Integrand size = 10, antiderivative size = 103 \[ \int x \cot ^{-1}\left (e^{a+b x}\right ) \, dx=-\frac {i x \operatorname {PolyLog}\left (2,-i e^{-a-b x}\right )}{2 b}+\frac {i x \operatorname {PolyLog}\left (2,i e^{-a-b x}\right )}{2 b}-\frac {i \operatorname {PolyLog}\left (3,-i e^{-a-b x}\right )}{2 b^2}+\frac {i \operatorname {PolyLog}\left (3,i e^{-a-b x}\right )}{2 b^2} \]
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Time = 0.05 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5252, 2611, 2320, 6724} \[ \int x \cot ^{-1}\left (e^{a+b x}\right ) \, dx=-\frac {i \operatorname {PolyLog}\left (3,-i e^{-a-b x}\right )}{2 b^2}+\frac {i \operatorname {PolyLog}\left (3,i e^{-a-b x}\right )}{2 b^2}-\frac {i x \operatorname {PolyLog}\left (2,-i e^{-a-b x}\right )}{2 b}+\frac {i x \operatorname {PolyLog}\left (2,i e^{-a-b x}\right )}{2 b} \]
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Rule 2320
Rule 2611
Rule 5252
Rule 6724
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} i \int x \log \left (1-i e^{-a-b x}\right ) \, dx-\frac {1}{2} i \int x \log \left (1+i e^{-a-b x}\right ) \, dx \\ & = -\frac {i x \operatorname {PolyLog}\left (2,-i e^{-a-b x}\right )}{2 b}+\frac {i x \operatorname {PolyLog}\left (2,i e^{-a-b x}\right )}{2 b}+\frac {i \int \operatorname {PolyLog}\left (2,-i e^{-a-b x}\right ) \, dx}{2 b}-\frac {i \int \operatorname {PolyLog}\left (2,i e^{-a-b x}\right ) \, dx}{2 b} \\ & = -\frac {i x \operatorname {PolyLog}\left (2,-i e^{-a-b x}\right )}{2 b}+\frac {i x \operatorname {PolyLog}\left (2,i e^{-a-b x}\right )}{2 b}-\frac {i \text {Subst}\left (\int \frac {\operatorname {PolyLog}(2,-i x)}{x} \, dx,x,e^{-a-b x}\right )}{2 b^2}+\frac {i \text {Subst}\left (\int \frac {\operatorname {PolyLog}(2,i x)}{x} \, dx,x,e^{-a-b x}\right )}{2 b^2} \\ & = -\frac {i x \operatorname {PolyLog}\left (2,-i e^{-a-b x}\right )}{2 b}+\frac {i x \operatorname {PolyLog}\left (2,i e^{-a-b x}\right )}{2 b}-\frac {i \operatorname {PolyLog}\left (3,-i e^{-a-b x}\right )}{2 b^2}+\frac {i \operatorname {PolyLog}\left (3,i e^{-a-b x}\right )}{2 b^2} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.81 \[ \int x \cot ^{-1}\left (e^{a+b x}\right ) \, dx=-\frac {i \left (b x \operatorname {PolyLog}\left (2,-i e^{-a-b x}\right )-b x \operatorname {PolyLog}\left (2,i e^{-a-b x}\right )+\operatorname {PolyLog}\left (3,-i e^{-a-b x}\right )-\operatorname {PolyLog}\left (3,i e^{-a-b x}\right )\right )}{2 b^2} \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 354 vs. \(2 (83 ) = 166\).
