Integrand size = 62, antiderivative size = 22 \[ \int \frac {e^{-\frac {48+48 x}{-5+x}} \left (288 e^2+100 x^3+248 x^4+4 x^5+e \left (-100 x-536 x^2-4 x^3\right )\right )}{25-10 x+x^2} \, dx=e^{-\frac {48 (1+x)}{-5+x}} \left (-e+x^2\right )^2 \]
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\[ \int \frac {e^{-\frac {48+48 x}{-5+x}} \left (288 e^2+100 x^3+248 x^4+4 x^5+e \left (-100 x-536 x^2-4 x^3\right )\right )}{25-10 x+x^2} \, dx=\int \frac {e^{-\frac {48+48 x}{-5+x}} \left (288 e^2+100 x^3+248 x^4+4 x^5+e \left (-100 x-536 x^2-4 x^3\right )\right )}{25-10 x+x^2} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{-\frac {48+48 x}{-5+x}} \left (288 e^2+100 x^3+248 x^4+4 x^5+e \left (-100 x-536 x^2-4 x^3\right )\right )}{(-5+x)^2} \, dx \\ & = \int \frac {4 e^{-\frac {48+48 x}{-5+x}} \left (72 e^2-25 e x-134 e x^2+25 \left (1-\frac {e}{25}\right ) x^3+62 x^4+x^5\right )}{(5-x)^2} \, dx \\ & = 4 \int \frac {e^{-\frac {48+48 x}{-5+x}} \left (72 e^2-25 e x-134 e x^2+25 \left (1-\frac {e}{25}\right ) x^3+62 x^4+x^5\right )}{(5-x)^2} \, dx \\ & = 4 \int \left (-72 e^{-\frac {48+48 x}{-5+x}} (-75+2 e)+\frac {72 (-25+e)^2 e^{-\frac {48+48 x}{-5+x}}}{(-5+x)^2}-\frac {1440 (-25+e) e^{-\frac {48+48 x}{-5+x}}}{-5+x}-(-720+e) e^{-\frac {48+48 x}{-5+x}} x+72 e^{-\frac {48+48 x}{-5+x}} x^2+e^{-\frac {48+48 x}{-5+x}} x^3\right ) \, dx \\ & = 4 \int e^{-\frac {48+48 x}{-5+x}} x^3 \, dx+288 \int e^{-\frac {48+48 x}{-5+x}} x^2 \, dx+(288 (75-2 e)) \int e^{-\frac {48+48 x}{-5+x}} \, dx+(5760 (25-e)) \int \frac {e^{-\frac {48+48 x}{-5+x}}}{-5+x} \, dx+\left (288 (25-e)^2\right ) \int \frac {e^{-\frac {48+48 x}{-5+x}}}{(-5+x)^2} \, dx+(4 (720-e)) \int e^{-\frac {48+48 x}{-5+x}} x \, dx \\ & = 4 \int e^{-\frac {48+48 x}{-5+x}} x^3 \, dx+288 \int e^{-\frac {48+48 x}{-5+x}} x^2 \, dx+(288 (75-2 e)) \int e^{-\frac {48+48 x}{-5+x}} \, dx+(5760 (25-e)) \int \frac {e^{-48-\frac {288}{-5+x}}}{-5+x} \, dx+\left (288 (25-e)^2\right ) \int \frac {e^{-48-\frac {288}{-5+x}}}{(-5+x)^2} \, dx+(4 (720-e)) \int e^{-\frac {48+48 x}{-5+x}} x \, dx \\ & = (25-e)^2 e^{-48+\frac {288}{5-x}}-\frac {5760 (25-e) \text {Ei}\left (\frac {288}{5-x}\right )}{e^{48}}+4 \int e^{-\frac {48+48 x}{-5+x}} x^3 \, dx+288 \int e^{-\frac {48+48 x}{-5+x}} x^2 \, dx+(288 (75-2 e)) \int e^{-\frac {48+48 x}{-5+x}} \, dx+(4 (720-e)) \int e^{-\frac {48+48 x}{-5+x}} x \, dx \\ \end{align*}
Time = 4.54 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-\frac {48+48 x}{-5+x}} \left (288 e^2+100 x^3+248 x^4+4 x^5+e \left (-100 x-536 x^2-4 x^3\right )\right )}{25-10 x+x^2} \, dx=e^{-\frac {48 (1+x)}{-5+x}} \left (e-x^2\right )^2 \]
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Time = 0.66 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18
method | result | size |
