Integrand size = 95, antiderivative size = 23 \[ \int \frac {e^{2 x} \left (18+4 x-4 x^2+(3-2 x) \log (3)+2 \log (5)\right )}{81-36 x^2+4 x^4+\left (18-18 x-4 x^2+4 x^3\right ) \log (3)+\left (1-2 x+x^2\right ) \log ^2(3)+\left (18-4 x^2+(2-2 x) \log (3)\right ) \log (5)+\log ^2(5)} \, dx=\frac {e^{2 x}}{9+\log (3)-x (2 x+\log (3))+\log (5)} \]
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Time = 0.51 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.021, Rules used = {6820, 2327} \[ \int \frac {e^{2 x} \left (18+4 x-4 x^2+(3-2 x) \log (3)+2 \log (5)\right )}{81-36 x^2+4 x^4+\left (18-18 x-4 x^2+4 x^3\right ) \log (3)+\left (1-2 x+x^2\right ) \log ^2(3)+\left (18-4 x^2+(2-2 x) \log (3)\right ) \log (5)+\log ^2(5)} \, dx=\frac {e^{2 x}}{-2 x^2-x \log (3)+9+\log (15)} \]
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Rule 2327
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{2 x} \left (18-4 x^2+x (4-\log (9))+\log (675)\right )}{\left (9-2 x^2-x \log (3)+\log (15)\right )^2} \, dx \\ & = \frac {e^{2 x}}{9-2 x^2-x \log (3)+\log (15)} \\ \end{align*}
Time = 0.82 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {e^{2 x} \left (18+4 x-4 x^2+(3-2 x) \log (3)+2 \log (5)\right )}{81-36 x^2+4 x^4+\left (18-18 x-4 x^2+4 x^3\right ) \log (3)+\left (1-2 x+x^2\right ) \log ^2(3)+\left (18-4 x^2+(2-2 x) \log (3)\right ) \log (5)+\log ^2(5)} \, dx=\frac {e^{2 x}}{9-2 x^2-x \log (3)+\log (15)} \]
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Time = 0.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04
method | result | size |
gosper | \(\frac {{\mathrm e}^{2 x}}{-x \ln \left (3\right )-2 x^{2}+\ln \left (5\right )+\ln \left (3\right )+9}\) | \(24\) |
norman | \(\frac {{\mathrm e}^{2 x}}{-x \ln \left (3\right )-2 x^{2}+\ln \left (5\right )+\ln \left (3\right )+9}\) | \(24\) |
parallelrisch | \(\frac {{\mathrm e}^{2 x}}{-x \ln \left (3\right )-2 x^{2}+\ln \left (5\right )+\ln \left (3\right )+9}\) | \(24\) |
default | \(-\frac {36 \,{\mathrm e}^{2 x} \ln \left (3\right )}{\left (\ln \left (3\right )^{2}+8 \ln \left (5\right )+8 \ln \left (3\right )+72\right ) \left (-x \ln \left (3\right )-2 x^{2}+\ln \left (5\right )+\ln \left (3\right )+9\right )}-\frac {{\mathrm e}^{2 x} \ln \left (3\right )^{2}}{\left (\ln \left (3\right )^{2}+8 \ln \left (5\right )+8 \ln \left (3\right )+72\right ) \left (-x \ln \left (3\right )-2 x^{2}+\ln \left (5\right )+\ln \left (3\right )+9\right )}+\frac {2 \,{\mathrm e}^{2 x} \left (-x \ln \left (3\right )^{2}+\ln \left (3\right ) \ln \left (5\right )-4 x \ln \left (5\right )+\ln \left (3\right )^{2}-4 x \ln \left (3\right )+9 \ln \left (3\right )-36 x \right )}{\left (\ln \left (3\right )^{2}+8 \ln \left (5\right )+8 \ln \left (3\right )+72\right ) \left (-x \ln \left (3\right )-2 x^{2}+\ln \left (5\right )+\ln \left (3\right )+9\right )}+\frac {4 \,{\mathrm e}^{2 x} \left (-x \ln \left (3\right )+2 \ln \left (5\right )+2 \ln \left (3\right )+18\right )}{\left (\ln \left (3\right )^{2}+8 \ln \left (5\right )+8 \ln \left (3\right )+72\right ) \left (-x \ln \left (3\right )-2 x^{2}+\ln \left (5\right )+\ln \left (3\right )+9\right )}+\frac {18 \,{\mathrm e}^{2 x} \left (\ln \left (3\right )+4 x \right )}{\left (\ln \left (3\right )^{2}+8 \ln \left (5\right )+8 \ln \left (3\right )+72\right ) \left (-x \ln \left (3\right )-2 x^{2}+\ln \left (5\right )+\ln \left (3\right )+9\right )}+\frac {2 \ln \left (3\right )^{2} {\mathrm e}^{2 x} x}{\left (\ln \left (3\right )^{2}+8 \ln \left (5\right )+8 \ln \left (3\right )+72\right ) \left (-x \ln \left (3\right )-2 x^{2}+\ln \left (5\right )+\ln \left (3\right )+9\right )}-\frac {2 \ln \left (5\right ) {\mathrm e}^{2 x} \ln \left (3\right )}{\left (\ln \left (3\right )^{2}+8 \ln \left (5\right )+8 \ln \left (3\right )+72\right ) \left (-x \ln \left (3\right )-2 x^{2}+\ln \left (5\right )+\ln \left (3\right )+9\right )}+\frac {8 \ln \left (5\right ) {\mathrm e}^{2 x} x}{\left (\ln \left (3\right )^{2}+8 \ln \left (5\right )+8 \ln \left (3\right )+72\right ) \left (-x \ln \left (3\right )-2 x^{2}+\ln \left (5\right )+\ln \left (3\right )+9\right )}+\frac {12 \ln \left (3\right ) {\mathrm e}^{2 x} x}{\left (\ln \left (3\right )^{2}+8 \ln \left (5\right )+8 \ln \left (3\right )+72\right ) \left (-x \ln \left (3\right )-2 x^{2}+\ln \left (5\right )+\ln \left (3\right )+9\right )}\) | \(438\) |
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Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {e^{2 x} \left (18+4 x-4 x^2+(3-2 x) \log (3)+2 \log (5)\right )}{81-36 x^2+4 x^4+\left (18-18 x-4 x^2+4 x^3\right ) \log (3)+\left (1-2 x+x^2\right ) \log ^2(3)+\left (18-4 x^2+(2-2 x) \log (3)\right ) \log (5)+\log ^2(5)} \, dx=-\frac {e^{\left (2 \, x\right )}}{2 \, x^{2} + {\left (x - 1\right )} \log \left (3\right ) - \log \left (5\right ) - 9} \]
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Time = 0.15 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {e^{2 x} \left (18+4 x-4 x^2+(3-2 x) \log (3)+2 \log (5)\right )}{81-36 x^2+4 x^4+\left (18-18 x-4 x^2+4 x^3\right ) \log (3)+\left (1-2 x+x^2\right ) \log ^2(3)+\left (18-4 x^2+(2-2 x) \log (3)\right ) \log (5)+\log ^2(5)} \, dx=- \frac {e^{2 x}}{2 x^{2} + x \log {\left (3 \right )} - 9 - \log {\left (5 \right )} - \log {\left (3 \right )}} \]
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Time = 0.32 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {e^{2 x} \left (18+4 x-4 x^2+(3-2 x) \log (3)+2 \log (5)\right )}{81-36 x^2+4 x^4+\left (18-18 x-4 x^2+4 x^3\right ) \log (3)+\left (1-2 x+x^2\right ) \log ^2(3)+\left (18-4 x^2+(2-2 x) \log (3)\right ) \log (5)+\log ^2(5)} \, dx=-\frac {e^{\left (2 \, x\right )}}{2 \, x^{2} + x \log \left (3\right ) - \log \left (5\right ) - \log \left (3\right ) - 9} \]
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Time = 0.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {e^{2 x} \left (18+4 x-4 x^2+(3-2 x) \log (3)+2 \log (5)\right )}{81-36 x^2+4 x^4+\left (18-18 x-4 x^2+4 x^3\right ) \log (3)+\left (1-2 x+x^2\right ) \log ^2(3)+\left (18-4 x^2+(2-2 x) \log (3)\right ) \log (5)+\log ^2(5)} \, dx=-\frac {e^{\left (2 \, x\right )}}{2 \, x^{2} + x \log \left (3\right ) - \log \left (5\right ) - \log \left (3\right ) - 9} \]
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Timed out. \[ \int \frac {e^{2 x} \left (18+4 x-4 x^2+(3-2 x) \log (3)+2 \log (5)\right )}{81-36 x^2+4 x^4+\left (18-18 x-4 x^2+4 x^3\right ) \log (3)+\left (1-2 x+x^2\right ) \log ^2(3)+\left (18-4 x^2+(2-2 x) \log (3)\right ) \log (5)+\log ^2(5)} \, dx=\int \frac {{\mathrm {e}}^{2\,x}\,\left (4\,x+2\,\ln \left (5\right )-\ln \left (3\right )\,\left (2\,x-3\right )-4\,x^2+18\right )}{{\ln \left (3\right )}^2\,\left (x^2-2\,x+1\right )-\ln \left (5\right )\,\left (\ln \left (3\right )\,\left (2\,x-2\right )+4\,x^2-18\right )-\ln \left (3\right )\,\left (-4\,x^3+4\,x^2+18\,x-18\right )+{\ln \left (5\right )}^2-36\,x^2+4\,x^4+81} \,d x \]
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