Time = 0.80 (sec) , antiderivative size = 355, normalized size of antiderivative = 3.45
method | result | size |
risch | \(-\frac {i \operatorname {polylog}\left (3, i {\mathrm e}^{b x +a}\right )}{2 b^{2}}+\frac {\pi \,x^{2}}{4}+\frac {i \operatorname {polylog}\left (3, -i {\mathrm e}^{b x +a}\right )}{2 b^{2}}-\frac {i \ln \left (-i \left ({\mathrm e}^{b x +a}+i\right )\right ) a^{2}}{2 b^{2}}+\frac {i a^{2} \ln \left (1-i {\mathrm e}^{b x +a}\right )}{2 b^{2}}+\frac {i x \operatorname {polylog}\left (2, i {\mathrm e}^{b x +a}\right )}{2 b}+\frac {i \ln \left (1-i {\mathrm e}^{b x +a}\right ) a x}{2 b}-\frac {i \operatorname {dilog}\left (-i \left ({\mathrm e}^{b x +a}+i\right )\right ) a}{2 b^{2}}-\frac {i \operatorname {polylog}\left (2, -i {\mathrm e}^{b x +a}\right ) a}{2 b^{2}}+\frac {i \operatorname {polylog}\left (2, i {\mathrm e}^{b x +a}\right ) a}{2 b^{2}}-\frac {i \ln \left (1+i {\mathrm e}^{b x +a}\right ) a x}{2 b}-\frac {i \operatorname {dilog}\left (-i {\mathrm e}^{b x +a}\right ) a}{2 b^{2}}+\frac {i \ln \left (-i \left (-{\mathrm e}^{b x +a}+i\right )\right ) a x}{2 b}-\frac {i \ln \left (-i \left ({\mathrm e}^{b x +a}+i\right )\right ) a x}{2 b}-\frac {i a^{2} \ln \left (1+i {\mathrm e}^{b x +a}\right )}{2 b^{2}}-\frac {i \ln \left (-i {\mathrm e}^{b x +a}\right ) \ln \left (-i \left (-{\mathrm e}^{b x +a}+i\right )\right ) a}{2 b^{2}}-\frac {i x \operatorname {polylog}\left (2, -i {\mathrm e}^{b x +a}\right )}{2 b}+\frac {i \ln \left (-i \left (-{\mathrm e}^{b x +a}+i\right )\right ) a^{2}}{2 b^{2}}\) | \(355\) |
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 151 vs. \(2 (73) = 146\).
Time = 0.36 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.47 \[ \int x \cot ^{-1}\left (e^{a+b x}\right ) \, dx=\frac {2 \, b^{2} x^{2} \operatorname {arccot}\left (e^{\left (b x + a\right )}\right ) + 2 i \, b x {\rm Li}_2\left (i \, e^{\left (b x + a\right )}\right ) - 2 i \, b x {\rm Li}_2\left (-i \, e^{\left (b x + a\right )}\right ) + i \, a^{2} \log \left (e^{\left (b x + a\right )} + i\right ) - i \, a^{2} \log \left (e^{\left (b x + a\right )} - i\right ) + {\left (-i \, b^{2} x^{2} + i \, a^{2}\right )} \log \left (i \, e^{\left (b x + a\right )} + 1\right ) + {\left (i \, b^{2} x^{2} - i \, a^{2}\right )} \log \left (-i \, e^{\left (b x + a\right )} + 1\right ) - 2 i \, {\rm polylog}\left (3, i \, e^{\left (b x + a\right )}\right ) + 2 i \, {\rm polylog}\left (3, -i \, e^{\left (b x + a\right )}\right )}{4 \, b^{2}} \]
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\[ \int x \cot ^{-1}\left (e^{a+b x}\right ) \, dx=\int x \operatorname {acot}{\left (e^{a} e^{b x} \right )}\, dx \]
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\[ \int x \cot ^{-1}\left (e^{a+b x}\right ) \, dx=\int { x \operatorname {arccot}\left (e^{\left (b x + a\right )}\right ) \,d x } \]
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\[ \int x \cot ^{-1}\left (e^{a+b x}\right ) \, dx=\int { x \operatorname {arccot}\left (e^{\left (b x + a\right )}\right ) \,d x } \]
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Timed out. \[ \int x \cot ^{-1}\left (e^{a+b x}\right ) \, dx=\int x\,\mathrm {acot}\left ({\mathrm {e}}^{a+b\,x}\right ) \,d x \]
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