Optimal. Leaf size=24 \[ \frac {9 e^{5-x} (2+\log (4))}{(5+x)^2 \log \left (x^2\right )} \]
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Rubi [F] time = 1.86, 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^{-x} \left (e^5 (-180-36 x)+e^5 (-90-18 x) \log (4)+\left (e^5 \left (-126 x-18 x^2\right )+e^5 \left (-63 x-9 x^2\right ) \log (4)\right ) \log \left (x^2\right )\right )}{\left (125 x+75 x^2+15 x^3+x^4\right ) \log ^2\left (x^2\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {9 e^{5-x} (2+\log (4)) \left (-2 (5+x)-x (7+x) \log \left (x^2\right )\right )}{x (5+x)^3 \log ^2\left (x^2\right )} \, dx\\ &=(9 (2+\log (4))) \int \frac {e^{5-x} \left (-2 (5+x)-x (7+x) \log \left (x^2\right )\right )}{x (5+x)^3 \log ^2\left (x^2\right )} \, dx\\ &=(9 (2+\log (4))) \int \left (-\frac {2 e^{5-x}}{x (5+x)^2 \log ^2\left (x^2\right )}+\frac {e^{5-x} (-7-x)}{(5+x)^3 \log \left (x^2\right )}\right ) \, dx\\ &=(9 (2+\log (4))) \int \frac {e^{5-x} (-7-x)}{(5+x)^3 \log \left (x^2\right )} \, dx-(18 (2+\log (4))) \int \frac {e^{5-x}}{x (5+x)^2 \log ^2\left (x^2\right )} \, dx\\ &=(9 (2+\log (4))) \int \left (-\frac {2 e^{5-x}}{(5+x)^3 \log \left (x^2\right )}-\frac {e^{5-x}}{(5+x)^2 \log \left (x^2\right )}\right ) \, dx-(18 (2+\log (4))) \int \left (\frac {e^{5-x}}{25 x \log ^2\left (x^2\right )}-\frac {e^{5-x}}{5 (5+x)^2 \log ^2\left (x^2\right )}-\frac {e^{5-x}}{25 (5+x) \log ^2\left (x^2\right )}\right ) \, dx\\ &=-\left (\frac {1}{25} (18 (2+\log (4))) \int \frac {e^{5-x}}{x \log ^2\left (x^2\right )} \, dx\right )+\frac {1}{25} (18 (2+\log (4))) \int \frac {e^{5-x}}{(5+x) \log ^2\left (x^2\right )} \, dx+\frac {1}{5} (18 (2+\log (4))) \int \frac {e^{5-x}}{(5+x)^2 \log ^2\left (x^2\right )} \, dx-(9 (2+\log (4))) \int \frac {e^{5-x}}{(5+x)^2 \log \left (x^2\right )} \, dx-(18 (2+\log (4))) \int \frac {e^{5-x}}{(5+x)^3 \log \left (x^2\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.22, size = 24, normalized size = 1.00 \begin {gather*} \frac {9 e^{5-x} (2+\log (4))}{(5+x)^2 \log \left (x^2\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.87, size = 30, normalized size = 1.25 \begin {gather*} \frac {18 \, {\left (e^{5} \log \relax (2) + e^{5}\right )} e^{\left (-x\right )}}{{\left (x^{2} + 10 \, x + 25\right )} \log \left (x^{2}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.20, size = 42, normalized size = 1.75 \begin {gather*} \frac {18 \, {\left (e^{\left (-x + 5\right )} \log \relax (2) + e^{\left (-x + 5\right )}\right )}}{x^{2} \log \left (x^{2}\right ) + 10 \, x \log \left (x^{2}\right ) + 25 \, \log \left (x^{2}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.09, size = 76, normalized size = 3.17
method | result | size |
risch | \(\frac {36 i \left (1+\ln \relax (2)\right ) {\mathrm e}^{5-x}}{\left (4 i \ln \relax (x )+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}\right ) \left (x^{2}+10 x +25\right )}\) | \(76\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.51, size = 26, normalized size = 1.08 \begin {gather*} \frac {9 \, {\left (\log \relax (2) + 1\right )} e^{\left (-x + 5\right )}}{{\left (x^{2} + 10 \, x + 25\right )} \log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.49, size = 87, normalized size = 3.62 \begin {gather*} \frac {9\,\left (\ln \relax (2)+1\right )\,\left (7\,x\,{\mathrm {e}}^{5-x}+x^2\,{\mathrm {e}}^{5-x}-7\,x\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^5-x^2\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^5\right )}{{\left (x+5\right )}^3}+\frac {9\,\left (\ln \relax (2)+1\right )\,\left (10\,{\mathrm {e}}^{5-x}+2\,x\,{\mathrm {e}}^{5-x}\right )}{\ln \left (x^2\right )\,{\left (x+5\right )}^3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.39, size = 39, normalized size = 1.62 \begin {gather*} \frac {\left (18 e^{5} \log {\relax (2 )} + 18 e^{5}\right ) e^{- x}}{x^{2} \log {\left (x^{2} \right )} + 10 x \log {\left (x^{2} \right )} + 25 \log {\left (x^{2} \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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