risch | \(\left (x^{4}-2 x^{2} {\mathrm e}+{\mathrm e}^{2}\right ) {\mathrm e}^{-\frac {48 \left (1+x \right )}{-5+x}}\) | \(26\) |
gosper | \(\left (x^{4}-2 x^{2} {\mathrm e}+{\mathrm e}^{2}\right ) {\mathrm e}^{-\frac {48 \left (1+x \right )}{-5+x}}\) | \(30\) |
parallelrisch | \(\frac {\left (x^{5}+{\mathrm e}^{2} x -5 x^{4}-5 \,{\mathrm e}^{2}+10 x^{2} {\mathrm e}-2 x^{3} {\mathrm e}\right ) {\mathrm e}^{-\frac {48 \left (1+x \right )}{-5+x}}}{-5+x}\) | \(55\) |
norman | \(\frac {\left (x^{5}+{\mathrm e}^{2} x -5 x^{4}-5 \,{\mathrm e}^{2}+10 x^{2} {\mathrm e}-2 x^{3} {\mathrm e}\right ) {\mathrm e}^{-\frac {48 x +48}{-5+x}}}{-5+x}\) | \(56\) |
derivativedivides | \(43700 \,{\mathrm e}^{-48-\frac {288}{-5+x}} \left (-5+x \right )-\frac {{\mathrm e}^{-48-\frac {288}{-5+x}} \left (\left (48+\frac {288}{-5+x}\right )^{3}-145 \left (48+\frac {288}{-5+x}\right )^{2}+223482+\frac {2018880}{-5+x}\right ) \left (-5+x \right )^{4}}{6}+58 \,{\mathrm e}^{-48-\frac {288}{-5+x}} \left (\left (48+\frac {288}{-5+x}\right )^{2}-2302-\frac {27936}{-5+x}\right ) \left (-5+x \right )^{3}-3030 \,{\mathrm e}^{-48-\frac {288}{-5+x}} \left (-1+\frac {288}{-5+x}\right ) \left (-5+x \right )^{2}+625 \,{\mathrm e}^{-48-\frac {288}{-5+x}}+10368 \,{\mathrm e} \left (-\frac {{\mathrm e}^{-48-\frac {288}{-5+x}} \left (-5+x \right )}{288}+{\mathrm e}^{-48} \operatorname {Ei}_{1}\left (\frac {288}{-5+x}\right )\right )-2304 \,{\mathrm e}^{2} \left (-\frac {{\mathrm e}^{-48-\frac {288}{-5+x}} \left (-5+x \right )}{288}+{\mathrm e}^{-48} \operatorname {Ei}_{1}\left (\frac {288}{-5+x}\right )\right )+331776 \,{\mathrm e} \left (\frac {{\mathrm e}^{-48-\frac {288}{-5+x}} \left (-1+\frac {288}{-5+x}\right ) \left (-5+x \right )^{2}}{165888}-\frac {{\mathrm e}^{-48} \operatorname {Ei}_{1}\left (\frac {288}{-5+x}\right )}{2}\right )+960 \,{\mathrm e} \left (-\frac {{\mathrm e}^{-48-\frac {288}{-5+x}} \left (-5+x \right )}{6}+47 \,{\mathrm e}^{-48} \operatorname {Ei}_{1}\left (\frac {288}{-5+x}\right )\right )+50 \,{\mathrm e} \left (-{\mathrm e}^{-48-\frac {288}{-5+x}}-8 \,{\mathrm e}^{-48-\frac {288}{-5+x}} \left (-5+x \right )+2208 \,{\mathrm e}^{-48} \operatorname {Ei}_{1}\left (\frac {288}{-5+x}\right )\right )+96 \,{\mathrm e}^{2} \left (-\frac {{\mathrm e}^{-48-\frac {288}{-5+x}} \left (-5+x \right )}{6}+47 \,{\mathrm e}^{-48} \operatorname {Ei}_{1}\left (\frac {288}{-5+x}\right )\right )-{\mathrm e}^{2} \left (-{\mathrm e}^{-48-\frac {288}{-5+x}}-8 \,{\mathrm e}^{-48-\frac {288}{-5+x}} \left (-5+x \right )+2208 \,{\mathrm e}^{-48} \operatorname {Ei}_{1}\left (\frac {288}{-5+x}\right )\right )\) | \(412\) |
