Integrand size = 82, antiderivative size = 25 \[ \int \frac {e^{1-x+x^2} (1536-1024 x)+e^x \left (-512+448 x+32 x^2\right )}{256 e^{2-4 x+2 x^2}+256 x^2+32 x^3+x^4+e^{1-2 x+x^2} \left (512 x+32 x^2\right )} \, dx=\frac {2 e^x}{e^{(1-x)^2}+x+\frac {x^2}{16}} \]
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\[ \int \frac {e^{1-x+x^2} (1536-1024 x)+e^x \left (-512+448 x+32 x^2\right )}{256 e^{2-4 x+2 x^2}+256 x^2+32 x^3+x^4+e^{1-2 x+x^2} \left (512 x+32 x^2\right )} \, dx=\int \frac {e^{1-x+x^2} (1536-1024 x)+e^x \left (-512+448 x+32 x^2\right )}{256 e^{2-4 x+2 x^2}+256 x^2+32 x^3+x^4+e^{1-2 x+x^2} \left (512 x+32 x^2\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {-512 e^{1+3 x+x^2} (-3+2 x)+32 e^{5 x} \left (-16+14 x+x^2\right )}{\left (16 e^{1+x^2}+e^{2 x} x (16+x)\right )^2} \, dx \\ & = \int \left (-\frac {32 e^{3 x} (-3+2 x)}{16 e^{1+x^2}+16 e^{2 x} x+e^{2 x} x^2}+\frac {64 e^{5 x} \left (-8-17 x+15 x^2+x^3\right )}{\left (16 e^{1+x^2}+16 e^{2 x} x+e^{2 x} x^2\right )^2}\right ) \, dx \\ & = -\left (32 \int \frac {e^{3 x} (-3+2 x)}{16 e^{1+x^2}+16 e^{2 x} x+e^{2 x} x^2} \, dx\right )+64 \int \frac {e^{5 x} \left (-8-17 x+15 x^2+x^3\right )}{\left (16 e^{1+x^2}+16 e^{2 x} x+e^{2 x} x^2\right )^2} \, dx \\ & = -\left (32 \int \left (-\frac {3 e^{3 x}}{16 e^{1+x^2}+16 e^{2 x} x+e^{2 x} x^2}+\frac {2 e^{3 x} x}{16 e^{1+x^2}+16 e^{2 x} x+e^{2 x} x^2}\right ) \, dx\right )+64 \int \left (-\frac {8 e^{5 x}}{\left (16 e^{1+x^2}+16 e^{2 x} x+e^{2 x} x^2\right )^2}-\frac {17 e^{5 x} x}{\left (16 e^{1+x^2}+16 e^{2 x} x+e^{2 x} x^2\right )^2}+\frac {15 e^{5 x} x^2}{\left (16 e^{1+x^2}+16 e^{2 x} x+e^{2 x} x^2\right )^2}+\frac {e^{5 x} x^3}{\left (16 e^{1+x^2}+16 e^{2 x} x+e^{2 x} x^2\right )^2}\right ) \, dx \\ & = 64 \int \frac {e^{5 x} x^3}{\left (16 e^{1+x^2}+16 e^{2 x} x+e^{2 x} x^2\right )^2} \, dx-64 \int \frac {e^{3 x} x}{16 e^{1+x^2}+16 e^{2 x} x+e^{2 x} x^2} \, dx+96 \int \frac {e^{3 x}}{16 e^{1+x^2}+16 e^{2 x} x+e^{2 x} x^2} \, dx-512 \int \frac {e^{5 x}}{\left (16 e^{1+x^2}+16 e^{2 x} x+e^{2 x} x^2\right )^2} \, dx+960 \int \frac {e^{5 x} x^2}{\left (16 e^{1+x^2}+16 e^{2 x} x+e^{2 x} x^2\right )^2} \, dx-1088 \int \frac {e^{5 x} x}{\left (16 e^{1+x^2}+16 e^{2 x} x+e^{2 x} x^2\right )^2} \, dx \\ \end{align*}
Time = 5.79 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {e^{1-x+x^2} (1536-1024 x)+e^x \left (-512+448 x+32 x^2\right )}{256 e^{2-4 x+2 x^2}+256 x^2+32 x^3+x^4+e^{1-2 x+x^2} \left (512 x+32 x^2\right )} \, dx=\frac {32 e^{3 x}}{16 e^{1+x^2}+e^{2 x} x (16+x)} \]
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Time = 0.14 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88
method | result | size |
risch | \(\frac {32 \,{\mathrm e}^{x}}{x^{2}+16 \,{\mathrm e}^{\left (-1+x \right )^{2}}+16 x}\) | \(22\) |
norman | \(\frac {32 \,{\mathrm e}^{x}}{x^{2}+16 \,{\mathrm e}^{x^{2}-2 x +1}+16 x}\) | \(25\) |
parallelrisch | \(\frac {32 \,{\mathrm e}^{x}}{x^{2}+16 \,{\mathrm e}^{x^{2}-2 x +1}+16 x}\) | \(25\) |
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Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (21) = 42\).
