Optimal. Leaf size=25 \[ \frac {e^{8-x} \left (\frac {e^4}{x}+x+\log (2)\right )}{5 x^2} \]
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Rubi [C] Result contains higher order function than in optimal. Order 4 vs. order 3 in
optimal.
time = 0.28, antiderivative size = 155, normalized size of antiderivative = 6.20, number of steps
used = 13, number of rules used = 4, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {12, 2230,
2208, 2209} \begin {gather*} \frac {1}{10} e^{12} \text {ExpIntegralEi}(-x)-\frac {1}{5} e^8 \text {ExpIntegralEi}(-x)-\frac {1}{10} e^8 \left (e^4+\log (4)\right ) \text {ExpIntegralEi}(-x)+\frac {1}{5} e^8 (1+\log (2)) \text {ExpIntegralEi}(-x)+\frac {e^{12-x}}{5 x^3}-\frac {e^{12-x}}{10 x^2}+\frac {e^{8-x} \left (e^4+\log (4)\right )}{10 x^2}+\frac {e^{12-x}}{10 x}-\frac {e^{8-x} \left (e^4+\log (4)\right )}{10 x}+\frac {e^{8-x} (1+\log (2))}{5 x} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 2208
Rule 2209
Rule 2230
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {e^{8-x} \left (e^4 (-3-x)-x^2-x^3+\left (-2 x-x^2\right ) \log (2)\right )}{x^4} \, dx\\ &=\frac {1}{5} \int \left (-\frac {3 e^{12-x}}{x^4}-\frac {e^{8-x}}{x}+\frac {e^{8-x} (-1-\log (2))}{x^2}+\frac {e^{8-x} \left (-e^4-\log (4)\right )}{x^3}\right ) \, dx\\ &=-\left (\frac {1}{5} \int \frac {e^{8-x}}{x} \, dx\right )-\frac {3}{5} \int \frac {e^{12-x}}{x^4} \, dx+\frac {1}{5} (-1-\log (2)) \int \frac {e^{8-x}}{x^2} \, dx+\frac {1}{5} \left (-e^4-\log (4)\right ) \int \frac {e^{8-x}}{x^3} \, dx\\ &=\frac {e^{12-x}}{5 x^3}-\frac {e^8 \text {Ei}(-x)}{5}+\frac {e^{8-x} (1+\log (2))}{5 x}+\frac {e^{8-x} \left (e^4+\log (4)\right )}{10 x^2}+\frac {1}{5} \int \frac {e^{12-x}}{x^3} \, dx+\frac {1}{5} (1+\log (2)) \int \frac {e^{8-x}}{x} \, dx+\frac {1}{10} \left (e^4+\log (4)\right ) \int \frac {e^{8-x}}{x^2} \, dx\\ &=\frac {e^{12-x}}{5 x^3}-\frac {e^{12-x}}{10 x^2}-\frac {e^8 \text {Ei}(-x)}{5}+\frac {e^{8-x} (1+\log (2))}{5 x}+\frac {1}{5} e^8 \text {Ei}(-x) (1+\log (2))+\frac {e^{8-x} \left (e^4+\log (4)\right )}{10 x^2}-\frac {e^{8-x} \left (e^4+\log (4)\right )}{10 x}-\frac {1}{10} \int \frac {e^{12-x}}{x^2} \, dx+\frac {1}{10} \left (-e^4-\log (4)\right ) \int \frac {e^{8-x}}{x} \, dx\\ &=\frac {e^{12-x}}{5 x^3}-\frac {e^{12-x}}{10 x^2}+\frac {e^{12-x}}{10 x}-\frac {e^8 \text {Ei}(-x)}{5}+\frac {e^{8-x} (1+\log (2))}{5 x}+\frac {1}{5} e^8 \text {Ei}(-x) (1+\log (2))+\frac {e^{8-x} \left (e^4+\log (4)\right )}{10 x^2}-\frac {e^{8-x} \left (e^4+\log (4)\right )}{10 x}-\frac {1}{10} e^8 \text {Ei}(-x) \left (e^4+\log (4)\right )+\frac {1}{10} \int \frac {e^{12-x}}{x} \, dx\\ &=\frac {e^{12-x}}{5 x^3}-\frac {e^{12-x}}{10 x^2}+\frac {e^{12-x}}{10 x}-\frac {e^8 \text {Ei}(-x)}{5}+\frac {e^{12} \text {Ei}(-x)}{10}+\frac {e^{8-x} (1+\log (2))}{5 x}+\frac {1}{5} e^8 \text {Ei}(-x) (1+\log (2))+\frac {e^{8-x} \left (e^4+\log (4)\right )}{10 x^2}-\frac {e^{8-x} \left (e^4+\log (4)\right )}{10 x}-\frac {1}{10} e^8 \text {Ei}(-x) \left (e^4+\log (4)\right )\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.05, size = 28, normalized size = 1.12 \begin {gather*} \frac {e^{8-x} \left (2 e^4+x (2 x+\log (4))\right )}{10 x^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order
3.
