6820, 6843, 32, 2339, 29} \[ \int \frac {4 x^2+4 x^2 \log (5)+x^2 \log ^2(5)+e^{2 x} \left (729+486 \log (5)+81 \log ^2(5)\right )+e^x \left (108 x+90 x \log (5)+18 x \log ^2(5)\right )+e^x \left (e \left (-27 x+27 x^2\right )+e \left (-9 x+9 x^2\right ) \log (5)\right ) \log (x)}{\left (4 x^3+4 x^3 \log (5)+x^3 \log ^2(5)+e^{2 x} \left (729 x+486 x \log (5)+81 x \log ^2(5)\right )+e^x \left (108 x^2+90 x^2 \log (5)+18 x^2 \log ^2(5)\right )\right ) \log (x)} \, dx=\log (\log (x))+\frac {9 e (3+\log (5))}{(2+\log (5)) \left (e^{-x} x (2+\log (5))+9 (3+\log (5))\right )} \]
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Rule 6
Rule 29
Rule 32
Rule 2339
Rule 6820
Rule 6843
Rubi steps \begin{align*} \text {integral}& = \int \frac {4 x^2+4 x^2 \log (5)+x^2 \log ^2(5)+e^{2 x} \left (729+486 \log (5)+81 \log ^2(5)\right )+e^x \left (108 x+90 x \log (5)+18 x \log ^2(5)\right )+e^x \left (e \left (-27 x+27 x^2\right )+e \left (-9 x+9 x^2\right ) \log (5)\right ) \log (x)}{\left (x^3 \log ^2(5)+x^3 (4+4 \log (5))+e^{2 x} \left (729 x+486 x \log (5)+81 x \log ^2(5)\right )+e^x \left (108 x^2+90 x^2 \log (5)+18 x^2 \log ^2(5)\right )\right ) \log (x)} \, dx \\ & = \int \frac {4 x^2+4 x^2 \log (5)+x^2 \log ^2(5)+e^{2 x} \left (729+486 \log (5)+81 \log ^2(5)\right )+e^x \left (108 x+90 x \log (5)+18 x \log ^2(5)\right )+e^x \left (e \left (-27 x+27 x^2\right )+e \left (-9 x+9 x^2\right ) \log (5)\right ) \log (x)}{\left (x^3 \left (4+4 \log (5)+\log ^2(5)\right )+e^{2 x} \left (729 x+486 x \log (5)+81 x \log ^2(5)\right )+e^x \left (108 x^2+90 x^2 \log (5)+18 x^2 \log ^2(5)\right )\right ) \log (x)} \, dx \\ & = \int \frac {x^2 \log ^2(5)+x^2 (4+4 \log (5))+e^{2 x} \left (729+486 \log (5)+81 \log ^2(5)\right )+e^x \left (108 x+90 x \log (5)+18 x \log ^2(5)\right )+e^x \left (e \left (-27 x+27 x^2\right )+e \left (-9 x+9 x^2\right ) \log (5)\right ) \log (x)}{\left (x^3 \left (4+4 \log (5)+\log ^2(5)\right )+e^{2 x} \left (729 x+486 x \log (5)+81 x \log ^2(5)\right )+e^x \left (108 x^2+90 x^2 \log (5)+18 x^2 \log ^2(5)\right )\right ) \log (x)} \, dx \\ & = \int \frac {x^2 \left (4+4 \log (5)+\log ^2(5)\right )+e^{2 x} \left (729+486 \log (5)+81 \log ^2(5)\right )+e^x \left (108 x+90 x \log (5)+18 x \log ^2(5)\right )+e^x \left (e \left (-27 x+27 x^2\right )+e \left (-9 x+9 x^2\right ) \log (5)\right ) \log (x)}{\left (x^3 \left (4+4 \log (5)+\log ^2(5)\right )+e^{2 x} \left (729 x+486 x \log (5)+81 x \log ^2(5)\right )+e^x \left (108 x^2+90 x^2 \log (5)+18 x^2 \log ^2(5)\right )\right ) \log (x)} \, dx \\ & = \int \left (\frac {9 e^{1+x} (-1+x) (3+\log (5))}{\left (x (2+\log (5))+9 e^x (3+\log (5))\right )^2}+\frac {1}{x \log (x)}\right ) \, dx \\ & = (9 (3+\log (5))) \int \frac {e^{1+x} (-1+x)}{\left (x (2+\log (5))+9 e^x (3+\log (5))\right )^2} \, dx+\int \frac {1}{x \log (x)} \, dx \\ & = -\left ((9 e (3+\log (5))) \text {Subst}\left (\int \frac {1}{(x (2+\log (5))+9 (3+\log (5)))^2} \, dx,x,e^{-x} x\right )\right )+\text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right ) \\ & = \frac {9 e (3+\log (5))}{(2+\log (5)) \left (e^{-x} x (2+\log (5))+9 (3+\log (5))\right )}+\log (\log (x)) \\ \end{align*}
Time = 0.82 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {4 x^2+4 x^2 \log (5)+x^2 \log ^2(5)+e^{2 x} \left (729+486 \log (5)+81 \log ^2(5)\right )+e^x \left (108 x+90 x \log (5)+18 x \log ^2(5)\right )+e^x \left (e \left (-27 x+27 x^2\right )+e \left (-9 x+9 x^2\right ) \log (5)\right ) \log (x)}{\left (4 x^3+4 x^3 \log (5)+x^3 \log ^2(5)+e^{2 x} \left (729 x+486 x \log (5)+81 x \log ^2(5)\right )+e^x \left (108 x^2+90 x^2 \log (5)+18 x^2 \log ^2(5)\right )\right ) \log (x)} \, dx=-\frac {e x}{27 e^x+2 x+9 e^x \log (5)+x \log (5)}+\log (\log (x)) \]
