Integrand size = 72, antiderivative size = 23 \[ \int \frac {x-\log \left (-130 e^{-4-x}\right )+4 \log ^2\left (-130 e^{-4-x}\right )}{1+(-8-4 x) \log \left (-130 e^{-4-x}\right )+\left (16+16 x+4 x^2\right ) \log ^2\left (-130 e^{-4-x}\right )} \, dx=\frac {x}{4+2 x-\frac {1}{\log \left (-130 e^{-4-x}\right )}} \]
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\[ \int \frac {x-\log \left (-130 e^{-4-x}\right )+4 \log ^2\left (-130 e^{-4-x}\right )}{1+(-8-4 x) \log \left (-130 e^{-4-x}\right )+\left (16+16 x+4 x^2\right ) \log ^2\left (-130 e^{-4-x}\right )} \, dx=\int \frac {x-\log \left (-130 e^{-4-x}\right )+4 \log ^2\left (-130 e^{-4-x}\right )}{1+(-8-4 x) \log \left (-130 e^{-4-x}\right )+\left (16+16 x+4 x^2\right ) \log ^2\left (-130 e^{-4-x}\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {x-\log \left (-130 e^{-4-x}\right )+4 \log ^2\left (-130 e^{-4-x}\right )}{\left (1-2 (2+x) \log \left (-130 e^{-4-x}\right )\right )^2} \, dx \\ & = \int \left (\frac {1}{(2+x)^2}+\frac {x \left (7+8 x+2 x^2\right )}{2 (2+x)^2 \left (-1+4 \log \left (-130 e^{-4-x}\right )+2 x \log \left (-130 e^{-4-x}\right )\right )^2}+\frac {2-x}{2 (2+x)^2 \left (-1+4 \log \left (-130 e^{-4-x}\right )+2 x \log \left (-130 e^{-4-x}\right )\right )}\right ) \, dx \\ & = -\frac {1}{2+x}+\frac {1}{2} \int \frac {x \left (7+8 x+2 x^2\right )}{(2+x)^2 \left (-1+4 \log \left (-130 e^{-4-x}\right )+2 x \log \left (-130 e^{-4-x}\right )\right )^2} \, dx+\frac {1}{2} \int \frac {2-x}{(2+x)^2 \left (-1+4 \log \left (-130 e^{-4-x}\right )+2 x \log \left (-130 e^{-4-x}\right )\right )} \, dx \\ & = -\frac {1}{2+x}+\frac {1}{2} \int \frac {x \left (7+8 x+2 x^2\right )}{(2+x)^2 \left (1-2 (2+x) \log \left (-130 e^{-4-x}\right )\right )^2} \, dx+\frac {1}{2} \int \frac {-2+x}{(2+x)^2 \left (1-2 (2+x) \log \left (-130 e^{-4-x}\right )\right )} \, dx \\ & = -\frac {1}{2+x}+\frac {1}{2} \int \left (\frac {2 x}{\left (-1+4 \log \left (-130 e^{-4-x}\right )+2 x \log \left (-130 e^{-4-x}\right )\right )^2}+\frac {2}{(2+x)^2 \left (-1+4 \log \left (-130 e^{-4-x}\right )+2 x \log \left (-130 e^{-4-x}\right )\right )^2}-\frac {1}{(2+x) \left (-1+4 \log \left (-130 e^{-4-x}\right )+2 x \log \left (-130 e^{-4-x}\right )\right )^2}\right ) \, dx+\frac {1}{2} \int \left (\frac {4}{(2+x)^2 \left (-1+4 \log \left (-130 e^{-4-x}\right )+2 x \log \left (-130 e^{-4-x}\right )\right )}-\frac {1}{(2+x) \left (-1+4 \log \left (-130 e^{-4-x}\right )+2 x \log \left (-130 e^{-4-x}\right )\right )}\right ) \, dx \\ & = -\frac {1}{2+x}-\frac {1}{2} \int \frac {1}{(2+x) \left (-1+4 \log \left (-130 e^{-4-x}\right )+2 x \log \left (-130 e^{-4-x}\right )\right )^2} \, dx-\frac {1}{2} \int \frac {1}{(2+x) \left (-1+4 \log \left (-130 e^{-4-x}\right )+2 x \log \left (-130 e^{-4-x}\right )\right )} \, dx+2 \int \frac {1}{(2+x)^2 \left (-1+4 \log \left (-130 e^{-4-x}\right )+2 x \log \left (-130 e^{-4-x}\right )\right )} \, dx+\int \frac {x}{\left (-1+4 \log \left (-130 e^{-4-x}\right )+2 x \log \left (-130 e^{-4-x}\right )\right )^2} \, dx+\int \frac {1}{(2+x)^2 \left (-1+4 \log \left (-130 e^{-4-x}\right )+2 x \log \left (-130 e^{-4-x}\right )\right )^2} \, dx \\ & = -\frac {1}{2+x}-\frac {1}{2} \int \frac {1}{(2+x) \left (-1+4 \log \left (-130 e^{-4-x}\right )+2 x \log \left (-130 e^{-4-x}\right )\right )} \, dx-\frac {1}{2} \int \frac {1}{(2+x) \left (1-2 (2+x) \log \left (-130 e^{-4-x}\right )\right )^2} \, dx+2 \int \frac {1}{(2+x)^2 \left (-1+2 (2+x) \log \left (-130 e^{-4-x}\right )\right )} \, dx+\int \frac {x}{\left (1-2 (2+x) \log \left (-130 e^{-4-x}\right )\right )^2} \, dx+\int \frac {1}{(2+x)^2 \left (1-2 (2+x) \log \left (-130 e^{-4-x}\right )\right )^2} \, dx \\ \end{align*}
Time = 0.17 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48 \[ \int \frac {x-\log \left (-130 e^{-4-x}\right )+4 \log ^2\left (-130 e^{-4-x}\right )}{1+(-8-4 x) \log \left (-130 e^{-4-x}\right )+\left (16+16 x+4 x^2\right ) \log ^2\left (-130 e^{-4-x}\right )} \, dx=\frac {1-4 \log \left (-130 e^{-4-x}\right )}{-2+4 (2+x) \log \left (-130 e^{-4-x}\right )} \]
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Time = 0.43 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.83
method | result | size |
norman | \(\frac {\frac {1}{2}-2 \ln \left (-130 \,{\mathrm e}^{-4-x}\right )}{2 \ln \left (-130 \,{\mathrm e}^{-4-x}\right ) x +4 \ln \left (-130 \,{\mathrm e}^{-4-x}\right )-1}\) | \(42\) |
parallelrisch | \(\frac {-8 \ln \left (-130 \,{\mathrm e}^{-4-x}\right )+2}{8 \ln \left (-130 \,{\mathrm e}^{-4-x}\right ) x +16 \ln \left (-130 \,{\mathrm e}^{-4-x}\right )-4}\) | \(43\) |
default | \(\frac {-\frac {1}{4}+\ln \left (-130 \,{\mathrm e}^{-4-x}\right )}{x^{2}+x \left (\ln \left ({\mathrm e}^{4+x}\right )-4-x \right )-x \left (\ln \left (-130 \,{\mathrm e}^{-4-x}\right )+\ln \left ({\mathrm e}^{4+x}\right )\right )+4 x +\frac {1}{2}-2 \ln \left (-130 \,{\mathrm e}^{-4-x}\right )}\) | \(64\) |
risch | \(-\frac {1}{2+x}-\frac {i x}{2 \left (2+x \right ) \left (-2 x \pi \operatorname {csgn}\left (i {\mathrm e}^{-4-x}\right )^{2}+2 x \pi \operatorname {csgn}\left (i {\mathrm e}^{-4-x}\right )^{3}-4 \pi \operatorname {csgn}\left (i {\mathrm e}^{-4-x}\right )^{2}+4 \pi \operatorname {csgn}\left (i {\mathrm e}^{-4-x}\right )^{3}+2 \pi x +4 \pi -4 i \ln \left (13\right )+4 i \ln \left ({\mathrm e}^{4+x}\right )-2 i x \ln \left (13\right )-2 i x \ln \left (2\right )-2 i x \ln \left (5\right )-4 i \ln \left (2\right )-4 i \ln \left (5\right )+2 i x \ln \left ({\mathrm e}^{4+x}\right )+i\right )}\) | \(142\) |
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Result contains complex when optimal does not.
