Integrand size = 98, antiderivative size = 31 \[ \int \frac {e^{-3/x} \left (180-15 x-50 x^2+28 e^{3/x} x^2+e^{\frac {3}{x}+x} \left (16 x^2+8 x^3+x^4\right )+e^{\frac {3}{x}+x^2} \left (32 x^3+16 x^4+2 x^5\right )\right )}{16 x^2+8 x^3+x^4} \, dx=4+e^x+e^{x^2}-\frac {\left (-4+5 e^{-3/x}\right ) (-3+x)}{4+x} \]
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Time = 1.64 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.26, number of steps used = 48, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.102, Rules used = {1608, 27, 6874, 2225, 2240, 2255, 2241, 2254, 2260, 2209} \[ \int \frac {e^{-3/x} \left (180-15 x-50 x^2+28 e^{3/x} x^2+e^{\frac {3}{x}+x} \left (16 x^2+8 x^3+x^4\right )+e^{\frac {3}{x}+x^2} \left (32 x^3+16 x^4+2 x^5\right )\right )}{16 x^2+8 x^3+x^4} \, dx=e^{x^2}-5 e^{-3/x}+e^x+\frac {35 e^{-3/x}}{x+4}-\frac {28}{x+4} \]
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Rule 27
Rule 1608
Rule 2209
Rule 2225
Rule 2240
Rule 2241
Rule 2254
Rule 2255
Rule 2260
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{-3/x} \left (180-15 x-50 x^2+28 e^{3/x} x^2+e^{\frac {3}{x}+x} \left (16 x^2+8 x^3+x^4\right )+e^{\frac {3}{x}+x^2} \left (32 x^3+16 x^4+2 x^5\right )\right )}{x^2 \left (16+8 x+x^2\right )} \, dx \\ & = \int \frac {e^{-3/x} \left (180-15 x-50 x^2+28 e^{3/x} x^2+e^{\frac {3}{x}+x} \left (16 x^2+8 x^3+x^4\right )+e^{\frac {3}{x}+x^2} \left (32 x^3+16 x^4+2 x^5\right )\right )}{x^2 (4+x)^2} \, dx \\ & = \int \left (e^x+2 e^{x^2} x+\frac {28}{(4+x)^2}-\frac {50 e^{-3/x}}{(4+x)^2}+\frac {180 e^{-3/x}}{x^2 (4+x)^2}-\frac {15 e^{-3/x}}{x (4+x)^2}\right ) \, dx \\ & = -\frac {28}{4+x}+2 \int e^{x^2} x \, dx-15 \int \frac {e^{-3/x}}{x (4+x)^2} \, dx-50 \int \frac {e^{-3/x}}{(4+x)^2} \, dx+180 \int \frac {e^{-3/x}}{x^2 (4+x)^2} \, dx+\int e^x \, dx \\ & = e^x+e^{x^2}-\frac {28}{4+x}+\frac {50 e^{-3/x}}{4+x}-15 \int \left (\frac {e^{-3/x}}{16 x}-\frac {e^{-3/x}}{4 (4+x)^2}-\frac {e^{-3/x}}{16 (4+x)}\right ) \, dx-150 \int \frac {e^{-3/x}}{x^2 (4+x)} \, dx+180 \int \left (\frac {e^{-3/x}}{16 x^2}-\frac {e^{-3/x}}{32 x}+\frac {e^{-3/x}}{16 (4+x)^2}+\frac {e^{-3/x}}{32 (4+x)}\right ) \, dx \\ & = e^x+e^{x^2}-\frac {28}{4+x}+\frac {50 e^{-3/x}}{4+x}-\frac {15}{16} \int \frac {e^{-3/x}}{x} \, dx+\frac {15}{16} \int \frac {e^{-3/x}}{4+x} \, dx+\frac {15}{4} \int \frac {e^{-3/x}}{(4+x)^2} \, dx-\frac {45}{8} \int \frac {e^{-3/x}}{x} \, dx+\frac {45}{8} \int \frac {e^{-3/x}}{4+x} \, dx+\frac {45}{4} \int \frac {e^{-3/x}}{x^2} \, dx+\frac {45}{4} \int \frac {e^{-3/x}}{(4+x)^2} \, dx-150 \int \left (\frac {e^{-3/x}}{4 x^2}-\frac {e^{-3/x}}{16 x}+\frac {e^{-3/x}}{16 (4+x)}\right ) \, dx \\ & = \frac {15 e^{-3/x}}{4}+e^x+e^{x^2}-\frac {28}{4+x}+\frac {35 e^{-3/x}}{4+x}+\frac {105 \operatorname {ExpIntegralEi}\left (-\frac {3}{x}\right )}{16}+\frac {15}{16} \int \frac {e^{-3/x}}{x} \, dx-\frac {15}{4} \int \frac {e^{-3/x}}{x (4+x)} \, dx+\frac {45}{8} \int \frac {e^{-3/x}}{x} \, dx+\frac {75}{8} \int \frac {e^{-3/x}}{x} \, dx-\frac {75}{8} \int \frac {e^{-3/x}}{4+x} \, dx+\frac {45}{4} \int \frac {e^{-3/x}}{x^2 (4+x)} \, dx-\frac {45}{2} \int \frac {e^{-3/x}}{x (4+x)} \, dx+\frac {135}{4} \int \frac {e^{-3/x}}{x^2 (4+x)} \, dx-\frac {75}{2} \int \frac {e^{-3/x}}{x^2} \, dx \\ & = -\frac {35}{4} e^{-3/x}+e^x+e^{x^2}-\frac {28}{4+x}+\frac {35 e^{-3/x}}{4+x}-\frac {75 \operatorname {ExpIntegralEi}\left (-\frac {3}{x}\right )}{8}+\frac {15}{16} \text {Subst}\left (\int \frac {e^{\frac {3}{4}-\frac {3 x}{4}}}{x} \, dx,x,\frac {4+x}{x}\right )+\frac {45}{8} \text {Subst}\left (\int \frac {e^{\frac {3}{4}-\frac {3 x}{4}}}{x} \, dx,x,\frac {4+x}{x}\right )-\frac {75}{8} \int \frac {e^{-3/x}}{x} \, dx+\frac {45}{4} \int \left (\frac {e^{-3/x}}{4 x^2}-\frac {e^{-3/x}}{16 x}+\frac {e^{-3/x}}{16 (4+x)}\right ) \, dx+\frac {135}{4} \int \left (\frac {e^{-3/x}}{4 x^2}-\frac {e^{-3/x}}{16 x}+\frac {e^{-3/x}}{16 (4+x)}\right ) \, dx+\frac {75}{2} \int \frac {e^{-3/x}}{x (4+x)} \, dx \\ & = -\frac {35}{4} e^{-3/x}+e^x+e^{x^2}-\frac {28}{4+x}+\frac {35 e^{-3/x}}{4+x}+\frac {105}{16} e^{3/4} \operatorname {ExpIntegralEi}\left (-\frac {3 (4+x)}{4 x}\right )-\frac {45}{64} \int \frac {e^{-3/x}}{x} \, dx+\frac {45}{64} \int \frac {e^{-3/x}}{4+x} \, dx-\frac {135}{64} \int \frac {e^{-3/x}}{x} \, dx+\frac {135}{64} \int \frac {e^{-3/x}}{4+x} \, dx+\frac {45}{16} \int \frac {e^{-3/x}}{x^2} \, dx+\frac {135}{16} \int \frac {e^{-3/x}}{x^2} \, dx-\frac {75}{8} \text {Subst}\left (\int \frac {e^{\frac {3}{4}-\frac {3 x}{4}}}{x} \, dx,x,\frac {4+x}{x}\right ) \\ & = -5 e^{-3/x}+e^x+e^{x^2}-\frac {28}{4+x}+\frac {35 e^{-3/x}}{4+x}+\frac {45 \operatorname {ExpIntegralEi}\left (-\frac {3}{x}\right )}{16}-\frac {45}{16} e^{3/4} \operatorname {ExpIntegralEi}\left (-\frac {3 (4+x)}{4 x}\right )+\frac {45}{64} \int \frac {e^{-3/x}}{x} \, dx+\frac {135}{64} \int \frac {e^{-3/x}}{x} \, dx-\frac {45}{16} \int \frac {e^{-3/x}}{x (4+x)} \, dx-\frac {135}{16} \int \frac {e^{-3/x}}{x (4+x)} \, dx \\ & = -5 e^{-3/x}+e^x+e^{x^2}-\frac {28}{4+x}+\frac {35 e^{-3/x}}{4+x}-\frac {45}{16} e^{3/4} \operatorname {ExpIntegralEi}\left (-\frac {3 (4+x)}{4 x}\right )+\frac {45}{64} \text {Subst}\left (\int \frac {e^{\frac {3}{4}-\frac {3 x}{4}}}{x} \, dx,x,\frac {4+x}{x}\right )+\frac {135}{64} \text {Subst}\left (\int \frac {e^{\frac {3}{4}-\frac {3 x}{4}}}{x} \, dx,x,\frac {4+x}{x}\right ) \\ & = -5 e^{-3/x}+e^x+e^{x^2}-\frac {28}{4+x}+\frac {35 e^{-3/x}}{4+x} \\ \end{align*}
Time = 5.08 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06 \[ \int \frac {e^{-3/x} \left (180-15 x-50 x^2+28 e^{3/x} x^2+e^{\frac {3}{x}+x} \left (16 x^2+8 x^3+x^4\right )+e^{\frac {3}{x}+x^2} \left (32 x^3+16 x^4+2 x^5\right )\right )}{16 x^2+8 x^3+x^4} \, dx=e^x+e^{x^2}-\frac {28}{4+x}+e^{-3/x} \left (-5+\frac {35}{4+x}\right ) \]
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Time = 8.06 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00
method | result | size |
risch | \(-\frac {28}{4+x}+{\mathrm e}^{x}+{\mathrm e}^{x^{2}}-\frac {5 \left (-3+x \right ) {\mathrm e}^{-\frac {3}{x}}}{4+x}\) | \(31\) |