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Time = 0.37 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20
method | result | size |
risch | \(-\frac {x \,{\mathrm e}}{9 \,{\mathrm e}^{x} \ln \left (5\right )+x \ln \left (5\right )+27 \,{\mathrm e}^{x}+2 x}+\ln \left (\ln \left (x \right )\right )\) | \(30\) |
parallelrisch | \(\frac {\ln \left (5\right )^{2} \ln \left (\ln \left (x \right )\right ) x +9 \ln \left (5\right )^{2} \ln \left (\ln \left (x \right )\right ) {\mathrm e}^{x}+9 \,{\mathrm e} \ln \left (5\right ) {\mathrm e}^{x}+4 \ln \left (5\right ) \ln \left (\ln \left (x \right )\right ) x +45 \ln \left (5\right ) {\mathrm e}^{x} \ln \left (\ln \left (x \right )\right )+27 \,{\mathrm e} \,{\mathrm e}^{x}+4 x \ln \left (\ln \left (x \right )\right )+54 \,{\mathrm e}^{x} \ln \left (\ln \left (x \right )\right )}{\left (2+\ln \left (5\right )\right ) \left (9 \,{\mathrm e}^{x} \ln \left (5\right )+x \ln \left (5\right )+27 \,{\mathrm e}^{x}+2 x \right )}\) | \(93\) |
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Time = 0.25 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.84 \[ \int \frac {4 x^2+4 x^2 \log (5)+x^2 \log ^2(5)+e^{2 x} \left (729+486 \log (5)+81 \log ^2(5)\right )+e^x \left (108 x+90 x \log (5)+18 x \log ^2(5)\right )+e^x \left (e \left (-27 x+27 x^2\right )+e \left (-9 x+9 x^2\right ) \log (5)\right ) \log (x)}{\left (4 x^3+4 x^3 \log (5)+x^3 \log ^2(5)+e^{2 x} \left (729 x+486 x \log (5)+81 x \log ^2(5)\right )+e^x \left (108 x^2+90 x^2 \log (5)+18 x^2 \log ^2(5)\right )\right ) \log (x)} \, dx=-\frac {x e - {\left (9 \, {\left (\log \left (5\right ) + 3\right )} e^{x} + x \log \left (5\right ) + 2 \, x\right )} \log \left (\log \left (x\right )\right )}{9 \, {\left (\log \left (5\right ) + 3\right )} e^{x} + x \log \left (5\right ) + 2 \, x} \]
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Time = 0.15 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {4 x^2+4 x^2 \log (5)+x^2 \log ^2(5)+e^{2 x} \left (729+486 \log (5)+81 \log ^2(5)\right )+e^x \left (108 x+90 x \log (5)+18 x \log ^2(5)\right )+e^x \left (e \left (-27 x+27 x^2\right )+e \left (-9 x+9 x^2\right ) \log (5)\right ) \log (x)}{\left (4 x^3+4 x^3 \log (5)+x^3 \log ^2(5)+e^{2 x} \left (729 x+486 x \log (5)+81 x \log ^2(5)\right )+e^x \left (108 x^2+90 x^2 \log (5)+18 x^2 \log ^2(5)\right )\right ) \log (x)} \, dx=- \frac {e x}{x \log {\left (5 \right )} + 2 x + \left (9 \log {\left (5 \right )} + 27\right ) e^{x}} + \log {\left (\log {\left (x \right )} \right )} \]
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Time = 0.32 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {4 x^2+4 x^2 \log (5)+x^2 \log ^2(5)+e^{2 x} \left (729+486 \log (5)+81 \log ^2(5)\right )+e^x \left (108 x+90 x \log (5)+18 x \log ^2(5)\right )+e^x \left (e \left (-27 x+27 x^2\right )+e \left (-9 x+9 x^2\right ) \log (5)\right ) \log (x)}{\left (4 x^3+4 x^3 \log (5)+x^3 \log ^2(5)+e^{2 x} \left (729 x+486 x \log (5)+81 x \log ^2(5)\right )+e^x \left (108 x^2+90 x^2 \log (5)+18 x^2 \log ^2(5)\right )\right ) \log (x)} \, dx=-\frac {x e}{x {\left (\log \left (5\right ) + 2\right )} + 9 \, {\left (\log \left (5\right ) + 3\right )} e^{x}} + \log \left (\log \left (x\right )\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (25) = 50\).
