Integrand size = 59, antiderivative size = 19 \[ \int e^{-320-340 x-20 x^2+e^{1+x} \left (80 x+85 x^2+5 x^3\right )} \left (-340-40 x+e^{1+x} \left (80+250 x+100 x^2+5 x^3\right )\right ) \, dx=e^{5 (1+x) (16+x) \left (-4+e^{1+x} x\right )} \]
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\[ \int e^{-320-340 x-20 x^2+e^{1+x} \left (80 x+85 x^2+5 x^3\right )} \left (-340-40 x+e^{1+x} \left (80+250 x+100 x^2+5 x^3\right )\right ) \, dx=\int \exp \left (-320-340 x-20 x^2+e^{1+x} \left (80 x+85 x^2+5 x^3\right )\right ) \left (-340-40 x+e^{1+x} \left (80+250 x+100 x^2+5 x^3\right )\right ) \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int e^{5 \left (-4+e^{1+x} x\right ) \left (16+17 x+x^2\right )} \left (-340-40 x+e^{1+x} \left (80+250 x+100 x^2+5 x^3\right )\right ) \, dx \\ & = \int \left (-340 e^{5 \left (-4+e^{1+x} x\right ) \left (16+17 x+x^2\right )}-40 e^{5 \left (-4+e^{1+x} x\right ) \left (16+17 x+x^2\right )} x+5 e^{1+x+5 \left (-4+e^{1+x} x\right ) \left (16+17 x+x^2\right )} \left (16+50 x+20 x^2+x^3\right )\right ) \, dx \\ & = 5 \int e^{1+x+5 \left (-4+e^{1+x} x\right ) \left (16+17 x+x^2\right )} \left (16+50 x+20 x^2+x^3\right ) \, dx-40 \int e^{5 \left (-4+e^{1+x} x\right ) \left (16+17 x+x^2\right )} x \, dx-340 \int e^{5 \left (-4+e^{1+x} x\right ) \left (16+17 x+x^2\right )} \, dx \\ & = 5 \int \left (16 e^{1+x+5 \left (-4+e^{1+x} x\right ) \left (16+17 x+x^2\right )}+50 e^{1+x+5 \left (-4+e^{1+x} x\right ) \left (16+17 x+x^2\right )} x+20 e^{1+x+5 \left (-4+e^{1+x} x\right ) \left (16+17 x+x^2\right )} x^2+e^{1+x+5 \left (-4+e^{1+x} x\right ) \left (16+17 x+x^2\right )} x^3\right ) \, dx-40 \int e^{5 \left (-4+e^{1+x} x\right ) \left (16+17 x+x^2\right )} x \, dx-340 \int e^{5 \left (-4+e^{1+x} x\right ) \left (16+17 x+x^2\right )} \, dx \\ & = 5 \int e^{1+x+5 \left (-4+e^{1+x} x\right ) \left (16+17 x+x^2\right )} x^3 \, dx-40 \int e^{5 \left (-4+e^{1+x} x\right ) \left (16+17 x+x^2\right )} x \, dx+80 \int e^{1+x+5 \left (-4+e^{1+x} x\right ) \left (16+17 x+x^2\right )} \, dx+100 \int e^{1+x+5 \left (-4+e^{1+x} x\right ) \left (16+17 x+x^2\right )} x^2 \, dx+250 \int e^{1+x+5 \left (-4+e^{1+x} x\right ) \left (16+17 x+x^2\right )} x \, dx-340 \int e^{5 \left (-4+e^{1+x} x\right ) \left (16+17 x+x^2\right )} \, dx \\ \end{align*}
Time = 1.81 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11 \[ \int e^{-320-340 x-20 x^2+e^{1+x} \left (80 x+85 x^2+5 x^3\right )} \left (-340-40 x+e^{1+x} \left (80+250 x+100 x^2+5 x^3\right )\right ) \, dx=e^{5 \left (-4+e^{1+x} x\right ) \left (16+17 x+x^2\right )} \]
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Time = 0.08 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95
method | result | size |
risch | \({\mathrm e}^{5 \left (x \,{\mathrm e}^{1+x}-4\right ) \left (1+x \right ) \left (x +16\right )}\) | \(18\) |
norman | \({\mathrm e}^{\left (5 x^{3}+85 x^{2}+80 x \right ) {\mathrm e}^{1+x}-20 x^{2}-340 x -320}\) | \(31\) |
parallelrisch | \({\mathrm e}^{\left (5 x^{3}+85 x^{2}+80 x \right ) {\mathrm e}^{1+x}-20 x^{2}-340 x -320}\) | \(31\) |
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Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.53 \[ \int e^{-320-340 x-20 x^2+e^{1+x} \left (80 x+85 x^2+5 x^3\right )} \left (-340-40 x+e^{1+x} \left (80+250 x+100 x^2+5 x^3\right )\right ) \, dx=e^{\left (-20 \, x^{2} + 5 \, {\left (x^{3} + 17 \, x^{2} + 16 \, x\right )} e^{\left (x + 1\right )} - 340 \, x - 320\right )} \]
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Time = 0.13 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.53 \[ \int e^{-320-340 x-20 x^2+e^{1+x} \left (80 x+85 x^2+5 x^3\right )} \left (-340-40 x+e^{1+x} \left (80+250 x+100 x^2+5 x^3\right )\right ) \, dx=e^{- 20 x^{2} - 340 x + \left (5 x^{3} + 85 x^{2} + 80 x\right ) e^{x + 1} - 320} \]
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Leaf count of result is larger than twice the leaf count of optimal. 36 vs. \(2 (17) = 34\).
Time = 0.32 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.89 \[ \int e^{-320-340 x-20 x^2+e^{1+x} \left (80 x+85 x^2+5 x^3\right )} \left (-340-40 x+e^{1+x} \left (80+250 x+100 x^2+5 x^3\right )\right ) \, dx=e^{\left (5 \, x^{3} e^{\left (x + 1\right )} + 85 \, x^{2} e^{\left (x + 1\right )} - 20 \, x^{2} + 80 \, x e^{\left (x + 1\right )} - 340 \, x - 320\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 36 vs. \(2 (17) = 34\).
Time = 0.30 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.89 \[ \int e^{-320-340 x-20 x^2+e^{1+x} \left (80 x+85 x^2+5 x^3\right )} \left (-340-40 x+e^{1+x} \left (80+250 x+100 x^2+5 x^3\right )\right ) \, dx=e^{\left (5 \, x^{3} e^{\left (x + 1\right )} + 85 \, x^{2} e^{\left (x + 1\right )} - 20 \, x^{2} + 80 \, x e^{\left (x + 1\right )} - 340 \, x - 320\right )} \]
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Time = 10.30 (sec) , antiderivative size = 41, normalized size of antiderivative = 2.16 \[ \int e^{-320-340 x-20 x^2+e^{1+x} \left (80 x+85 x^2+5 x^3\right )} \left (-340-40 x+e^{1+x} \left (80+250 x+100 x^2+5 x^3\right )\right ) \, dx={\mathrm {e}}^{-340\,x}\,{\mathrm {e}}^{-320}\,{\mathrm {e}}^{80\,x\,\mathrm {e}\,{\mathrm {e}}^x}\,{\mathrm {e}}^{-20\,x^2}\,{\mathrm {e}}^{5\,x^3\,\mathrm {e}\,{\mathrm {e}}^x}\,{\mathrm {e}}^{85\,x^2\,\mathrm {e}\,{\mathrm {e}}^x} \]
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