Integrand size = 81, antiderivative size = 24 \[ \int \frac {-10-20 e^{2 x} x^3-10 e^{4 x} x^6+e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} \left (-12 x^2-8 x^3\right )}{5+10 e^{2 x} x^3+5 e^{4 x} x^6} \, dx=e^{\frac {4 x}{5 \left (x+e^{2 x} x^4\right )}}-2 x \]
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\[ \int \frac {-10-20 e^{2 x} x^3-10 e^{4 x} x^6+e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} \left (-12 x^2-8 x^3\right )}{5+10 e^{2 x} x^3+5 e^{4 x} x^6} \, dx=\int \frac {-10-20 e^{2 x} x^3-10 e^{4 x} x^6+e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} \left (-12 x^2-8 x^3\right )}{5+10 e^{2 x} x^3+5 e^{4 x} x^6} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {-10-20 e^{2 x} x^3-10 e^{4 x} x^6+e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} \left (-12 x^2-8 x^3\right )}{5 \left (1+e^{2 x} x^3\right )^2} \, dx \\ & = \frac {1}{5} \int \frac {-10-20 e^{2 x} x^3-10 e^{4 x} x^6+e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} \left (-12 x^2-8 x^3\right )}{\left (1+e^{2 x} x^3\right )^2} \, dx \\ & = \frac {1}{5} \int \left (-10-\frac {4 e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} x^2 (3+2 x)}{\left (1+e^{2 x} x^3\right )^2}\right ) \, dx \\ & = -2 x-\frac {4}{5} \int \frac {e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} x^2 (3+2 x)}{\left (1+e^{2 x} x^3\right )^2} \, dx \\ & = -2 x-\frac {4}{5} \int \left (\frac {3 e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} x^2}{\left (1+e^{2 x} x^3\right )^2}+\frac {2 e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} x^3}{\left (1+e^{2 x} x^3\right )^2}\right ) \, dx \\ & = -2 x-\frac {8}{5} \int \frac {e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} x^3}{\left (1+e^{2 x} x^3\right )^2} \, dx-\frac {12}{5} \int \frac {e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} x^2}{\left (1+e^{2 x} x^3\right )^2} \, dx \\ \end{align*}
Time = 0.85 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {-10-20 e^{2 x} x^3-10 e^{4 x} x^6+e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} \left (-12 x^2-8 x^3\right )}{5+10 e^{2 x} x^3+5 e^{4 x} x^6} \, dx=-\frac {2}{5} \left (-\frac {5}{2} e^{\frac {4}{5 \left (1+e^{2 x} x^3\right )}}+5 x\right ) \]
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Time = 7.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83
method | result | size |
risch | \({\mathrm e}^{\frac {4}{5 \left ({\mathrm e}^{2 x} x^{3}+1\right )}}-2 x\) | \(20\) |
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Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (20) = 40\).
Time = 0.27 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.88 \[ \int \frac {-10-20 e^{2 x} x^3-10 e^{4 x} x^6+e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} \left (-12 x^2-8 x^3\right )}{5+10 e^{2 x} x^3+5 e^{4 x} x^6} \, dx=-{\left (2 \, x e^{\left (2 \, x\right )} - e^{\left (\frac {2 \, {\left (5 \, x^{4} e^{\left (2 \, x\right )} + 5 \, x + 2\right )}}{5 \, {\left (x^{3} e^{\left (2 \, x\right )} + 1\right )}}\right )}\right )} e^{\left (-2 \, x\right )} \]
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Time = 0.18 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {-10-20 e^{2 x} x^3-10 e^{4 x} x^6+e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} \left (-12 x^2-8 x^3\right )}{5+10 e^{2 x} x^3+5 e^{4 x} x^6} \, dx=- 2 x + e^{\frac {4}{5 x^{3} e^{2 x} + 5}} \]
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none
Time = 0.22 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {-10-20 e^{2 x} x^3-10 e^{4 x} x^6+e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} \left (-12 x^2-8 x^3\right )}{5+10 e^{2 x} x^3+5 e^{4 x} x^6} \, dx=-2 \, x + e^{\left (\frac {4}{5 \, {\left (x^{3} e^{\left (2 \, x\right )} + 1\right )}}\right )} \]
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\[ \int \frac {-10-20 e^{2 x} x^3-10 e^{4 x} x^6+e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} \left (-12 x^2-8 x^3\right )}{5+10 e^{2 x} x^3+5 e^{4 x} x^6} \, dx=\int { -\frac {2 \, {\left (5 \, x^{6} e^{\left (4 \, x\right )} + 10 \, x^{3} e^{\left (2 \, x\right )} + 2 \, {\left (2 \, x^{3} + 3 \, x^{2}\right )} e^{\left (2 \, x + \frac {4}{5 \, {\left (x^{3} e^{\left (2 \, x\right )} + 1\right )}}\right )} + 5\right )}}{5 \, {\left (x^{6} e^{\left (4 \, x\right )} + 2 \, x^{3} e^{\left (2 \, x\right )} + 1\right )}} \,d x } \]
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Time = 0.23 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {-10-20 e^{2 x} x^3-10 e^{4 x} x^6+e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} \left (-12 x^2-8 x^3\right )}{5+10 e^{2 x} x^3+5 e^{4 x} x^6} \, dx={\mathrm {e}}^{\frac {4}{5\,x^3\,{\mathrm {e}}^{2\,x}+5}}-2\,x \]
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