parts | \(-\frac {28}{4+x}+{\mathrm e}^{x^{2}}-\frac {105 \,{\mathrm e}^{-\frac {3}{x}}}{4 \left (3+\frac {12}{x}\right )}+\frac {15 \,{\mathrm e}^{-\frac {3}{x}}}{4}+{\mathrm e}^{x}\) | \(40\) |
parallelrisch | \(-\frac {\left (-12 x^{2} {\mathrm e}^{\frac {3}{x}} {\mathrm e}^{x}-12 \,{\mathrm e}^{\frac {3}{x}} {\mathrm e}^{x^{2}} x^{2}-48 \,{\mathrm e}^{\frac {3}{x}} {\mathrm e}^{x} x -48 \,{\mathrm e}^{\frac {3}{x}} {\mathrm e}^{x^{2}} x +60 x^{2}+336 x \,{\mathrm e}^{\frac {3}{x}}-180 x \right ) {\mathrm e}^{-\frac {3}{x}}}{12 x \left (4+x \right )}\) | \(89\) |
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Time = 0.31 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.71 \[ \int \frac {e^{-3/x} \left (180-15 x-50 x^2+28 e^{3/x} x^2+e^{\frac {3}{x}+x} \left (16 x^2+8 x^3+x^4\right )+e^{\frac {3}{x}+x^2} \left (32 x^3+16 x^4+2 x^5\right )\right )}{16 x^2+8 x^3+x^4} \, dx=\frac {{\left ({\left (x + 4\right )} e^{\left (\frac {x^{3} + 3}{x}\right )} + {\left (x + 4\right )} e^{\left (\frac {x^{2} + 3}{x}\right )} - 5 \, x - 28 \, e^{\frac {3}{x}} + 15\right )} e^{\left (-\frac {3}{x}\right )}}{x + 4} \]
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Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {e^{-3/x} \left (180-15 x-50 x^2+28 e^{3/x} x^2+e^{\frac {3}{x}+x} \left (16 x^2+8 x^3+x^4\right )+e^{\frac {3}{x}+x^2} \left (32 x^3+16 x^4+2 x^5\right )\right )}{16 x^2+8 x^3+x^4} \, dx=\frac {\left (15 - 5 x\right ) e^{- \frac {3}{x}}}{x + 4} + e^{x} + e^{x^{2}} - \frac {28}{x + 4} \]
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Time = 0.23 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06 \[ \int \frac {e^{-3/x} \left (180-15 x-50 x^2+28 e^{3/x} x^2+e^{\frac {3}{x}+x} \left (16 x^2+8 x^3+x^4\right )+e^{\frac {3}{x}+x^2} \left (32 x^3+16 x^4+2 x^5\right )\right )}{16 x^2+8 x^3+x^4} \, dx=\frac {{\left (x + 4\right )} e^{\left (x^{2}\right )} + {\left (x + 4\right )} e^{x} - 5 \, {\left (x - 3\right )} e^{\left (-\frac {3}{x}\right )} - 28}{x + 4} \]
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Time = 0.28 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.45 \[ \int \frac {e^{-3/x} \left (180-15 x-50 x^2+28 e^{3/x} x^2+e^{\frac {3}{x}+x} \left (16 x^2+8 x^3+x^4\right )+e^{\frac {3}{x}+x^2} \left (32 x^3+16 x^4+2 x^5\right )\right )}{16 x^2+8 x^3+x^4} \, dx=\frac {x e^{\left (x^{2}\right )} + x e^{x} - 5 \, x e^{\left (-\frac {3}{x}\right )} + 4 \, e^{\left (x^{2}\right )} + 4 \, e^{x} + 15 \, e^{\left (-\frac {3}{x}\right )} - 28}{x + 4} \]
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Time = 12.52 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {e^{-3/x} \left (180-15 x-50 x^2+28 e^{3/x} x^2+e^{\frac {3}{x}+x} \left (16 x^2+8 x^3+x^4\right )+e^{\frac {3}{x}+x^2} \left (32 x^3+16 x^4+2 x^5\right )\right )}{16 x^2+8 x^3+x^4} \, dx={\mathrm {e}}^{x^2}+{\mathrm {e}}^x-\frac {28}{x+4}-\frac {{\mathrm {e}}^{-\frac {3}{x}}\,\left (5\,x-15\right )}{x+4} \]
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