Optimal. Leaf size=303 \[ \frac {\text {Li}_4\left (\frac {(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{8 b^3}-\frac {\text {Li}_4\left (\frac {(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{8 b^3}-\frac {x \text {Li}_3\left (\frac {(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{4 b^2}+\frac {x \text {Li}_3\left (\frac {(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{4 b^2}+\frac {x^2 \text {Li}_2\left (\frac {(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{4 b}-\frac {x^2 \text {Li}_2\left (\frac {(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{4 b}+\frac {1}{6} x^3 \log \left (1-\frac {(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )-\frac {1}{6} x^3 \log \left (1-\frac {(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )+\frac {1}{3} x^3 \coth ^{-1}(d \coth (a+b x)+c) \]
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Rubi [A] time = 0.47, antiderivative size = 303, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6246, 2190, 2531, 6609, 2282, 6589} \[ -\frac {x \text {PolyLog}\left (3,\frac {(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{4 b^2}+\frac {x \text {PolyLog}\left (3,\frac {(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{4 b^2}+\frac {\text {PolyLog}\left (4,\frac {(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{8 b^3}-\frac {\text {PolyLog}\left (4,\frac {(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{8 b^3}+\frac {x^2 \text {PolyLog}\left (2,\frac {(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{4 b}-\frac {x^2 \text {PolyLog}\left (2,\frac {(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{4 b}+\frac {1}{6} x^3 \log \left (1-\frac {(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )-\frac {1}{6} x^3 \log \left (1-\frac {(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )+\frac {1}{3} x^3 \coth ^{-1}(d \coth (a+b x)+c) \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2282
Rule 2531
Rule 6246
Rule 6589
Rule 6609
Rubi steps
\begin {align*} \int x^2 \coth ^{-1}(c+d \coth (a+b x)) \, dx &=\frac {1}{3} x^3 \coth ^{-1}(c+d \coth (a+b x))-\frac {1}{3} (b (1-c-d)) \int \frac {e^{2 a+2 b x} x^3}{1-c+d+(-1+c+d) e^{2 a+2 b x}} \, dx+\frac {1}{3} (b (1+c+d)) \int \frac {e^{2 a+2 b x} x^3}{1+c-d+(-1-c-d) e^{2 a+2 b x}} \, dx\\ &=\frac {1}{3} x^3 \coth ^{-1}(c+d \coth (a+b x))+\frac {1}{6} x^3 \log \left (1-\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )-\frac {1}{6} x^3 \log \left (1-\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )+\frac {1}{2} \int x^2 \log \left (1+\frac {(-1-c-d) e^{2 a+2 b x}}{1+c-d}\right ) \, dx-\frac {1}{2} \int x^2 \log \left (1+\frac {(-1+c+d) e^{2 a+2 b x}}{1-c+d}\right ) \, dx\\ &=\frac {1}{3} x^3 \coth ^{-1}(c+d \coth (a+b x))+\frac {1}{6} x^3 \log \left (1-\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )-\frac {1}{6} x^3 \log \left (1-\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )+\frac {x^2 \text {Li}_2\left (\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b}-\frac {x^2 \text {Li}_2\left (\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b}+\frac {\int x \text {Li}_2\left (-\frac {(-1-c-d) e^{2 a+2 b x}}{1+c-d}\right ) \, dx}{2 b}-\frac {\int x \text {Li}_2\left (-\frac {(-1+c+d) e^{2 a+2 b x}}{1-c+d}\right ) \, dx}{2 b}\\ &=\frac {1}{3} x^3 \coth ^{-1}(c+d \coth (a+b x))+\frac {1}{6} x^3 \log \left (1-\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )-\frac {1}{6} x^3 \log \left (1-\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )+\frac {x^2 \text {Li}_2\left (\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b}-\frac {x^2 \text {Li}_2\left (\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b}-\frac {x \text {Li}_3\left (\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b^2}+\frac {x \text {Li}_3\left (\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b^2}-\frac {\int \text {Li}_3\left (-\frac {(-1-c-d) e^{2 a+2 b x}}{1+c-d}\right ) \, dx}{4 b^2}+\frac {\int \text {Li}_3\left (-\frac {(-1+c+d) e^{2 a+2 b x}}{1-c+d}\right ) \, dx}{4 b^2}\\ &=\frac {1}{3} x^3 \coth ^{-1}(c+d \coth (a+b x))+\frac {1}{6} x^3 \log \left (1-\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )-\frac {1}{6} x^3 \log \left (1-\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )+\frac {x^2 \text {Li}_2\left (\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b}-\frac {x^2 \text {Li}_2\left (\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b}-\frac {x \text {Li}_3\left (\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b^2}+\frac {x \text {Li}_3\left (\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b^2}+\frac {\operatorname {Subst}\left (\int \frac {\text {Li}_3\left (\frac {(-1+c+d) x}{-1+c-d}\right )}{x} \, dx,x,e^{2 a+2 b x}\right )}{8 b^3}-\frac {\operatorname {Subst}\left (\int \frac {\text {Li}_3\left (\frac {(1+c+d) x}{1+c-d}\right )}{x} \, dx,x,e^{2 a+2 b x}\right )}{8 b^3}\\ &=\frac {1}{3} x^3 \coth ^{-1}(c+d \coth (a+b x))+\frac {1}{6} x^3 \log \left (1-\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )-\frac {1}{6} x^3 \log \left (1-\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )+\frac {x^2 \text {Li}_2\left (\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b}-\frac {x^2 \text {Li}_2\left (\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b}-\frac {x \text {Li}_3\left (\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b^2}+\frac {x \text {Li}_3\left (\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b^2}+\frac {\text {Li}_4\left (\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{8 b^3}-\frac {\text {Li}_4\left (\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{8 b^3}\\ \end {align*}
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Mathematica [A] time = 0.45, size = 353, normalized size = 1.17 \[ \frac {4 b^3 x^3 \log \left (\frac {2 (\cosh (a+b x)-\sinh (a+b x)) ((c-1) \sinh (a+b x)+d \cosh (a+b x))}{c+d-1}\right )-4 b^3 x^3 \log \left (\frac {(c-d+1) (\sinh (2 (a+b x))-\cosh (2 (a+b x)))}{c+d+1}+1\right )-6 b^2 x^2 \text {Li}_2\left (\frac {(c-d-1) (\cosh (2 (a+b x))-\sinh (2 (a+b x)))}{c+d-1}\right )+6 b^2 x^2 \text {Li}_2\left (\frac {(c-d+1) (\cosh (2 (a+b x))-\sinh (2 (a+b x)))}{c+d+1}\right )-6 b x \text {Li}_3\left (\frac {(c-d-1) (\cosh (2 (a+b x))-\sinh (2 (a+b x)))}{c+d-1}\right )+6 b x \text {Li}_3\left (\frac {(c-d+1) (\cosh (2 (a+b x))-\sinh (2 (a+b x)))}{c+d+1}\right )-3 \text {Li}_4\left (\frac {(c-d-1) (\cosh (2 (a+b x))-\sinh (2 (a+b x)))}{c+d-1}\right )+3 \text {Li}_4\left (\frac {(c-d+1) (\cosh (2 (a+b x))-\sinh (2 (a+b x)))}{c+d+1}\right )}{24 b^3}+\frac {1}{3} x^3 \coth ^{-1}(d \coth (a+b x)+c) \]
Warning: Unable to verify antiderivative.
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fricas [C] time = 0.56, size = 879, normalized size = 2.90 \[ \text {result too large to display} \]
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 \[ \int x^{2} \operatorname {arcoth}\left (d \coth \left (b x + a\right ) + c\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 10.36, size = 5222, normalized size = 17.23 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.70, size = 277, normalized size = 0.91 \[ \frac {1}{3} \, x^{3} \operatorname {arcoth}\left (d \coth \left (b x + a\right ) + c\right ) - \frac {1}{18} \, b d {\left (\frac {4 \, b^{3} x^{3} \log \left (-\frac {{\left (c + d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d + 1} + 1\right ) + 6 \, b^{2} x^{2} {\rm Li}_2\left (\frac {{\left (c + d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d + 1}\right ) - 6 \, b x {\rm Li}_{3}(\frac {{\left (c + d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d + 1}) + 3 \, {\rm Li}_{4}(\frac {{\left (c + d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d + 1})}{b^{4} d} - \frac {4 \, b^{3} x^{3} \log \left (-\frac {{\left (c + d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d - 1} + 1\right ) + 6 \, b^{2} x^{2} {\rm Li}_2\left (\frac {{\left (c + d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d - 1}\right ) - 6 \, b x {\rm Li}_{3}(\frac {{\left (c + d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d - 1}) + 3 \, {\rm Li}_{4}(\frac {{\left (c + d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d - 1})}{b^{4} d}\right )} \]
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 \[ \int x^2\,\mathrm {acoth}\left (c+d\,\mathrm {coth}\left (a+b\,x\right )\right ) \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \operatorname {acoth}{\left (c + d \coth {\left (a + b x \right )} \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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