Optimal. Leaf size=30 \[ -e^x x+\frac {4 x \log \left (5 e^{-x} x^2\right )}{5 (16+2 x)} \]
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Rubi [A] time = 0.24, antiderivative size = 43, normalized size of antiderivative = 1.43, number of steps used = 13, number of rules used = 8, integrand size = 55, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {27, 12, 6742, 2176, 2194, 43, 2551, 72} \begin {gather*} -\frac {16 \log \left (5 e^{-x} x^2\right )}{5 (x+8)}-\frac {2 x}{5}+e^x-e^x (x+1)+\frac {4 \log (x)}{5} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 27
Rule 43
Rule 72
Rule 2176
Rule 2194
Rule 2551
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {32-12 x-2 x^2+e^x \left (-320-400 x-85 x^2-5 x^3\right )+16 \log \left (5 e^{-x} x^2\right )}{5 (8+x)^2} \, dx\\ &=\frac {1}{5} \int \frac {32-12 x-2 x^2+e^x \left (-320-400 x-85 x^2-5 x^3\right )+16 \log \left (5 e^{-x} x^2\right )}{(8+x)^2} \, dx\\ &=\frac {1}{5} \int \left (-5 e^x (1+x)+\frac {32}{(8+x)^2}-\frac {12 x}{(8+x)^2}-\frac {2 x^2}{(8+x)^2}+\frac {16 \log \left (5 e^{-x} x^2\right )}{(8+x)^2}\right ) \, dx\\ &=-\frac {32}{5 (8+x)}-\frac {2}{5} \int \frac {x^2}{(8+x)^2} \, dx-\frac {12}{5} \int \frac {x}{(8+x)^2} \, dx+\frac {16}{5} \int \frac {\log \left (5 e^{-x} x^2\right )}{(8+x)^2} \, dx-\int e^x (1+x) \, dx\\ &=-e^x (1+x)-\frac {32}{5 (8+x)}-\frac {16 \log \left (5 e^{-x} x^2\right )}{5 (8+x)}-\frac {2}{5} \int \left (1+\frac {64}{(8+x)^2}-\frac {16}{8+x}\right ) \, dx-\frac {12}{5} \int \left (-\frac {8}{(8+x)^2}+\frac {1}{8+x}\right ) \, dx+\frac {16}{5} \int \frac {2-x}{x (8+x)} \, dx+\int e^x \, dx\\ &=e^x-\frac {2 x}{5}-e^x (1+x)-\frac {16 \log \left (5 e^{-x} x^2\right )}{5 (8+x)}+4 \log (8+x)+\frac {16}{5} \int \left (\frac {1}{4 x}-\frac {5}{4 (8+x)}\right ) \, dx\\ &=e^x-\frac {2 x}{5}-e^x (1+x)+\frac {4 \log (x)}{5}-\frac {16 \log \left (5 e^{-x} x^2\right )}{5 (8+x)}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.14, size = 36, normalized size = 1.20 \begin {gather*} \frac {1}{5} \left (-2 x-5 e^x x+4 \log (x)-\frac {16 \log \left (5 e^{-x} x^2\right )}{8+x}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.71, size = 32, normalized size = 1.07 \begin {gather*} -\frac {5 \, {\left (x^{2} + 8 \, x\right )} e^{x} - 2 \, x \log \left (5 \, x^{2} e^{\left (-x\right )}\right )}{5 \, {\left (x + 8\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.24, size = 46, normalized size = 1.53 \begin {gather*} -\frac {5 \, x^{2} e^{x} + 2 \, x^{2} + 40 \, x e^{x} - 4 \, x \log \relax (x) + 16 \, x + 16 \, \log \left (5 \, x^{2}\right ) - 32 \, \log \relax (x) + 128}{5 \, {\left (x + 8\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.49, size = 31, normalized size = 1.03
method | result | size |
default | \(-{\mathrm e}^{x} x -\frac {2 x}{5}-\frac {16 \ln \left (5 x^{2} {\mathrm e}^{-x}\right )}{5 \left (x +8\right )}+\frac {4 \ln \relax (x )}{5}\) | \(31\) |
norman | \(\frac {-8 \,{\mathrm e}^{x} x -{\mathrm e}^{x} x^{2}+\frac {2 \ln \left (5 x^{2} {\mathrm e}^{-x}\right ) x}{5}}{x +8}\) | \(33\) |
risch | \(\frac {16 \ln \left ({\mathrm e}^{x}\right )}{5 \left (x +8\right )}-\frac {-8 i \pi \,\mathrm {csgn}\left (i x^{2} {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x}\right )-8 i \pi \mathrm {csgn}\left (i x^{2} {\mathrm e}^{-x}\right )^{3}+8 i \pi \mathrm {csgn}\left (i x^{2} {\mathrm e}^{-x}\right )^{2} \mathrm {csgn}\left (i x^{2}\right )+16 i \pi \mathrm {csgn}\left (i x^{2}\right )^{2} \mathrm {csgn}\left (i x \right )-8 i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+8 i \pi \mathrm {csgn}\left (i x^{2} {\mathrm e}^{-x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{-x}\right )-8 i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )^{2}+5 \,{\mathrm e}^{x} x^{2}-4 x \ln \relax (x )+2 x^{2}+40 \,{\mathrm e}^{x} x +16 \ln \relax (5)+16 x}{5 \left (x +8\right )}\) | \(194\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {2}{5} \, x + \frac {64 \, e^{\left (-8\right )} E_{2}\left (-x - 8\right )}{x + 8} + \frac {4 \, {\left (x \log \relax (x) - 4 \, \log \relax (5) - 32\right )}}{5 \, {\left (x + 8\right )}} - \frac {1}{5} \, \int \frac {5 \, {\left (x^{3} + 17 \, x^{2} + 80 \, x\right )} e^{x}}{x^{2} + 16 \, x + 64}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.45, size = 47, normalized size = 1.57 \begin {gather*} \frac {4\,\ln \relax (x)}{5}-\frac {2\,x}{5}-\frac {16\,\ln \relax (5)}{5\,\left (x+8\right )}+\frac {16\,x}{5\,\left (x+8\right )}-x\,{\mathrm {e}}^x-\frac {16\,\ln \left (x^2\right )}{5\,\left (x+8\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.26, size = 32, normalized size = 1.07 \begin {gather*} - x e^{x} - \frac {2 x}{5} + \frac {4 \log {\relax (x )}}{5} - \frac {16 \log {\left (5 x^{2} e^{- x} \right )}}{5 x + 40} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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