default | \(43700 \,{\mathrm e}^{-48-\frac {288}{-5+x}} \left (-5+x \right )-\frac {{\mathrm e}^{-48-\frac {288}{-5+x}} \left (\left (48+\frac {288}{-5+x}\right )^{3}-145 \left (48+\frac {288}{-5+x}\right )^{2}+223482+\frac {2018880}{-5+x}\right ) \left (-5+x \right )^{4}}{6}+58 \,{\mathrm e}^{-48-\frac {288}{-5+x}} \left (\left (48+\frac {288}{-5+x}\right )^{2}-2302-\frac {27936}{-5+x}\right ) \left (-5+x \right )^{3}-3030 \,{\mathrm e}^{-48-\frac {288}{-5+x}} \left (-1+\frac {288}{-5+x}\right ) \left (-5+x \right )^{2}+625 \,{\mathrm e}^{-48-\frac {288}{-5+x}}+10368 \,{\mathrm e} \left (-\frac {{\mathrm e}^{-48-\frac {288}{-5+x}} \left (-5+x \right )}{288}+{\mathrm e}^{-48} \operatorname {Ei}_{1}\left (\frac {288}{-5+x}\right )\right )-2304 \,{\mathrm e}^{2} \left (-\frac {{\mathrm e}^{-48-\frac {288}{-5+x}} \left (-5+x \right )}{288}+{\mathrm e}^{-48} \operatorname {Ei}_{1}\left (\frac {288}{-5+x}\right )\right )+331776 \,{\mathrm e} \left (\frac {{\mathrm e}^{-48-\frac {288}{-5+x}} \left (-1+\frac {288}{-5+x}\right ) \left (-5+x \right )^{2}}{165888}-\frac {{\mathrm e}^{-48} \operatorname {Ei}_{1}\left (\frac {288}{-5+x}\right )}{2}\right )+960 \,{\mathrm e} \left (-\frac {{\mathrm e}^{-48-\frac {288}{-5+x}} \left (-5+x \right )}{6}+47 \,{\mathrm e}^{-48} \operatorname {Ei}_{1}\left (\frac {288}{-5+x}\right )\right )+50 \,{\mathrm e} \left (-{\mathrm e}^{-48-\frac {288}{-5+x}}-8 \,{\mathrm e}^{-48-\frac {288}{-5+x}} \left (-5+x \right )+2208 \,{\mathrm e}^{-48} \operatorname {Ei}_{1}\left (\frac {288}{-5+x}\right )\right )+96 \,{\mathrm e}^{2} \left (-\frac {{\mathrm e}^{-48-\frac {288}{-5+x}} \left (-5+x \right )}{6}+47 \,{\mathrm e}^{-48} \operatorname {Ei}_{1}\left (\frac {288}{-5+x}\right )\right )-{\mathrm e}^{2} \left (-{\mathrm e}^{-48-\frac {288}{-5+x}}-8 \,{\mathrm e}^{-48-\frac {288}{-5+x}} \left (-5+x \right )+2208 \,{\mathrm e}^{-48} \operatorname {Ei}_{1}\left (\frac {288}{-5+x}\right )\right )\) | \(412\) |
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Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {e^{-\frac {48+48 x}{-5+x}} \left (288 e^2+100 x^3+248 x^4+4 x^5+e \left (-100 x-536 x^2-4 x^3\right )\right )}{25-10 x+x^2} \, dx={\left (x^{4} - 2 \, x^{2} e + e^{2}\right )} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5}\right )} \]
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Time = 0.22 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {e^{-\frac {48+48 x}{-5+x}} \left (288 e^2+100 x^3+248 x^4+4 x^5+e \left (-100 x-536 x^2-4 x^3\right )\right )}{25-10 x+x^2} \, dx=\left (x^{4} - 2 e x^{2} + e^{2}\right ) e^{- \frac {48 x + 48}{x - 5}} \]
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\[ \int \frac {e^{-\frac {48+48 x}{-5+x}} \left (288 e^2+100 x^3+248 x^4+4 x^5+e \left (-100 x-536 x^2-4 x^3\right )\right )}{25-10 x+x^2} \, dx=\int { \frac {4 \, {\left (x^{5} + 62 \, x^{4} + 25 \, x^{3} - {\left (x^{3} + 134 \, x^{2} + 25 \, x\right )} e + 72 \, e^{2}\right )} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5}\right )}}{x^{2} - 10 \, x + 25} \,d x } \]
