Integrand size = 136, antiderivative size = 20 \[ \int \frac {-4 x^2+e^{\frac {1}{x}} x^2+\left (8 x+4 x^2+e^{\frac {1}{x}} \left (-2-3 x-x^2\right )\right ) \log (2+x)}{2 x^3+x^4+\left (16 x^2+8 x^3+e^{\frac {1}{x}} \left (-4 x^2-2 x^3\right )\right ) \log (2+x)+\left (32 x+16 x^2+e^{\frac {1}{x}} \left (-16 x-8 x^2\right )+e^{2/x} \left (2 x+x^2\right )\right ) \log ^2(2+x)} \, dx=\frac {x}{x+\left (4-e^{\frac {1}{x}}\right ) \log (2+x)} \]
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Time = 0.67 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.022, Rules used = {6820, 6843, 32} \[ \int \frac {-4 x^2+e^{\frac {1}{x}} x^2+\left (8 x+4 x^2+e^{\frac {1}{x}} \left (-2-3 x-x^2\right )\right ) \log (2+x)}{2 x^3+x^4+\left (16 x^2+8 x^3+e^{\frac {1}{x}} \left (-4 x^2-2 x^3\right )\right ) \log (2+x)+\left (32 x+16 x^2+e^{\frac {1}{x}} \left (-16 x-8 x^2\right )+e^{2/x} \left (2 x+x^2\right )\right ) \log ^2(2+x)} \, dx=-\frac {1}{1-\frac {x}{\left (e^{\frac {1}{x}}-4\right ) \log (x+2)}} \]
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Rule 32
Rule 6820
Rule 6843
Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (-4+e^{\frac {1}{x}}\right ) x^2-(2+x) \left (-4 x+e^{\frac {1}{x}} (1+x)\right ) \log (2+x)}{x (2+x) \left (x-\left (-4+e^{\frac {1}{x}}\right ) \log (2+x)\right )^2} \, dx \\ & = \text {Subst}\left (\int \frac {1}{(1+x)^2} \, dx,x,-\frac {x}{\left (-4+e^{\frac {1}{x}}\right ) \log (2+x)}\right ) \\ & = -\frac {1}{1-\frac {x}{\left (-4+e^{\frac {1}{x}}\right ) \log (2+x)}} \\ \end{align*}
Time = 0.72 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {-4 x^2+e^{\frac {1}{x}} x^2+\left (8 x+4 x^2+e^{\frac {1}{x}} \left (-2-3 x-x^2\right )\right ) \log (2+x)}{2 x^3+x^4+\left (16 x^2+8 x^3+e^{\frac {1}{x}} \left (-4 x^2-2 x^3\right )\right ) \log (2+x)+\left (32 x+16 x^2+e^{\frac {1}{x}} \left (-16 x-8 x^2\right )+e^{2/x} \left (2 x+x^2\right )\right ) \log ^2(2+x)} \, dx=\frac {x}{x-\left (-4+e^{\frac {1}{x}}\right ) \log (2+x)} \]
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Time = 0.99 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15
method | result | size |
risch | \(\frac {x}{-\ln \left (2+x \right ) {\mathrm e}^{\frac {1}{x}}+x +4 \ln \left (2+x \right )}\) | \(23\) |
parallelrisch | \(\frac {x}{-\ln \left (2+x \right ) {\mathrm e}^{\frac {1}{x}}+x +4 \ln \left (2+x \right )}\) | \(23\) |
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Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {-4 x^2+e^{\frac {1}{x}} x^2+\left (8 x+4 x^2+e^{\frac {1}{x}} \left (-2-3 x-x^2\right )\right ) \log (2+x)}{2 x^3+x^4+\left (16 x^2+8 x^3+e^{\frac {1}{x}} \left (-4 x^2-2 x^3\right )\right ) \log (2+x)+\left (32 x+16 x^2+e^{\frac {1}{x}} \left (-16 x-8 x^2\right )+e^{2/x} \left (2 x+x^2\right )\right ) \log ^2(2+x)} \, dx=-\frac {x}{{\left (e^{\frac {1}{x}} - 4\right )} \log \left (x + 2\right ) - x} \]
