Integrand size = 201, antiderivative size = 33 \[ \int \frac {e^x \left (-2-2 x+2 x^2+2 x^3\right )+e^{\frac {5 e^{e^x} x-2 x^2-2 x^3}{2+2 x}} \left (e^x \left (2+8 x+10 x^2+4 x^3\right )+e^{e^x} \left (-5 e^x+e^{2 x} \left (-5 x-5 x^2\right )\right )\right )}{8 x^2+16 x^3+8 x^4+e^{\frac {2 \left (5 e^{e^x} x-2 x^2-2 x^3\right )}{2+2 x}} \left (8+16 x+8 x^2\right )+e^{\frac {5 e^{e^x} x-2 x^2-2 x^3}{2+2 x}} \left (16 x+32 x^2+16 x^3\right )} \, dx=\frac {e^x}{4 \left (e^{x \left (-x+\frac {5 e^{e^x}}{2 (1+x)}\right )}+x\right )} \]
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\[ \int \frac {e^x \left (-2-2 x+2 x^2+2 x^3\right )+e^{\frac {5 e^{e^x} x-2 x^2-2 x^3}{2+2 x}} \left (e^x \left (2+8 x+10 x^2+4 x^3\right )+e^{e^x} \left (-5 e^x+e^{2 x} \left (-5 x-5 x^2\right )\right )\right )}{8 x^2+16 x^3+8 x^4+e^{\frac {2 \left (5 e^{e^x} x-2 x^2-2 x^3\right )}{2+2 x}} \left (8+16 x+8 x^2\right )+e^{\frac {5 e^{e^x} x-2 x^2-2 x^3}{2+2 x}} \left (16 x+32 x^2+16 x^3\right )} \, dx=\int \frac {e^x \left (-2-2 x+2 x^2+2 x^3\right )+e^{\frac {5 e^{e^x} x-2 x^2-2 x^3}{2+2 x}} \left (e^x \left (2+8 x+10 x^2+4 x^3\right )+e^{e^x} \left (-5 e^x+e^{2 x} \left (-5 x-5 x^2\right )\right )\right )}{8 x^2+16 x^3+8 x^4+e^{\frac {2 \left (5 e^{e^x} x-2 x^2-2 x^3\right )}{2+2 x}} \left (8+16 x+8 x^2\right )+e^{\frac {5 e^{e^x} x-2 x^2-2 x^3}{2+2 x}} \left (16 x+32 x^2+16 x^3\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{x+x^2} \left (-5 e^{e^x+\frac {5 e^{e^x} x}{2+2 x}}-5 e^{e^x+x+\frac {5 e^{e^x} x}{2+2 x}} x (1+x)+2 e^{x^2} (-1+x) (1+x)^2+2 e^{\frac {5 e^{e^x} x}{2+2 x}} (1+x)^2 (1+2 x)\right )}{8 (1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2} \, dx \\ & = \frac {1}{8} \int \frac {e^{x+x^2} \left (-5 e^{e^x+\frac {5 e^{e^x} x}{2+2 x}}-5 e^{e^x+x+\frac {5 e^{e^x} x}{2+2 x}} x (1+x)+2 e^{x^2} (-1+x) (1+x)^2+2 e^{\frac {5 e^{e^x} x}{2+2 x}} (1+x)^2 (1+2 x)\right )}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2} \, dx \\ & = \frac {1}{8} \int \left (\frac {e^{x+x^2} \left (2-5 e^{e^x}+8 x-5 e^{e^x+x} x+10 x^2-5 e^{e^x+x} x^2+4 x^3\right )}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )}-\frac {e^{x+2 x^2} \left (2+4 x-5 e^{e^x} x+6 x^2-5 e^{e^x+x} x^2+8 x^3-5 e^{e^x+x} x^3+4 x^4\right )}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2}\right ) \, dx \\ & = \frac {1}{8} \int \frac {e^{x+x^2} \left (2-5 e^{e^x}+8 x-5 e^{e^x+x} x+10 x^2-5 e^{e^x+x} x^2+4 x^3\right )}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )} \, dx-\frac {1}{8} \int \frac {e^{x+2 x^2} \left (2+4 x-5 e^{e^x} x+6 x^2-5 e^{e^x+x} x^2+8 x^3-5 e^{e^x+x} x^3+4 x^4\right )}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2} \, dx \\ & = -\left (\frac {1}{8} \int \left (\frac {2 e^{x+2 x^2}}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2}+\frac {4 e^{x+2 x^2} x}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2}-\frac {5 e^{e^x+x+2 x^2} x}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2}+\frac {6 e^{x+2 x^2} x^2}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2}-\frac {5 e^{e^x+2 x+2 x^2} x^2}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2}+\frac {8 e^{x+2 x^2} x^3}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2}-\frac {5 e^{e^x+2 x+2 x^2} x^3}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2}+\frac {4 e^{x+2 x^2} x^4}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2}\right ) \, dx\right )+\frac {1}{8} \int \left (\frac {2 e^{x+x^2}}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )}-\frac {5 e^{e^x+x+x^2}}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )}+\frac {8 e^{x+x^2} x}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )}-\frac {5 e^{e^x+2 x+x^2} x}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )}+\frac {10 e^{x+x^2} x^2}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )}-\frac {5 e^{e^x+2 x+x^2} x^2}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )}+\frac {4 e^{x+x^2} x^3}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )}\right ) \, dx \\ & = -\left (\frac {1}{4} \int \frac {e^{x+2 x^2}}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2} \, dx\right )+\frac {1}{4} \int \frac {e^{x+x^2}}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )} \, dx-\frac {1}{2} \int \frac {e^{x+2 x^2} x}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2} \, dx-\frac {1}{2} \int \frac {e^{x+2 x^2} x^4}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2} \, dx+\frac {1}{2} \int \frac {e^{x+x^2} x^3}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )} \, dx+\frac {5}{8} \int \frac {e^{e^x+x+2 x^2} x}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2} \, dx+\frac {5}{8} \int \frac {e^{e^x+2 x+2 x^2} x^2}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2} \, dx+\frac {5}{8} \int \frac {e^{e^x+2 x+2 x^2} x^3}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2} \, dx-\frac {5}{8} \int \frac {e^{e^x+x+x^2}}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )} \, dx-\frac {5}{8} \int \frac {e^{e^x+2 x+x^2} x}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )} \, dx-\frac {5}{8} \int \frac {e^{e^x+2 x+x^2} x^2}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )} \, dx-\frac {3}{4} \int \frac {e^{x+2 x^2} x^2}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2} \, dx+\frac {5}{4} \int \frac {e^{x+x^2} x^2}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )} \, dx-\int \frac {e^{x+2 x^2} x^3}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2} \, dx+\int \frac {e^{x+x^2} x}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )} \, dx \\ & = \text {Too large to display} \\ \end{align*}
