Integrand size = 168, antiderivative size = 30 \[ \int \frac {-12+e^x \left (-3-3 x-x^2\right )+\left (e^{3 x} \left (6 x+2 x^2\right )+e^{2 x} \left (24 x+8 x^2\right )\right ) \log \left (\frac {16+4 e^x}{3+x}\right )+\left (e^{3 x} \left (6 x+2 x^2\right )+e^{2 x} \left (24 x+8 x^2\right )\right ) \log \left (\frac {x}{\log (3)}\right )}{\left (12 x+4 x^2+e^x \left (3 x+x^2\right )\right ) \log \left (\frac {16+4 e^x}{3+x}\right )+\left (12 x+4 x^2+e^x \left (3 x+x^2\right )\right ) \log \left (\frac {x}{\log (3)}\right )} \, dx=e^{2 x}-\log \left (\log \left (\frac {4 \left (4+e^x\right )}{3+x}\right )+\log \left (\frac {x}{\log (3)}\right )\right ) \]
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\[ \int \frac {-12+e^x \left (-3-3 x-x^2\right )+\left (e^{3 x} \left (6 x+2 x^2\right )+e^{2 x} \left (24 x+8 x^2\right )\right ) \log \left (\frac {16+4 e^x}{3+x}\right )+\left (e^{3 x} \left (6 x+2 x^2\right )+e^{2 x} \left (24 x+8 x^2\right )\right ) \log \left (\frac {x}{\log (3)}\right )}{\left (12 x+4 x^2+e^x \left (3 x+x^2\right )\right ) \log \left (\frac {16+4 e^x}{3+x}\right )+\left (12 x+4 x^2+e^x \left (3 x+x^2\right )\right ) \log \left (\frac {x}{\log (3)}\right )} \, dx=\int \frac {-12+e^x \left (-3-3 x-x^2\right )+\left (e^{3 x} \left (6 x+2 x^2\right )+e^{2 x} \left (24 x+8 x^2\right )\right ) \log \left (\frac {16+4 e^x}{3+x}\right )+\left (e^{3 x} \left (6 x+2 x^2\right )+e^{2 x} \left (24 x+8 x^2\right )\right ) \log \left (\frac {x}{\log (3)}\right )}{\left (12 x+4 x^2+e^x \left (3 x+x^2\right )\right ) \log \left (\frac {16+4 e^x}{3+x}\right )+\left (12 x+4 x^2+e^x \left (3 x+x^2\right )\right ) \log \left (\frac {x}{\log (3)}\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {-12-3 e^x-3 e^x x-e^x x^2+2 e^{2 x} \left (4+e^x\right ) x (3+x) \log \left (\frac {4 \left (4+e^x\right )}{3+x}\right )+2 e^{2 x} \left (4+e^x\right ) x (3+x) \log \left (\frac {x}{\log (3)}\right )}{\left (4+e^x\right ) x (3+x) \left (\log \left (\frac {4+e^x}{3+x}\right )+\log \left (\frac {4 x}{\log (3)}\right )\right )} \, dx \\ & = \int \left (2 e^{2 x}+\frac {4}{\left (4+e^x\right ) \left (\log \left (\frac {4+e^x}{3+x}\right )+\log \left (\frac {4 x}{\log (3)}\right )\right )}+\frac {-3-3 x-x^2}{x (3+x) \left (\log \left (\frac {4+e^x}{3+x}\right )+\log \left (\frac {4 x}{\log (3)}\right )\right )}\right ) \, dx \\ & = 2 \int e^{2 x} \, dx+4 \int \frac {1}{\left (4+e^x\right ) \left (\log \left (\frac {4+e^x}{3+x}\right )+\log \left (\frac {4 x}{\log (3)}\right )\right )} \, dx+\int \frac {-3-3 x-x^2}{x (3+x) \left (\log \left (\frac {4+e^x}{3+x}\right )+\log \left (\frac {4 x}{\log (3)}\right )\right )} \, dx \\ & = e^{2 x}+4 \int \frac {1}{\left (4+e^x\right ) \left (\log \left (\frac {4+e^x}{3+x}\right )+\log \left (\frac {4 x}{\log (3)}\right )\right )} \, dx+\int \left (\frac {1}{-\log \left (\frac {4+e^x}{3+x}\right )-\log \left (\frac {4 x}{\log (3)}\right )}-\frac {1}{x \left (\log \left (\frac {4+e^x}{3+x}\right )+\log \left (\frac {4 x}{\log (3)}\right )\right )}+\frac {1}{(3+x) \left (\log \left (\frac {4+e^x}{3+x}\right )+\log \left (\frac {4 x}{\log (3)}\right )\right )}\right ) \, dx \\ & = e^{2 x}+4 \int \frac {1}{\left (4+e^x\right ) \left (\log \left (\frac {4+e^x}{3+x}\right )+\log \left (\frac {4 x}{\log (3)}\right )\right )} \, dx+\int \frac {1}{-\log \left (\frac {4+e^x}{3+x}\right )-\log \left (\frac {4 x}{\log (3)}\right )} \, dx-\int \frac {1}{x \left (\log \left (\frac {4+e^x}{3+x}\right )+\log \left (\frac {4 x}{\log (3)}\right )\right )} \, dx+\int \frac {1}{(3+x) \left (\log \left (\frac {4+e^x}{3+x}\right )+\log \left (\frac {4 x}{\log (3)}\right )\right )} \, dx \\ \end{align*}