Time = 0.32 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.24 \[ \int \frac {4 x^2+4 x^2 \log (5)+x^2 \log ^2(5)+e^{2 x} \left (729+486 \log (5)+81 \log ^2(5)\right )+e^x \left (108 x+90 x \log (5)+18 x \log ^2(5)\right )+e^x \left (e \left (-27 x+27 x^2\right )+e \left (-9 x+9 x^2\right ) \log (5)\right ) \log (x)}{\left (4 x^3+4 x^3 \log (5)+x^3 \log ^2(5)+e^{2 x} \left (729 x+486 x \log (5)+81 x \log ^2(5)\right )+e^x \left (108 x^2+90 x^2 \log (5)+18 x^2 \log ^2(5)\right )\right ) \log (x)} \, dx=\frac {x \log \left (5\right ) \log \left (\log \left (x\right )\right ) + 9 \, e^{x} \log \left (5\right ) \log \left (\log \left (x\right )\right ) - x e + 2 \, x \log \left (\log \left (x\right )\right ) + 27 \, e^{x} \log \left (\log \left (x\right )\right )}{x \log \left (5\right ) + 9 \, e^{x} \log \left (5\right ) + 2 \, x + 27 \, e^{x}} \]
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Timed out. \[ \int \frac {4 x^2+4 x^2 \log (5)+x^2 \log ^2(5)+e^{2 x} \left (729+486 \log (5)+81 \log ^2(5)\right )+e^x \left (108 x+90 x \log (5)+18 x \log ^2(5)\right )+e^x \left (e \left (-27 x+27 x^2\right )+e \left (-9 x+9 x^2\right ) \log (5)\right ) \log (x)}{\left (4 x^3+4 x^3 \log (5)+x^3 \log ^2(5)+e^{2 x} \left (729 x+486 x \log (5)+81 x \log ^2(5)\right )+e^x \left (108 x^2+90 x^2 \log (5)+18 x^2 \log ^2(5)\right )\right ) \log (x)} \, dx=\int \frac {x^2\,{\ln \left (5\right )}^2+{\mathrm {e}}^x\,\left (108\,x+90\,x\,\ln \left (5\right )+18\,x\,{\ln \left (5\right )}^2\right )+4\,x^2\,\ln \left (5\right )+4\,x^2+{\mathrm {e}}^{2\,x}\,\left (486\,\ln \left (5\right )+81\,{\ln \left (5\right )}^2+729\right )-{\mathrm {e}}^x\,\ln \left (x\right )\,\left (\mathrm {e}\,\left (27\,x-27\,x^2\right )+\mathrm {e}\,\ln \left (5\right )\,\left (9\,x-9\,x^2\right )\right )}{\ln \left (x\right )\,\left (x^3\,{\ln \left (5\right )}^2+{\mathrm {e}}^x\,\left (18\,x^2\,{\ln \left (5\right )}^2+90\,x^2\,\ln \left (5\right )+108\,x^2\right )+4\,x^3\,\ln \left (5\right )+{\mathrm {e}}^{2\,x}\,\left (729\,x+486\,x\,\ln \left (5\right )+81\,x\,{\ln \left (5\right )}^2\right )+4\,x^3\right )} \,d x \]
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