Integrand size = 44, antiderivative size = 31 \[ \int \frac {1}{320} e^{-x} \left (-10000+10400 x+\left (400 x-204 x^2\right ) \log \left (x^2\right )+\left (-3 x^2+x^3\right ) \log ^2\left (x^2\right )\right ) \, dx=5+\frac {1}{4} \left (-4-\frac {1}{5} e^{-x} x \left (-25+\frac {1}{4} x \log \left (x^2\right )\right )^2\right ) \]
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\[ \int \frac {1}{320} e^{-x} \left (-10000+10400 x+\left (400 x-204 x^2\right ) \log \left (x^2\right )+\left (-3 x^2+x^3\right ) \log ^2\left (x^2\right )\right ) \, dx=\int \frac {1}{320} e^{-x} \left (-10000+10400 x+\left (400 x-204 x^2\right ) \log \left (x^2\right )+\left (-3 x^2+x^3\right ) \log ^2\left (x^2\right )\right ) \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {1}{320} \int e^{-x} \left (-10000+10400 x+\left (400 x-204 x^2\right ) \log \left (x^2\right )+\left (-3 x^2+x^3\right ) \log ^2\left (x^2\right )\right ) \, dx \\ & = \frac {1}{320} \int \left (-10000 e^{-x}+10400 e^{-x} x-4 e^{-x} x (-100+51 x) \log \left (x^2\right )+e^{-x} (-3+x) x^2 \log ^2\left (x^2\right )\right ) \, dx \\ & = \frac {1}{320} \int e^{-x} (-3+x) x^2 \log ^2\left (x^2\right ) \, dx-\frac {1}{80} \int e^{-x} x (-100+51 x) \log \left (x^2\right ) \, dx-\frac {125}{4} \int e^{-x} \, dx+\frac {65}{2} \int e^{-x} x \, dx \\ & = \frac {125 e^{-x}}{4}-\frac {65 e^{-x} x}{2}+\frac {1}{40} e^{-x} \log \left (x^2\right )+\frac {1}{40} e^{-x} x \log \left (x^2\right )+\frac {51}{80} e^{-x} x^2 \log \left (x^2\right )+\frac {1}{320} \int \left (-3 e^{-x} x^2 \log ^2\left (x^2\right )+e^{-x} x^3 \log ^2\left (x^2\right )\right ) \, dx+\frac {1}{80} \int \frac {2 e^{-x} \left (-2-2 x-51 x^2\right )}{x} \, dx+\frac {65}{2} \int e^{-x} \, dx \\ & = -\frac {5 e^{-x}}{4}-\frac {65 e^{-x} x}{2}+\frac {1}{40} e^{-x} \log \left (x^2\right )+\frac {1}{40} e^{-x} x \log \left (x^2\right )+\frac {51}{80} e^{-x} x^2 \log \left (x^2\right )+\frac {1}{320} \int e^{-x} x^3 \log ^2\left (x^2\right ) \, dx-\frac {3}{320} \int e^{-x} x^2 \log ^2\left (x^2\right ) \, dx+\frac {1}{40} \int \frac {e^{-x} \left (-2-2 x-51 x^2\right )}{x} \, dx \\ & = -\frac {5 e^{-x}}{4}-\frac {65 e^{-x} x}{2}+\frac {1}{40} e^{-x} \log \left (x^2\right )+\frac {1}{40} e^{-x} x \log \left (x^2\right )+\frac {51}{80} e^{-x} x^2 \log \left (x^2\right )+\frac {1}{320} \int e^{-x} x^3 \log ^2\left (x^2\right ) \, dx-\frac {3}{320} \int e^{-x} x^2 \log ^2\left (x^2\right ) \, dx+\frac {1}{40} \int \left (-2 e^{-x}-\frac {2 e^{-x}}{x}-51 e^{-x} x\right ) \, dx \\ & = -\frac {5 e^{-x}}{4}-\frac {65 e^{-x} x}{2}+\frac {1}{40} e^{-x} \log \left (x^2\right )+\frac {1}{40} e^{-x} x \log \left (x^2\right )+\frac {51}{80} e^{-x} x^2 \log \left (x^2\right )+\frac {1}{320} \int e^{-x} x^3 \log ^2\left (x^2\right ) \, dx-\frac {3}{320} \int e^{-x} x^2 \log ^2\left (x^2\right ) \, dx-\frac {1}{20} \int e^{-x} \, dx-\frac {1}{20} \int \frac {e^{-x}}{x} \, dx-\frac {51}{40} \int e^{-x} x \, dx \\ & = -\frac {6 e^{-x}}{5}-\frac {1249 e^{-x} x}{40}-\frac {\operatorname {ExpIntegralEi}(-x)}{20}+\frac {1}{40} e^{-x} \log \left (x^2\right )+\frac {1}{40} e^{-x} x \log \left (x^2\right )+\frac {51}{80} e^{-x} x^2 \log \left (x^2\right )+\frac {1}{320} \int e^{-x} x^3 \log ^2\left (x^2\right ) \, dx-\frac {3}{320} \int e^{-x} x^2 \log ^2\left (x^2\right ) \, dx-\frac {51}{40} \int e^{-x} \, dx \\ & = \frac {3 e^{-x}}{40}-\frac {1249 e^{-x} x}{40}-\frac {\operatorname {ExpIntegralEi}(-x)}{20}+\frac {1}{40} e^{-x} \log \left (x^2\right )+\frac {1}{40} e^{-x} x \log \left (x^2\right )+\frac {51}{80} e^{-x} x^2 \log \left (x^2\right )+\frac {1}{320} \int e^{-x} x^3 \log ^2\left (x^2\right ) \, dx-\frac {3}{320} \int e^{-x} x^2 \log ^2\left (x^2\right ) \, dx \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.65 \[ \int \frac {1}{320} e^{-x} \left (-10000+10400 x+\left (400 x-204 x^2\right ) \log \left (x^2\right )+\left (-3 x^2+x^3\right ) \log ^2\left (x^2\right )\right ) \, dx=-\frac {1}{320} e^{-x} x \left (-100+x \log \left (x^2\right )\right )^2 \]
