Optimal. Leaf size=265 \[ \frac {i \text {Li}_3\left (\frac {(-c-d+i) e^{2 a+2 b x}}{-c+d+i}\right )}{8 b^2}-\frac {i \text {Li}_3\left (\frac {(c+d+i) e^{2 a+2 b x}}{c-d+i}\right )}{8 b^2}-\frac {i x \text {Li}_2\left (\frac {(-c-d+i) e^{2 a+2 b x}}{-c+d+i}\right )}{4 b}+\frac {i x \text {Li}_2\left (\frac {(c+d+i) e^{2 a+2 b x}}{c-d+i}\right )}{4 b}-\frac {1}{4} i x^2 \log \left (1-\frac {(-c-d+i) e^{2 a+2 b x}}{-c+d+i}\right )+\frac {1}{4} i x^2 \log \left (1-\frac {(c+d+i) e^{2 a+2 b x}}{c-d+i}\right )+\frac {1}{2} x^2 \cot ^{-1}(d \coth (a+b x)+c) \]
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Rubi [A] time = 0.38, antiderivative size = 265, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {5202, 2190, 2531, 2282, 6589} \[ \frac {i \text {PolyLog}\left (3,\frac {(-c-d+i) e^{2 a+2 b x}}{-c+d+i}\right )}{8 b^2}-\frac {i \text {PolyLog}\left (3,\frac {(c+d+i) e^{2 a+2 b x}}{c-d+i}\right )}{8 b^2}-\frac {i x \text {PolyLog}\left (2,\frac {(-c-d+i) e^{2 a+2 b x}}{-c+d+i}\right )}{4 b}+\frac {i x \text {PolyLog}\left (2,\frac {(c+d+i) e^{2 a+2 b x}}{c-d+i}\right )}{4 b}-\frac {1}{4} i x^2 \log \left (1-\frac {(-c-d+i) e^{2 a+2 b x}}{-c+d+i}\right )+\frac {1}{4} i x^2 \log \left (1-\frac {(c+d+i) e^{2 a+2 b x}}{c-d+i}\right )+\frac {1}{2} x^2 \cot ^{-1}(d \coth (a+b x)+c) \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2282
Rule 2531
Rule 5202
Rule 6589
Rubi steps
\begin {align*} \int x \cot ^{-1}(c+d \coth (a+b x)) \, dx &=\frac {1}{2} x^2 \cot ^{-1}(c+d \coth (a+b x))+\frac {1}{2} (b (1-i (c+d))) \int \frac {e^{2 a+2 b x} x^2}{i+c-d+(-i-c-d) e^{2 a+2 b x}} \, dx-\frac {1}{2} (b (1+i (c+d))) \int \frac {e^{2 a+2 b x} x^2}{i-c+d+(-i+c+d) e^{2 a+2 b x}} \, dx\\ &=\frac {1}{2} x^2 \cot ^{-1}(c+d \coth (a+b x))-\frac {1}{4} i x^2 \log \left (1-\frac {(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )+\frac {1}{4} i x^2 \log \left (1-\frac {(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )-\frac {1}{2} i \int x \log \left (1+\frac {(-i-c-d) e^{2 a+2 b x}}{i+c-d}\right ) \, dx+\frac {1}{2} i \int x \log \left (1+\frac {(-i+c+d) e^{2 a+2 b x}}{i-c+d}\right ) \, dx\\ &=\frac {1}{2} x^2 \cot ^{-1}(c+d \coth (a+b x))-\frac {1}{4} i x^2 \log \left (1-\frac {(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )+\frac {1}{4} i x^2 \log \left (1-\frac {(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )-\frac {i x \text {Li}_2\left (\frac {(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )}{4 b}+\frac {i x \text {Li}_2\left (\frac {(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )}{4 b}-\frac {i \int \text {Li}_2\left (-\frac {(-i-c-d) e^{2 a+2 b x}}{i+c-d}\right ) \, dx}{4 b}+\frac {i \int \text {Li}_2\left (-\frac {(-i+c+d) e^{2 a+2 b x}}{i-c+d}\right ) \, dx}{4 b}\\ &=\frac {1}{2} x^2 \cot ^{-1}(c+d \coth (a+b x))-\frac {1}{4} i x^2 \log \left (1-\frac {(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )+\frac {1}{4} i x^2 \log \left (1-\frac {(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )-\frac {i x \text {Li}_2\left (\frac {(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )}{4 b}+\frac {i x \text {Li}_2\left (\frac {(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )}{4 b}+\frac {i \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {(-i+c+d) x}{-i+c-d}\right )}{x} \, dx,x,e^{2 a+2 b x}\right )}{8 b^2}-\frac {i \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {(i+c+d) x}{i+c-d}\right )}{x} \, dx,x,e^{2 a+2 b x}\right )}{8 b^2}\\ &=\frac {1}{2} x^2 \cot ^{-1}(c+d \coth (a+b x))-\frac {1}{4} i x^2 \log \left (1-\frac {(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )+\frac {1}{4} i x^2 \log \left (1-\frac {(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )-\frac {i x \text {Li}_2\left (\frac {(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )}{4 b}+\frac {i x \text {Li}_2\left (\frac {(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )}{4 b}+\frac {i \text {Li}_3\left (\frac {(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )}{8 b^2}-\frac {i \text {Li}_3\left (\frac {(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )}{8 b^2}\\ \end {align*}
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Mathematica [A] time = 4.51, size = 225, normalized size = 0.85 \[ \frac {1}{2} x^2 \cot ^{-1}(d \coth (a+b x)+c)-\frac {i \left (2 b^2 x^2 \log \left (1+\frac {(c+d-i) e^{2 (a+b x)}}{-c+d+i}\right )-2 b^2 x^2 \log \left (1+\frac {(c+d+i) e^{2 (a+b x)}}{-c+d-i}\right )+2 b x \text {Li}_2\left (\frac {(c+d-i) e^{2 (a+b x)}}{c-d-i}\right )-2 b x \text {Li}_2\left (\frac {(c+d+i) e^{2 (a+b x)}}{c-d+i}\right )-\text {Li}_3\left (\frac {(c+d-i) e^{2 (a+b x)}}{c-d-i}\right )+\text {Li}_3\left (\frac {(c+d+i) e^{2 (a+b x)}}{c-d+i}\right )\right )}{8 b^2} \]
Antiderivative was successfully verified.
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fricas [C] time = 1.08, size = 1087, normalized size = 4.10 \[ \text {result too large to display} \]
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 \[ \int x \operatorname {arccot}\left (d \coth \left (b x + a\right ) + c\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 6.46, size = 6508, normalized size = 24.56 \[ \text {output too large to display} \]
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 \[ \frac {1}{2} \, x^{2} \arctan \left (e^{\left (2 \, b x + 2 \, a\right )} - 1, {\left (c e^{\left (2 \, a\right )} + d e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )} - c + d\right ) - 2 \, b d \int \frac {x^{2} e^{\left (2 \, b x + 2 \, a\right )}}{c^{2} - 2 \, c d + d^{2} + {\left (c^{2} e^{\left (4 \, a\right )} + 2 \, c d e^{\left (4 \, a\right )} + d^{2} e^{\left (4 \, a\right )} + e^{\left (4 \, a\right )}\right )} e^{\left (4 \, b x\right )} - 2 \, {\left (c^{2} e^{\left (2 \, a\right )} - d^{2} e^{\left (2 \, a\right )} + e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )} + 1}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int x\,\mathrm {acot}\left (c+d\,\mathrm {coth}\left (a+b\,x\right )\right ) \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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