Optimal. Leaf size=31 \[ e^{\frac {16 \left (e^{\frac {5}{5-x}}-x\right )^2}{25 (-3+x) x^7}} \]
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Rubi [F] time = 38.52, 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 {\exp \left (\frac {16 e^{-\frac {10}{-5+x}}-32 e^{-\frac {5}{-5+x}} x+16 x^2}{-75 x^7+25 x^8}\right ) \left (6000 x^2-4800 x^3+1200 x^4-96 x^5+e^{-\frac {10}{-5+x}} \left (8400-7040 x+1776 x^2-128 x^3\right )+e^{-\frac {5}{-5+x}} \left (-14400 x+11840 x^2-2976 x^3+224 x^4\right )\right )}{5625 x^8-6000 x^9+2350 x^{10}-400 x^{11}+25 x^{12}} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (\frac {16 e^{-\frac {10}{-5+x}} \left (-1+e^{\frac {5}{-5+x}} x\right )^2}{25 (-3+x) x^7}\right ) \left (6000 x^2-4800 x^3+1200 x^4-96 x^5+e^{-\frac {10}{-5+x}} \left (8400-7040 x+1776 x^2-128 x^3\right )+e^{-\frac {5}{-5+x}} \left (-14400 x+11840 x^2-2976 x^3+224 x^4\right )\right )}{25 x^8 \left (15-8 x+x^2\right )^2} \, dx\\ &=\frac {1}{25} \int \frac {\exp \left (\frac {16 e^{-\frac {10}{-5+x}} \left (-1+e^{\frac {5}{-5+x}} x\right )^2}{25 (-3+x) x^7}\right ) \left (6000 x^2-4800 x^3+1200 x^4-96 x^5+e^{-\frac {10}{-5+x}} \left (8400-7040 x+1776 x^2-128 x^3\right )+e^{-\frac {5}{-5+x}} \left (-14400 x+11840 x^2-2976 x^3+224 x^4\right )\right )}{x^8 \left (15-8 x+x^2\right )^2} \, dx\\ &=\frac {1}{25} \int \left (\frac {6000 \exp \left (\frac {16 e^{-\frac {10}{-5+x}} \left (-1+e^{\frac {5}{-5+x}} x\right )^2}{25 (-3+x) x^7}\right )}{(-5+x)^2 (-3+x)^2 x^6}-\frac {4800 \exp \left (\frac {16 e^{-\frac {10}{-5+x}} \left (-1+e^{\frac {5}{-5+x}} x\right )^2}{25 (-3+x) x^7}\right )}{(-5+x)^2 (-3+x)^2 x^5}+\frac {1200 \exp \left (\frac {16 e^{-\frac {10}{-5+x}} \left (-1+e^{\frac {5}{-5+x}} x\right )^2}{25 (-3+x) x^7}\right )}{(-5+x)^2 (-3+x)^2 x^4}-\frac {96 \exp \left (\frac {16 e^{-\frac {10}{-5+x}} \left (-1+e^{\frac {5}{-5+x}} x\right )^2}{25 (-3+x) x^7}\right )}{(-5+x)^2 (-3+x)^2 x^3}+\frac {32 \exp \left (-\frac {5}{-5+x}+\frac {16 e^{-\frac {10}{-5+x}} \left (-1+e^{\frac {5}{-5+x}} x\right )^2}{25 (-3+x) x^7}\right ) \left (-450+370 x-93 x^2+7 x^3\right )}{(-5+x)^2 (-3+x)^2 x^7}-\frac {16 \exp \left (-\frac {10}{-5+x}+\frac {16 e^{-\frac {10}{-5+x}} \left (-1+e^{\frac {5}{-5+x}} x\right )^2}{25 (-3+x) x^7}\right ) \left (-525+440 x-111 x^2+8 x^3\right )}{(-5+x)^2 (-3+x)^2 x^8}\right ) \, dx\\ &=-\left (\frac {16}{25} \int \frac {\exp \left (-\frac {10}{-5+x}+\frac {16 e^{-\frac {10}{-5+x}} \left (-1+e^{\frac {5}{-5+x}} x\right )^2}{25 (-3+x) x^7}\right ) \left (-525+440 x-111 x^2+8 x^3\right )}{(-5+x)^2 (-3+x)^2 x^8} \, dx\right )+\frac {32}{25} \int \frac {\exp \left (-\frac {5}{-5+x}+\frac {16 e^{-\frac {10}{-5+x}} \left (-1+e^{\frac {5}{-5+x}} x\right )^2}{25 (-3+x) x^7}\right ) \left (-450+370 x-93 x^2+7 x^3\right )}{(-5+x)^2 (-3+x)^2 x^7} \, dx-\frac {96}{25} \int \frac {\exp \left (\frac {16 e^{-\frac {10}{-5+x}} \left (-1+e^{\frac {5}{-5+x}} x\right )^2}{25 (-3+x) x^7}\right )}{(-5+x)^2 (-3+x)^2 x^3} \, dx+48 \int \frac {\exp \left (\frac {16 e^{-\frac {10}{-5+x}} \left (-1+e^{\frac {5}{-5+x}} x\right )^2}{25 (-3+x) x^7}\right )}{(-5+x)^2 (-3+x)^2 x^4} \, dx-192 \int \frac {\exp \left (\frac {16 e^{-\frac {10}{-5+x}} \left (-1+e^{\frac {5}{-5+x}} x\right )^2}{25 (-3+x) x^7}\right )}{(-5+x)^2 (-3+x)^2 x^5} \, dx+240 \int \frac {\exp \left (\frac {16 e^{-\frac {10}{-5+x}} \left (-1+e^{\frac {5}{-5+x}} x\right )^2}{25 (-3+x) x^7}\right )}{(-5+x)^2 (-3+x)^2 x^6} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [A] time = 0.36, size = 29, normalized size = 0.94 \begin {gather*} e^{\frac {16 \left (e^{-\frac {5}{-5+x}}-x\right )^2}{25 (-3+x) x^7}} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.82, size = 37, normalized size = 1.19 \begin {gather*} e^{\left (\frac {16 \, {\left (x^{2} - 2 \, x e^{\left (-\frac {5}{x - 5}\right )} + e^{\left (-\frac {10}{x - 5}\right )}\right )}}{25 \, {\left (x^{8} - 3 \, x^{7}\right )}}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 1.18, size = 61, normalized size = 1.97 \begin {gather*} e^{\left (\frac {16 \, x^{2}}{25 \, {\left (x^{8} - 3 \, x^{7}\right )}} - \frac {32 \, x e^{\left (-\frac {5}{x - 5}\right )}}{25 \, {\left (x^{8} - 3 \, x^{7}\right )}} + \frac {16 \, e^{\left (-\frac {10}{x - 5}\right )}}{25 \, {\left (x^{8} - 3 \, x^{7}\right )}}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.12, size = 35, normalized size = 1.13
method | result | size |
risch | \({\mathrm e}^{\frac {-\frac {32 x \,{\mathrm e}^{-\frac {5}{x -5}}}{25}+\frac {16 x^{2}}{25}+\frac {16 \,{\mathrm e}^{-\frac {10}{x -5}}}{25}}{x^{7} \left (x -3\right )}}\) | \(35\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 1.16, size = 233, normalized size = 7.52 \begin {gather*} e^{\left (-\frac {32 \, e^{\left (-\frac {5}{x - 5}\right )}}{18225 \, {\left (x - 3\right )}} + \frac {32 \, e^{\left (-\frac {5}{x - 5}\right )}}{18225 \, x} + \frac {16 \, e^{\left (-\frac {10}{x - 5}\right )}}{54675 \, {\left (x - 3\right )}} - \frac {16 \, e^{\left (-\frac {10}{x - 5}\right )}}{54675 \, x} + \frac {16}{6075 \, {\left (x - 3\right )}} - \frac {16}{6075 \, x} + \frac {32 \, e^{\left (-\frac {5}{x - 5}\right )}}{6075 \, x^{2}} - \frac {16 \, e^{\left (-\frac {10}{x - 5}\right )}}{18225 \, x^{2}} - \frac {16}{2025 \, x^{2}} + \frac {32 \, e^{\left (-\frac {5}{x - 5}\right )}}{2025 \, x^{3}} - \frac {16 \, e^{\left (-\frac {10}{x - 5}\right )}}{6075 \, x^{3}} - \frac {16}{675 \, x^{3}} + \frac {32 \, e^{\left (-\frac {5}{x - 5}\right )}}{675 \, x^{4}} - \frac {16 \, e^{\left (-\frac {10}{x - 5}\right )}}{2025 \, x^{4}} - \frac {16}{225 \, x^{4}} + \frac {32 \, e^{\left (-\frac {5}{x - 5}\right )}}{225 \, x^{5}} - \frac {16 \, e^{\left (-\frac {10}{x - 5}\right )}}{675 \, x^{5}} - \frac {16}{75 \, x^{5}} + \frac {32 \, e^{\left (-\frac {5}{x - 5}\right )}}{75 \, x^{6}} - \frac {16 \, e^{\left (-\frac {10}{x - 5}\right )}}{225 \, x^{6}} - \frac {16 \, e^{\left (-\frac {10}{x - 5}\right )}}{75 \, x^{7}}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 6.03, size = 69, normalized size = 2.23 \begin {gather*} {\mathrm {e}}^{\frac {32\,x\,{\mathrm {e}}^{-\frac {5}{x-5}}}{75\,x^7-25\,x^8}}\,{\mathrm {e}}^{-\frac {16\,x^2}{75\,x^7-25\,x^8}}\,{\mathrm {e}}^{-\frac {16\,{\mathrm {e}}^{-\frac {10}{x-5}}}{75\,x^7-25\,x^8}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.93, size = 34, normalized size = 1.10 \begin {gather*} e^{\frac {16 x^{2} - 32 x e^{- \frac {5}{x - 5}} + 16 e^{- \frac {10}{x - 5}}}{25 x^{8} - 75 x^{7}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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