Time = 0.25 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.76 \[ \int \frac {e^{1-x+x^2} (1536-1024 x)+e^x \left (-512+448 x+32 x^2\right )}{256 e^{2-4 x+2 x^2}+256 x^2+32 x^3+x^4+e^{1-2 x+x^2} \left (512 x+32 x^2\right )} \, dx=\frac {32 \, e^{\left (x^{2} - x + 1\right )}}{{\left (x^{2} + 16 \, x\right )} e^{\left (x^{2} - 2 \, x + 1\right )} + 16 \, e^{\left (2 \, x^{2} - 4 \, x + 2\right )}} \]
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Time = 0.08 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {e^{1-x+x^2} (1536-1024 x)+e^x \left (-512+448 x+32 x^2\right )}{256 e^{2-4 x+2 x^2}+256 x^2+32 x^3+x^4+e^{1-2 x+x^2} \left (512 x+32 x^2\right )} \, dx=\frac {2 e^{x}}{\frac {x^{2}}{16} + x + e^{x^{2} - 2 x + 1}} \]
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Time = 0.23 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {e^{1-x+x^2} (1536-1024 x)+e^x \left (-512+448 x+32 x^2\right )}{256 e^{2-4 x+2 x^2}+256 x^2+32 x^3+x^4+e^{1-2 x+x^2} \left (512 x+32 x^2\right )} \, dx=\frac {32 \, e^{\left (3 \, x\right )}}{{\left (x^{2} + 16 \, x\right )} e^{\left (2 \, x\right )} + 16 \, e^{\left (x^{2} + 1\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 255 vs. \(2 (21) = 42\).
Time = 0.27 (sec) , antiderivative size = 255, normalized size of antiderivative = 10.20 \[ \int \frac {e^{1-x+x^2} (1536-1024 x)+e^x \left (-512+448 x+32 x^2\right )}{256 e^{2-4 x+2 x^2}+256 x^2+32 x^3+x^4+e^{1-2 x+x^2} \left (512 x+32 x^2\right )} \, dx=\frac {32 \, {\left (x^{5} e^{\left (2 \, x\right )} + 31 \, x^{4} e^{\left (2 \, x\right )} + 16 \, x^{3} e^{\left (x^{2} + 1\right )} + 223 \, x^{3} e^{\left (2 \, x\right )} + 240 \, x^{2} e^{\left (x^{2} + 1\right )} - 280 \, x^{2} e^{\left (2 \, x\right )} - 272 \, x e^{\left (x^{2} + 1\right )} - 128 \, x e^{\left (2 \, x\right )} - 128 \, e^{\left (x^{2} + 1\right )}\right )}}{x^{7} e^{x} + 47 \, x^{6} e^{x} + 32 \, x^{5} e^{\left (x^{2} - x + 1\right )} + 719 \, x^{5} e^{x} + 992 \, x^{4} e^{\left (x^{2} - x + 1\right )} + 3288 \, x^{4} e^{x} + 256 \, x^{3} e^{\left (2 \, x^{2} - 3 \, x + 2\right )} + 7136 \, x^{3} e^{\left (x^{2} - x + 1\right )} - 4608 \, x^{3} e^{x} + 3840 \, x^{2} e^{\left (2 \, x^{2} - 3 \, x + 2\right )} - 8960 \, x^{2} e^{\left (x^{2} - x + 1\right )} - 2048 \, x^{2} e^{x} - 4352 \, x e^{\left (2 \, x^{2} - 3 \, x + 2\right )} - 4096 \, x e^{\left (x^{2} - x + 1\right )} - 2048 \, e^{\left (2 \, x^{2} - 3 \, x + 2\right )}} \]
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Timed out. \[ \int \frac {e^{1-x+x^2} (1536-1024 x)+e^x \left (-512+448 x+32 x^2\right )}{256 e^{2-4 x+2 x^2}+256 x^2+32 x^3+x^4+e^{1-2 x+x^2} \left (512 x+32 x^2\right )} \, dx=\int -\frac {{\mathrm {e}}^{x^2-x+1}\,\left (1024\,x-1536\right )-{\mathrm {e}}^x\,\left (32\,x^2+448\,x-512\right )}{256\,{\mathrm {e}}^{2\,x^2-4\,x+2}+{\mathrm {e}}^{x^2-2\,x+1}\,\left (32\,x^2+512\,x\right )+256\,x^2+32\,x^3+x^4} \,d x \]
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