Time = 0.24 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.61 \[ \int \frac {x-\log \left (-130 e^{-4-x}\right )+4 \log ^2\left (-130 e^{-4-x}\right )}{1+(-8-4 x) \log \left (-130 e^{-4-x}\right )+\left (16+16 x+4 x^2\right ) \log ^2\left (-130 e^{-4-x}\right )} \, dx=\frac {-4 i \, \pi + 4 \, x - 4 \, \log \left (130\right ) + 17}{2 \, {\left (2 \, {\left (i \, \pi + \log \left (130\right )\right )} {\left (x + 2\right )} - 2 \, x^{2} - 12 \, x - 17\right )}} \]
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Result contains complex when optimal does not.
Time = 1.09 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.91 \[ \int \frac {x-\log \left (-130 e^{-4-x}\right )+4 \log ^2\left (-130 e^{-4-x}\right )}{1+(-8-4 x) \log \left (-130 e^{-4-x}\right )+\left (16+16 x+4 x^2\right ) \log ^2\left (-130 e^{-4-x}\right )} \, dx=\frac {- 4 x - 17 + 4 \log {\left (130 \right )} + 4 i \pi }{4 x^{2} + x \left (- 4 \log {\left (130 \right )} + 24 - 4 i \pi \right ) - 8 \log {\left (130 \right )} + 34 - 8 i \pi } \]
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Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (22) = 44\).
Time = 0.33 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.61 \[ \int \frac {x-\log \left (-130 e^{-4-x}\right )+4 \log ^2\left (-130 e^{-4-x}\right )}{1+(-8-4 x) \log \left (-130 e^{-4-x}\right )+\left (16+16 x+4 x^2\right ) \log ^2\left (-130 e^{-4-x}\right )} \, dx=-\frac {4 \, \log \left (13\right ) + 4 \, \log \left (5\right ) + 4 \, \log \left (2\right ) - 4 \, \log \left (-e^{x}\right ) - 17}{2 \, {\left (2 \, x {\left (\log \left (13\right ) + \log \left (5\right ) + \log \left (2\right ) - 4\right )} - 2 \, {\left (x + 2\right )} \log \left (-e^{x}\right ) + 4 \, \log \left (13\right ) + 4 \, \log \left (5\right ) + 4 \, \log \left (2\right ) - 17\right )}} \]
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Result contains complex when optimal does not.
Time = 0.33 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.78 \[ \int \frac {x-\log \left (-130 e^{-4-x}\right )+4 \log ^2\left (-130 e^{-4-x}\right )}{1+(-8-4 x) \log \left (-130 e^{-4-x}\right )+\left (16+16 x+4 x^2\right ) \log ^2\left (-130 e^{-4-x}\right )} \, dx=\frac {4 i \, \pi - 4 \, x + 4 \, \log \left (130\right ) - 17}{-8 i \, \pi - 4 i \, \pi x + 4 \, x^{2} - 4 \, x \log \left (130\right ) + 24 \, x - 8 \, \log \left (130\right ) + 34} \]
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Time = 14.42 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.87 \[ \int \frac {x-\log \left (-130 e^{-4-x}\right )+4 \log ^2\left (-130 e^{-4-x}\right )}{1+(-8-4 x) \log \left (-130 e^{-4-x}\right )+\left (16+16 x+4 x^2\right ) \log ^2\left (-130 e^{-4-x}\right )} \, dx=\frac {2\,x-2\,\ln \left (130\right )+\frac {17}{2}-\pi \,2{}\mathrm {i}}{-2\,x^2+\left (2\,\ln \left (130\right )-12+\pi \,2{}\mathrm {i}\right )\,x+\pi \,4{}\mathrm {i}+4\,\ln \left (130\right )-17} \]
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