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Leaf count of result is larger than twice the leaf count of optimal. 375 vs. \(2 (22) = 44\).
Time = 0.39 (sec) , antiderivative size = 375, normalized size of antiderivative = 17.05 \[ \int \frac {e^{-\frac {48+48 x}{-5+x}} \left (288 e^2+100 x^3+248 x^4+4 x^5+e \left (-100 x-536 x^2-4 x^3\right )\right )}{25-10 x+x^2} \, dx=\frac {\frac {{\left (x + 1\right )}^{4} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5} + 2\right )}}{{\left (x - 5\right )}^{4}} - \frac {4 \, {\left (x + 1\right )}^{3} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5} + 2\right )}}{{\left (x - 5\right )}^{3}} + \frac {6 \, {\left (x + 1\right )}^{2} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5} + 2\right )}}{{\left (x - 5\right )}^{2}} - \frac {4 \, {\left (x + 1\right )} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5} + 2\right )}}{x - 5} - \frac {50 \, {\left (x + 1\right )}^{4} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5} + 1\right )}}{{\left (x - 5\right )}^{4}} + \frac {80 \, {\left (x + 1\right )}^{3} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5} + 1\right )}}{{\left (x - 5\right )}^{3}} - \frac {12 \, {\left (x + 1\right )}^{2} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5} + 1\right )}}{{\left (x - 5\right )}^{2}} - \frac {16 \, {\left (x + 1\right )} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5} + 1\right )}}{x - 5} + \frac {625 \, {\left (x + 1\right )}^{4} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5}\right )}}{{\left (x - 5\right )}^{4}} + \frac {500 \, {\left (x + 1\right )}^{3} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5}\right )}}{{\left (x - 5\right )}^{3}} + \frac {150 \, {\left (x + 1\right )}^{2} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5}\right )}}{{\left (x - 5\right )}^{2}} + \frac {20 \, {\left (x + 1\right )} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5}\right )}}{x - 5} + e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5} + 2\right )} - 2 \, e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5} + 1\right )} + e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5}\right )}}{\frac {{\left (x + 1\right )}^{4}}{{\left (x - 5\right )}^{4}} - \frac {4 \, {\left (x + 1\right )}^{3}}{{\left (x - 5\right )}^{3}} + \frac {6 \, {\left (x + 1\right )}^{2}}{{\left (x - 5\right )}^{2}} - \frac {4 \, {\left (x + 1\right )}}{x - 5} + 1} \]
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Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27 \[ \int \frac {e^{-\frac {48+48 x}{-5+x}} \left (288 e^2+100 x^3+248 x^4+4 x^5+e \left (-100 x-536 x^2-4 x^3\right )\right )}{25-10 x+x^2} \, dx={\mathrm {e}}^{-\frac {48\,x}{x-5}-\frac {48}{x-5}}\,{\left (\mathrm {e}-x^2\right )}^2 \]
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