Integrand size = 98, antiderivative size = 27 \[ \int \frac {e^8 \left (5+8 x-2 x^2\right )+\left (5+8 x-2 x^2\right ) \log (5)}{-15 x^3+3 x^4+\left (-30 x^2+6 x^3\right ) \log \left (e^{5+x} \left (5 x-x^2\right )\right )+\left (-15 x+3 x^2\right ) \log ^2\left (e^{5+x} \left (5 x-x^2\right )\right )} \, dx=\frac {e^8+\log (5)}{3 \left (x+\log \left (e^{5+x} (5-x) x\right )\right )} \]
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\[ \int \frac {e^8 \left (5+8 x-2 x^2\right )+\left (5+8 x-2 x^2\right ) \log (5)}{-15 x^3+3 x^4+\left (-30 x^2+6 x^3\right ) \log \left (e^{5+x} \left (5 x-x^2\right )\right )+\left (-15 x+3 x^2\right ) \log ^2\left (e^{5+x} \left (5 x-x^2\right )\right )} \, dx=\int \frac {e^8 \left (5+8 x-2 x^2\right )+\left (5+8 x-2 x^2\right ) \log (5)}{-15 x^3+3 x^4+\left (-30 x^2+6 x^3\right ) \log \left (e^{5+x} \left (5 x-x^2\right )\right )+\left (-15 x+3 x^2\right ) \log ^2\left (e^{5+x} \left (5 x-x^2\right )\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (5+8 x-2 x^2\right ) \left (e^8+\log (5)\right )}{-15 x^3+3 x^4+\left (-30 x^2+6 x^3\right ) \log \left (e^{5+x} \left (5 x-x^2\right )\right )+\left (-15 x+3 x^2\right ) \log ^2\left (e^{5+x} \left (5 x-x^2\right )\right )} \, dx \\ & = \left (e^8+\log (5)\right ) \int \frac {5+8 x-2 x^2}{-15 x^3+3 x^4+\left (-30 x^2+6 x^3\right ) \log \left (e^{5+x} \left (5 x-x^2\right )\right )+\left (-15 x+3 x^2\right ) \log ^2\left (e^{5+x} \left (5 x-x^2\right )\right )} \, dx \\ & = \left (e^8+\log (5)\right ) \int \frac {-5-8 x+2 x^2}{3 (5-x) x \left (x+\log \left (-e^{5+x} (-5+x) x\right )\right )^2} \, dx \\ & = \frac {1}{3} \left (e^8+\log (5)\right ) \int \frac {-5-8 x+2 x^2}{(5-x) x \left (x+\log \left (-e^{5+x} (-5+x) x\right )\right )^2} \, dx \\ & = \frac {1}{3} \left (e^8+\log (5)\right ) \int \left (-\frac {2}{\left (x+\log \left (-e^{5+x} (-5+x) x\right )\right )^2}-\frac {1}{(-5+x) \left (x+\log \left (-e^{5+x} (-5+x) x\right )\right )^2}-\frac {1}{x \left (x+\log \left (-e^{5+x} (-5+x) x\right )\right )^2}\right ) \, dx \\ & = \frac {1}{3} \left (-e^8-\log (5)\right ) \int \frac {1}{(-5+x) \left (x+\log \left (-e^{5+x} (-5+x) x\right )\right )^2} \, dx+\frac {1}{3} \left (-e^8-\log (5)\right ) \int \frac {1}{x \left (x+\log \left (-e^{5+x} (-5+x) x\right )\right )^2} \, dx-\frac {1}{3} \left (2 \left (e^8+\log (5)\right )\right ) \int \frac {1}{\left (x+\log \left (-e^{5+x} (-5+x) x\right )\right )^2} \, dx \\ \end{align*}
Time = 0.12 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {e^8 \left (5+8 x-2 x^2\right )+\left (5+8 x-2 x^2\right ) \log (5)}{-15 x^3+3 x^4+\left (-30 x^2+6 x^3\right ) \log \left (e^{5+x} \left (5 x-x^2\right )\right )+\left (-15 x+3 x^2\right ) \log ^2\left (e^{5+x} \left (5 x-x^2\right )\right )} \, dx=\frac {e^8+\log (5)}{3 \left (x+\log \left (-e^{5+x} (-5+x) x\right )\right )} \]
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Time = 0.56 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07
method | result | size |
parallelrisch | \(\frac {\ln \left (5\right )+{\mathrm e}^{8}}{3 x +3 \ln \left (\left (-x^{2}+5 x \right ) {\mathrm e}^{5+x}\right )}\) | \(29\) |
risch | \(-\frac {2 i \ln \left (5\right )}{3 \left (-2 \pi \operatorname {csgn}\left (i x \left (-5+x \right ) {\mathrm e}^{5+x}\right )^{2}-\pi \,\operatorname {csgn}\left (i \left (-5+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{5+x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{5+x} \left (-5+x \right )\right )+\pi \,\operatorname {csgn}\left (i \left (-5+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{5+x} \left (-5+x \right )\right )^{2}+\pi \,\operatorname {csgn}\left (i {\mathrm e}^{5+x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{5+x} \left (-5+x \right )\right )^{2}-\pi \operatorname {csgn}\left (i {\mathrm e}^{5+x} \left (-5+x \right )\right )^{3}+\pi \,\operatorname {csgn}\left (i {\mathrm e}^{5+x} \left (-5+x \right )\right ) \operatorname {csgn}\left (i x \left (-5+x \right ) {\mathrm e}^{5+x}\right )^{2}-\pi \,\operatorname {csgn}\left (i {\mathrm e}^{5+x} \left (-5+x \right )\right ) \operatorname {csgn}\left (i x \left (-5+x \right ) {\mathrm e}^{5+x}\right ) \operatorname {csgn}\left (i x \right )+\pi \operatorname {csgn}\left (i x \left (-5+x \right ) {\mathrm e}^{5+x}\right )^{3}+\pi \operatorname {csgn}\left (i x \left (-5+x \right ) {\mathrm e}^{5+x}\right )^{2} \operatorname {csgn}\left (i x \right )+2 \pi -2 i \ln \left (x \right )-2 i x -2 i \ln \left ({\mathrm e}^{5+x}\right )-2 i \ln \left (-5+x \right )\right )}-\frac {2 i {\mathrm e}^{8}}{3 \left (-2 \pi \operatorname {csgn}\left (i x \left (-5+x \right ) {\mathrm e}^{5+x}\right )^{2}-\pi \,\operatorname {csgn}\left (i \left (-5+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{5+x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{5+x} \left (-5+x \right )\right )+\pi \,\operatorname {csgn}\left (i \left (-5+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{5+x} \left (-5+x \right )\right )^{2}+\pi \,\operatorname {csgn}\left (i {\mathrm e}^{5+x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{5+x} \left (-5+x \right )\right )^{2}-\pi \operatorname {csgn}\left (i {\mathrm e}^{5+x} \left (-5+x \right )\right )^{3}+\pi \,\operatorname {csgn}\left (i {\mathrm e}^{5+x} \left (-5+x \right )\right ) \operatorname {csgn}\left (i x \left (-5+x \right ) {\mathrm e}^{5+x}\right )^{2}-\pi \,\operatorname {csgn}\left (i {\mathrm e}^{5+x} \left (-5+x \right )\right ) \operatorname {csgn}\left (i x \left (-5+x \right ) {\mathrm e}^{5+x}\right ) \operatorname {csgn}\left (i x \right )+\pi \operatorname {csgn}\left (i x \left (-5+x \right ) {\mathrm e}^{5+x}\right )^{3}+\pi \operatorname {csgn}\left (i x \left (-5+x \right ) {\mathrm e}^{5+x}\right )^{2} \operatorname {csgn}\left (i x \right )+2 \pi -2 i \ln \left (x \right )-2 i x -2 i \ln \left ({\mathrm e}^{5+x}\right )-2 i \ln \left (-5+x \right )\right )}\) | \(476\) |
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Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {e^8 \left (5+8 x-2 x^2\right )+\left (5+8 x-2 x^2\right ) \log (5)}{-15 x^3+3 x^4+\left (-30 x^2+6 x^3\right ) \log \left (e^{5+x} \left (5 x-x^2\right )\right )+\left (-15 x+3 x^2\right ) \log ^2\left (e^{5+x} \left (5 x-x^2\right )\right )} \, dx=\frac {e^{8} + \log \left (5\right )}{3 \, {\left (x + \log \left (-{\left (x^{2} - 5 \, x\right )} e^{\left (x + 5\right )}\right )\right )}} \]
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Time = 0.12 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {e^8 \left (5+8 x-2 x^2\right )+\left (5+8 x-2 x^2\right ) \log (5)}{-15 x^3+3 x^4+\left (-30 x^2+6 x^3\right ) \log \left (e^{5+x} \left (5 x-x^2\right )\right )+\left (-15 x+3 x^2\right ) \log ^2\left (e^{5+x} \left (5 x-x^2\right )\right )} \, dx=\frac {\log {\left (5 \right )} + e^{8}}{3 x + 3 \log {\left (\left (- x^{2} + 5 x\right ) e^{x + 5} \right )}} \]
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Time = 0.37 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {e^8 \left (5+8 x-2 x^2\right )+\left (5+8 x-2 x^2\right ) \log (5)}{-15 x^3+3 x^4+\left (-30 x^2+6 x^3\right ) \log \left (e^{5+x} \left (5 x-x^2\right )\right )+\left (-15 x+3 x^2\right ) \log ^2\left (e^{5+x} \left (5 x-x^2\right )\right )} \, dx=\frac {e^{8} + \log \left (5\right )}{3 \, {\left (2 \, x + \log \left (x\right ) + \log \left (-x + 5\right ) + 5\right )}} \]
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Time = 0.36 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {e^8 \left (5+8 x-2 x^2\right )+\left (5+8 x-2 x^2\right ) \log (5)}{-15 x^3+3 x^4+\left (-30 x^2+6 x^3\right ) \log \left (e^{5+x} \left (5 x-x^2\right )\right )+\left (-15 x+3 x^2\right ) \log ^2\left (e^{5+x} \left (5 x-x^2\right )\right )} \, dx=\frac {e^{8} + \log \left (5\right )}{3 \, {\left (2 \, x + \log \left (-x^{2} + 5 \, x\right ) + 5\right )}} \]
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Time = 14.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {e^8 \left (5+8 x-2 x^2\right )+\left (5+8 x-2 x^2\right ) \log (5)}{-15 x^3+3 x^4+\left (-30 x^2+6 x^3\right ) \log \left (e^{5+x} \left (5 x-x^2\right )\right )+\left (-15 x+3 x^2\right ) \log ^2\left (e^{5+x} \left (5 x-x^2\right )\right )} \, dx=\frac {\frac {{\mathrm {e}}^8}{3}+\frac {\ln \left (5\right )}{3}}{x+\ln \left ({\mathrm {e}}^5\,{\mathrm {e}}^x\,\left (5\,x-x^2\right )\right )} \]
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