Integrand size = 18, antiderivative size = 69 \[ \int \frac {e^x x^3}{1-e^{2 x}} \, dx=x^3 \text {arctanh}\left (e^x\right )+\frac {3}{2} x^2 \operatorname {PolyLog}\left (2,-e^x\right )-\frac {3}{2} x^2 \operatorname {PolyLog}\left (2,e^x\right )-3 x \operatorname {PolyLog}\left (3,-e^x\right )+3 x \operatorname {PolyLog}\left (3,e^x\right )+3 \operatorname {PolyLog}\left (4,-e^x\right )-3 \operatorname {PolyLog}\left (4,e^x\right ) \]
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Time = 0.10 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {2281, 212, 2277, 6348, 2611, 6744, 2320, 6724} \[ \int \frac {e^x x^3}{1-e^{2 x}} \, dx=x^3 \text {arctanh}\left (e^x\right )+\frac {3}{2} x^2 \operatorname {PolyLog}\left (2,-e^x\right )-\frac {3}{2} x^2 \operatorname {PolyLog}\left (2,e^x\right )-3 x \operatorname {PolyLog}\left (3,-e^x\right )+3 x \operatorname {PolyLog}\left (3,e^x\right )+3 \operatorname {PolyLog}\left (4,-e^x\right )-3 \operatorname {PolyLog}\left (4,e^x\right ) \]
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Rule 212
Rule 2277
Rule 2281
Rule 2320
Rule 2611
Rule 6348
Rule 6724
Rule 6744
Rubi steps \begin{align*} \text {integral}& = x^3 \tanh ^{-1}\left (e^x\right )-3 \int x^2 \tanh ^{-1}\left (e^x\right ) \, dx \\ & = x^3 \tanh ^{-1}\left (e^x\right )+\frac {3}{2} \int x^2 \log \left (1-e^x\right ) \, dx-\frac {3}{2} \int x^2 \log \left (1+e^x\right ) \, dx \\ & = x^3 \tanh ^{-1}\left (e^x\right )+\frac {3}{2} x^2 \text {Li}_2\left (-e^x\right )-\frac {3}{2} x^2 \text {Li}_2\left (e^x\right )-3 \int x \text {Li}_2\left (-e^x\right ) \, dx+3 \int x \text {Li}_2\left (e^x\right ) \, dx \\ & = x^3 \tanh ^{-1}\left (e^x\right )+\frac {3}{2} x^2 \text {Li}_2\left (-e^x\right )-\frac {3}{2} x^2 \text {Li}_2\left (e^x\right )-3 x \text {Li}_3\left (-e^x\right )+3 x \text {Li}_3\left (e^x\right )+3 \int \text {Li}_3\left (-e^x\right ) \, dx-3 \int \text {Li}_3\left (e^x\right ) \, dx \\ & = x^3 \tanh ^{-1}\left (e^x\right )+\frac {3}{2} x^2 \text {Li}_2\left (-e^x\right )-\frac {3}{2} x^2 \text {Li}_2\left (e^x\right )-3 x \text {Li}_3\left (-e^x\right )+3 x \text {Li}_3\left (e^x\right )+3 \text {Subst}\left (\int \frac {\text {Li}_3(-x)}{x} \, dx,x,e^x\right )-3 \text {Subst}\left (\int \frac {\text {Li}_3(x)}{x} \, dx,x,e^x\right ) \\ & = x^3 \tanh ^{-1}\left (e^x\right )+\frac {3}{2} x^2 \text {Li}_2\left (-e^x\right )-\frac {3}{2} x^2 \text {Li}_2\left (e^x\right )-3 x \text {Li}_3\left (-e^x\right )+3 x \text {Li}_3\left (e^x\right )+3 \text {Li}_4\left (-e^x\right )-3 \text {Li}_4\left (e^x\right ) \\ \end{align*}
Time = 0.09 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.29 \[ \int \frac {e^x x^3}{1-e^{2 x}} \, dx=-\frac {1}{2} x^3 \log \left (1-e^x\right )+\frac {1}{2} x^3 \log \left (1+e^x\right )+\frac {3}{2} x^2 \operatorname {PolyLog}\left (2,-e^x\right )-\frac {3}{2} x^2 \operatorname {PolyLog}\left (2,e^x\right )-3 x \operatorname {PolyLog}\left (3,-e^x\right )+3 x \operatorname {PolyLog}\left (3,e^x\right )+3 \operatorname {PolyLog}\left (4,-e^x\right )-3 \operatorname {PolyLog}\left (4,e^x\right ) \]
