Integrand size = 23, antiderivative size = 29 \[ \int \frac {1}{3} e^{2 x} \left (4-50 x-6 x^2\right ) \log ^4(4) \, dx=-5+e^{2 x} \left (\frac {5}{3} (5-2 x)-(2+x)^2\right ) \log ^4(4) \]
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Time = 0.03 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.45, number of steps used = 9, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {12, 2227, 2225, 2207} \[ \int \frac {1}{3} e^{2 x} \left (4-50 x-6 x^2\right ) \log ^4(4) \, dx=-e^{2 x} x^2 \log ^4(4)-\frac {22}{3} e^{2 x} x \log ^4(4)+\frac {13}{3} e^{2 x} \log ^4(4) \]
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Rule 12
Rule 2207
Rule 2225
Rule 2227
Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \log ^4(4) \int e^{2 x} \left (4-50 x-6 x^2\right ) \, dx \\ & = \frac {1}{3} \log ^4(4) \int \left (4 e^{2 x}-50 e^{2 x} x-6 e^{2 x} x^2\right ) \, dx \\ & = \frac {1}{3} \left (4 \log ^4(4)\right ) \int e^{2 x} \, dx-\left (2 \log ^4(4)\right ) \int e^{2 x} x^2 \, dx-\frac {1}{3} \left (50 \log ^4(4)\right ) \int e^{2 x} x \, dx \\ & = \frac {2}{3} e^{2 x} \log ^4(4)-\frac {25}{3} e^{2 x} x \log ^4(4)-e^{2 x} x^2 \log ^4(4)+\left (2 \log ^4(4)\right ) \int e^{2 x} x \, dx+\frac {1}{3} \left (25 \log ^4(4)\right ) \int e^{2 x} \, dx \\ & = \frac {29}{6} e^{2 x} \log ^4(4)-\frac {22}{3} e^{2 x} x \log ^4(4)-e^{2 x} x^2 \log ^4(4)-\log ^4(4) \int e^{2 x} \, dx \\ & = \frac {13}{3} e^{2 x} \log ^4(4)-\frac {22}{3} e^{2 x} x \log ^4(4)-e^{2 x} x^2 \log ^4(4) \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {1}{3} e^{2 x} \left (4-50 x-6 x^2\right ) \log ^4(4) \, dx=-\frac {2}{3} e^{2 x} \left (-\frac {13}{2}+11 x+\frac {3 x^2}{2}\right ) \log ^4(4) \]
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Time = 0.04 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.72
method | result | size |
gosper | \(-\frac {16 \,{\mathrm e}^{2 x} \ln \left (2\right )^{4} \left (3 x^{2}+22 x -13\right )}{3}\) | \(21\) |
risch | \(\frac {16 \ln \left (2\right )^{4} \left (-3 x^{2}-22 x +13\right ) {\mathrm e}^{2 x}}{3}\) | \(21\) |
default | \(\frac {32 \ln \left (2\right )^{4} \left (\frac {13 \,{\mathrm e}^{2 x}}{2}-11 x \,{\mathrm e}^{2 x}-\frac {3 \,{\mathrm e}^{2 x} x^{2}}{2}\right )}{3}\) | \(30\) |
parallelrisch | \(\frac {16 \ln \left (2\right )^{4} \left (-3 \,{\mathrm e}^{2 x} x^{2}-22 x \,{\mathrm e}^{2 x}+13 \,{\mathrm e}^{2 x}\right )}{3}\) | \(30\) |
norman | \(\frac {208 \ln \left (2\right )^{4} {\mathrm e}^{2 x}}{3}-\frac {352 x \ln \left (2\right )^{4} {\mathrm e}^{2 x}}{3}-16 x^{2} \ln \left (2\right )^{4} {\mathrm e}^{2 x}\) | \(36\) |
meijerg | \(-\frac {32 \ln \left (2\right )^{4} \left (1-{\mathrm e}^{2 x}\right )}{3}+4 \ln \left (2\right )^{4} \left (2-\frac {\left (12 x^{2}-12 x +6\right ) {\mathrm e}^{2 x}}{3}\right )-\frac {200 \ln \left (2\right )^{4} \left (1-\frac {\left (-4 x +2\right ) {\mathrm e}^{2 x}}{2}\right )}{3}\) | \(59\) |
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Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69 \[ \int \frac {1}{3} e^{2 x} \left (4-50 x-6 x^2\right ) \log ^4(4) \, dx=-\frac {16}{3} \, {\left (3 \, x^{2} + 22 \, x - 13\right )} e^{\left (2 \, x\right )} \log \left (2\right )^{4} \]
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Time = 0.05 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {1}{3} e^{2 x} \left (4-50 x-6 x^2\right ) \log ^4(4) \, dx=\frac {\left (- 48 x^{2} \log {\left (2 \right )}^{4} - 352 x \log {\left (2 \right )}^{4} + 208 \log {\left (2 \right )}^{4}\right ) e^{2 x}}{3} \]
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Time = 0.22 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.38 \[ \int \frac {1}{3} e^{2 x} \left (4-50 x-6 x^2\right ) \log ^4(4) \, dx=-\frac {8}{3} \, {\left (3 \, {\left (2 \, x^{2} - 2 \, x + 1\right )} e^{\left (2 \, x\right )} + 25 \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} - 4 \, e^{\left (2 \, x\right )}\right )} \log \left (2\right )^{4} \]
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Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69 \[ \int \frac {1}{3} e^{2 x} \left (4-50 x-6 x^2\right ) \log ^4(4) \, dx=-\frac {16}{3} \, {\left (3 \, x^{2} + 22 \, x - 13\right )} e^{\left (2 \, x\right )} \log \left (2\right )^{4} \]
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Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69 \[ \int \frac {1}{3} e^{2 x} \left (4-50 x-6 x^2\right ) \log ^4(4) \, dx=-\frac {16\,{\mathrm {e}}^{2\,x}\,{\ln \left (2\right )}^4\,\left (3\,x^2+22\,x-13\right )}{3} \]
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