Integrand size = 86, antiderivative size = 24 \[ \int \frac {e^{-x} \left (e^5 (-180-36 x)+e^5 (-90-18 x) \log (4)+\left (e^5 \left (-126 x-18 x^2\right )+e^5 \left (-63 x-9 x^2\right ) \log (4)\right ) \log \left (x^2\right )\right )}{\left (125 x+75 x^2+15 x^3+x^4\right ) \log ^2\left (x^2\right )} \, dx=\frac {9 e^{5-x} (2+\log (4))}{(5+x)^2 \log \left (x^2\right )} \]
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\[ \int \frac {e^{-x} \left (e^5 (-180-36 x)+e^5 (-90-18 x) \log (4)+\left (e^5 \left (-126 x-18 x^2\right )+e^5 \left (-63 x-9 x^2\right ) \log (4)\right ) \log \left (x^2\right )\right )}{\left (125 x+75 x^2+15 x^3+x^4\right ) \log ^2\left (x^2\right )} \, dx=\int \frac {e^{-x} \left (e^5 (-180-36 x)+e^5 (-90-18 x) \log (4)+\left (e^5 \left (-126 x-18 x^2\right )+e^5 \left (-63 x-9 x^2\right ) \log (4)\right ) \log \left (x^2\right )\right )}{\left (125 x+75 x^2+15 x^3+x^4\right ) \log ^2\left (x^2\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {9 e^{5-x} (2+\log (4)) \left (-2 (5+x)-x (7+x) \log \left (x^2\right )\right )}{x (5+x)^3 \log ^2\left (x^2\right )} \, dx \\ & = (9 (2+\log (4))) \int \frac {e^{5-x} \left (-2 (5+x)-x (7+x) \log \left (x^2\right )\right )}{x (5+x)^3 \log ^2\left (x^2\right )} \, dx \\ & = (9 (2+\log (4))) \int \left (-\frac {2 e^{5-x}}{x (5+x)^2 \log ^2\left (x^2\right )}+\frac {e^{5-x} (-7-x)}{(5+x)^3 \log \left (x^2\right )}\right ) \, dx \\ & = (9 (2+\log (4))) \int \frac {e^{5-x} (-7-x)}{(5+x)^3 \log \left (x^2\right )} \, dx-(18 (2+\log (4))) \int \frac {e^{5-x}}{x (5+x)^2 \log ^2\left (x^2\right )} \, dx \\ & = (9 (2+\log (4))) \int \left (-\frac {2 e^{5-x}}{(5+x)^3 \log \left (x^2\right )}-\frac {e^{5-x}}{(5+x)^2 \log \left (x^2\right )}\right ) \, dx-(18 (2+\log (4))) \int \left (\frac {e^{5-x}}{25 x \log ^2\left (x^2\right )}-\frac {e^{5-x}}{5 (5+x)^2 \log ^2\left (x^2\right )}-\frac {e^{5-x}}{25 (5+x) \log ^2\left (x^2\right )}\right ) \, dx \\ & = -\left (\frac {1}{25} (18 (2+\log (4))) \int \frac {e^{5-x}}{x \log ^2\left (x^2\right )} \, dx\right )+\frac {1}{25} (18 (2+\log (4))) \int \frac {e^{5-x}}{(5+x) \log ^2\left (x^2\right )} \, dx+\frac {1}{5} (18 (2+\log (4))) \int \frac {e^{5-x}}{(5+x)^2 \log ^2\left (x^2\right )} \, dx-(9 (2+\log (4))) \int \frac {e^{5-x}}{(5+x)^2 \log \left (x^2\right )} \, dx-(18 (2+\log (4))) \int \frac {e^{5-x}}{(5+x)^3 \log \left (x^2\right )} \, dx \\ \end{align*}
Time = 0.22 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-x} \left (e^5 (-180-36 x)+e^5 (-90-18 x) \log (4)+\left (e^5 \left (-126 x-18 x^2\right )+e^5 \left (-63 x-9 x^2\right ) \log (4)\right ) \log \left (x^2\right )\right )}{\left (125 x+75 x^2+15 x^3+x^4\right ) \log ^2\left (x^2\right )} \, dx=\frac {9 e^{5-x} (2+\log (4))}{(5+x)^2 \log \left (x^2\right )} \]
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Time = 7.97 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42
method | result | size |
parallelrisch | \(\frac {\left (36 \,{\mathrm e}^{5} \ln \left (2\right )+36 \,{\mathrm e}^{5}\right ) {\mathrm e}^{-x}}{2 \ln \left (x^{2}\right ) \left (x^{2}+10 x +25\right )}\) | \(34\) |
