Integrand size = 16, antiderivative size = 259 \[ \int \frac {x^2}{-1+e^x+e^{2 x}} \, dx=\frac {2 x^3}{3 \sqrt {5} \left (1-\sqrt {5}\right )}-\frac {2 x^3}{3 \sqrt {5} \left (1+\sqrt {5}\right )}-\frac {2 x^2 \log \left (1+\frac {2 e^x}{1-\sqrt {5}}\right )}{\sqrt {5} \left (1-\sqrt {5}\right )}+\frac {2 x^2 \log \left (1+\frac {2 e^x}{1+\sqrt {5}}\right )}{\sqrt {5} \left (1+\sqrt {5}\right )}-\frac {4 x \operatorname {PolyLog}\left (2,-\frac {2 e^x}{1-\sqrt {5}}\right )}{\sqrt {5} \left (1-\sqrt {5}\right )}+\frac {4 x \operatorname {PolyLog}\left (2,-\frac {2 e^x}{1+\sqrt {5}}\right )}{\sqrt {5} \left (1+\sqrt {5}\right )}+\frac {4 \operatorname {PolyLog}\left (3,-\frac {2 e^x}{1-\sqrt {5}}\right )}{\sqrt {5} \left (1-\sqrt {5}\right )}-\frac {4 \operatorname {PolyLog}\left (3,-\frac {2 e^x}{1+\sqrt {5}}\right )}{\sqrt {5} \left (1+\sqrt {5}\right )} \]
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Time = 0.19 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {2295, 2215, 2221, 2611, 2320, 6724} \[ \int \frac {x^2}{-1+e^x+e^{2 x}} \, dx=-\frac {4 x \operatorname {PolyLog}\left (2,-\frac {2 e^x}{1-\sqrt {5}}\right )}{\sqrt {5} \left (1-\sqrt {5}\right )}+\frac {4 x \operatorname {PolyLog}\left (2,-\frac {2 e^x}{1+\sqrt {5}}\right )}{\sqrt {5} \left (1+\sqrt {5}\right )}+\frac {4 \operatorname {PolyLog}\left (3,-\frac {2 e^x}{1-\sqrt {5}}\right )}{\sqrt {5} \left (1-\sqrt {5}\right )}-\frac {4 \operatorname {PolyLog}\left (3,-\frac {2 e^x}{1+\sqrt {5}}\right )}{\sqrt {5} \left (1+\sqrt {5}\right )}-\frac {2 x^3}{3 \sqrt {5} \left (1+\sqrt {5}\right )}+\frac {2 x^3}{3 \sqrt {5} \left (1-\sqrt {5}\right )}-\frac {2 x^2 \log \left (\frac {2 e^x}{1-\sqrt {5}}+1\right )}{\sqrt {5} \left (1-\sqrt {5}\right )}+\frac {2 x^2 \log \left (\frac {2 e^x}{1+\sqrt {5}}+1\right )}{\sqrt {5} \left (1+\sqrt {5}\right )} \]
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Rule 2215
Rule 2221
Rule 2295
Rule 2320
Rule 2611
Rule 6724
Rubi steps \begin{align*} \text {integral}& = \frac {2 \int \frac {x^2}{1-\sqrt {5}+2 e^x} \, dx}{\sqrt {5}}-\frac {2 \int \frac {x^2}{1+\sqrt {5}+2 e^x} \, dx}{\sqrt {5}} \\ & = \frac {2 x^3}{3 \sqrt {5} \left (1-\sqrt {5}\right )}-\frac {2 x^3}{3 \sqrt {5} \left (1+\sqrt {5}\right )}-\frac {4 \int \frac {e^x x^2}{1-\sqrt {5}+2 e^x} \, dx}{\sqrt {5} \left (1-\sqrt {5}\right )}+\frac {4 \int \frac {e^x x^2}{1+\sqrt {5}+2 e^x} \, dx}{\sqrt {5} \left (1+\sqrt {5}\right )} \\ & = \frac {2 x^3}{3 \sqrt {5} \left (1-\sqrt {5}\right )}-\frac {2 x^3}{3 \sqrt {5} \left (1+\sqrt {5}\right )}-\frac {2 x^2 \log \left (1+\frac {2 e^x}{1-\sqrt {5}}\right )}{\sqrt {5} \left (1-\sqrt {5}\right )}+\frac {2 x^2 \log \left (1+\frac {2 e^x}{1+\sqrt {5}}\right )}{\sqrt {5} \left (1+\sqrt {5}\right )}+\frac {4 \int x \log \left (1+\frac {2 e^x}{1-\sqrt {5}}\right ) \, dx}{\sqrt {5} \left (1-\sqrt {5}\right )}-\frac {4 \int x \log \left (1+\frac {2 e^x}{1+\sqrt {5}}\right ) \, dx}{\sqrt {5} \left (1+\sqrt {5}\right )} \\ & = \frac {2 x^3}{3 \sqrt {5} \left (1-\sqrt {5}\right )}-\frac {2 x^3}{3 \sqrt {5} \left (1+\sqrt {5}\right )}-\frac {2 x^2 \log \left (1+\frac {2 e^x}{1-\sqrt {5}}\right )}{\sqrt {5} \left (1-\sqrt {5}\right )}+\frac {2 x^2 \log \left (1+\frac {2 e^x}{1+\sqrt {5}}\right )}{\sqrt {5} \left (1+\sqrt {5}\right )}-\frac {4 x \text {Li}_2\left (-\frac {2 e^x}{1-\sqrt {5}}\right )}{\sqrt {5} \left (1-\sqrt {5}\right )}+\frac {4 x \text {Li}_2\left (-\frac {2 e^x}{1+\sqrt {5}}\right )}{\sqrt {5} \left (1+\sqrt {5}\right )}+\frac {4 \int \text {Li}_2\left (-\frac {2 e^x}{1-\sqrt {5}}\right ) \, dx}{\sqrt {5} \left (1-\sqrt {5}\right )}-\frac {4 \int \text {Li}_2\left (-\frac {2 e^x}{1+\sqrt {5}}\right ) \, dx}{\sqrt {5} \left (1+\sqrt {5}\right )} \\ & = \frac {2 x^3}{3 \sqrt {5} \left (1-\sqrt {5}\right )}-\frac {2 x^3}{3 \sqrt {5} \left (1+\sqrt {5}\right )}-\frac {2 x^2 \log \left (1+\frac {2 e^x}{1-\sqrt {5}}\right )}{\sqrt {5} \left (1-\sqrt {5}\right )}+\frac {2 x^2 \log \left (1+\frac {2 e^x}{1+\sqrt {5}}\right )}{\sqrt {5} \left (1+\sqrt {5}\right )}-\frac {4 x \text {Li}_2\left (-\frac {2 e^x}{1-\sqrt {5}}\right )}{\sqrt {5} \left (1-\sqrt {5}\right )}+\frac {4 x \text {Li}_2\left (-\frac {2 e^x}{1+\sqrt {5}}\right )}{\sqrt {5} \left (1+\sqrt {5}\right )}+\frac {4 \text {Subst}\left (\int \frac {\text {Li}_2\left (\frac {2 x}{-1+\sqrt {5}}\right )}{x} \, dx,x,e^x\right )}{\sqrt {5} \left (1-\sqrt {5}\right )}-\frac {4 \text {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {2 x}{1+\sqrt {5}}\right )}{x} \, dx,x,e^x\right )}{\sqrt {5} \left (1+\sqrt {5}\right )} \\ & = \frac {2 x^3}{3 \sqrt {5} \left (1-\sqrt {5}\right )}-\frac {2 x^3}{3 \sqrt {5} \left (1+\sqrt {5}\right )}-\frac {2 x^2 \log \left (1+\frac {2 e^x}{1-\sqrt {5}}\right )}{\sqrt {5} \left (1-\sqrt {5}\right )}+\frac {2 x^2 \log \left (1+\frac {2 e^x}{1+\sqrt {5}}\right )}{\sqrt {5} \left (1+\sqrt {5}\right )}-\frac {4 x \text {Li}_2\left (-\frac {2 e^x}{1-\sqrt {5}}\right )}{\sqrt {5} \left (1-\sqrt {5}\right )}+\frac {4 x \text {Li}_2\left (-\frac {2 e^x}{1+\sqrt {5}}\right )}{\sqrt {5} \left (1+\sqrt {5}\right )}+\frac {4 \text {Li}_3\left (-\frac {2 e^x}{1-\sqrt {5}}\right )}{\sqrt {5} \left (1-\sqrt {5}\right )}-\frac {4 \text {Li}_3\left (-\frac {2 e^x}{1+\sqrt {5}}\right )}{\sqrt {5} \left (1+\sqrt {5}\right )} \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.66 \[ \int \frac {x^2}{-1+e^x+e^{2 x}} \, dx=\frac {2 \left (\frac {x^2 \log \left (1-\frac {1}{2} \left (-1+\sqrt {5}\right ) e^{-x}\right )}{-1+\sqrt {5}}+\frac {x^2 \log \left (1+\frac {1}{2} \left (1+\sqrt {5}\right ) e^{-x}\right )}{1+\sqrt {5}}-\frac {2 \left (x \operatorname {PolyLog}\left (2,\frac {1}{2} \left (-1+\sqrt {5}\right ) e^{-x}\right )+\operatorname {PolyLog}\left (3,\frac {1}{2} \left (-1+\sqrt {5}\right ) e^{-x}\right )\right )}{-1+\sqrt {5}}-\frac {2 \left (x \operatorname {PolyLog}\left (2,-\frac {1}{2} \left (1+\sqrt {5}\right ) e^{-x}\right )+\operatorname {PolyLog}\left (3,-\frac {1}{2} \left (1+\sqrt {5}\right ) e^{-x}\right )\right )}{1+\sqrt {5}}\right )}{\sqrt {5}} \]
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\[\int \frac {x^{2}}{-1+{\mathrm e}^{x}+{\mathrm e}^{2 x}}d x\]
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Time = 0.31 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.53 \[ \int \frac {x^2}{-1+e^x+e^{2 x}} \, dx=-\frac {1}{3} \, x^{3} + \frac {1}{5} \, {\left (\sqrt {5} x + 5 \, x\right )} {\rm Li}_2\left (\frac {1}{2} \, {\left (\sqrt {5} + 1\right )} e^{x}\right ) - \frac {1}{5} \, {\left (\sqrt {5} x - 5 \, x\right )} {\rm Li}_2\left (-\frac {1}{2} \, {\left (\sqrt {5} - 1\right )} e^{x}\right ) + \frac {1}{10} \, {\left (\sqrt {5} x^{2} + 5 \, x^{2}\right )} \log \left (-\frac {1}{2} \, {\left (\sqrt {5} + 1\right )} e^{x} + 1\right ) - \frac {1}{10} \, {\left (\sqrt {5} x^{2} - 5 \, x^{2}\right )} \log \left (\frac {1}{2} \, {\left (\sqrt {5} - 1\right )} e^{x} + 1\right ) - \frac {1}{5} \, {\left (\sqrt {5} + 5\right )} {\rm polylog}\left (3, \frac {1}{2} \, {\left (\sqrt {5} + 1\right )} e^{x}\right ) + \frac {1}{5} \, {\left (\sqrt {5} - 5\right )} {\rm polylog}\left (3, -\frac {1}{2} \, {\left (\sqrt {5} - 1\right )} e^{x}\right ) \]
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\[ \int \frac {x^2}{-1+e^x+e^{2 x}} \, dx=\int \frac {x^{2}}{e^{2 x} + e^{x} - 1}\, dx \]
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\[ \int \frac {x^2}{-1+e^x+e^{2 x}} \, dx=\int { \frac {x^{2}}{e^{\left (2 \, x\right )} + e^{x} - 1} \,d x } \]
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\[ \int \frac {x^2}{-1+e^x+e^{2 x}} \, dx=\int { \frac {x^{2}}{e^{\left (2 \, x\right )} + e^{x} - 1} \,d x } \]
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Timed out. \[ \int \frac {x^2}{-1+e^x+e^{2 x}} \, dx=\int \frac {x^2}{{\mathrm {e}}^{2\,x}+{\mathrm {e}}^x-1} \,d x \]
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