Integrand size = 15, antiderivative size = 303 \[ \int x^2 \coth ^{-1}(c+d \coth (a+b x)) \, dx=\frac {1}{3} x^3 \coth ^{-1}(c+d \coth (a+b x))+\frac {1}{6} x^3 \log \left (1-\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )-\frac {1}{6} x^3 \log \left (1-\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )+\frac {x^2 \operatorname {PolyLog}\left (2,\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b}-\frac {x^2 \operatorname {PolyLog}\left (2,\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b}-\frac {x \operatorname {PolyLog}\left (3,\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b^2}+\frac {x \operatorname {PolyLog}\left (3,\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b^2}+\frac {\operatorname {PolyLog}\left (4,\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{8 b^3}-\frac {\operatorname {PolyLog}\left (4,\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{8 b^3} \]
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Time = 0.41 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6381, 2221, 2611, 6744, 2320, 6724} \[ \int x^2 \coth ^{-1}(c+d \coth (a+b x)) \, dx=\frac {\operatorname {PolyLog}\left (4,\frac {(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{8 b^3}-\frac {\operatorname {PolyLog}\left (4,\frac {(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{8 b^3}-\frac {x \operatorname {PolyLog}\left (3,\frac {(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{4 b^2}+\frac {x \operatorname {PolyLog}\left (3,\frac {(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{4 b^2}+\frac {x^2 \operatorname {PolyLog}\left (2,\frac {(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{4 b}-\frac {x^2 \operatorname {PolyLog}\left (2,\frac {(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{4 b}+\frac {1}{6} x^3 \log \left (1-\frac {(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )-\frac {1}{6} x^3 \log \left (1-\frac {(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )+\frac {1}{3} x^3 \coth ^{-1}(d \coth (a+b x)+c) \]
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Rule 2221
Rule 2320
Rule 2611
Rule 6381
Rule 6724
Rule 6744
Rubi steps \begin{align*} \text {integral}& = \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 \operatorname {PolyLog}\left (2,\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b}-\frac {x^2 \operatorname {PolyLog}\left (2,\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b}+\frac {\int x \operatorname {PolyLog}\left (2,-\frac {(-1-c-d) e^{2 a+2 b x}}{1+c-d}\right ) \, dx}{2 b}-\frac {\int x \operatorname {PolyLog}\left (2,-\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 \operatorname {PolyLog}\left (2,\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b}-\frac {x^2 \operatorname {PolyLog}\left (2,\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b}-\frac {x \operatorname {PolyLog}\left (3,\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b^2}+\frac {x \operatorname {PolyLog}\left (3,\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b^2}-\frac {\int \operatorname {PolyLog}\left (3,-\frac {(-1-c-d) e^{2 a+2 b x}}{1+c-d}\right ) \, dx}{4 b^2}+\frac {\int \operatorname {PolyLog}\left (3,-\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 \operatorname {PolyLog}\left (2,\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b}-\frac {x^2 \operatorname {PolyLog}\left (2,\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b}-\frac {x \operatorname {PolyLog}\left (3,\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b^2}+\frac {x \operatorname {PolyLog}\left (3,\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b^2}+\frac {\text {Subst}\left (\int \frac {\operatorname {PolyLog}\left (3,\frac {(-1+c+d) x}{-1+c-d}\right )}{x} \, dx,x,e^{2 a+2 b x}\right )}{8 b^3}-\frac {\text {Subst}\left (\int \frac {\operatorname {PolyLog}\left (3,\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 \operatorname {PolyLog}\left (2,\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b}-\frac {x^2 \operatorname {PolyLog}\left (2,\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b}-\frac {x \operatorname {PolyLog}\left (3,\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b^2}+\frac {x \operatorname {PolyLog}\left (3,\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b^2}+\frac {\operatorname {PolyLog}\left (4,\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{8 b^3}-\frac {\operatorname {PolyLog}\left (4,\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{8 b^3} \\ \end{align*}
Time = 0.20 (sec) , antiderivative size = 265, normalized size of antiderivative = 0.87 \[ \int x^2 \coth ^{-1}(c+d \coth (a+b x)) \, dx=\frac {1}{3} x^3 \coth ^{-1}(c+d \coth (a+b x))-\frac {-4 b^3 x^3 \log \left (1+\frac {(1-c+d) e^{-2 (a+b x)}}{-1+c+d}\right )+4 b^3 x^3 \log \left (1+\frac {(-1-c+d) e^{-2 (a+b x)}}{1+c+d}\right )+6 b^2 x^2 \operatorname {PolyLog}\left (2,\frac {(-1+c-d) e^{-2 (a+b x)}}{-1+c+d}\right )-6 b^2 x^2 \operatorname {PolyLog}\left (2,\frac {(1+c-d) e^{-2 (a+b x)}}{1+c+d}\right )+6 b x \operatorname {PolyLog}\left (3,\frac {(-1+c-d) e^{-2 (a+b x)}}{-1+c+d}\right )-6 b x \operatorname {PolyLog}\left (3,\frac {(1+c-d) e^{-2 (a+b x)}}{1+c+d}\right )+3 \operatorname {PolyLog}\left (4,\frac {(-1+c-d) e^{-2 (a+b x)}}{-1+c+d}\right )-3 \operatorname {PolyLog}\left (4,\frac {(1+c-d) e^{-2 (a+b x)}}{1+c+d}\right )}{24 b^3} \]
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 7.20 (sec) , antiderivative size = 5185, normalized size of antiderivative = 17.11
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Leaf count of result is larger than twice the leaf count of optimal. 879 vs. \(2 (259) = 518\).
Time = 0.29 (sec) , antiderivative size = 879, normalized size of antiderivative = 2.90 \[ \int x^2 \coth ^{-1}(c+d \coth (a+b x)) \, dx=\text {Too large to display} \]
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\[ \int x^2 \coth ^{-1}(c+d \coth (a+b x)) \, dx=\int x^{2} \operatorname {acoth}{\left (c + d \coth {\left (a + b x \right )} \right )}\, dx \]
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Time = 0.49 (sec) , antiderivative size = 277, normalized size of antiderivative = 0.91 \[ \int x^2 \coth ^{-1}(c+d \coth (a+b x)) \, dx=\frac {1}{3} \, x^{3} \operatorname {arcoth}\left (d \coth \left (b x + a\right ) + c\right ) - \frac {1}{18} \, b d {\left (\frac {4 \, b^{3} x^{3} \log \left (-\frac {{\left (c + d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d + 1} + 1\right ) + 6 \, b^{2} x^{2} {\rm Li}_2\left (\frac {{\left (c + d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d + 1}\right ) - 6 \, b x {\rm Li}_{3}(\frac {{\left (c + d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d + 1}) + 3 \, {\rm Li}_{4}(\frac {{\left (c + d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d + 1})}{b^{4} d} - \frac {4 \, b^{3} x^{3} \log \left (-\frac {{\left (c + d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d - 1} + 1\right ) + 6 \, b^{2} x^{2} {\rm Li}_2\left (\frac {{\left (c + d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d - 1}\right ) - 6 \, b x {\rm Li}_{3}(\frac {{\left (c + d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d - 1}) + 3 \, {\rm Li}_{4}(\frac {{\left (c + d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d - 1})}{b^{4} d}\right )} \]
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\[ \int x^2 \coth ^{-1}(c+d \coth (a+b x)) \, dx=\int { x^{2} \operatorname {arcoth}\left (d \coth \left (b x + a\right ) + c\right ) \,d x } \]
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Timed out. \[ \int x^2 \coth ^{-1}(c+d \coth (a+b x)) \, dx=\int x^2\,\mathrm {acoth}\left (c+d\,\mathrm {coth}\left (a+b\,x\right )\right ) \,d x \]
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