Integrand size = 49, antiderivative size = 23 \[ \int \frac {e^{\frac {256 e^x x}{\log ^4(2)}} \left (e^x \left (256 x-4096 x^2-4352 x^3\right )+(1-34 x) \log ^4(2)\right )}{5 \log ^4(2)} \, dx=\frac {1}{5} e^{\frac {256 e^x x}{\log ^4(2)}} \left (x-17 x^2\right ) \]
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Time = 0.03 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.78, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.041, Rules used = {12, 2326} \[ \int \frac {e^{\frac {256 e^x x}{\log ^4(2)}} \left (e^x \left (256 x-4096 x^2-4352 x^3\right )+(1-34 x) \log ^4(2)\right )}{5 \log ^4(2)} \, dx=\frac {\left (-17 x^3-16 x^2+x\right ) e^{x+\frac {256 e^x x}{\log ^4(2)}}}{5 \left (e^x x+e^x\right )} \]
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Rule 12
Rule 2326
Rubi steps \begin{align*} \text {integral}& = \frac {\int e^{\frac {256 e^x x}{\log ^4(2)}} \left (e^x \left (256 x-4096 x^2-4352 x^3\right )+(1-34 x) \log ^4(2)\right ) \, dx}{5 \log ^4(2)} \\ & = \frac {e^{x+\frac {256 e^x x}{\log ^4(2)}} \left (x-16 x^2-17 x^3\right )}{5 \left (e^x+e^x x\right )} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {256 e^x x}{\log ^4(2)}} \left (e^x \left (256 x-4096 x^2-4352 x^3\right )+(1-34 x) \log ^4(2)\right )}{5 \log ^4(2)} \, dx=-\frac {1}{5} e^{\frac {256 e^x x}{\log ^4(2)}} x (-1+17 x) \]
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Time = 0.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83
method | result | size |
risch | \(-\frac {\left (17 x -1\right ) x \,{\mathrm e}^{\frac {256 x \,{\mathrm e}^{x}}{\ln \left (2\right )^{4}}}}{5}\) | \(19\) |
norman | \(\frac {\frac {x \ln \left (2\right )^{3} {\mathrm e}^{\frac {256 x \,{\mathrm e}^{x}}{\ln \left (2\right )^{4}}}}{5}-\frac {17 x^{2} \ln \left (2\right )^{3} {\mathrm e}^{\frac {256 x \,{\mathrm e}^{x}}{\ln \left (2\right )^{4}}}}{5}}{\ln \left (2\right )^{3}}\) | \(43\) |
parallelrisch | \(\frac {\ln \left (2\right )^{4} {\mathrm e}^{\frac {256 x \,{\mathrm e}^{x}}{\ln \left (2\right )^{4}}} x -17 \ln \left (2\right )^{4} {\mathrm e}^{\frac {256 x \,{\mathrm e}^{x}}{\ln \left (2\right )^{4}}} x^{2}}{5 \ln \left (2\right )^{4}}\) | \(43\) |
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Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {e^{\frac {256 e^x x}{\log ^4(2)}} \left (e^x \left (256 x-4096 x^2-4352 x^3\right )+(1-34 x) \log ^4(2)\right )}{5 \log ^4(2)} \, dx=-\frac {1}{5} \, {\left (17 \, x^{2} - x\right )} e^{\left (\frac {256 \, x e^{x}}{\log \left (2\right )^{4}}\right )} \]
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Time = 0.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {e^{\frac {256 e^x x}{\log ^4(2)}} \left (e^x \left (256 x-4096 x^2-4352 x^3\right )+(1-34 x) \log ^4(2)\right )}{5 \log ^4(2)} \, dx=\frac {\left (- 17 x^{2} + x\right ) e^{\frac {256 x e^{x}}{\log {\left (2 \right )}^{4}}}}{5} \]
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Time = 0.31 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.43 \[ \int \frac {e^{\frac {256 e^x x}{\log ^4(2)}} \left (e^x \left (256 x-4096 x^2-4352 x^3\right )+(1-34 x) \log ^4(2)\right )}{5 \log ^4(2)} \, dx=-\frac {{\left (17 \, x^{2} \log \left (2\right )^{4} - x \log \left (2\right )^{4}\right )} e^{\left (\frac {256 \, x e^{x}}{\log \left (2\right )^{4}}\right )}}{5 \, \log \left (2\right )^{4}} \]
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\[ \int \frac {e^{\frac {256 e^x x}{\log ^4(2)}} \left (e^x \left (256 x-4096 x^2-4352 x^3\right )+(1-34 x) \log ^4(2)\right )}{5 \log ^4(2)} \, dx=\int { -\frac {{\left ({\left (34 \, x - 1\right )} \log \left (2\right )^{4} + 256 \, {\left (17 \, x^{3} + 16 \, x^{2} - x\right )} e^{x}\right )} e^{\left (\frac {256 \, x e^{x}}{\log \left (2\right )^{4}}\right )}}{5 \, \log \left (2\right )^{4}} \,d x } \]
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Time = 0.16 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {e^{\frac {256 e^x x}{\log ^4(2)}} \left (e^x \left (256 x-4096 x^2-4352 x^3\right )+(1-34 x) \log ^4(2)\right )}{5 \log ^4(2)} \, dx=-\frac {x\,{\mathrm {e}}^{\frac {256\,x\,{\mathrm {e}}^x}{{\ln \left (2\right )}^4}}\,\left (17\,x-1\right )}{5} \]
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