Optimal. Leaf size=103 \[ -\frac {i \text {Li}_3\left (-i e^{-a-b x}\right )}{2 b^2}+\frac {i \text {Li}_3\left (i e^{-a-b x}\right )}{2 b^2}-\frac {i x \text {Li}_2\left (-i e^{-a-b x}\right )}{2 b}+\frac {i x \text {Li}_2\left (i e^{-a-b x}\right )}{2 b} \]
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Rubi [A] time = 0.06, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5144, 2531, 2282, 6589} \[ -\frac {i \text {PolyLog}\left (3,-i e^{-a-b x}\right )}{2 b^2}+\frac {i \text {PolyLog}\left (3,i e^{-a-b x}\right )}{2 b^2}-\frac {i x \text {PolyLog}\left (2,-i e^{-a-b x}\right )}{2 b}+\frac {i x \text {PolyLog}\left (2,i e^{-a-b x}\right )}{2 b} \]
Antiderivative was successfully verified.
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Rule 2282
Rule 2531
Rule 5144
Rule 6589
Rubi steps
\begin {align*} \int x \cot ^{-1}\left (e^{a+b x}\right ) \, dx &=\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 \text {Li}_2\left (-i e^{-a-b x}\right )}{2 b}+\frac {i x \text {Li}_2\left (i e^{-a-b x}\right )}{2 b}+\frac {i \int \text {Li}_2\left (-i e^{-a-b x}\right ) \, dx}{2 b}-\frac {i \int \text {Li}_2\left (i e^{-a-b x}\right ) \, dx}{2 b}\\ &=-\frac {i x \text {Li}_2\left (-i e^{-a-b x}\right )}{2 b}+\frac {i x \text {Li}_2\left (i e^{-a-b x}\right )}{2 b}-\frac {i \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{-a-b x}\right )}{2 b^2}+\frac {i \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{-a-b x}\right )}{2 b^2}\\ &=-\frac {i x \text {Li}_2\left (-i e^{-a-b x}\right )}{2 b}+\frac {i x \text {Li}_2\left (i e^{-a-b x}\right )}{2 b}-\frac {i \text {Li}_3\left (-i e^{-a-b x}\right )}{2 b^2}+\frac {i \text {Li}_3\left (i e^{-a-b x}\right )}{2 b^2}\\ \end {align*}
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Mathematica [A] time = 0.02, size = 83, normalized size = 0.81 \[ -\frac {i \left (b x \text {Li}_2\left (-i e^{-a-b x}\right )-b x \text {Li}_2\left (i e^{-a-b x}\right )+\text {Li}_3\left (-i e^{-a-b x}\right )-\text {Li}_3\left (i e^{-a-b x}\right )\right )}{2 b^2} \]
Antiderivative was successfully verified.
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fricas [C] time = 0.66, size = 151, normalized size = 1.47 \[ \frac {2 \, b^{2} x^{2} \operatorname {arccot}\left (e^{\left (b x + a\right )}\right ) + 2 i \, b x {\rm Li}_2\left (i \, e^{\left (b x + a\right )}\right ) - 2 i \, b x {\rm Li}_2\left (-i \, e^{\left (b x + a\right )}\right ) + i \, a^{2} \log \left (e^{\left (b x + a\right )} + i\right ) - i \, a^{2} \log \left (e^{\left (b x + a\right )} - i\right ) + {\left (-i \, b^{2} x^{2} + i \, a^{2}\right )} \log \left (i \, e^{\left (b x + a\right )} + 1\right ) + {\left (i \, b^{2} x^{2} - i \, a^{2}\right )} \log \left (-i \, e^{\left (b x + a\right )} + 1\right ) - 2 i \, {\rm polylog}\left (3, i \, e^{\left (b x + a\right )}\right ) + 2 i \, {\rm polylog}\left (3, -i \, e^{\left (b x + a\right )}\right )}{4 \, b^{2}} \]
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 \[ \int x \operatorname {arccot}\left (e^{\left (b x + a\right )}\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.34, size = 355, normalized size = 3.45 \[ \frac {\pi \,x^{2}}{4}+\frac {i \ln \left (-i \left (-{\mathrm e}^{b x +a}+i\right )\right ) a^{2}}{2 b^{2}}-\frac {i \polylog \left (3, i {\mathrm e}^{b x +a}\right )}{2 b^{2}}-\frac {i \dilog \left (-i {\mathrm e}^{b x +a}\right ) a}{2 b^{2}}-\frac {i \ln \left (-i \left ({\mathrm e}^{b x +a}+i\right )\right ) a^{2}}{2 b^{2}}-\frac {i \ln \left (-i \left ({\mathrm e}^{b x +a}+i\right )\right ) x a}{2 b}-\frac {i a^{2} \ln \left (1+i {\mathrm e}^{b x +a}\right )}{2 b^{2}}-\frac {i \ln \left (1+i {\mathrm e}^{b x +a}\right ) x a}{2 b}+\frac {i \ln \left (1-i {\mathrm e}^{b x +a}\right ) x a}{2 b}+\frac {i a^{2} \ln \left (1-i {\mathrm e}^{b x +a}\right )}{2 b^{2}}+\frac {i \polylog \left (2, i {\mathrm e}^{b x +a}\right ) a}{2 b^{2}}-\frac {i \ln \left (-i {\mathrm e}^{b x +a}\right ) \ln \left (-i \left (-{\mathrm e}^{b x +a}+i\right )\right ) a}{2 b^{2}}+\frac {i \polylog \left (3, -i {\mathrm e}^{b x +a}\right )}{2 b^{2}}+\frac {i \ln \left (-i \left (-{\mathrm e}^{b x +a}+i\right )\right ) x a}{2 b}-\frac {i \polylog \left (2, -i {\mathrm e}^{b x +a}\right ) a}{2 b^{2}}+\frac {i \polylog \left (2, i {\mathrm e}^{b x +a}\right ) x}{2 b}-\frac {i \polylog \left (2, -i {\mathrm e}^{b x +a}\right ) x}{2 b}-\frac {i \dilog \left (-i \left ({\mathrm e}^{b x +a}+i\right )\right ) a}{2 b^{2}} \]
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 \[ \frac {1}{2} \, x^{2} \arctan \left (e^{\left (-b x - a\right )}\right ) + b \int \frac {x^{2} e^{\left (b x + a\right )}}{2 \, {\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}}\,{d x} \]
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 \[ \int x\,\mathrm {acot}\left ({\mathrm {e}}^{a+b\,x}\right ) \,d x \]
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 \[ \int x \operatorname {acot}{\left (e^{a} e^{b x} \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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