Integrand size = 34, antiderivative size = 12 \[ \int \frac {25+10 x+e^x \left (5-3 x-x^2\right )}{25+10 e^x+e^{2 x}} \, dx=\frac {x (5+x)}{5+e^x} \]
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Time = 0.45 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.83, number of steps used = 41, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {6873, 6874, 2216, 2215, 2221, 2317, 2438, 2222, 2320, 36, 29, 31, 2611, 6724} \[ \int \frac {25+10 x+e^x \left (5-3 x-x^2\right )}{25+10 e^x+e^{2 x}} \, dx=\frac {x^2}{e^x+5}+\frac {5 x}{e^x+5} \]
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Rule 29
Rule 31
Rule 36
Rule 2215
Rule 2216
Rule 2221
Rule 2222
Rule 2317
Rule 2320
Rule 2438
Rule 2611
Rule 6724
Rule 6873
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {25+10 x+e^x \left (5-3 x-x^2\right )}{\left (5+e^x\right )^2} \, dx \\ & = \int \left (\frac {5 x (5+x)}{\left (5+e^x\right )^2}-\frac {-5+3 x+x^2}{5+e^x}\right ) \, dx \\ & = 5 \int \frac {x (5+x)}{\left (5+e^x\right )^2} \, dx-\int \frac {-5+3 x+x^2}{5+e^x} \, dx \\ & = 5 \int \left (\frac {5 x}{\left (5+e^x\right )^2}+\frac {x^2}{\left (5+e^x\right )^2}\right ) \, dx-\int \left (-\frac {5}{5+e^x}+\frac {3 x}{5+e^x}+\frac {x^2}{5+e^x}\right ) \, dx \\ & = -\left (3 \int \frac {x}{5+e^x} \, dx\right )+5 \int \frac {1}{5+e^x} \, dx+5 \int \frac {x^2}{\left (5+e^x\right )^2} \, dx+25 \int \frac {x}{\left (5+e^x\right )^2} \, dx-\int \frac {x^2}{5+e^x} \, dx \\ & = -\frac {3 x^2}{10}-\frac {x^3}{15}+\frac {1}{5} \int \frac {e^x x^2}{5+e^x} \, dx+\frac {3}{5} \int \frac {e^x x}{5+e^x} \, dx-5 \int \frac {e^x x}{\left (5+e^x\right )^2} \, dx+5 \int \frac {x}{5+e^x} \, dx+5 \text {Subst}\left (\int \frac {1}{x (5+x)} \, dx,x,e^x\right )-\int \frac {e^x x^2}{\left (5+e^x\right )^2} \, dx+\int \frac {x^2}{5+e^x} \, dx \\ & = \frac {5 x}{5+e^x}+\frac {x^2}{5}+\frac {x^2}{5+e^x}+\frac {3}{5} x \log \left (1+\frac {e^x}{5}\right )+\frac {1}{5} x^2 \log \left (1+\frac {e^x}{5}\right )-\frac {1}{5} \int \frac {e^x x^2}{5+e^x} \, dx-\frac {2}{5} \int x \log \left (1+\frac {e^x}{5}\right ) \, dx-\frac {3}{5} \int \log \left (1+\frac {e^x}{5}\right ) \, dx-2 \int \frac {x}{5+e^x} \, dx-5 \int \frac {1}{5+e^x} \, dx-\int \frac {e^x x}{5+e^x} \, dx+\text {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )-\text {Subst}\left (\int \frac {1}{5+x} \, dx,x,e^x\right ) \\ & = x+\frac {5 x}{5+e^x}+\frac {x^2}{5+e^x}-\frac {2}{5} x \log \left (1+\frac {e^x}{5}\right )-\log \left (5+e^x\right )+\frac {2}{5} x \operatorname {PolyLog}\left (2,-\frac {e^x}{5}\right )+\frac {2}{5} \int \frac {e^x x}{5+e^x} \, dx+\frac {2}{5} \int x \log \left (1+\frac {e^x}{5}\right ) \, dx-\frac {2}{5} \int \operatorname {PolyLog}\left (2,-\frac {e^x}{5}\right ) \, dx-\frac {3}{5} \text {Subst}\left (\int \frac {\log \left (1+\frac {x}{5}\right )}{x} \, dx,x,e^x\right )-5 \text {Subst}\left (\int \frac {1}{x (5+x)} \, dx,x,e^x\right )+\int \log \left (1+\frac {e^x}{5}\right ) \, dx \\ & = x+\frac {5 x}{5+e^x}+\frac {x^2}{5+e^x}-\log \left (5+e^x\right )+\frac {3}{5} \operatorname {PolyLog}\left (2,-\frac {e^x}{5}\right )-\frac {2}{5} \int \log \left (1+\frac {e^x}{5}\right ) \, dx+\frac {2}{5} \int \operatorname {PolyLog}\left (2,-\frac {e^x}{5}\right ) \, dx-\frac {2}{5} \text {Subst}\left (\int \frac {\operatorname {PolyLog}\left (2,-\frac {x}{5}\right )}{x} \, dx,x,e^x\right )-\text {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )+\text {Subst}\left (\int \frac {1}{5+x} \, dx,x,e^x\right )+\text {Subst}\left (\int \frac {\log \left (1+\frac {x}{5}\right )}{x} \, dx,x,e^x\right ) \\ & = \frac {5 x}{5+e^x}+\frac {x^2}{5+e^x}-\frac {2}{5} \operatorname {PolyLog}\left (2,-\frac {e^x}{5}\right )-\frac {2}{5} \operatorname {PolyLog}\left (3,-\frac {e^x}{5}\right )-\frac {2}{5} \text {Subst}\left (\int \frac {\log \left (1+\frac {x}{5}\right )}{x} \, dx,x,e^x\right )+\frac {2}{5} \text {Subst}\left (\int \frac {\operatorname {PolyLog}\left (2,-\frac {x}{5}\right )}{x} \, dx,x,e^x\right ) \\ & = \frac {5 x}{5+e^x}+\frac {x^2}{5+e^x} \\ \end{align*}
Time = 0.30 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {25+10 x+e^x \left (5-3 x-x^2\right )}{25+10 e^x+e^{2 x}} \, dx=\frac {x (5+x)}{5+e^x} \]
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Time = 0.18 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00
method | result | size |
risch | \(\frac {\left (5+x \right ) x}{{\mathrm e}^{x}+5}\) | \(12\) |
norman | \(\frac {x^{2}+5 x}{{\mathrm e}^{x}+5}\) | \(15\) |
parallelrisch | \(\frac {x^{2}+5 x}{{\mathrm e}^{x}+5}\) | \(15\) |
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Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.17 \[ \int \frac {25+10 x+e^x \left (5-3 x-x^2\right )}{25+10 e^x+e^{2 x}} \, dx=\frac {x^{2} + 5 \, x}{e^{x} + 5} \]
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Time = 0.05 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \frac {25+10 x+e^x \left (5-3 x-x^2\right )}{25+10 e^x+e^{2 x}} \, dx=\frac {x^{2} + 5 x}{e^{x} + 5} \]
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Leaf count of result is larger than twice the leaf count of optimal. 27 vs. \(2 (11) = 22\).
Time = 0.21 (sec) , antiderivative size = 27, normalized size of antiderivative = 2.25 \[ \int \frac {25+10 x+e^x \left (5-3 x-x^2\right )}{25+10 e^x+e^{2 x}} \, dx=x + \frac {x^{2} - x e^{x} - 5}{e^{x} + 5} + \frac {5}{e^{x} + 5} \]
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Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.17 \[ \int \frac {25+10 x+e^x \left (5-3 x-x^2\right )}{25+10 e^x+e^{2 x}} \, dx=\frac {x^{2} + 5 \, x}{e^{x} + 5} \]
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Time = 13.81 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.92 \[ \int \frac {25+10 x+e^x \left (5-3 x-x^2\right )}{25+10 e^x+e^{2 x}} \, dx=\frac {x\,\left (x+5\right )}{{\mathrm {e}}^x+5} \]
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