Integrand size = 68, antiderivative size = 28 \[ \int \frac {-2 x^2+e^x \left (4 x-2 x^2\right )+e^{1+x} \left (e^x-x^2\right )}{5 e^{5+2 x} x^2+20 e^{4+x} x^3+20 e^3 x^4} \, dx=\frac {-e^x+x}{5 e^3 x \left (e^{1+x}+2 x\right )} \]
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\[ \int \frac {-2 x^2+e^x \left (4 x-2 x^2\right )+e^{1+x} \left (e^x-x^2\right )}{5 e^{5+2 x} x^2+20 e^{4+x} x^3+20 e^3 x^4} \, dx=\int \frac {-2 x^2+e^x \left (4 x-2 x^2\right )+e^{1+x} \left (e^x-x^2\right )}{5 e^{5+2 x} x^2+20 e^{4+x} x^3+20 e^3 x^4} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {-2 x^2+e^x \left (4 x-2 x^2\right )+e^{1+x} \left (e^x-x^2\right )}{5 e^3 x^2 \left (e^{1+x}+2 x\right )^2} \, dx \\ & = \frac {\int \frac {-2 x^2+e^x \left (4 x-2 x^2\right )+e^{1+x} \left (e^x-x^2\right )}{x^2 \left (e^{1+x}+2 x\right )^2} \, dx}{5 e^3} \\ & = \frac {\int \left (\frac {1}{e x^2}+\frac {2 (2+e) (-1+x)}{e \left (e^{1+x}+2 x\right )^2}-\frac {2+e}{e \left (e^{1+x}+2 x\right )}\right ) \, dx}{5 e^3} \\ & = -\frac {1}{5 e^4 x}-\frac {(2+e) \int \frac {1}{e^{1+x}+2 x} \, dx}{5 e^4}+\frac {(2 (2+e)) \int \frac {-1+x}{\left (e^{1+x}+2 x\right )^2} \, dx}{5 e^4} \\ & = -\frac {1}{5 e^4 x}-\frac {(2+e) \int \frac {1}{e^{1+x}+2 x} \, dx}{5 e^4}+\frac {(2 (2+e)) \int \left (-\frac {1}{\left (e^{1+x}+2 x\right )^2}+\frac {x}{\left (e^{1+x}+2 x\right )^2}\right ) \, dx}{5 e^4} \\ & = -\frac {1}{5 e^4 x}-\frac {(2+e) \int \frac {1}{e^{1+x}+2 x} \, dx}{5 e^4}-\frac {(2 (2+e)) \int \frac {1}{\left (e^{1+x}+2 x\right )^2} \, dx}{5 e^4}+\frac {(2 (2+e)) \int \frac {x}{\left (e^{1+x}+2 x\right )^2} \, dx}{5 e^4} \\ \end{align*}
Time = 1.02 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {-2 x^2+e^x \left (4 x-2 x^2\right )+e^{1+x} \left (e^x-x^2\right )}{5 e^{5+2 x} x^2+20 e^{4+x} x^3+20 e^3 x^4} \, dx=\frac {-e^x+x}{5 e^3 x \left (e^{1+x}+2 x\right )} \]
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Time = 0.79 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93
method | result | size |
parallelrisch | \(\frac {{\mathrm e}^{-3} \left (x -{\mathrm e}^{x}\right )}{5 x \left (2 x +{\mathrm e}^{1+x}\right )}\) | \(26\) |
norman | \(\frac {\frac {{\mathrm e}^{-3} x}{5}-\frac {{\mathrm e}^{-3} {\mathrm e}^{x}}{5}}{x \left ({\mathrm e} \,{\mathrm e}^{x}+2 x \right )}\) | \(32\) |
risch | \(-\frac {{\mathrm e}^{-4}}{5 x}+\frac {{\mathrm e}^{-4} {\mathrm e}}{10 x +5 \,{\mathrm e}^{1+x}}+\frac {2 \,{\mathrm e}^{-4}}{5 \left (2 x +{\mathrm e}^{1+x}\right )}\) | \(39\) |
parts | \(\frac {{\mathrm e}^{-3}}{10 x +5 \,{\mathrm e}^{1+x}}-\frac {{\mathrm e}^{-3} {\mathrm e}^{x}}{5 x \left ({\mathrm e} \,{\mathrm e}^{x}+2 x \right )}\) | \(40\) |
