Optimal. Leaf size=23 \[ \frac {9}{2+x \left (5 e^4+e^x-5 \log (3)+\log (5)\right )} \]
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Rubi [A] time = 0.31, antiderivative size = 25, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 4, integrand size = 135, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.030, Rules used = {6, 6688, 12, 6686} \begin {gather*} \frac {9}{e^x x+x \left (5 e^4-\log \left (\frac {243}{5}\right )\right )+2} \end {gather*}
Antiderivative was successfully verified.
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Rule 6
Rule 12
Rule 6686
Rule 6688
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-45 e^4+e^x (-9-9 x)+45 \log (3)-9 \log (5)}{4+20 e^4 x+e^{2 x} x^2+\left (-20 x-50 e^4 x^2\right ) \log (3)+25 x^2 \left (e^8+\log ^2(3)\right )+\left (4 x+10 e^4 x^2-10 x^2 \log (3)\right ) \log (5)+x^2 \log ^2(5)+e^x \left (4 x+10 e^4 x^2-10 x^2 \log (3)+2 x^2 \log (5)\right )} \, dx\\ &=\int \frac {-45 e^4+e^x (-9-9 x)+45 \log (3)-9 \log (5)}{4+20 e^4 x+e^{2 x} x^2+\left (-20 x-50 e^4 x^2\right ) \log (3)+\left (4 x+10 e^4 x^2-10 x^2 \log (3)\right ) \log (5)+e^x \left (4 x+10 e^4 x^2-10 x^2 \log (3)+2 x^2 \log (5)\right )+x^2 \left (25 \left (e^8+\log ^2(3)\right )+\log ^2(5)\right )} \, dx\\ &=\int \frac {9 \left (-e^x (1+x)-5 e^4 \left (1-\frac {\log (3) \log \left (\frac {243}{5}\right )}{e^4 \log (243)}\right )\right )}{\left (2+e^x x+5 e^4 x \left (1-\frac {\log \left (\frac {243}{5}\right )}{5 e^4}\right )\right )^2} \, dx\\ &=9 \int \frac {-e^x (1+x)-5 e^4 \left (1-\frac {\log (3) \log \left (\frac {243}{5}\right )}{e^4 \log (243)}\right )}{\left (2+e^x x+5 e^4 x \left (1-\frac {\log \left (\frac {243}{5}\right )}{5 e^4}\right )\right )^2} \, dx\\ &=\frac {9}{2+e^x x+x \left (5 e^4-\log \left (\frac {243}{5}\right )\right )}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.05, size = 25, normalized size = 1.09 \begin {gather*} \frac {9}{2+e^x x+x \left (5 e^4-\log \left (\frac {243}{5}\right )\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.56, size = 24, normalized size = 1.04 \begin {gather*} \frac {9}{5 \, x e^{4} + x e^{x} + x \log \relax (5) - 5 \, x \log \relax (3) + 2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.36, size = 24, normalized size = 1.04 \begin {gather*} \frac {9}{5 \, x e^{4} + x e^{x} + x \log \relax (5) - 5 \, x \log \relax (3) + 2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.20, size = 27, normalized size = 1.17
method | result | size |
norman | \(-\frac {9}{5 x \ln \relax (3)-x \ln \relax (5)-5 x \,{\mathrm e}^{4}-{\mathrm e}^{x} x -2}\) | \(27\) |
risch | \(-\frac {9}{5 x \ln \relax (3)-x \ln \relax (5)-5 x \,{\mathrm e}^{4}-{\mathrm e}^{x} x -2}\) | \(27\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.50, size = 23, normalized size = 1.00 \begin {gather*} \frac {9}{x {\left (5 \, e^{4} + \log \relax (5) - 5 \, \log \relax (3)\right )} + x e^{x} + 2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.39, size = 45, normalized size = 1.96 \begin {gather*} -\frac {\frac {9\,x\,{\mathrm {e}}^x}{2}+x\,\left (\frac {45\,{\mathrm {e}}^4}{2}-\ln \left (\frac {31381059609\,\sqrt {15}}{3125}\right )\right )}{5\,x\,{\mathrm {e}}^4-5\,x\,\ln \relax (3)+x\,\ln \relax (5)+x\,{\mathrm {e}}^x+2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.21, size = 26, normalized size = 1.13 \begin {gather*} \frac {9}{x e^{x} - 5 x \log {\relax (3 )} + x \log {\relax (5 )} + 5 x e^{4} + 2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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