Time = 0.64 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.15 \[ \int \frac {e^x \left (-2-2 x+2 x^2+2 x^3\right )+e^{\frac {5 e^{e^x} x-2 x^2-2 x^3}{2+2 x}} \left (e^x \left (2+8 x+10 x^2+4 x^3\right )+e^{e^x} \left (-5 e^x+e^{2 x} \left (-5 x-5 x^2\right )\right )\right )}{8 x^2+16 x^3+8 x^4+e^{\frac {2 \left (5 e^{e^x} x-2 x^2-2 x^3\right )}{2+2 x}} \left (8+16 x+8 x^2\right )+e^{\frac {5 e^{e^x} x-2 x^2-2 x^3}{2+2 x}} \left (16 x+32 x^2+16 x^3\right )} \, dx=\frac {e^{x+x^2}}{4 \left (e^{\frac {5 e^{e^x} x}{2 (1+x)}}+e^{x^2} x\right )} \]
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Time = 5.57 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97
method | result | size |
risch | \(\frac {{\mathrm e}^{x}}{4 x +4 \,{\mathrm e}^{-\frac {x \left (2 x^{2}-5 \,{\mathrm e}^{{\mathrm e}^{x}}+2 x \right )}{2 \left (1+x \right )}}}\) | \(32\) |
parallelrisch | \(\frac {{\mathrm e}^{x}}{4 \,{\mathrm e}^{\frac {x \left (-2 x^{2}+5 \,{\mathrm e}^{{\mathrm e}^{x}}-2 x \right )}{2+2 x}}+4 x}\) | \(32\) |
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Time = 0.24 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00 \[ \int \frac {e^x \left (-2-2 x+2 x^2+2 x^3\right )+e^{\frac {5 e^{e^x} x-2 x^2-2 x^3}{2+2 x}} \left (e^x \left (2+8 x+10 x^2+4 x^3\right )+e^{e^x} \left (-5 e^x+e^{2 x} \left (-5 x-5 x^2\right )\right )\right )}{8 x^2+16 x^3+8 x^4+e^{\frac {2 \left (5 e^{e^x} x-2 x^2-2 x^3\right )}{2+2 x}} \left (8+16 x+8 x^2\right )+e^{\frac {5 e^{e^x} x-2 x^2-2 x^3}{2+2 x}} \left (16 x+32 x^2+16 x^3\right )} \, dx=\frac {e^{x}}{4 \, {\left (x + e^{\left (-\frac {2 \, x^{3} + 2 \, x^{2} - 5 \, x e^{\left (e^{x}\right )}}{2 \, {\left (x + 1\right )}}\right )}\right )}} \]
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Time = 0.35 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int \frac {e^x \left (-2-2 x+2 x^2+2 x^3\right )+e^{\frac {5 e^{e^x} x-2 x^2-2 x^3}{2+2 x}} \left (e^x \left (2+8 x+10 x^2+4 x^3\right )+e^{e^x} \left (-5 e^x+e^{2 x} \left (-5 x-5 x^2\right )\right )\right )}{8 x^2+16 x^3+8 x^4+e^{\frac {2 \left (5 e^{e^x} x-2 x^2-2 x^3\right )}{2+2 x}} \left (8+16 x+8 x^2\right )+e^{\frac {5 e^{e^x} x-2 x^2-2 x^3}{2+2 x}} \left (16 x+32 x^2+16 x^3\right )} \, dx=\frac {e^{x}}{4 x + 4 e^{\frac {- 2 x^{3} - 2 x^{2} + 5 x e^{e^{x}}}{2 x + 2}}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 161 vs. \(2 (26) = 52\).
