Optimal. Leaf size=21 \[ 3+\frac {9 e^{-2 x} x^2 \log (3)}{16 (-5+x)^2} \]
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Rubi [A] time = 0.10, antiderivative size = 38, normalized size of antiderivative = 1.81, number of steps used = 3, number of rules used = 3, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {12, 1594, 2288} \begin {gather*} \frac {9 e^{-2 x} x \left (5 x-x^2\right ) \log (3)}{16 \left (-x^3+15 x^2-75 x+125\right )} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 1594
Rule 2288
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\log (3) \int \frac {e^{-2 x} \left (-45 x+45 x^2-9 x^3\right )}{-1000+600 x-120 x^2+8 x^3} \, dx\\ &=\log (3) \int \frac {e^{-2 x} x \left (-45+45 x-9 x^2\right )}{-1000+600 x-120 x^2+8 x^3} \, dx\\ &=\frac {9 e^{-2 x} x \left (5 x-x^2\right ) \log (3)}{16 \left (125-75 x+15 x^2-x^3\right )}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.01, size = 19, normalized size = 0.90 \begin {gather*} \frac {9 e^{-2 x} x^2 \log (3)}{16 (-5+x)^2} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.21, size = 21, normalized size = 1.00 \begin {gather*} \frac {9 \, x^{2} e^{\left (-2 \, x\right )} \log \relax (3)}{16 \, {\left (x^{2} - 10 \, x + 25\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.18, size = 21, normalized size = 1.00 \begin {gather*} \frac {9 \, x^{2} e^{\left (-2 \, x\right )} \log \relax (3)}{16 \, {\left (x^{2} - 10 \, x + 25\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.07, size = 17, normalized size = 0.81
method | result | size |
norman | \(\frac {9 x^{2} {\mathrm e}^{-2 x} \ln \relax (3)}{16 \left (x -5\right )^{2}}\) | \(17\) |
risch | \(\frac {9 x^{2} {\mathrm e}^{-2 x} \ln \relax (3)}{16 \left (x -5\right )^{2}}\) | \(17\) |
gosper | \(\frac {9 x^{2} \ln \relax (3) {\mathrm e}^{-2 x}}{16 \left (x^{2}-10 x +25\right )}\) | \(22\) |
default | \(\frac {9 \ln \relax (3) \left (-\frac {5 \,{\mathrm e}^{-2 x} \left (8 x -45\right )}{2 \left (x^{2}-10 x +25\right )}+\frac {25 \,{\mathrm e}^{-2 x} \left (6 x -35\right )}{2 \left (x^{2}-10 x +25\right )}+\frac {{\mathrm e}^{-2 x}}{2}-\frac {25 \,{\mathrm e}^{-2 x} \left (4 x -25\right )}{2 \left (x^{2}-10 x +25\right )}\right )}{8}\) | \(75\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.42, size = 21, normalized size = 1.00 \begin {gather*} \frac {9 \, x^{2} e^{\left (-2 \, x\right )} \log \relax (3)}{16 \, {\left (x^{2} - 10 \, x + 25\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.16, size = 16, normalized size = 0.76 \begin {gather*} \frac {9\,x^2\,{\mathrm {e}}^{-2\,x}\,\ln \relax (3)}{16\,{\left (x-5\right )}^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.16, size = 22, normalized size = 1.05 \begin {gather*} \frac {9 x^{2} e^{- 2 x} \log {\relax (3 )}}{16 x^{2} - 160 x + 400} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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