Integrand size = 53, antiderivative size = 25 \[ \int \frac {e^{-x} \left (-5600+525 x+20500 x^5-750 x^6-12500 x^{10}\right )}{3675-630 x+27 x^2-10500 x^5+900 x^6+7500 x^{10}} \, dx=\frac {5 e^{-x}}{3-\frac {9 x}{5 \left (7-10 x^5\right )}} \]
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\[ \int \frac {e^{-x} \left (-5600+525 x+20500 x^5-750 x^6-12500 x^{10}\right )}{3675-630 x+27 x^2-10500 x^5+900 x^6+7500 x^{10}} \, dx=\int \frac {e^{-x} \left (-5600+525 x+20500 x^5-750 x^6-12500 x^{10}\right )}{3675-630 x+27 x^2-10500 x^5+900 x^6+7500 x^{10}} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {25 e^{-x} \left (-224+21 x+820 x^5-30 x^6-500 x^{10}\right )}{3 \left (35-3 x-50 x^5\right )^2} \, dx \\ & = \frac {25}{3} \int \frac {e^{-x} \left (-224+21 x+820 x^5-30 x^6-500 x^{10}\right )}{\left (35-3 x-50 x^5\right )^2} \, dx \\ & = \frac {25}{3} \int \left (-\frac {e^{-x}}{5}-\frac {3 e^{-x} (-175+12 x)}{5 \left (-35+3 x+50 x^5\right )^2}+\frac {3 e^{-x} (4+x)}{5 \left (-35+3 x+50 x^5\right )}\right ) \, dx \\ & = -\left (\frac {5}{3} \int e^{-x} \, dx\right )-5 \int \frac {e^{-x} (-175+12 x)}{\left (-35+3 x+50 x^5\right )^2} \, dx+5 \int \frac {e^{-x} (4+x)}{-35+3 x+50 x^5} \, dx \\ & = \frac {5 e^{-x}}{3}-5 \int \left (-\frac {175 e^{-x}}{\left (-35+3 x+50 x^5\right )^2}+\frac {12 e^{-x} x}{\left (-35+3 x+50 x^5\right )^2}\right ) \, dx+5 \int \left (\frac {4 e^{-x}}{-35+3 x+50 x^5}+\frac {e^{-x} x}{-35+3 x+50 x^5}\right ) \, dx \\ & = \frac {5 e^{-x}}{3}+5 \int \frac {e^{-x} x}{-35+3 x+50 x^5} \, dx+20 \int \frac {e^{-x}}{-35+3 x+50 x^5} \, dx-60 \int \frac {e^{-x} x}{\left (-35+3 x+50 x^5\right )^2} \, dx+875 \int \frac {e^{-x}}{\left (-35+3 x+50 x^5\right )^2} \, dx \\ \end{align*}
Time = 3.04 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {e^{-x} \left (-5600+525 x+20500 x^5-750 x^6-12500 x^{10}\right )}{3675-630 x+27 x^2-10500 x^5+900 x^6+7500 x^{10}} \, dx=-\frac {25}{3} e^{-x} \left (-\frac {1}{5}+\frac {3 x}{5 \left (-35+3 x+50 x^5\right )}\right ) \]
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Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00
method | result | size |
norman | \(\frac {\left (-\frac {175}{3}+\frac {250 x^{5}}{3}\right ) {\mathrm e}^{-x}}{50 x^{5}+3 x -35}\) | \(25\) |
gosper | \(\frac {25 \left (10 x^{5}-7\right ) {\mathrm e}^{-x}}{3 \left (50 x^{5}+3 x -35\right )}\) | \(26\) |
risch | \(\frac {25 \left (10 x^{5}-7\right ) {\mathrm e}^{-x}}{3 \left (50 x^{5}+3 x -35\right )}\) | \(26\) |
parallelrisch | \(\frac {\left (12500 x^{5}-8750\right ) {\mathrm e}^{-x}}{7500 x^{5}+450 x -5250}\) | \(26\) |
default | \(\frac {5600 \,{\mathrm e}^{-x} \left (216000 x^{4}+3150000 x^{3}+45937500 x^{2}+669921875 x +10368\right )}{351709077687 \left (50 x^{5}+3 x -35\right )}-\frac {5600 \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (50 \textit {\_Z}^{5}+3 \textit {\_Z} -35\right )}{\sum }\frac {\left (216000 \textit {\_R1}^{4}+3366000 \textit {\_R1}^{3}+52237500 \textit {\_R1}^{2}+807734375 \textit {\_R1} +2679697868\right ) {\mathrm e}^{-\textit {\_R1}} \operatorname {Ei}_{1}\left (x -\textit {\_R1} \right )}{250 \textit {\_R1}^{4}+3}\right )}{351709077687}-\frac {175 \,{\mathrm e}^{-x} \left (3150000 x^{4}+45937500 x^{3}+669921875 x^{2}-2592 x +151200\right )}{117236359229 \left (50 x^{5}+3 x -35\right )}+\frac {175 \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (50 \textit {\_Z}^{5}+3 \textit {\_Z} -35\right )}{\sum }\frac {\left (3150000 \textit {\_R1}^{4}+49087500 \textit {\_R1}^{3}+761796875 \textit {\_R1}^{2}+2009763033 \textit {\_R1} +140832\right ) {\mathrm e}^{-\textit {\_R1}} \operatorname {Ei}_{1}\left (x -\textit {\_R1} \right )}{250 \textit {\_R1}^{4}+3}\right )}{117236359229}+\frac {410 \,{\mathrm e}^{-x} \left (1890000 x^{4}+27562500 x^{3}+401953125 x^{2}-23447273401 x +90720\right )}{351709077687 \left (50 x^{5}+3 x -35\right )}-\frac {410 \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (50 \textit {\_Z}^{5}+3 \textit {\_Z} -35\right )}{\sum }\frac {\left (1890000 \textit {\_R1}^{4}+29452500 \textit {\_R1}^{3}+457078125 \textit {\_R1}^{2}-22241414026 \textit {\_R1} +23447356345\right ) {\mathrm e}^{-\textit {\_R1}} \operatorname {Ei}_{1}\left (x -\textit {\_R1} \right )}{250 \textit {\_R1}^{4}+3}\right )}{351709077687}-\frac {5 \,{\mathrm e}^{-x} \left (27562500 x^{4}+401953125 x^{3}-23447273401 x^{2}-22680 x +1323000\right )}{117236359229 \left (50 x^{5}+3 x -35\right )}+\frac {5 \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (50 \textit {\_Z}^{5}+3 \textit {\_Z} -35\right )}{\sum }\frac {\left (27562500 \textit {\_R1}^{4}+429515625 \textit {\_R1}^{3}-22643367151 \textit {\_R1}^{2}+46894516346 \textit {\_R1} +1232280\right ) {\mathrm e}^{-\textit {\_R1}} \operatorname {Ei}_{1}\left (x -\textit {\_R1} \right )}{250 \textit {\_R1}^{4}+3}\right )}{117236359229}+\frac {5 \,{\mathrm e}^{-x}}{3}+\frac {5 \,{\mathrm e}^{-x} \left (16537500 x^{4}+241171875 x^{3}-84410179578 x^{2}+820654500995 x +793800\right )}{351709077687 \left (50 x^{5}+3 x -35\right )}-\frac {5 \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (50 \textit {\_Z}^{5}+3 \textit {\_Z} -35\right )}{\sum }\frac {\left (16537500 \textit {\_R1}^{4}+257709375 \textit {\_R1}^{3}-83927835828 \textit {\_R1}^{2}+1270842117635 \textit {\_R1} -4923926348250\right ) {\mathrm e}^{-\textit {\_R1}} \operatorname {Ei}_{1}\left (x -\textit {\_R1} \right )}{250 \textit {\_R1}^{4}+3}\right )}{351709077687}\) | \(483\) |
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Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-x} \left (-5600+525 x+20500 x^5-750 x^6-12500 x^{10}\right )}{3675-630 x+27 x^2-10500 x^5+900 x^6+7500 x^{10}} \, dx=\frac {25 \, {\left (10 \, x^{5} - 7\right )} e^{\left (-x\right )}}{3 \, {\left (50 \, x^{5} + 3 \, x - 35\right )}} \]
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Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {e^{-x} \left (-5600+525 x+20500 x^5-750 x^6-12500 x^{10}\right )}{3675-630 x+27 x^2-10500 x^5+900 x^6+7500 x^{10}} \, dx=\frac {\left (250 x^{5} - 175\right ) e^{- x}}{150 x^{5} + 9 x - 105} \]
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Time = 0.22 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-x} \left (-5600+525 x+20500 x^5-750 x^6-12500 x^{10}\right )}{3675-630 x+27 x^2-10500 x^5+900 x^6+7500 x^{10}} \, dx=\frac {25 \, {\left (10 \, x^{5} - 7\right )} e^{\left (-x\right )}}{3 \, {\left (50 \, x^{5} + 3 \, x - 35\right )}} \]
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Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {e^{-x} \left (-5600+525 x+20500 x^5-750 x^6-12500 x^{10}\right )}{3675-630 x+27 x^2-10500 x^5+900 x^6+7500 x^{10}} \, dx=\frac {25 \, {\left (10 \, x^{5} e^{\left (-x\right )} - 7 \, e^{\left (-x\right )}\right )}}{3 \, {\left (50 \, x^{5} + 3 \, x - 35\right )}} \]
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Time = 13.50 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-x} \left (-5600+525 x+20500 x^5-750 x^6-12500 x^{10}\right )}{3675-630 x+27 x^2-10500 x^5+900 x^6+7500 x^{10}} \, dx=\frac {{\mathrm {e}}^{-x}\,\left (\frac {5\,x^5}{3}-\frac {7}{6}\right )}{x^5+\frac {3\,x}{50}-\frac {7}{10}} \]
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