Time = 0.38 (sec) , antiderivative size = 161, normalized size of antiderivative = 4.88 \[ \int \frac {e^x \left (-2-2 x+2 x^2+2 x^3\right )+e^{\frac {5 e^{e^x} x-2 x^2-2 x^3}{2+2 x}} \left (e^x \left (2+8 x+10 x^2+4 x^3\right )+e^{e^x} \left (-5 e^x+e^{2 x} \left (-5 x-5 x^2\right )\right )\right )}{8 x^2+16 x^3+8 x^4+e^{\frac {2 \left (5 e^{e^x} x-2 x^2-2 x^3\right )}{2+2 x}} \left (8+16 x+8 x^2\right )+e^{\frac {5 e^{e^x} x-2 x^2-2 x^3}{2+2 x}} \left (16 x+32 x^2+16 x^3\right )} \, dx=\frac {{\left (4 \, x^{4} + 8 \, x^{3} + 6 \, x^{2} - 5 \, {\left ({\left (x^{3} + x^{2}\right )} e^{x} + x\right )} e^{\left (e^{x}\right )} + 4 \, x + 2\right )} e^{\left (x^{2} + x + \frac {5 \, e^{\left (e^{x}\right )}}{2 \, {\left (x + 1\right )}}\right )}}{4 \, {\left ({\left (4 \, x^{5} + 8 \, x^{4} + 6 \, x^{3} + 4 \, x^{2} - 5 \, {\left (x^{2} + {\left (x^{4} + x^{3}\right )} e^{x}\right )} e^{\left (e^{x}\right )} + 2 \, x\right )} e^{\left (x^{2} + \frac {5 \, e^{\left (e^{x}\right )}}{2 \, {\left (x + 1\right )}}\right )} + {\left (4 \, x^{4} + 8 \, x^{3} + 6 \, x^{2} - 5 \, {\left ({\left (x^{3} + x^{2}\right )} e^{x} + x\right )} e^{\left (e^{x}\right )} + 4 \, x + 2\right )} e^{\left (\frac {5}{2} \, e^{\left (e^{x}\right )}\right )}\right )}} \]
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\[ \int \frac {e^x \left (-2-2 x+2 x^2+2 x^3\right )+e^{\frac {5 e^{e^x} x-2 x^2-2 x^3}{2+2 x}} \left (e^x \left (2+8 x+10 x^2+4 x^3\right )+e^{e^x} \left (-5 e^x+e^{2 x} \left (-5 x-5 x^2\right )\right )\right )}{8 x^2+16 x^3+8 x^4+e^{\frac {2 \left (5 e^{e^x} x-2 x^2-2 x^3\right )}{2+2 x}} \left (8+16 x+8 x^2\right )+e^{\frac {5 e^{e^x} x-2 x^2-2 x^3}{2+2 x}} \left (16 x+32 x^2+16 x^3\right )} \, dx=\int { \frac {2 \, {\left (x^{3} + x^{2} - x - 1\right )} e^{x} + {\left (2 \, {\left (2 \, x^{3} + 5 \, x^{2} + 4 \, x + 1\right )} e^{x} - 5 \, {\left ({\left (x^{2} + x\right )} e^{\left (2 \, x\right )} + e^{x}\right )} e^{\left (e^{x}\right )}\right )} e^{\left (-\frac {2 \, x^{3} + 2 \, x^{2} - 5 \, x e^{\left (e^{x}\right )}}{2 \, {\left (x + 1\right )}}\right )}}{8 \, {\left (x^{4} + 2 \, x^{3} + x^{2} + 2 \, {\left (x^{3} + 2 \, x^{2} + x\right )} e^{\left (-\frac {2 \, x^{3} + 2 \, x^{2} - 5 \, x e^{\left (e^{x}\right )}}{2 \, {\left (x + 1\right )}}\right )} + {\left (x^{2} + 2 \, x + 1\right )} e^{\left (-\frac {2 \, x^{3} + 2 \, x^{2} - 5 \, x e^{\left (e^{x}\right )}}{x + 1}\right )}\right )}} \,d x } \]
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Time = 11.49 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.45 \[ \int \frac {e^x \left (-2-2 x+2 x^2+2 x^3\right )+e^{\frac {5 e^{e^x} x-2 x^2-2 x^3}{2+2 x}} \left (e^x \left (2+8 x+10 x^2+4 x^3\right )+e^{e^x} \left (-5 e^x+e^{2 x} \left (-5 x-5 x^2\right )\right )\right )}{8 x^2+16 x^3+8 x^4+e^{\frac {2 \left (5 e^{e^x} x-2 x^2-2 x^3\right )}{2+2 x}} \left (8+16 x+8 x^2\right )+e^{\frac {5 e^{e^x} x-2 x^2-2 x^3}{2+2 x}} \left (16 x+32 x^2+16 x^3\right )} \, dx=\frac {{\mathrm {e}}^x}{4\,x}-\frac {{\mathrm {e}}^x}{x\,\left (4\,x\,{\mathrm {e}}^{\frac {2\,x^2-5\,x\,{\mathrm {e}}^{{\mathrm {e}}^x}+2\,x^3}{2\,x+2}}+4\right )} \]
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