Integrand size = 41, antiderivative size = 19 \[ \int 4 e^{-8+\frac {4 \left (25 x-4 x^3-x^3 \log (4)\right )}{e^8}} \left (25-12 x^2-3 x^2 \log (4)\right ) \, dx=e^{\frac {4 x \left (25-x^2 (4+\log (4))\right )}{e^8}} \]
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Time = 0.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.42, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.073, Rules used = {6, 12, 6838} \[ \int 4 e^{-8+\frac {4 \left (25 x-4 x^3-x^3 \log (4)\right )}{e^8}} \left (25-12 x^2-3 x^2 \log (4)\right ) \, dx=4^{-\frac {4 x^3}{e^8}} e^{\frac {4 \left (25 x-4 x^3\right )}{e^8}} \]
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Rule 6
Rule 12
Rule 6838
Rubi steps \begin{align*} \text {integral}& = \int 4 e^{-8+\frac {4 \left (25 x-4 x^3-x^3 \log (4)\right )}{e^8}} \left (25+x^2 (-12-3 \log (4))\right ) \, dx \\ & = 4 \int e^{-8+\frac {4 \left (25 x-4 x^3-x^3 \log (4)\right )}{e^8}} \left (25+x^2 (-12-3 \log (4))\right ) \, dx \\ & = 4^{-\frac {4 x^3}{e^8}} e^{\frac {4 \left (25 x-4 x^3\right )}{e^8}} \\ \end{align*}
Time = 0.39 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int 4 e^{-8+\frac {4 \left (25 x-4 x^3-x^3 \log (4)\right )}{e^8}} \left (25-12 x^2-3 x^2 \log (4)\right ) \, dx=e^{-\frac {4 x \left (-25+x^2 (4+\log (4))\right )}{e^8}} \]
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Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11
method | result | size |
risch | \({\mathrm e}^{-4 x \left (2 x^{2} \ln \left (2\right )+4 x^{2}-25\right ) {\mathrm e}^{-8}}\) | \(21\) |
gosper | \({\mathrm e}^{-x \left (2 x^{2} \ln \left (2\right )+4 x^{2}-25\right ) {\mathrm e}^{2 \ln \left (2\right )-8}}\) | \(26\) |
derivativedivides | \({\mathrm e}^{\left (-2 x^{3} \ln \left (2\right )-4 x^{3}+25 x \right ) {\mathrm e}^{2 \ln \left (2\right )-8}}\) | \(26\) |
default | \({\mathrm e}^{\left (-2 x^{3} \ln \left (2\right )-4 x^{3}+25 x \right ) {\mathrm e}^{2 \ln \left (2\right )-8}}\) | \(26\) |
parallelrisch | \({\mathrm e}^{\left (-2 x^{3} \ln \left (2\right )-4 x^{3}+25 x \right ) {\mathrm e}^{2 \ln \left (2\right )-8}}\) | \(26\) |
norman | \({\mathrm e}^{-8} {\mathrm e}^{8} {\mathrm e}^{\left (-2 x^{3} \ln \left (2\right )-4 x^{3}+25 x \right ) {\mathrm e}^{2 \ln \left (2\right )-8}}\) | \(35\) |
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Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.37 \[ \int 4 e^{-8+\frac {4 \left (25 x-4 x^3-x^3 \log (4)\right )}{e^8}} \left (25-12 x^2-3 x^2 \log (4)\right ) \, dx=e^{\left (-{\left (2 \, x^{3} \log \left (2\right ) + 4 \, x^{3} - 25 \, x\right )} e^{\left (2 \, \log \left (2\right ) - 8\right )}\right )} \]
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Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int 4 e^{-8+\frac {4 \left (25 x-4 x^3-x^3 \log (4)\right )}{e^8}} \left (25-12 x^2-3 x^2 \log (4)\right ) \, dx=e^{\frac {4 \left (- 4 x^{3} - 2 x^{3} \log {\left (2 \right )} + 25 x\right )}{e^{8}}} \]
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Time = 0.20 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11 \[ \int 4 e^{-8+\frac {4 \left (25 x-4 x^3-x^3 \log (4)\right )}{e^8}} \left (25-12 x^2-3 x^2 \log (4)\right ) \, dx=e^{\left (-4 \, {\left (2 \, x^{3} \log \left (2\right ) + 4 \, x^{3} - 25 \, x\right )} e^{\left (-8\right )}\right )} \]
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Time = 0.27 (sec) , antiderivative size = 44, normalized size of antiderivative = 2.32 \[ \int 4 e^{-8+\frac {4 \left (25 x-4 x^3-x^3 \log (4)\right )}{e^8}} \left (25-12 x^2-3 x^2 \log (4)\right ) \, dx=\frac {1}{4} \, e^{\left (-2 \, x^{3} e^{\left (2 \, \log \left (2\right ) - 8\right )} \log \left (2\right ) - 4 \, x^{3} e^{\left (2 \, \log \left (2\right ) - 8\right )} + 25 \, x e^{\left (2 \, \log \left (2\right ) - 8\right )} + 2 \, \log \left (2\right )\right )} \]
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Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.37 \[ \int 4 e^{-8+\frac {4 \left (25 x-4 x^3-x^3 \log (4)\right )}{e^8}} \left (25-12 x^2-3 x^2 \log (4)\right ) \, dx=\frac {{\mathrm {e}}^{-16\,x^3\,{\mathrm {e}}^{-8}}\,{\mathrm {e}}^{100\,x\,{\mathrm {e}}^{-8}}}{2^{8\,x^3\,{\mathrm {e}}^{-8}}} \]
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