Integrand size = 112, antiderivative size = 21 \[ \int \frac {28 x^2-144 x^3+e^{20/x} \left (-60-20 x-10 x^2\right )+\left (-2 x^2-e^{20/x} x^2-16 x^3\right ) \log \left (2+e^{20/x}+16 x\right )}{18 x^2+156 x^3+98 x^4+16 x^5+e^{20/x} \left (9 x^2+6 x^3+x^4\right )} \, dx=\frac {10+\log \left (2+e^{20/x}+16 x\right )}{3+x} \]
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Time = 1.86 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29, number of steps used = 16, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {6820, 6874, 907, 2631, 46} \[ \int \frac {28 x^2-144 x^3+e^{20/x} \left (-60-20 x-10 x^2\right )+\left (-2 x^2-e^{20/x} x^2-16 x^3\right ) \log \left (2+e^{20/x}+16 x\right )}{18 x^2+156 x^3+98 x^4+16 x^5+e^{20/x} \left (9 x^2+6 x^3+x^4\right )} \, dx=\frac {10}{x+3}+\frac {\log \left (16 x+e^{20/x}+2\right )}{x+3} \]
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Rule 46
Rule 907
Rule 2631
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {-4 x^2 (-7+36 x)-10 e^{20/x} \left (6+2 x+x^2\right )-x^2 \left (2+e^{20/x}+16 x\right ) \log \left (2+e^{20/x}+16 x\right )}{x^2 (3+x)^2 \left (2+e^{20/x}+16 x\right )} \, dx \\ & = \int \left (\frac {8 \left (5+40 x+2 x^2\right )}{x^2 (3+x) \left (2+e^{20/x}+16 x\right )}+\frac {-60-20 x-10 x^2-x^2 \log \left (2+e^{20/x}+16 x\right )}{x^2 (3+x)^2}\right ) \, dx \\ & = 8 \int \frac {5+40 x+2 x^2}{x^2 (3+x) \left (2+e^{20/x}+16 x\right )} \, dx+\int \frac {-60-20 x-10 x^2-x^2 \log \left (2+e^{20/x}+16 x\right )}{x^2 (3+x)^2} \, dx \\ & = 8 \int \left (\frac {5}{3 x^2 \left (2+e^{20/x}+16 x\right )}+\frac {115}{9 x \left (2+e^{20/x}+16 x\right )}-\frac {97}{9 (3+x) \left (2+e^{20/x}+16 x\right )}\right ) \, dx+\int \left (-\frac {10 \left (6+2 x+x^2\right )}{x^2 (3+x)^2}-\frac {\log \left (2+e^{20/x}+16 x\right )}{(3+x)^2}\right ) \, dx \\ & = -\left (10 \int \frac {6+2 x+x^2}{x^2 (3+x)^2} \, dx\right )+\frac {40}{3} \int \frac {1}{x^2 \left (2+e^{20/x}+16 x\right )} \, dx-\frac {776}{9} \int \frac {1}{(3+x) \left (2+e^{20/x}+16 x\right )} \, dx+\frac {920}{9} \int \frac {1}{x \left (2+e^{20/x}+16 x\right )} \, dx-\int \frac {\log \left (2+e^{20/x}+16 x\right )}{(3+x)^2} \, dx \\ & = \frac {\log \left (2+e^{20/x}+16 x\right )}{3+x}-10 \int \left (\frac {2}{3 x^2}-\frac {2}{9 x}+\frac {1}{(3+x)^2}+\frac {2}{9 (3+x)}\right ) \, dx+\frac {40}{3} \int \frac {1}{x^2 \left (2+e^{20/x}+16 x\right )} \, dx-\frac {776}{9} \int \frac {1}{(3+x) \left (2+e^{20/x}+16 x\right )} \, dx+\frac {920}{9} \int \frac {1}{x \left (2+e^{20/x}+16 x\right )} \, dx-\int \frac {16-\frac {20 e^{20/x}}{x^2}}{(3+x) \left (2+e^{20/x}+16 x\right )} \, dx \\ & = \frac {20}{3 x}+\frac {10}{3+x}+\frac {20 \log (x)}{9}-\frac {20}{9} \log (3+x)+\frac {\log \left (2+e^{20/x}+16 x\right )}{3+x}+\frac {40}{3} \int \frac {1}{x^2 \left (2+e^{20/x}+16 x\right )} \, dx-\frac {776}{9} \int \frac {1}{(3+x) \left (2+e^{20/x}+16 x\right )} \, dx+\frac {920}{9} \int \frac {1}{x \left (2+e^{20/x}+16 x\right )} \, dx-\int \left (-\frac {20}{x^2 (3+x)}+\frac {8 \left (5+40 x+2 x^2\right )}{x^2 (3+x) \left (2+e^{20/x}+16 x\right )}\right ) \, dx \\ & = \frac {20}{3 x}+\frac {10}{3+x}+\frac {20 \log (x)}{9}-\frac {20}{9} \log (3+x)+\frac {\log \left (2+e^{20/x}+16 x\right )}{3+x}-8 \int \frac {5+40 x+2 x^2}{x^2 (3+x) \left (2+e^{20/x}+16 x\right )} \, dx+\frac {40}{3} \int \frac {1}{x^2 \left (2+e^{20/x}+16 x\right )} \, dx+20 \int \frac {1}{x^2 (3+x)} \, dx-\frac {776}{9} \int \frac {1}{(3+x) \left (2+e^{20/x}+16 x\right )} \, dx+\frac {920}{9} \int \frac {1}{x \left (2+e^{20/x}+16 x\right )} \, dx \\ & = \frac {20}{3 x}+\frac {10}{3+x}+\frac {20 \log (x)}{9}-\frac {20}{9} \log (3+x)+\frac {\log \left (2+e^{20/x}+16 x\right )}{3+x}-8 \int \left (\frac {5}{3 x^2 \left (2+e^{20/x}+16 x\right )}+\frac {115}{9 x \left (2+e^{20/x}+16 x\right )}-\frac {97}{9 (3+x) \left (2+e^{20/x}+16 x\right )}\right ) \, dx+\frac {40}{3} \int \frac {1}{x^2 \left (2+e^{20/x}+16 x\right )} \, dx+20 \int \left (\frac {1}{3 x^2}-\frac {1}{9 x}+\frac {1}{9 (3+x)}\right ) \, dx-\frac {776}{9} \int \frac {1}{(3+x) \left (2+e^{20/x}+16 x\right )} \, dx+\frac {920}{9} \int \frac {1}{x \left (2+e^{20/x}+16 x\right )} \, dx \\ & = \frac {10}{3+x}+\frac {\log \left (2+e^{20/x}+16 x\right )}{3+x} \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {28 x^2-144 x^3+e^{20/x} \left (-60-20 x-10 x^2\right )+\left (-2 x^2-e^{20/x} x^2-16 x^3\right ) \log \left (2+e^{20/x}+16 x\right )}{18 x^2+156 x^3+98 x^4+16 x^5+e^{20/x} \left (9 x^2+6 x^3+x^4\right )} \, dx=\frac {10+\log \left (2+e^{20/x}+16 x\right )}{3+x} \]
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Time = 0.51 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24
method | result | size |
parallelrisch | \(\frac {-40 x +12 \ln \left ({\mathrm e}^{\frac {20}{x}}+16 x +2\right )}{12 x +36}\) | \(26\) |
risch | \(\frac {\ln \left ({\mathrm e}^{\frac {20}{x}}+16 x +2\right )}{3+x}+\frac {10}{3+x}\) | \(27\) |
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Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {28 x^2-144 x^3+e^{20/x} \left (-60-20 x-10 x^2\right )+\left (-2 x^2-e^{20/x} x^2-16 x^3\right ) \log \left (2+e^{20/x}+16 x\right )}{18 x^2+156 x^3+98 x^4+16 x^5+e^{20/x} \left (9 x^2+6 x^3+x^4\right )} \, dx=\frac {\log \left (16 \, x + e^{\frac {20}{x}} + 2\right ) + 10}{x + 3} \]
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Time = 0.16 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {28 x^2-144 x^3+e^{20/x} \left (-60-20 x-10 x^2\right )+\left (-2 x^2-e^{20/x} x^2-16 x^3\right ) \log \left (2+e^{20/x}+16 x\right )}{18 x^2+156 x^3+98 x^4+16 x^5+e^{20/x} \left (9 x^2+6 x^3+x^4\right )} \, dx=\frac {\log {\left (16 x + e^{\frac {20}{x}} + 2 \right )}}{x + 3} + \frac {10}{x + 3} \]
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Time = 0.22 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {28 x^2-144 x^3+e^{20/x} \left (-60-20 x-10 x^2\right )+\left (-2 x^2-e^{20/x} x^2-16 x^3\right ) \log \left (2+e^{20/x}+16 x\right )}{18 x^2+156 x^3+98 x^4+16 x^5+e^{20/x} \left (9 x^2+6 x^3+x^4\right )} \, dx=\frac {\log \left (16 \, x + e^{\frac {20}{x}} + 2\right ) + 10}{x + 3} \]
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Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {28 x^2-144 x^3+e^{20/x} \left (-60-20 x-10 x^2\right )+\left (-2 x^2-e^{20/x} x^2-16 x^3\right ) \log \left (2+e^{20/x}+16 x\right )}{18 x^2+156 x^3+98 x^4+16 x^5+e^{20/x} \left (9 x^2+6 x^3+x^4\right )} \, dx=\frac {\log \left (16 \, x + e^{\frac {20}{x}} + 2\right ) + 10}{x + 3} \]
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Time = 13.49 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {28 x^2-144 x^3+e^{20/x} \left (-60-20 x-10 x^2\right )+\left (-2 x^2-e^{20/x} x^2-16 x^3\right ) \log \left (2+e^{20/x}+16 x\right )}{18 x^2+156 x^3+98 x^4+16 x^5+e^{20/x} \left (9 x^2+6 x^3+x^4\right )} \, dx=\frac {\ln \left (16\,x+{\mathrm {e}}^{20/x}+2\right )+10}{x+3} \]
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