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Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {-2 x^2+e^x \left (4 x-2 x^2\right )+e^{1+x} \left (e^x-x^2\right )}{5 e^{5+2 x} x^2+20 e^{4+x} x^3+20 e^3 x^4} \, dx=\frac {x e^{4} - e^{\left (x + 4\right )}}{5 \, {\left (2 \, x^{2} e^{7} + x e^{\left (x + 8\right )}\right )}} \]
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Time = 0.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {-2 x^2+e^x \left (4 x-2 x^2\right )+e^{1+x} \left (e^x-x^2\right )}{5 e^{5+2 x} x^2+20 e^{4+x} x^3+20 e^3 x^4} \, dx=\frac {2 + e}{10 x e^{4} + 5 e^{5} e^{x}} - \frac {1}{5 x e^{4}} \]
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Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {-2 x^2+e^x \left (4 x-2 x^2\right )+e^{1+x} \left (e^x-x^2\right )}{5 e^{5+2 x} x^2+20 e^{4+x} x^3+20 e^3 x^4} \, dx=\frac {x - e^{x}}{5 \, {\left (2 \, x^{2} e^{3} + x e^{\left (x + 4\right )}\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 211 vs. \(2 (23) = 46\).
Time = 0.29 (sec) , antiderivative size = 211, normalized size of antiderivative = 7.54 \[ \int \frac {-2 x^2+e^x \left (4 x-2 x^2\right )+e^{1+x} \left (e^x-x^2\right )}{5 e^{5+2 x} x^2+20 e^{4+x} x^3+20 e^3 x^4} \, dx=\frac {2 \, {\left (x + 4\right )}^{3} e^{7} - 26 \, {\left (x + 4\right )}^{2} e^{7} + {\left (x + 4\right )}^{2} e^{\left (x + 8\right )} - 2 \, {\left (x + 4\right )}^{2} e^{\left (x + 7\right )} + 112 \, {\left (x + 4\right )} e^{7} - {\left (x + 4\right )} e^{\left (2 \, x + 8\right )} - 9 \, {\left (x + 4\right )} e^{\left (x + 8\right )} + 18 \, {\left (x + 4\right )} e^{\left (x + 7\right )} - 160 \, e^{7} + 5 \, e^{\left (2 \, x + 8\right )} + 20 \, e^{\left (x + 8\right )} - 40 \, e^{\left (x + 7\right )}}{5 \, {\left (4 \, {\left (x + 4\right )}^{4} e^{10} - 68 \, {\left (x + 4\right )}^{3} e^{10} + 4 \, {\left (x + 4\right )}^{3} e^{\left (x + 11\right )} + 432 \, {\left (x + 4\right )}^{2} e^{10} + {\left (x + 4\right )}^{2} e^{\left (2 \, x + 12\right )} - 52 \, {\left (x + 4\right )}^{2} e^{\left (x + 11\right )} - 1216 \, {\left (x + 4\right )} e^{10} - 9 \, {\left (x + 4\right )} e^{\left (2 \, x + 12\right )} + 224 \, {\left (x + 4\right )} e^{\left (x + 11\right )} + 1280 \, e^{10} + 20 \, e^{\left (2 \, x + 12\right )} - 320 \, e^{\left (x + 11\right )}\right )}} \]
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Time = 13.49 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82 \[ \int \frac {-2 x^2+e^x \left (4 x-2 x^2\right )+e^{1+x} \left (e^x-x^2\right )}{5 e^{5+2 x} x^2+20 e^{4+x} x^3+20 e^3 x^4} \, dx=\frac {x-{\mathrm {e}}^x}{5\,x\,\left ({\mathrm {e}}^{x+4}+2\,x\,{\mathrm {e}}^3\right )} \]
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