Optimal. Leaf size=180 \[ -\frac {2 \text {Li}_2\left (-\frac {2 e^x}{1-\sqrt {5}}\right )}{\sqrt {5} \left (1-\sqrt {5}\right )}+\frac {2 \text {Li}_2\left (-\frac {2 e^x}{1+\sqrt {5}}\right )}{\sqrt {5} \left (1+\sqrt {5}\right )}-\frac {x^2}{\sqrt {5} \left (1+\sqrt {5}\right )}+\frac {x^2}{\sqrt {5} \left (1-\sqrt {5}\right )}-\frac {2 x \log \left (\frac {2 e^x}{1-\sqrt {5}}+1\right )}{\sqrt {5} \left (1-\sqrt {5}\right )}+\frac {2 x \log \left (\frac {2 e^x}{1+\sqrt {5}}+1\right )}{\sqrt {5} \left (1+\sqrt {5}\right )} \]
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Rubi [A] time = 0.19, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {2263, 2184, 2190, 2279, 2391} \[ -\frac {2 \text {PolyLog}\left (2,-\frac {2 e^x}{1-\sqrt {5}}\right )}{\sqrt {5} \left (1-\sqrt {5}\right )}+\frac {2 \text {PolyLog}\left (2,-\frac {2 e^x}{1+\sqrt {5}}\right )}{\sqrt {5} \left (1+\sqrt {5}\right )}-\frac {x^2}{\sqrt {5} \left (1+\sqrt {5}\right )}+\frac {x^2}{\sqrt {5} \left (1-\sqrt {5}\right )}-\frac {2 x \log \left (\frac {2 e^x}{1-\sqrt {5}}+1\right )}{\sqrt {5} \left (1-\sqrt {5}\right )}+\frac {2 x \log \left (\frac {2 e^x}{1+\sqrt {5}}+1\right )}{\sqrt {5} \left (1+\sqrt {5}\right )} \]
Antiderivative was successfully verified.
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Rule 2184
Rule 2190
Rule 2263
Rule 2279
Rule 2391
Rubi steps
\begin {align*} \int \frac {x}{-1+e^x+e^{2 x}} \, dx &=\frac {2 \int \frac {x}{1-\sqrt {5}+2 e^x} \, dx}{\sqrt {5}}-\frac {2 \int \frac {x}{1+\sqrt {5}+2 e^x} \, dx}{\sqrt {5}}\\ &=\frac {x^2}{\sqrt {5} \left (1-\sqrt {5}\right )}-\frac {x^2}{\sqrt {5} \left (1+\sqrt {5}\right )}-\frac {4 \int \frac {e^x x}{1-\sqrt {5}+2 e^x} \, dx}{\sqrt {5} \left (1-\sqrt {5}\right )}+\frac {4 \int \frac {e^x x}{1+\sqrt {5}+2 e^x} \, dx}{\sqrt {5} \left (1+\sqrt {5}\right )}\\ &=\frac {x^2}{\sqrt {5} \left (1-\sqrt {5}\right )}-\frac {x^2}{\sqrt {5} \left (1+\sqrt {5}\right )}-\frac {2 x \log \left (1+\frac {2 e^x}{1-\sqrt {5}}\right )}{\sqrt {5} \left (1-\sqrt {5}\right )}+\frac {2 x \log \left (1+\frac {2 e^x}{1+\sqrt {5}}\right )}{\sqrt {5} \left (1+\sqrt {5}\right )}+\frac {2 \int \log \left (1+\frac {2 e^x}{1-\sqrt {5}}\right ) \, dx}{\sqrt {5} \left (1-\sqrt {5}\right )}-\frac {2 \int \log \left (1+\frac {2 e^x}{1+\sqrt {5}}\right ) \, dx}{\sqrt {5} \left (1+\sqrt {5}\right )}\\ &=\frac {x^2}{\sqrt {5} \left (1-\sqrt {5}\right )}-\frac {x^2}{\sqrt {5} \left (1+\sqrt {5}\right )}-\frac {2 x \log \left (1+\frac {2 e^x}{1-\sqrt {5}}\right )}{\sqrt {5} \left (1-\sqrt {5}\right )}+\frac {2 x \log \left (1+\frac {2 e^x}{1+\sqrt {5}}\right )}{\sqrt {5} \left (1+\sqrt {5}\right )}+\frac {2 \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {2 x}{1-\sqrt {5}}\right )}{x} \, dx,x,e^x\right )}{\sqrt {5} \left (1-\sqrt {5}\right )}-\frac {2 \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {2 x}{1+\sqrt {5}}\right )}{x} \, dx,x,e^x\right )}{\sqrt {5} \left (1+\sqrt {5}\right )}\\ &=\frac {x^2}{\sqrt {5} \left (1-\sqrt {5}\right )}-\frac {x^2}{\sqrt {5} \left (1+\sqrt {5}\right )}-\frac {2 x \log \left (1+\frac {2 e^x}{1-\sqrt {5}}\right )}{\sqrt {5} \left (1-\sqrt {5}\right )}+\frac {2 x \log \left (1+\frac {2 e^x}{1+\sqrt {5}}\right )}{\sqrt {5} \left (1+\sqrt {5}\right )}-\frac {2 \text {Li}_2\left (-\frac {2 e^x}{1-\sqrt {5}}\right )}{\sqrt {5} \left (1-\sqrt {5}\right )}+\frac {2 \text {Li}_2\left (-\frac {2 e^x}{1+\sqrt {5}}\right )}{\sqrt {5} \left (1+\sqrt {5}\right )}\\ \end {align*}
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Mathematica [A] time = 0.11, size = 120, normalized size = 0.67 \[ \frac {-\left (1+\sqrt {5}\right ) \text {Li}_2\left (\frac {1}{2} \left (-1+\sqrt {5}\right ) e^{-x}\right )-\left (\sqrt {5}-1\right ) \text {Li}_2\left (-\frac {1}{2} \left (1+\sqrt {5}\right ) e^{-x}\right )+\left (1+\sqrt {5}\right ) x \log \left (1-\frac {1}{2} \left (\sqrt {5}-1\right ) e^{-x}\right )+\left (\sqrt {5}-1\right ) x \log \left (\frac {1}{2} \left (1+\sqrt {5}\right ) e^{-x}+1\right )}{2 \sqrt {5}} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.45, size = 86, normalized size = 0.48 \[ -\frac {1}{2} \, x^{2} + \frac {1}{10} \, {\left (\sqrt {5} + 5\right )} {\rm Li}_2\left (\frac {1}{2} \, {\left (\sqrt {5} + 1\right )} e^{x}\right ) - \frac {1}{10} \, {\left (\sqrt {5} - 5\right )} {\rm Li}_2\left (-\frac {1}{2} \, {\left (\sqrt {5} - 1\right )} e^{x}\right ) + \frac {1}{10} \, {\left (\sqrt {5} x + 5 \, x\right )} \log \left (-\frac {1}{2} \, {\left (\sqrt {5} + 1\right )} e^{x} + 1\right ) - \frac {1}{10} \, {\left (\sqrt {5} x - 5 \, x\right )} \log \left (\frac {1}{2} \, {\left (\sqrt {5} - 1\right )} e^{x} + 1\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{e^{\left (2 \, x\right )} + e^{x} - 1}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.02, size = 183, normalized size = 1.02 \[ -\frac {x^{2}}{2}-\frac {\sqrt {5}\, x \ln \left (\frac {2 \,{\mathrm e}^{x}+1+\sqrt {5}}{\sqrt {5}+1}\right )}{10}+\frac {x \ln \left (\frac {2 \,{\mathrm e}^{x}+1+\sqrt {5}}{\sqrt {5}+1}\right )}{2}+\frac {\sqrt {5}\, x \ln \left (\frac {-2 \,{\mathrm e}^{x}+\sqrt {5}-1}{\sqrt {5}-1}\right )}{10}+\frac {x \ln \left (\frac {-2 \,{\mathrm e}^{x}+\sqrt {5}-1}{\sqrt {5}-1}\right )}{2}-\frac {\sqrt {5}\, \dilog \left (\frac {2 \,{\mathrm e}^{x}+1+\sqrt {5}}{\sqrt {5}+1}\right )}{10}+\frac {\dilog \left (\frac {2 \,{\mathrm e}^{x}+1+\sqrt {5}}{\sqrt {5}+1}\right )}{2}+\frac {\sqrt {5}\, \dilog \left (\frac {-2 \,{\mathrm e}^{x}+\sqrt {5}-1}{\sqrt {5}-1}\right )}{10}+\frac {\dilog \left (\frac {-2 \,{\mathrm e}^{x}+\sqrt {5}-1}{\sqrt {5}-1}\right )}{2} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{e^{\left (2 \, x\right )} + e^{x} - 1}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x}{{\mathrm {e}}^{2\,x}+{\mathrm {e}}^x-1} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{e^{2 x} + e^{x} - 1}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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