Optimal. Leaf size=87 \[ \frac {1}{250} e^{2 x} \cos (4 x)+\frac {2}{25} e^{2 x} x \cos (4 x)-\frac {1}{5} e^{2 x} x^2 \cos (4 x)-\frac {11}{500} e^{2 x} \sin (4 x)+\frac {3}{50} e^{2 x} x \sin (4 x)+\frac {1}{10} e^{2 x} x^2 \sin (4 x) \]
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Rubi [A]
time = 0.11, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps
used = 11, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {4517, 4553,
14, 4518, 4554} \begin {gather*} \frac {1}{10} e^{2 x} x^2 \sin (4 x)-\frac {1}{5} e^{2 x} x^2 \cos (4 x)+\frac {3}{50} e^{2 x} x \sin (4 x)-\frac {11}{500} e^{2 x} \sin (4 x)+\frac {2}{25} e^{2 x} x \cos (4 x)+\frac {1}{250} e^{2 x} \cos (4 x) \end {gather*}
Antiderivative was successfully verified.
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Rule 14
Rule 4517
Rule 4518
Rule 4553
Rule 4554
Rubi steps
\begin {align*} \int e^{2 x} x^2 \sin (4 x) \, dx &=-\frac {1}{5} e^{2 x} x^2 \cos (4 x)+\frac {1}{10} e^{2 x} x^2 \sin (4 x)-2 \int x \left (-\frac {1}{5} e^{2 x} \cos (4 x)+\frac {1}{10} e^{2 x} \sin (4 x)\right ) \, dx\\ &=-\frac {1}{5} e^{2 x} x^2 \cos (4 x)+\frac {1}{10} e^{2 x} x^2 \sin (4 x)-2 \int \left (-\frac {1}{5} e^{2 x} x \cos (4 x)+\frac {1}{10} e^{2 x} x \sin (4 x)\right ) \, dx\\ &=-\frac {1}{5} e^{2 x} x^2 \cos (4 x)+\frac {1}{10} e^{2 x} x^2 \sin (4 x)-\frac {1}{5} \int e^{2 x} x \sin (4 x) \, dx+\frac {2}{5} \int e^{2 x} x \cos (4 x) \, dx\\ &=\frac {2}{25} e^{2 x} x \cos (4 x)-\frac {1}{5} e^{2 x} x^2 \cos (4 x)+\frac {3}{50} e^{2 x} x \sin (4 x)+\frac {1}{10} e^{2 x} x^2 \sin (4 x)+\frac {1}{5} \int \left (-\frac {1}{5} e^{2 x} \cos (4 x)+\frac {1}{10} e^{2 x} \sin (4 x)\right ) \, dx-\frac {2}{5} \int \left (\frac {1}{10} e^{2 x} \cos (4 x)+\frac {1}{5} e^{2 x} \sin (4 x)\right ) \, dx\\ &=\frac {2}{25} e^{2 x} x \cos (4 x)-\frac {1}{5} e^{2 x} x^2 \cos (4 x)+\frac {3}{50} e^{2 x} x \sin (4 x)+\frac {1}{10} e^{2 x} x^2 \sin (4 x)+\frac {1}{50} \int e^{2 x} \sin (4 x) \, dx-2 \left (\frac {1}{25} \int e^{2 x} \cos (4 x) \, dx\right )-\frac {2}{25} \int e^{2 x} \sin (4 x) \, dx\\ &=\frac {3}{250} e^{2 x} \cos (4 x)+\frac {2}{25} e^{2 x} x \cos (4 x)-\frac {1}{5} e^{2 x} x^2 \cos (4 x)-\frac {3}{500} e^{2 x} \sin (4 x)+\frac {3}{50} e^{2 x} x \sin (4 x)+\frac {1}{10} e^{2 x} x^2 \sin (4 x)-2 \left (\frac {1}{250} e^{2 x} \cos (4 x)+\frac {1}{125} e^{2 x} \sin (4 x)\right )\\ \end {align*}
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Mathematica [A]
time = 0.06, size = 40, normalized size = 0.46 \begin {gather*} \frac {1}{500} e^{2 x} \left (\left (2+40 x-100 x^2\right ) \cos (4 x)+\left (-11+30 x+50 x^2\right ) \sin (4 x)\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.05, size = 40, normalized size = 0.46
method | result | size |
