Integrand size = 30, antiderivative size = 20 \[ \int \left (1+59049 e^{-e^{-3+4 x}+2 x} \left (-2+4 e^{-3+4 x}\right )\right ) \, dx=-1-59049 e^{-e^{-3+4 x}+2 x}+x \]
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Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2320, 2258, 2236, 2243} \[ \int \left (1+59049 e^{-e^{-3+4 x}+2 x} \left (-2+4 e^{-3+4 x}\right )\right ) \, dx=x-59049 e^{2 x-e^{4 x-3}} \]
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Rule 2236
Rule 2243
Rule 2258
Rule 2320
Rubi steps \begin{align*} \text {integral}& = x+59049 \int e^{-e^{-3+4 x}+2 x} \left (-2+4 e^{-3+4 x}\right ) \, dx \\ & = x+\frac {59049}{2} \text {Subst}\left (\int e^{-\frac {x^2}{e^3}} \left (-2+\frac {4 x^2}{e^3}\right ) \, dx,x,e^{2 x}\right ) \\ & = x+\frac {59049}{2} \text {Subst}\left (\int \left (-2 e^{-\frac {x^2}{e^3}}+4 e^{-3-\frac {x^2}{e^3}} x^2\right ) \, dx,x,e^{2 x}\right ) \\ & = x-59049 \text {Subst}\left (\int e^{-\frac {x^2}{e^3}} \, dx,x,e^{2 x}\right )+118098 \text {Subst}\left (\int e^{-3-\frac {x^2}{e^3}} x^2 \, dx,x,e^{2 x}\right ) \\ & = -59049 e^{-e^{-3+4 x}+2 x}+x-\frac {59049}{2} e^{3/2} \sqrt {\pi } \text {erf}\left (e^{-\frac {3}{2}+2 x}\right )+\left (59049 e^3\right ) \text {Subst}\left (\int e^{-3-\frac {x^2}{e^3}} \, dx,x,e^{2 x}\right ) \\ & = -59049 e^{-e^{-3+4 x}+2 x}+x \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \left (1+59049 e^{-e^{-3+4 x}+2 x} \left (-2+4 e^{-3+4 x}\right )\right ) \, dx=-59049 e^{-e^{-3+4 x}+2 x}+x \]
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Time = 0.40 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90
method | result | size |
risch | \(x -59049 \,{\mathrm e}^{-{\mathrm e}^{-3+4 x}+2 x}\) | \(18\) |
default | \(x -{\mathrm e}^{-{\mathrm e}^{-3+4 x}+10 \ln \left (3\right )+2 x}\) | \(22\) |
norman | \(x -{\mathrm e}^{-{\mathrm e}^{-3+4 x}+10 \ln \left (3\right )+2 x}\) | \(22\) |
parallelrisch | \(x -{\mathrm e}^{-{\mathrm e}^{-3+4 x}+10 \ln \left (3\right )+2 x}\) | \(22\) |
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Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \left (1+59049 e^{-e^{-3+4 x}+2 x} \left (-2+4 e^{-3+4 x}\right )\right ) \, dx=x - e^{\left (2 \, x - e^{\left (4 \, x - 3\right )} + 10 \, \log \left (3\right )\right )} \]
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Time = 0.09 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int \left (1+59049 e^{-e^{-3+4 x}+2 x} \left (-2+4 e^{-3+4 x}\right )\right ) \, dx=x - 59049 e^{2 x - e^{4 x - 3}} \]
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Time = 0.21 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \left (1+59049 e^{-e^{-3+4 x}+2 x} \left (-2+4 e^{-3+4 x}\right )\right ) \, dx=x - 59049 \, e^{\left (2 \, x - e^{\left (4 \, x - 3\right )}\right )} \]
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Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \left (1+59049 e^{-e^{-3+4 x}+2 x} \left (-2+4 e^{-3+4 x}\right )\right ) \, dx=x - e^{\left (2 \, x - e^{\left (4 \, x - 3\right )} + 10 \, \log \left (3\right )\right )} \]
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Time = 0.10 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \left (1+59049 e^{-e^{-3+4 x}+2 x} \left (-2+4 e^{-3+4 x}\right )\right ) \, dx=x-59049\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^{-3}} \]
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