Optimal. Leaf size=36 \[ \frac {e^{-\frac {1}{4} e^x (-4+x)^2} \left (-e^x+\frac {4}{5 x}+x\right )^2}{-1+x} \]
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Rubi [F] time = 18.85, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {e^{-\frac {1}{4} e^x \left (16-8 x+x^2\right )} \left (128-192 x-160 x^3-200 x^4+100 x^5+e^{3 x} \left (200 x^3-350 x^4+175 x^5-25 x^6\right )+e^{2 x} \left (-320 x^2+260 x^3-480 x^4+740 x^5-350 x^6+50 x^7\right )+e^x \left (-32 x+256 x^2+472 x^3-376 x^4+280 x^5-390 x^6+175 x^7-25 x^8\right )\right )}{100 x^3-200 x^4+100 x^5} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-\frac {1}{4} e^x \left (16-8 x+x^2\right )} \left (128-192 x-160 x^3-200 x^4+100 x^5+e^{3 x} \left (200 x^3-350 x^4+175 x^5-25 x^6\right )+e^{2 x} \left (-320 x^2+260 x^3-480 x^4+740 x^5-350 x^6+50 x^7\right )+e^x \left (-32 x+256 x^2+472 x^3-376 x^4+280 x^5-390 x^6+175 x^7-25 x^8\right )\right )}{x^3 \left (100-200 x+100 x^2\right )} \, dx\\ &=\int \frac {e^{-\frac {1}{4} e^x \left (16-8 x+x^2\right )} \left (128-192 x-160 x^3-200 x^4+100 x^5+e^{3 x} \left (200 x^3-350 x^4+175 x^5-25 x^6\right )+e^{2 x} \left (-320 x^2+260 x^3-480 x^4+740 x^5-350 x^6+50 x^7\right )+e^x \left (-32 x+256 x^2+472 x^3-376 x^4+280 x^5-390 x^6+175 x^7-25 x^8\right )\right )}{100 (-1+x)^2 x^3} \, dx\\ &=\frac {1}{100} \int \frac {e^{-\frac {1}{4} e^x \left (16-8 x+x^2\right )} \left (128-192 x-160 x^3-200 x^4+100 x^5+e^{3 x} \left (200 x^3-350 x^4+175 x^5-25 x^6\right )+e^{2 x} \left (-320 x^2+260 x^3-480 x^4+740 x^5-350 x^6+50 x^7\right )+e^x \left (-32 x+256 x^2+472 x^3-376 x^4+280 x^5-390 x^6+175 x^7-25 x^8\right )\right )}{(-1+x)^2 x^3} \, dx\\ &=\frac {1}{100} \int \left (-\frac {160 e^{-\frac {1}{4} e^x \left (16-8 x+x^2\right )}}{(-1+x)^2}-\frac {25 e^{3 x-\frac {1}{4} e^x \left (16-8 x+x^2\right )} (-4+x) (-2+x)}{-1+x}+\frac {128 e^{-\frac {1}{4} e^x \left (16-8 x+x^2\right )}}{(-1+x)^2 x^3}-\frac {192 e^{-\frac {1}{4} e^x \left (16-8 x+x^2\right )}}{(-1+x)^2 x^2}-\frac {200 e^{-\frac {1}{4} e^x \left (16-8 x+x^2\right )} x}{(-1+x)^2}+\frac {100 e^{-\frac {1}{4} e^x \left (16-8 x+x^2\right )} x^2}{(-1+x)^2}+\frac {10 e^{2 x-\frac {1}{4} e^x \left (16-8 x+x^2\right )} \left (-32+26 x-48 x^2+74 x^3-35 x^4+5 x^5\right )}{(-1+x)^2 x}-\frac {e^{x-\frac {1}{4} e^x \left (16-8 x+x^2\right )} \left (32-256 x-472 x^2+376 x^3-280 x^4+390 x^5-175 x^6+25 x^7\right )}{(-1+x)^2 x^2}\right ) \, dx\\ &=-\left (\frac {1}{100} \int \frac {e^{x-\frac {1}{4} e^x \left (16-8 x+x^2\right )} \left (32-256 x-472 x^2+376 x^3-280 x^4+390 x^5-175 x^6+25 x^7\right )}{(-1+x)^2 x^2} \, dx\right )+\frac {1}{10} \int \frac {e^{2 x-\frac {1}{4} e^x \left (16-8 x+x^2\right )} \left (-32+26 x-48 x^2+74 x^3-35 x^4+5 x^5\right )}{(-1+x)^2 x} \, dx-\frac {1}{4} \int \frac {e^{3 x-\frac {1}{4} e^x \left (16-8 x+x^2\right )} (-4+x) (-2+x)}{-1+x} \, dx+\frac {32}{25} \int \frac {e^{-\frac {1}{4} e^x \left (16-8 x+x^2\right )}}{(-1+x)^2 x^3} \, dx-\frac {8}{5} \int \frac {e^{-\frac {1}{4} e^x \left (16-8 x+x^2\right )}}{(-1+x)^2} \, dx-\frac {48}{25} \int \frac {e^{-\frac {1}{4} e^x \left (16-8 x+x^2\right )}}{(-1+x)^2 x^2} \, dx-2 \int \frac {e^{-\frac {1}{4} e^x \left (16-8 x+x^2\right )} x}{(-1+x)^2} \, dx+\int \frac {e^{-\frac {1}{4} e^x \left (16-8 x+x^2\right )} x^2}{(-1+x)^2} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [A] time = 0.13, size = 41, normalized size = 1.14 \begin {gather*} \frac {e^{-\frac {1}{4} e^x (-4+x)^2} \left (4-5 e^x x+5 x^2\right )^2}{25 (-1+x) x^2} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.86, size = 60, normalized size = 1.67 \begin {gather*} \frac {{\left (25 \, x^{4} + 25 \, x^{2} e^{\left (2 \, x\right )} + 40 \, x^{2} - 10 \, {\left (5 \, x^{3} + 4 \, x\right )} e^{x} + 16\right )} e^{\left (-\frac {1}{4} \, {\left (x^{2} - 8 \, x + 16\right )} e^{x}\right )}}{25 \, {\left (x^{3} - x^{2}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (100 \, x^{5} - 200 \, x^{4} - 160 \, x^{3} - 25 \, {\left (x^{6} - 7 \, x^{5} + 14 \, x^{4} - 8 \, x^{3}\right )} e^{\left (3 \, x\right )} + 10 \, {\left (5 \, x^{7} - 35 \, x^{6} + 74 \, x^{5} - 48 \, x^{4} + 26 \, x^{3} - 32 \, x^{2}\right )} e^{\left (2 \, x\right )} - {\left (25 \, x^{8} - 175 \, x^{7} + 390 \, x^{6} - 280 \, x^{5} + 376 \, x^{4} - 472 \, x^{3} - 256 \, x^{2} + 32 \, x\right )} e^{x} - 192 \, x + 128\right )} e^{\left (-\frac {1}{4} \, {\left (x^{2} - 8 \, x + 16\right )} e^{x}\right )}}{100 \, {\left (x^{5} - 2 \, x^{4} + x^{3}\right )}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 54, normalized size = 1.50
method | result | size |
risch | \(\frac {\left (25 x^{4}-50 \,{\mathrm e}^{x} x^{3}+25 \,{\mathrm e}^{2 x} x^{2}+40 x^{2}-40 \,{\mathrm e}^{x} x +16\right ) {\mathrm e}^{-\frac {\left (x -4\right )^{2} {\mathrm e}^{x}}{4}}}{25 x^{2} \left (x -1\right )}\) | \(54\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.86, size = 65, normalized size = 1.81 \begin {gather*} \frac {{\left (25 \, x^{4} + 25 \, x^{2} e^{\left (2 \, x\right )} + 40 \, x^{2} - 10 \, {\left (5 \, x^{3} + 4 \, x\right )} e^{x} + 16\right )} e^{\left (-\frac {1}{4} \, x^{2} e^{x} + 2 \, x e^{x} - 4 \, e^{x}\right )}}{25 \, {\left (x^{3} - x^{2}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.77, size = 61, normalized size = 1.69 \begin {gather*} -\frac {{\mathrm {e}}^{2\,x\,{\mathrm {e}}^x-\frac {x^2\,{\mathrm {e}}^x}{4}-4\,{\mathrm {e}}^x}\,\left (x^2\,{\mathrm {e}}^{2\,x}-2\,x^3\,{\mathrm {e}}^x-\frac {8\,x\,{\mathrm {e}}^x}{5}+\frac {8\,x^2}{5}+x^4+\frac {16}{25}\right )}{x^2-x^3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.51, size = 61, normalized size = 1.69 \begin {gather*} \frac {\left (25 x^{4} - 50 x^{3} e^{x} + 25 x^{2} e^{2 x} + 40 x^{2} - 40 x e^{x} + 16\right ) e^{- \left (\frac {x^{2}}{4} - 2 x + 4\right ) e^{x}}}{25 x^{3} - 25 x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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