\[ \int \frac {-12+e^x \left (-3-3 x-x^2\right )+\left (e^{3 x} \left (6 x+2 x^2\right )+e^{2 x} \left (24 x+8 x^2\right )\right ) \log \left (\frac {16+4 e^x}{3+x}\right )+\left (e^{3 x} \left (6 x+2 x^2\right )+e^{2 x} \left (24 x+8 x^2\right )\right ) \log \left (\frac {x}{\log (3)}\right )}{\left (12 x+4 x^2+e^x \left (3 x+x^2\right )\right ) \log \left (\frac {16+4 e^x}{3+x}\right )+\left (12 x+4 x^2+e^x \left (3 x+x^2\right )\right ) \log \left (\frac {x}{\log (3)}\right )} \, dx=\int \frac {-12+e^x \left (-3-3 x-x^2\right )+\left (e^{3 x} \left (6 x+2 x^2\right )+e^{2 x} \left (24 x+8 x^2\right )\right ) \log \left (\frac {16+4 e^x}{3+x}\right )+\left (e^{3 x} \left (6 x+2 x^2\right )+e^{2 x} \left (24 x+8 x^2\right )\right ) \log \left (\frac {x}{\log (3)}\right )}{\left (12 x+4 x^2+e^x \left (3 x+x^2\right )\right ) \log \left (\frac {16+4 e^x}{3+x}\right )+\left (12 x+4 x^2+e^x \left (3 x+x^2\right )\right ) \log \left (\frac {x}{\log (3)}\right )} \, dx \]
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Time = 8.04 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97
method | result | size |
parallelrisch | \({\mathrm e}^{2 x}-\ln \left (\ln \left (\frac {4 \,{\mathrm e}^{x}+16}{3+x}\right )+\ln \left (\frac {x}{\ln \left (3\right )}\right )\right )\) | \(29\) |
risch | \({\mathrm e}^{2 x}-\ln \left (\ln \left ({\mathrm e}^{x}+4\right )-\frac {i \left (\pi \,\operatorname {csgn}\left (\frac {i}{3+x}\right ) \operatorname {csgn}\left (i \left ({\mathrm e}^{x}+4\right )\right ) \operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}+4\right )}{3+x}\right )-\pi \,\operatorname {csgn}\left (\frac {i}{3+x}\right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}+4\right )}{3+x}\right )}^{2}-\pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{x}+4\right )\right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}+4\right )}{3+x}\right )}^{2}+\pi {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}+4\right )}{3+x}\right )}^{3}+4 i \ln \left (2\right )+2 i \ln \left (\frac {x}{\ln \left (3\right )}\right )-2 i \ln \left (3+x \right )\right )}{2}\right )\) | \(143\) |
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Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {-12+e^x \left (-3-3 x-x^2\right )+\left (e^{3 x} \left (6 x+2 x^2\right )+e^{2 x} \left (24 x+8 x^2\right )\right ) \log \left (\frac {16+4 e^x}{3+x}\right )+\left (e^{3 x} \left (6 x+2 x^2\right )+e^{2 x} \left (24 x+8 x^2\right )\right ) \log \left (\frac {x}{\log (3)}\right )}{\left (12 x+4 x^2+e^x \left (3 x+x^2\right )\right ) \log \left (\frac {16+4 e^x}{3+x}\right )+\left (12 x+4 x^2+e^x \left (3 x+x^2\right )\right ) \log \left (\frac {x}{\log (3)}\right )} \, dx=e^{\left (2 \, x\right )} - \log \left (\log \left (\frac {4 \, {\left (e^{x} + 4\right )}}{x + 3}\right ) + \log \left (\frac {x}{\log \left (3\right )}\right )\right ) \]