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Time = 0.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {-4 x^2+e^{\frac {1}{x}} x^2+\left (8 x+4 x^2+e^{\frac {1}{x}} \left (-2-3 x-x^2\right )\right ) \log (2+x)}{2 x^3+x^4+\left (16 x^2+8 x^3+e^{\frac {1}{x}} \left (-4 x^2-2 x^3\right )\right ) \log (2+x)+\left (32 x+16 x^2+e^{\frac {1}{x}} \left (-16 x-8 x^2\right )+e^{2/x} \left (2 x+x^2\right )\right ) \log ^2(2+x)} \, dx=- \frac {x}{- x + e^{\frac {1}{x}} \log {\left (x + 2 \right )} - 4 \log {\left (x + 2 \right )}} \]
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Time = 0.23 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20 \[ \int \frac {-4 x^2+e^{\frac {1}{x}} x^2+\left (8 x+4 x^2+e^{\frac {1}{x}} \left (-2-3 x-x^2\right )\right ) \log (2+x)}{2 x^3+x^4+\left (16 x^2+8 x^3+e^{\frac {1}{x}} \left (-4 x^2-2 x^3\right )\right ) \log (2+x)+\left (32 x+16 x^2+e^{\frac {1}{x}} \left (-16 x-8 x^2\right )+e^{2/x} \left (2 x+x^2\right )\right ) \log ^2(2+x)} \, dx=-\frac {x}{e^{\frac {1}{x}} \log \left (x + 2\right ) - x - 4 \, \log \left (x + 2\right )} \]
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Time = 0.31 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20 \[ \int \frac {-4 x^2+e^{\frac {1}{x}} x^2+\left (8 x+4 x^2+e^{\frac {1}{x}} \left (-2-3 x-x^2\right )\right ) \log (2+x)}{2 x^3+x^4+\left (16 x^2+8 x^3+e^{\frac {1}{x}} \left (-4 x^2-2 x^3\right )\right ) \log (2+x)+\left (32 x+16 x^2+e^{\frac {1}{x}} \left (-16 x-8 x^2\right )+e^{2/x} \left (2 x+x^2\right )\right ) \log ^2(2+x)} \, dx=-\frac {x}{e^{\frac {1}{x}} \log \left (x + 2\right ) - x - 4 \, \log \left (x + 2\right )} \]
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Time = 13.40 (sec) , antiderivative size = 178, normalized size of antiderivative = 8.90 \[ \int \frac {-4 x^2+e^{\frac {1}{x}} x^2+\left (8 x+4 x^2+e^{\frac {1}{x}} \left (-2-3 x-x^2\right )\right ) \log (2+x)}{2 x^3+x^4+\left (16 x^2+8 x^3+e^{\frac {1}{x}} \left (-4 x^2-2 x^3\right )\right ) \log (2+x)+\left (32 x+16 x^2+e^{\frac {1}{x}} \left (-16 x-8 x^2\right )+e^{2/x} \left (2 x+x^2\right )\right ) \log ^2(2+x)} \, dx=\frac {{\ln \left (x+2\right )}^2\,\left (4\,x+8\right )\,{\left (x^3+2\,x^2\right )}^2-x^3\,{\left (x^3+2\,x^2\right )}^2+\ln \left (x+2\right )\,{\left (x^3+2\,x^2\right )}^2\,\left (x^3+3\,x^2+2\,x\right )}{x\,\left (x+2\right )\,\left (x+4\,\ln \left (x+2\right )-\ln \left (x+2\right )\,{\mathrm {e}}^{1/x}\right )\,\left (x^6\,\ln \left (x+2\right )-x^6+5\,x^5\,\ln \left (x+2\right )-2\,x^5+4\,x^4\,{\ln \left (x+2\right )}^2+8\,x^4\,\ln \left (x+2\right )+16\,x^3\,{\ln \left (x+2\right )}^2+4\,x^3\,\ln \left (x+2\right )+16\,x^2\,{\ln \left (x+2\right )}^2\right )} \]
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