Integrand size = 45, antiderivative size = 15 \[ \int \frac {2 e x+e^x \left (2 x-x^2\right )}{e^2 \log (25)+e^{2 x} \log (25)+2 e^{1+x} \log (25)} \, dx=\frac {x^2}{\left (e+e^x\right ) \log (25)} \]
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Time = 0.44 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 26, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {6820, 12, 6874, 2215, 2221, 2317, 2438, 2611, 2320, 6724, 2216, 2222} \[ \int \frac {2 e x+e^x \left (2 x-x^2\right )}{e^2 \log (25)+e^{2 x} \log (25)+2 e^{1+x} \log (25)} \, dx=\frac {x^2}{\left (e^x+e\right ) \log (25)} \]
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Rule 12
Rule 2215
Rule 2216
Rule 2221
Rule 2222
Rule 2317
Rule 2320
Rule 2438
Rule 2611
Rule 6724
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {2 e x-e^x (-2+x) x}{\left (e+e^x\right )^2 \log (25)} \, dx \\ & = \frac {\int \frac {2 e x-e^x (-2+x) x}{\left (e+e^x\right )^2} \, dx}{\log (25)} \\ & = \frac {\int \left (-\frac {(-2+x) x}{e+e^x}+\frac {e x^2}{\left (e+e^x\right )^2}\right ) \, dx}{\log (25)} \\ & = -\frac {\int \frac {(-2+x) x}{e+e^x} \, dx}{\log (25)}+\frac {e \int \frac {x^2}{\left (e+e^x\right )^2} \, dx}{\log (25)} \\ & = -\frac {\int \frac {e^x x^2}{\left (e+e^x\right )^2} \, dx}{\log (25)}+\frac {\int \frac {x^2}{e+e^x} \, dx}{\log (25)}-\frac {\int \left (-\frac {2 x}{e+e^x}+\frac {x^2}{e+e^x}\right ) \, dx}{\log (25)} \\ & = \frac {x^2}{\left (e+e^x\right ) \log (25)}+\frac {x^3}{3 e \log (25)}-\frac {\int \frac {x^2}{e+e^x} \, dx}{\log (25)}-\frac {\int \frac {e^x x^2}{e+e^x} \, dx}{e \log (25)} \\ & = \frac {x^2}{\left (e+e^x\right ) \log (25)}-\frac {x^2 \log \left (1+e^{-1+x}\right )}{e \log (25)}+\frac {\int \frac {e^x x^2}{e+e^x} \, dx}{e \log (25)}+\frac {2 \int x \log \left (1+e^{-1+x}\right ) \, dx}{e \log (25)} \\ & = \frac {x^2}{\left (e+e^x\right ) \log (25)}-\frac {2 x \operatorname {PolyLog}\left (2,-e^{-1+x}\right )}{e \log (25)}-\frac {2 \int x \log \left (1+e^{-1+x}\right ) \, dx}{e \log (25)}+\frac {2 \int \operatorname {PolyLog}\left (2,-e^{-1+x}\right ) \, dx}{e \log (25)} \\ & = \frac {x^2}{\left (e+e^x\right ) \log (25)}-\frac {2 \int \operatorname {PolyLog}\left (2,-e^{-1+x}\right ) \, dx}{e \log (25)}+\frac {2 \text {Subst}\left (\int \frac {\operatorname {PolyLog}(2,-x)}{x} \, dx,x,e^{-1+x}\right )}{e \log (25)} \\ & = \frac {x^2}{\left (e+e^x\right ) \log (25)}+\frac {2 \operatorname {PolyLog}\left (3,-e^{-1+x}\right )}{e \log (25)}-\frac {2 \text {Subst}\left (\int \frac {\operatorname {PolyLog}(2,-x)}{x} \, dx,x,e^{-1+x}\right )}{e \log (25)} \\ & = \frac {x^2}{\left (e+e^x\right ) \log (25)} \\ \end{align*}
Time = 0.35 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {2 e x+e^x \left (2 x-x^2\right )}{e^2 \log (25)+e^{2 x} \log (25)+2 e^{1+x} \log (25)} \, dx=\frac {x^2}{\left (e+e^x\right ) \log (25)} \]
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Time = 0.19 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.13
method | result | size |
norman | \(\frac {x^{2}}{2 \ln \left (5\right ) \left ({\mathrm e}+{\mathrm e}^{x}\right )}\) | \(17\) |
risch | \(\frac {x^{2}}{2 \ln \left (5\right ) \left ({\mathrm e}+{\mathrm e}^{x}\right )}\) | \(17\) |
parallelrisch | \(\frac {x^{2}}{2 \ln \left (5\right ) \left ({\mathrm e}+{\mathrm e}^{x}\right )}\) | \(17\) |
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Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.47 \[ \int \frac {2 e x+e^x \left (2 x-x^2\right )}{e^2 \log (25)+e^{2 x} \log (25)+2 e^{1+x} \log (25)} \, dx=\frac {x^{2} e}{2 \, {\left (e^{2} \log \left (5\right ) + e^{\left (x + 1\right )} \log \left (5\right )\right )}} \]
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Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.27 \[ \int \frac {2 e x+e^x \left (2 x-x^2\right )}{e^2 \log (25)+e^{2 x} \log (25)+2 e^{1+x} \log (25)} \, dx=\frac {x^{2}}{2 e^{x} \log {\left (5 \right )} + 2 e \log {\left (5 \right )}} \]
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Time = 0.33 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.20 \[ \int \frac {2 e x+e^x \left (2 x-x^2\right )}{e^2 \log (25)+e^{2 x} \log (25)+2 e^{1+x} \log (25)} \, dx=\frac {x^{2}}{2 \, {\left (e \log \left (5\right ) + e^{x} \log \left (5\right )\right )}} \]
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Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.20 \[ \int \frac {2 e x+e^x \left (2 x-x^2\right )}{e^2 \log (25)+e^{2 x} \log (25)+2 e^{1+x} \log (25)} \, dx=\frac {x^{2}}{2 \, {\left (e \log \left (5\right ) + e^{x} \log \left (5\right )\right )}} \]
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Time = 0.17 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.07 \[ \int \frac {2 e x+e^x \left (2 x-x^2\right )}{e^2 \log (25)+e^{2 x} \log (25)+2 e^{1+x} \log (25)} \, dx=\frac {x^2}{2\,\ln \left (5\right )\,\left (\mathrm {e}+{\mathrm {e}}^x\right )} \]
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