risch | \(\frac {36 i \left (1+\ln \left (2\right )\right ) {\mathrm e}^{5-x}}{\left (\pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+\pi \operatorname {csgn}\left (i x^{2}\right )^{3}+4 i \ln \left (x \right )\right ) \left (x^{2}+10 x +25\right )}\) | \(76\) |
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Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {e^{-x} \left (e^5 (-180-36 x)+e^5 (-90-18 x) \log (4)+\left (e^5 \left (-126 x-18 x^2\right )+e^5 \left (-63 x-9 x^2\right ) \log (4)\right ) \log \left (x^2\right )\right )}{\left (125 x+75 x^2+15 x^3+x^4\right ) \log ^2\left (x^2\right )} \, dx=\frac {18 \, {\left (e^{5} \log \left (2\right ) + e^{5}\right )} e^{\left (-x\right )}}{{\left (x^{2} + 10 \, x + 25\right )} \log \left (x^{2}\right )} \]
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Time = 0.14 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62 \[ \int \frac {e^{-x} \left (e^5 (-180-36 x)+e^5 (-90-18 x) \log (4)+\left (e^5 \left (-126 x-18 x^2\right )+e^5 \left (-63 x-9 x^2\right ) \log (4)\right ) \log \left (x^2\right )\right )}{\left (125 x+75 x^2+15 x^3+x^4\right ) \log ^2\left (x^2\right )} \, dx=\frac {\left (18 e^{5} \log {\left (2 \right )} + 18 e^{5}\right ) e^{- x}}{x^{2} \log {\left (x^{2} \right )} + 10 x \log {\left (x^{2} \right )} + 25 \log {\left (x^{2} \right )}} \]
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Time = 0.31 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^{-x} \left (e^5 (-180-36 x)+e^5 (-90-18 x) \log (4)+\left (e^5 \left (-126 x-18 x^2\right )+e^5 \left (-63 x-9 x^2\right ) \log (4)\right ) \log \left (x^2\right )\right )}{\left (125 x+75 x^2+15 x^3+x^4\right ) \log ^2\left (x^2\right )} \, dx=\frac {9 \, {\left (\log \left (2\right ) + 1\right )} e^{\left (-x + 5\right )}}{{\left (x^{2} + 10 \, x + 25\right )} \log \left (x\right )} \]
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Time = 0.31 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.75 \[ \int \frac {e^{-x} \left (e^5 (-180-36 x)+e^5 (-90-18 x) \log (4)+\left (e^5 \left (-126 x-18 x^2\right )+e^5 \left (-63 x-9 x^2\right ) \log (4)\right ) \log \left (x^2\right )\right )}{\left (125 x+75 x^2+15 x^3+x^4\right ) \log ^2\left (x^2\right )} \, dx=\frac {18 \, {\left (e^{\left (-x + 5\right )} \log \left (2\right ) + e^{\left (-x + 5\right )}\right )}}{x^{2} \log \left (x^{2}\right ) + 10 \, x \log \left (x^{2}\right ) + 25 \, \log \left (x^{2}\right )} \]
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Time = 14.96 (sec) , antiderivative size = 87, normalized size of antiderivative = 3.62 \[ \int \frac {e^{-x} \left (e^5 (-180-36 x)+e^5 (-90-18 x) \log (4)+\left (e^5 \left (-126 x-18 x^2\right )+e^5 \left (-63 x-9 x^2\right ) \log (4)\right ) \log \left (x^2\right )\right )}{\left (125 x+75 x^2+15 x^3+x^4\right ) \log ^2\left (x^2\right )} \, dx=\frac {9\,\left (\ln \left (2\right )+1\right )\,\left (7\,x\,{\mathrm {e}}^{5-x}+x^2\,{\mathrm {e}}^{5-x}-7\,x\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^5-x^2\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^5\right )}{{\left (x+5\right )}^3}+\frac {9\,\left (\ln \left (2\right )+1\right )\,\left (10\,{\mathrm {e}}^{5-x}+2\,x\,{\mathrm {e}}^{5-x}\right )}{\ln \left (x^2\right )\,{\left (x+5\right )}^3} \]
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