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Time = 0.16 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97
method | result | size |
parallelrisch | \(-\frac {\left (x^{3} \ln \left (x^{2}\right )^{2}-200 x^{2} \ln \left (x^{2}\right )+10000 x \right ) {\mathrm e}^{-x}}{320}\) | \(30\) |
risch | \(-\frac {x^{3} {\mathrm e}^{-x} \ln \left (x \right )^{2}}{80}+\frac {i x^{2} \left (x \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 x \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+x \pi \operatorname {csgn}\left (i x^{2}\right )^{3}-200 i\right ) {\mathrm e}^{-x} \ln \left (x \right )}{160}+\frac {\left (-40000 x -400 i \pi \,x^{2} \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+800 i \pi \,x^{2} \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}-400 i \pi \,x^{2} \operatorname {csgn}\left (i x^{2}\right )^{3}+x^{3} \pi ^{2} \operatorname {csgn}\left (i x \right )^{4} \operatorname {csgn}\left (i x^{2}\right )^{2}-4 x^{3} \pi ^{2} \operatorname {csgn}\left (i x \right )^{3} \operatorname {csgn}\left (i x^{2}\right )^{3}+6 x^{3} \pi ^{2} \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )^{4}-4 x^{3} \pi ^{2} \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{5}+x^{3} \pi ^{2} \operatorname {csgn}\left (i x^{2}\right )^{6}\right ) {\mathrm e}^{-x}}{1280}\) | \(254\) |
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Time = 0.24 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16 \[ \int \frac {1}{320} e^{-x} \left (-10000+10400 x+\left (400 x-204 x^2\right ) \log \left (x^2\right )+\left (-3 x^2+x^3\right ) \log ^2\left (x^2\right )\right ) \, dx=-\frac {1}{320} \, x^{3} e^{\left (-x\right )} \log \left (x^{2}\right )^{2} + \frac {5}{8} \, x^{2} e^{\left (-x\right )} \log \left (x^{2}\right ) - \frac {125}{4} \, x e^{\left (-x\right )} \]
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Time = 0.14 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {1}{320} e^{-x} \left (-10000+10400 x+\left (400 x-204 x^2\right ) \log \left (x^2\right )+\left (-3 x^2+x^3\right ) \log ^2\left (x^2\right )\right ) \, dx=\frac {\left (- x^{3} \log {\left (x^{2} \right )}^{2} + 200 x^{2} \log {\left (x^{2} \right )} - 10000 x\right ) e^{- x}}{320} \]
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Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (19) = 38\).
Time = 0.23 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35 \[ \int \frac {1}{320} e^{-x} \left (-10000+10400 x+\left (400 x-204 x^2\right ) \log \left (x^2\right )+\left (-3 x^2+x^3\right ) \log ^2\left (x^2\right )\right ) \, dx=-\frac {1}{80} \, {\left (x^{3} \log \left (x\right )^{2} - 100 \, x^{2} \log \left (x\right ) - 100 \, x - 100\right )} e^{\left (-x\right )} - \frac {65}{2} \, {\left (x + 1\right )} e^{\left (-x\right )} + \frac {125}{4} \, e^{\left (-x\right )} \]
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Time = 0.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16 \[ \int \frac {1}{320} e^{-x} \left (-10000+10400 x+\left (400 x-204 x^2\right ) \log \left (x^2\right )+\left (-3 x^2+x^3\right ) \log ^2\left (x^2\right )\right ) \, dx=-\frac {1}{320} \, x^{3} e^{\left (-x\right )} \log \left (x^{2}\right )^{2} + \frac {5}{8} \, x^{2} e^{\left (-x\right )} \log \left (x^{2}\right ) - \frac {125}{4} \, x e^{\left (-x\right )} \]
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Time = 9.41 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.55 \[ \int \frac {1}{320} e^{-x} \left (-10000+10400 x+\left (400 x-204 x^2\right ) \log \left (x^2\right )+\left (-3 x^2+x^3\right ) \log ^2\left (x^2\right )\right ) \, dx=-\frac {x\,{\mathrm {e}}^{-x}\,{\left (x\,\ln \left (x^2\right )-100\right )}^2}{320} \]
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