time = 1.37, size = 116, normalized size = 4.64
method | result | size |
gosper | \(\frac {{\mathrm e}^{8-x} \left (x \ln \left (2\right )+x^{2}+{\mathrm e}^{4}\right )}{5 x^{3}}\) | \(22\) |
risch | \(\frac {{\mathrm e}^{8-x} \left (x \ln \left (2\right )+x^{2}+{\mathrm e}^{4}\right )}{5 x^{3}}\) | \(22\) |
norman | \(\frac {\frac {x^{2} {\mathrm e}^{8-x}}{5}+\frac {{\mathrm e}^{4} {\mathrm e}^{8-x}}{5}+\frac {x \ln \left (2\right ) {\mathrm e}^{8-x}}{5}}{x^{3}}\) | \(38\) |
derivativedivides | \(\frac {{\mathrm e}^{8-x}}{5 x}-\frac {{\mathrm e}^{4} \left (-\frac {11 \,{\mathrm e}^{8-x}}{6 x^{2}}+\frac {11 \,{\mathrm e}^{8-x}}{6 x}-\frac {11 \,{\mathrm e}^{8} \expIntegral \left (1, x\right )}{6}+\frac {8 \,{\mathrm e}^{8-x}}{3 x^{3}}\right )}{5}+\frac {11 \,{\mathrm e}^{4} \left (\frac {{\mathrm e}^{8-x}}{3 x^{3}}-\frac {{\mathrm e}^{8-x}}{6 x^{2}}+\frac {{\mathrm e}^{8-x}}{6 x}-\frac {{\mathrm e}^{8} \expIntegral \left (1, x\right )}{6}\right )}{5}+\frac {\ln \left (2\right ) {\mathrm e}^{8-x}}{5 x^{2}}\) | \(116\) |
default | \(\frac {{\mathrm e}^{8-x}}{5 x}-\frac {{\mathrm e}^{4} \left (-\frac {11 \,{\mathrm e}^{8-x}}{6 x^{2}}+\frac {11 \,{\mathrm e}^{8-x}}{6 x}-\frac {11 \,{\mathrm e}^{8} \expIntegral \left (1, x\right )}{6}+\frac {8 \,{\mathrm e}^{8-x}}{3 x^{3}}\right )}{5}+\frac {11 \,{\mathrm e}^{4} \left (\frac {{\mathrm e}^{8-x}}{3 x^{3}}-\frac {{\mathrm e}^{8-x}}{6 x^{2}}+\frac {{\mathrm e}^{8-x}}{6 x}-\frac {{\mathrm e}^{8} \expIntegral \left (1, x\right )}{6}\right )}{5}+\frac {\ln \left (2\right ) {\mathrm e}^{8-x}}{5 x^{2}}\) | \(116\) |
meijerg | \(\left (-\frac {\ln \left (2\right )}{5}-\frac {1}{5}\right ) {\mathrm e}^{8} \left (\frac {-2 x +2}{2 x}-\frac {{\mathrm e}^{-x}}{x}+\expIntegral \left (1, x\right )+1-\frac {1}{x}\right )+\left (-\frac {{\mathrm e}^{4}}{5}-\frac {2 \ln \left (2\right )}{5}\right ) {\mathrm e}^{8} \left (\frac {9 x^{2}-12 x +6}{12 x^{2}}-\frac {\left (-3 x +3\right ) {\mathrm e}^{-x}}{6 x^{2}}-\frac {\expIntegral \left (1, x\right )}{2}-\frac {3}{4}-\frac {1}{2 x^{2}}+\frac {1}{x}\right )+\frac {{\mathrm e}^{8} \expIntegral \left (1, x\right )}{5}-\frac {3 \,{\mathrm e}^{12} \left (\frac {-22 x^{3}+36 x^{2}-36 x +24}{72 x^{3}}-\frac {\left (4 x^{2}-4 x +8\right ) {\mathrm e}^{-x}}{24 x^{3}}+\frac {\expIntegral \left (1, x\right )}{6}+\frac {11}{36}-\frac {1}{3 x^{3}}+\frac {1}{2 x^{2}}-\frac {1}{2 x}\right )}{5}\) | \(168\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [C] Result contains higher order function than in optimal. Order 4 vs. order
3.
time = 0.31, size = 48, normalized size = 1.92 \begin {gather*} \frac {1}{5} \, e^{8} \Gamma \left (-1, x\right ) \log \left (2\right ) + \frac {2}{5} \, e^{8} \Gamma \left (-2, x\right ) \log \left (2\right ) - \frac {1}{5} \, {\rm Ei}\left (-x\right ) e^{8} + \frac {1}{5} \, e^{8} \Gamma \left (-1, x\right ) + \frac {1}{5} \, e^{12} \Gamma \left (-2, x\right ) + \frac {3}{5} \, e^{12} \Gamma \left (-3, x\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.44, size = 21, normalized size = 0.84 \begin {gather*} \frac {{\left (x^{2} + x \log \left (2\right ) + e^{4}\right )} e^{\left (-x + 8\right )}}{5 \, x^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.06, size = 20, normalized size = 0.80 \begin {gather*} \frac {\left (x^{2} + x \log {\left (2 \right )} + e^{4}\right ) e^{8 - x}}{5 x^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 81 vs.
\(2 (21) = 42\).
time = 0.39, size = 81, normalized size = 3.24 \begin {gather*} \frac {{\left (x - 8\right )}^{2} e^{\left (-x + 8\right )} + {\left (x - 8\right )} e^{\left (-x + 8\right )} \log \left (2\right ) + 16 \, {\left (x - 8\right )} e^{\left (-x + 8\right )} + 8 \, e^{\left (-x + 8\right )} \log \left (2\right ) + e^{\left (-x + 12\right )} + 64 \, e^{\left (-x + 8\right )}}{5 \, {\left ({\left (x - 8\right )}^{3} + 24 \, {\left (x - 8\right )}^{2} + 192 \, x - 1024\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 2.05, size = 36, normalized size = 1.44 \begin {gather*} \frac {{\mathrm {e}}^{-x}\,{\mathrm {e}}^8}{5\,x}+\frac {{\mathrm {e}}^{-x}\,{\mathrm {e}}^{12}}{5\,x^3}+\frac {{\mathrm {e}}^{-x}\,{\mathrm {e}}^8\,\ln \left (2\right )}{5\,x^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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