Integrand size = 99, antiderivative size = 28 \[ \int e^{3 x+4 e^{16 x-2 x \log (x)} x-8 e^{8 x-x \log (x)} x^2+4 x^3} \left (3+12 x^2+e^{16 x-2 x \log (x)} (4+56 x-8 x \log (x))+e^{8 x-x \log (x)} \left (-16 x-56 x^2+8 x^2 \log (x)\right )\right ) \, dx=e^{x+2 \left (x+2 \left (e^{8 x-x \log (x)}-x\right )^2 x\right )} \]
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Time = 1.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.010, Rules used = {6838} \[ \int e^{3 x+4 e^{16 x-2 x \log (x)} x-8 e^{8 x-x \log (x)} x^2+4 x^3} \left (3+12 x^2+e^{16 x-2 x \log (x)} (4+56 x-8 x \log (x))+e^{8 x-x \log (x)} \left (-16 x-56 x^2+8 x^2 \log (x)\right )\right ) \, dx=\exp \left (4 e^{16 x} x^{1-2 x}-8 e^{8 x} x^{2-x}+4 x^3+3 x\right ) \]
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Rule 6838
Rubi steps \begin{align*} \text {integral}& = \exp \left (3 x+4 x^3+4 e^{16 x} x^{1-2 x}-8 e^{8 x} x^{2-x}\right ) \\ \end{align*}
Time = 0.18 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int e^{3 x+4 e^{16 x-2 x \log (x)} x-8 e^{8 x-x \log (x)} x^2+4 x^3} \left (3+12 x^2+e^{16 x-2 x \log (x)} (4+56 x-8 x \log (x))+e^{8 x-x \log (x)} \left (-16 x-56 x^2+8 x^2 \log (x)\right )\right ) \, dx=e^{3 x+4 x^3+4 e^{16 x} x^{1-2 x}-8 e^{8 x} x^{2-x}} \]
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Time = 0.34 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29
method | result | size |
risch | \({\mathrm e}^{x \left (-8 x^{-x} {\mathrm e}^{8 x} x +4 x^{-2 x} {\mathrm e}^{16 x}+4 x^{2}+3\right )}\) | \(36\) |
parallelrisch | \({\mathrm e}^{4 x \,{\mathrm e}^{-2 x \ln \left (x \right )+16 x}-8 x^{2} {\mathrm e}^{-x \ln \left (x \right )+8 x}+4 x^{3}+3 x}\) | \(41\) |
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Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int e^{3 x+4 e^{16 x-2 x \log (x)} x-8 e^{8 x-x \log (x)} x^2+4 x^3} \left (3+12 x^2+e^{16 x-2 x \log (x)} (4+56 x-8 x \log (x))+e^{8 x-x \log (x)} \left (-16 x-56 x^2+8 x^2 \log (x)\right )\right ) \, dx=e^{\left (4 \, x^{3} - 8 \, x^{2} e^{\left (-x \log \left (x\right ) + 8 \, x\right )} + 4 \, x e^{\left (-2 \, x \log \left (x\right ) + 16 \, x\right )} + 3 \, x\right )} \]
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Time = 0.43 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int e^{3 x+4 e^{16 x-2 x \log (x)} x-8 e^{8 x-x \log (x)} x^2+4 x^3} \left (3+12 x^2+e^{16 x-2 x \log (x)} (4+56 x-8 x \log (x))+e^{8 x-x \log (x)} \left (-16 x-56 x^2+8 x^2 \log (x)\right )\right ) \, dx=e^{4 x^{3} - 8 x^{2} e^{- x \log {\left (x \right )} + 8 x} + 4 x e^{- 2 x \log {\left (x \right )} + 16 x} + 3 x} \]
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\[ \int e^{3 x+4 e^{16 x-2 x \log (x)} x-8 e^{8 x-x \log (x)} x^2+4 x^3} \left (3+12 x^2+e^{16 x-2 x \log (x)} (4+56 x-8 x \log (x))+e^{8 x-x \log (x)} \left (-16 x-56 x^2+8 x^2 \log (x)\right )\right ) \, dx=\int { {\left (12 \, x^{2} + 8 \, {\left (x^{2} \log \left (x\right ) - 7 \, x^{2} - 2 \, x\right )} e^{\left (-x \log \left (x\right ) + 8 \, x\right )} - 4 \, {\left (2 \, x \log \left (x\right ) - 14 \, x - 1\right )} e^{\left (-2 \, x \log \left (x\right ) + 16 \, x\right )} + 3\right )} e^{\left (4 \, x^{3} - 8 \, x^{2} e^{\left (-x \log \left (x\right ) + 8 \, x\right )} + 4 \, x e^{\left (-2 \, x \log \left (x\right ) + 16 \, x\right )} + 3 \, x\right )} \,d x } \]
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Time = 1.03 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int e^{3 x+4 e^{16 x-2 x \log (x)} x-8 e^{8 x-x \log (x)} x^2+4 x^3} \left (3+12 x^2+e^{16 x-2 x \log (x)} (4+56 x-8 x \log (x))+e^{8 x-x \log (x)} \left (-16 x-56 x^2+8 x^2 \log (x)\right )\right ) \, dx=e^{\left (4 \, x^{3} - 8 \, x^{2} e^{\left (-x \log \left (x\right ) + 8 \, x\right )} + 4 \, x e^{\left (-2 \, x \log \left (x\right ) + 16 \, x\right )} + 3 \, x\right )} \]
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Time = 12.64 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int e^{3 x+4 e^{16 x-2 x \log (x)} x-8 e^{8 x-x \log (x)} x^2+4 x^3} \left (3+12 x^2+e^{16 x-2 x \log (x)} (4+56 x-8 x \log (x))+e^{8 x-x \log (x)} \left (-16 x-56 x^2+8 x^2 \log (x)\right )\right ) \, dx={\mathrm {e}}^{-8\,x^{2-x}\,{\mathrm {e}}^{8\,x}}\,{\mathrm {e}}^{4\,x^{1-2\,x}\,{\mathrm {e}}^{16\,x}}\,{\mathrm {e}}^{3\,x}\,{\mathrm {e}}^{4\,x^3} \]
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