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Time = 0.03 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.07
method | result | size |
default | \(\frac {x^{3} \ln \left (1+{\mathrm e}^{x}\right )}{2}+\frac {3 x^{2} \operatorname {Li}_{2}\left (-{\mathrm e}^{x}\right )}{2}-3 x \,\operatorname {Li}_{3}\left (-{\mathrm e}^{x}\right )+3 \,\operatorname {Li}_{4}\left (-{\mathrm e}^{x}\right )-\frac {x^{3} \ln \left (1-{\mathrm e}^{x}\right )}{2}-\frac {3 x^{2} \operatorname {Li}_{2}\left ({\mathrm e}^{x}\right )}{2}+3 x \,\operatorname {Li}_{3}\left ({\mathrm e}^{x}\right )-3 \,\operatorname {Li}_{4}\left ({\mathrm e}^{x}\right )\) | \(74\) |
risch | \(\frac {x^{3} \ln \left (1+{\mathrm e}^{x}\right )}{2}+\frac {3 x^{2} \operatorname {Li}_{2}\left (-{\mathrm e}^{x}\right )}{2}-3 x \,\operatorname {Li}_{3}\left (-{\mathrm e}^{x}\right )+3 \,\operatorname {Li}_{4}\left (-{\mathrm e}^{x}\right )-\frac {x^{3} \ln \left (1-{\mathrm e}^{x}\right )}{2}-\frac {3 x^{2} \operatorname {Li}_{2}\left ({\mathrm e}^{x}\right )}{2}+3 x \,\operatorname {Li}_{3}\left ({\mathrm e}^{x}\right )-3 \,\operatorname {Li}_{4}\left ({\mathrm e}^{x}\right )\) | \(74\) |
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Time = 0.28 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.03 \[ \int \frac {e^x x^3}{1-e^{2 x}} \, dx=\frac {1}{2} \, x^{3} \log \left (e^{x} + 1\right ) - \frac {1}{2} \, x^{3} \log \left (-e^{x} + 1\right ) + \frac {3}{2} \, x^{2} {\rm Li}_2\left (-e^{x}\right ) - \frac {3}{2} \, x^{2} {\rm Li}_2\left (e^{x}\right ) - 3 \, x {\rm polylog}\left (3, -e^{x}\right ) + 3 \, x {\rm polylog}\left (3, e^{x}\right ) + 3 \, {\rm polylog}\left (4, -e^{x}\right ) - 3 \, {\rm polylog}\left (4, e^{x}\right ) \]
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\[ \int \frac {e^x x^3}{1-e^{2 x}} \, dx=- \int \frac {x^{3} e^{x}}{e^{2 x} - 1}\, dx \]
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Time = 0.17 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.03 \[ \int \frac {e^x x^3}{1-e^{2 x}} \, dx=\frac {1}{2} \, x^{3} \log \left (e^{x} + 1\right ) - \frac {1}{2} \, x^{3} \log \left (-e^{x} + 1\right ) + \frac {3}{2} \, x^{2} {\rm Li}_2\left (-e^{x}\right ) - \frac {3}{2} \, x^{2} {\rm Li}_2\left (e^{x}\right ) - 3 \, x {\rm Li}_{3}(-e^{x}) + 3 \, x {\rm Li}_{3}(e^{x}) + 3 \, {\rm Li}_{4}(-e^{x}) - 3 \, {\rm Li}_{4}(e^{x}) \]
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\[ \int \frac {e^x x^3}{1-e^{2 x}} \, dx=\int { -\frac {x^{3} e^{x}}{e^{\left (2 \, x\right )} - 1} \,d x } \]
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Timed out. \[ \int \frac {e^x x^3}{1-e^{2 x}} \, dx=-\int \frac {x^3\,{\mathrm {e}}^x}{{\mathrm {e}}^{2\,x}-1} \,d x \]
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