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Time = 0.41 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {-12+e^x \left (-3-3 x-x^2\right )+\left (e^{3 x} \left (6 x+2 x^2\right )+e^{2 x} \left (24 x+8 x^2\right )\right ) \log \left (\frac {16+4 e^x}{3+x}\right )+\left (e^{3 x} \left (6 x+2 x^2\right )+e^{2 x} \left (24 x+8 x^2\right )\right ) \log \left (\frac {x}{\log (3)}\right )}{\left (12 x+4 x^2+e^x \left (3 x+x^2\right )\right ) \log \left (\frac {16+4 e^x}{3+x}\right )+\left (12 x+4 x^2+e^x \left (3 x+x^2\right )\right ) \log \left (\frac {x}{\log (3)}\right )} \, dx=e^{2 x} - \log {\left (\log {\left (\frac {x}{\log {\left (3 \right )}} \right )} + \log {\left (\frac {4 e^{x} + 16}{x + 3} \right )} \right )} \]
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Time = 0.34 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {-12+e^x \left (-3-3 x-x^2\right )+\left (e^{3 x} \left (6 x+2 x^2\right )+e^{2 x} \left (24 x+8 x^2\right )\right ) \log \left (\frac {16+4 e^x}{3+x}\right )+\left (e^{3 x} \left (6 x+2 x^2\right )+e^{2 x} \left (24 x+8 x^2\right )\right ) \log \left (\frac {x}{\log (3)}\right )}{\left (12 x+4 x^2+e^x \left (3 x+x^2\right )\right ) \log \left (\frac {16+4 e^x}{3+x}\right )+\left (12 x+4 x^2+e^x \left (3 x+x^2\right )\right ) \log \left (\frac {x}{\log (3)}\right )} \, dx=e^{\left (2 \, x\right )} - \log \left (2 \, \log \left (2\right ) - \log \left (x + 3\right ) + \log \left (x\right ) + \log \left (e^{x} + 4\right ) - \log \left (\log \left (3\right )\right )\right ) \]
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Time = 0.37 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {-12+e^x \left (-3-3 x-x^2\right )+\left (e^{3 x} \left (6 x+2 x^2\right )+e^{2 x} \left (24 x+8 x^2\right )\right ) \log \left (\frac {16+4 e^x}{3+x}\right )+\left (e^{3 x} \left (6 x+2 x^2\right )+e^{2 x} \left (24 x+8 x^2\right )\right ) \log \left (\frac {x}{\log (3)}\right )}{\left (12 x+4 x^2+e^x \left (3 x+x^2\right )\right ) \log \left (\frac {16+4 e^x}{3+x}\right )+\left (12 x+4 x^2+e^x \left (3 x+x^2\right )\right ) \log \left (\frac {x}{\log (3)}\right )} \, dx=e^{\left (2 \, x\right )} - \log \left (-\log \left (x + 3\right ) + \log \left (x\right ) + \log \left (4 \, e^{x} + 16\right ) - \log \left (\log \left (3\right )\right )\right ) \]
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Time = 13.55 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {-12+e^x \left (-3-3 x-x^2\right )+\left (e^{3 x} \left (6 x+2 x^2\right )+e^{2 x} \left (24 x+8 x^2\right )\right ) \log \left (\frac {16+4 e^x}{3+x}\right )+\left (e^{3 x} \left (6 x+2 x^2\right )+e^{2 x} \left (24 x+8 x^2\right )\right ) \log \left (\frac {x}{\log (3)}\right )}{\left (12 x+4 x^2+e^x \left (3 x+x^2\right )\right ) \log \left (\frac {16+4 e^x}{3+x}\right )+\left (12 x+4 x^2+e^x \left (3 x+x^2\right )\right ) \log \left (\frac {x}{\log (3)}\right )} \, dx={\mathrm {e}}^{2\,x}-\ln \left (\ln \left (\frac {4\,{\mathrm {e}}^x+16}{\ln \left (3\right )\,\left (x+3\right )}\right )+\ln \left (x\right )\right ) \]
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