Integrand size = 39, antiderivative size = 20 \[ \int \left (2+2^{2 e^x x} e^x (2+2 x) \log (2)+2^{e^x x} e^x (8+8 x) \log (2)\right ) \, dx=1+\left (4+2^{e^x x}\right )^2+2 x-4 \log (5) \]
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Time = 0.18 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05, number of steps used = 3, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {6838} \[ \int \left (2+2^{2 e^x x} e^x (2+2 x) \log (2)+2^{e^x x} e^x (8+8 x) \log (2)\right ) \, dx=2 x+2^{2 e^x x}+2^{e^x x+3} \]
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Rule 6838
Rubi steps \begin{align*} \text {integral}& = 2 x+\log (2) \int 2^{2 e^x x} e^x (2+2 x) \, dx+\log (2) \int 2^{e^x x} e^x (8+8 x) \, dx \\ & = 2^{2 e^x x}+2^{3+e^x x}+2 x \\ \end{align*}
Time = 0.14 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \left (2+2^{2 e^x x} e^x (2+2 x) \log (2)+2^{e^x x} e^x (8+8 x) \log (2)\right ) \, dx=\left (4+2^{e^x x}\right )^2+2 x \]
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Time = 0.14 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05
method | result | size |
risch | \(2 x +2^{2 \,{\mathrm e}^{x} x}+8 \,2^{{\mathrm e}^{x} x}\) | \(21\) |
default | \(2 x +{\mathrm e}^{2 x \ln \left (2\right ) {\mathrm e}^{x}}+8 \,{\mathrm e}^{x \ln \left (2\right ) {\mathrm e}^{x}}\) | \(23\) |
norman | \(2 x +{\mathrm e}^{2 x \ln \left (2\right ) {\mathrm e}^{x}}+8 \,{\mathrm e}^{x \ln \left (2\right ) {\mathrm e}^{x}}\) | \(23\) |
parallelrisch | \(2 x +{\mathrm e}^{2 x \ln \left (2\right ) {\mathrm e}^{x}}+8 \,{\mathrm e}^{x \ln \left (2\right ) {\mathrm e}^{x}}\) | \(23\) |
parts | \(2 x +{\mathrm e}^{2 x \ln \left (2\right ) {\mathrm e}^{x}}+8 \,{\mathrm e}^{x \ln \left (2\right ) {\mathrm e}^{x}}\) | \(23\) |
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none
Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \left (2+2^{2 e^x x} e^x (2+2 x) \log (2)+2^{e^x x} e^x (8+8 x) \log (2)\right ) \, dx=2^{2 \, x e^{x}} + 8 \cdot 2^{x e^{x}} + 2 \, x \]
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Time = 0.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.30 \[ \int \left (2+2^{2 e^x x} e^x (2+2 x) \log (2)+2^{e^x x} e^x (8+8 x) \log (2)\right ) \, dx=2 x + e^{2 x e^{x} \log {\left (2 \right )}} + 8 e^{x e^{x} \log {\left (2 \right )}} \]
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none
Time = 0.32 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \left (2+2^{2 e^x x} e^x (2+2 x) \log (2)+2^{e^x x} e^x (8+8 x) \log (2)\right ) \, dx=2^{2 \, x e^{x}} + 8 \cdot 2^{x e^{x}} + 2 \, x \]
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none
Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \left (2+2^{2 e^x x} e^x (2+2 x) \log (2)+2^{e^x x} e^x (8+8 x) \log (2)\right ) \, dx=2^{2 \, x e^{x}} + 8 \cdot 2^{x e^{x}} + 2 \, x \]
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Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \left (2+2^{2 e^x x} e^x (2+2 x) \log (2)+2^{e^x x} e^x (8+8 x) \log (2)\right ) \, dx=2\,x+8\,2^{x\,{\mathrm {e}}^x}+2^{2\,x\,{\mathrm {e}}^x} \]
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