default | \(\left (-\frac {1}{5} x^{2}+\frac {2}{25} x +\frac {1}{250}\right ) {\mathrm e}^{2 x} \cos \left (4 x \right )+\left (\frac {1}{10} x^{2}+\frac {3}{50} x -\frac {11}{500}\right ) {\mathrm e}^{2 x} \sin \left (4 x \right )\) | \(40\) |
risch | \(\left (-\frac {1}{500}-\frac {i}{1000}\right ) \left (50 x^{2}+20 i x -10 x -4 i-3\right ) {\mathrm e}^{\left (2+4 i\right ) x}+\left (-\frac {1}{500}+\frac {i}{1000}\right ) \left (50 x^{2}-20 i x -10 x +4 i-3\right ) {\mathrm e}^{\left (2-4 i\right ) x}\) | \(54\) |
norman | \(\frac {\frac {2 \,{\mathrm e}^{2 x} x}{25}-\frac {{\mathrm e}^{2 x} x^{2}}{5}-\frac {11 \,{\mathrm e}^{2 x} \tan \left (2 x \right )}{250}-\frac {{\mathrm e}^{2 x} \left (\tan ^{2}\left (2 x \right )\right )}{250}+\frac {3 \,{\mathrm e}^{2 x} x \tan \left (2 x \right )}{25}-\frac {2 \,{\mathrm e}^{2 x} x \left (\tan ^{2}\left (2 x \right )\right )}{25}+\frac {{\mathrm e}^{2 x} x^{2} \tan \left (2 x \right )}{5}+\frac {{\mathrm e}^{2 x} x^{2} \left (\tan ^{2}\left (2 x \right )\right )}{5}+\frac {{\mathrm e}^{2 x}}{250}}{1+\tan ^{2}\left (2 x \right )}\) | \(109\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 2.61, size = 41, normalized size = 0.47 \begin {gather*} -\frac {1}{250} \, {\left (50 \, x^{2} - 20 \, x - 1\right )} \cos \left (4 \, x\right ) e^{\left (2 \, x\right )} + \frac {1}{500} \, {\left (50 \, x^{2} + 30 \, x - 11\right )} e^{\left (2 \, x\right )} \sin \left (4 \, x\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.69, size = 41, normalized size = 0.47 \begin {gather*} -\frac {1}{250} \, {\left (50 \, x^{2} - 20 \, x - 1\right )} \cos \left (4 \, x\right ) e^{\left (2 \, x\right )} + \frac {1}{500} \, {\left (50 \, x^{2} + 30 \, x - 11\right )} e^{\left (2 \, x\right )} \sin \left (4 \, x\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.33, size = 85, normalized size = 0.98 \begin {gather*} \frac {x^{2} e^{2 x} \sin {\left (4 x \right )}}{10} - \frac {x^{2} e^{2 x} \cos {\left (4 x \right )}}{5} + \frac {3 x e^{2 x} \sin {\left (4 x \right )}}{50} + \frac {2 x e^{2 x} \cos {\left (4 x \right )}}{25} - \frac {11 e^{2 x} \sin {\left (4 x \right )}}{500} + \frac {e^{2 x} \cos {\left (4 x \right )}}{250} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 1.61, size = 39, normalized size = 0.45 \begin {gather*} -\frac {1}{500} \, {\left (2 \, {\left (50 \, x^{2} - 20 \, x - 1\right )} \cos \left (4 \, x\right ) - {\left (50 \, x^{2} + 30 \, x - 11\right )} \sin \left (4 \, x\right )\right )} e^{\left (2 \, x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 0.35, size = 51, normalized size = 0.59 \begin {gather*} \frac {{\mathrm {e}}^{2\,x}\,\left (2\,\cos \left (4\,x\right )-11\,\sin \left (4\,x\right )+40\,x\,\cos \left (4\,x\right )+30\,x\,\sin \left (4\,x\right )-100\,x^2\,\cos \left (4\,x\right )+50\,x^2\,\sin \left (4\,x\right )\right )}{500} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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