Integrand size = 52, antiderivative size = 39 \[ \int \frac {3-2 x+\log (4)+4 e^{16+e^{16+4 x}+4 x} (i \pi +\log (5-\log (3)))}{i \pi +\log (5-\log (3))} \, dx=-e^5+e^{e^{4 (4+x)}}+\frac {x (3-x+\log (4))}{i \pi +\log (5-\log (3))} \]
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Time = 0.03 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.31, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.058, Rules used = {12, 2320, 2225} \[ \int \frac {3-2 x+\log (4)+4 e^{16+e^{16+4 x}+4 x} (i \pi +\log (5-\log (3)))}{i \pi +\log (5-\log (3))} \, dx=-\frac {x^2}{\log (5-\log (3))+i \pi }+e^{e^{4 x+16}}+\frac {x (3+\log (4))}{\log (5-\log (3))+i \pi } \]
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Rule 12
Rule 2225
Rule 2320
Rubi steps \begin{align*} \text {integral}& = \frac {\int \left (3-2 x+\log (4)+4 e^{16+e^{16+4 x}+4 x} (i \pi +\log (5-\log (3)))\right ) \, dx}{i \pi +\log (5-\log (3))} \\ & = -\frac {x^2}{i \pi +\log (5-\log (3))}+\frac {x (3+\log (4))}{i \pi +\log (5-\log (3))}+4 \int e^{16+e^{16+4 x}+4 x} \, dx \\ & = -\frac {x^2}{i \pi +\log (5-\log (3))}+\frac {x (3+\log (4))}{i \pi +\log (5-\log (3))}+\text {Subst}\left (\int e^{16+e^{16} x} \, dx,x,e^{4 x}\right ) \\ & = e^{e^{16+4 x}}-\frac {x^2}{i \pi +\log (5-\log (3))}+\frac {x (3+\log (4))}{i \pi +\log (5-\log (3))} \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.33 \[ \int \frac {3-2 x+\log (4)+4 e^{16+e^{16+4 x}+4 x} (i \pi +\log (5-\log (3)))}{i \pi +\log (5-\log (3))} \, dx=\frac {3 x-x^2+x \log (4)+e^{e^{4 (4+x)}} (i \pi +\log (5-\log (3)))}{i \pi +\log (5-\log (3))} \]
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Time = 0.14 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.79
method | result | size |
parts | \(\frac {2 x \ln \left (2\right )-x^{2}+3 x}{\ln \left (\ln \left (3\right )-5\right )}+{\mathrm e}^{{\mathrm e}^{4 x +16}}\) | \(31\) |
default | \(\frac {-x^{2}+3 x +\ln \left (\ln \left (3\right )-5\right ) {\mathrm e}^{{\mathrm e}^{4 x +16}}+2 x \ln \left (2\right )}{\ln \left (\ln \left (3\right )-5\right )}\) | \(36\) |
norman | \(\frac {\left (2 \ln \left (2\right )+3\right ) x}{\ln \left (\ln \left (3\right )-5\right )}-\frac {x^{2}}{\ln \left (\ln \left (3\right )-5\right )}+{\mathrm e}^{{\mathrm e}^{4 x +16}}\) | \(36\) |
parallelrisch | \(\frac {-x^{2}+\ln \left (\ln \left (3\right )-5\right ) {\mathrm e}^{{\mathrm e}^{4 x +16}}+x \left (2 \ln \left (2\right )+3\right )}{\ln \left (\ln \left (3\right )-5\right )}\) | \(36\) |
risch | \(-\frac {x^{2}}{\ln \left (\ln \left (3\right )-5\right )}+\frac {3 x}{\ln \left (\ln \left (3\right )-5\right )}+{\mathrm e}^{{\mathrm e}^{4 x +16}}+\frac {2 x \ln \left (2\right )}{\ln \left (\ln \left (3\right )-5\right )}\) | \(43\) |
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Time = 0.28 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.38 \[ \int \frac {3-2 x+\log (4)+4 e^{16+e^{16+4 x}+4 x} (i \pi +\log (5-\log (3)))}{i \pi +\log (5-\log (3))} \, dx=-\frac {{\left ({\left (x^{2} - 2 \, x \log \left (2\right ) - 3 \, x\right )} e^{\left (4 \, x + 16\right )} - e^{\left (4 \, x + e^{\left (4 \, x + 16\right )} + 16\right )} \log \left (\log \left (3\right ) - 5\right )\right )} e^{\left (-4 \, x - 16\right )}}{\log \left (\log \left (3\right ) - 5\right )} \]
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Time = 0.15 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.05 \[ \int \frac {3-2 x+\log (4)+4 e^{16+e^{16+4 x}+4 x} (i \pi +\log (5-\log (3)))}{i \pi +\log (5-\log (3))} \, dx=- \frac {x^{2}}{\log {\left (5 - \log {\left (3 \right )} \right )} + i \pi } + \frac {x \left (2 \log {\left (2 \right )} + 3\right )}{\log {\left (5 - \log {\left (3 \right )} \right )} + i \pi } + e^{e^{16} e^{4 x}} \]
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Time = 0.21 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.90 \[ \int \frac {3-2 x+\log (4)+4 e^{16+e^{16+4 x}+4 x} (i \pi +\log (5-\log (3)))}{i \pi +\log (5-\log (3))} \, dx=-\frac {x^{2} - 2 \, x \log \left (2\right ) - e^{\left (e^{\left (4 \, x + 16\right )}\right )} \log \left (\log \left (3\right ) - 5\right ) - 3 \, x}{\log \left (\log \left (3\right ) - 5\right )} \]
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Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.90 \[ \int \frac {3-2 x+\log (4)+4 e^{16+e^{16+4 x}+4 x} (i \pi +\log (5-\log (3)))}{i \pi +\log (5-\log (3))} \, dx=-\frac {x^{2} - 2 \, x \log \left (2\right ) - e^{\left (e^{\left (4 \, x + 16\right )}\right )} \log \left (\log \left (3\right ) - 5\right ) - 3 \, x}{\log \left (\log \left (3\right ) - 5\right )} \]
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Time = 0.16 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.90 \[ \int \frac {3-2 x+\log (4)+4 e^{16+e^{16+4 x}+4 x} (i \pi +\log (5-\log (3)))}{i \pi +\log (5-\log (3))} \, dx=\frac {3\,x+x\,\ln \left (4\right )-x^2+\ln \left (\ln \left (3\right )-5\right )\,{\mathrm {e}}^{{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^{16}}}{\ln \left (\ln \left (3\right )-5\right )} \]
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