Integrand size = 19, antiderivative size = 168 \[ \int x^3 \coth ^{-1}(1-d-d \tanh (a+b x)) \, dx=\frac {b x^5}{20}+\frac {1}{4} x^4 \coth ^{-1}(1-d-d \tanh (a+b x))-\frac {1}{8} x^4 \log \left (1+(1-d) e^{2 a+2 b x}\right )-\frac {x^3 \operatorname {PolyLog}\left (2,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{4 b}+\frac {3 x^2 \operatorname {PolyLog}\left (3,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{8 b^2}-\frac {3 x \operatorname {PolyLog}\left (4,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{8 b^3}+\frac {3 \operatorname {PolyLog}\left (5,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{16 b^4} \]
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Time = 0.26 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {6375, 2215, 2221, 2611, 6744, 2320, 6724} \[ \int x^3 \coth ^{-1}(1-d-d \tanh (a+b x)) \, dx=\frac {3 \operatorname {PolyLog}\left (5,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{16 b^4}-\frac {3 x \operatorname {PolyLog}\left (4,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{8 b^3}+\frac {3 x^2 \operatorname {PolyLog}\left (3,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{8 b^2}-\frac {x^3 \operatorname {PolyLog}\left (2,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{4 b}-\frac {1}{8} x^4 \log \left ((1-d) e^{2 a+2 b x}+1\right )+\frac {1}{4} x^4 \coth ^{-1}(d (-\tanh (a+b x))-d+1)+\frac {b x^5}{20} \]
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Rule 2215
Rule 2221
Rule 2320
Rule 2611
Rule 6375
Rule 6724
Rule 6744
Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} x^4 \coth ^{-1}(1-d-d \tanh (a+b x))+\frac {1}{4} b \int \frac {x^4}{1+(1-d) e^{2 a+2 b x}} \, dx \\ & = \frac {b x^5}{20}+\frac {1}{4} x^4 \coth ^{-1}(1-d-d \tanh (a+b x))-\frac {1}{4} (b (1-d)) \int \frac {e^{2 a+2 b x} x^4}{1+(1-d) e^{2 a+2 b x}} \, dx \\ & = \frac {b x^5}{20}+\frac {1}{4} x^4 \coth ^{-1}(1-d-d \tanh (a+b x))-\frac {1}{8} x^4 \log \left (1+(1-d) e^{2 a+2 b x}\right )+\frac {1}{2} \int x^3 \log \left (1+(1-d) e^{2 a+2 b x}\right ) \, dx \\ & = \frac {b x^5}{20}+\frac {1}{4} x^4 \coth ^{-1}(1-d-d \tanh (a+b x))-\frac {1}{8} x^4 \log \left (1+(1-d) e^{2 a+2 b x}\right )-\frac {x^3 \operatorname {PolyLog}\left (2,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{4 b}+\frac {3 \int x^2 \operatorname {PolyLog}\left (2,(-1+d) e^{2 a+2 b x}\right ) \, dx}{4 b} \\ & = \frac {b x^5}{20}+\frac {1}{4} x^4 \coth ^{-1}(1-d-d \tanh (a+b x))-\frac {1}{8} x^4 \log \left (1+(1-d) e^{2 a+2 b x}\right )-\frac {x^3 \operatorname {PolyLog}\left (2,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{4 b}+\frac {3 x^2 \operatorname {PolyLog}\left (3,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{8 b^2}-\frac {3 \int x \operatorname {PolyLog}\left (3,(-1+d) e^{2 a+2 b x}\right ) \, dx}{4 b^2} \\ & = \frac {b x^5}{20}+\frac {1}{4} x^4 \coth ^{-1}(1-d-d \tanh (a+b x))-\frac {1}{8} x^4 \log \left (1+(1-d) e^{2 a+2 b x}\right )-\frac {x^3 \operatorname {PolyLog}\left (2,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{4 b}+\frac {3 x^2 \operatorname {PolyLog}\left (3,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{8 b^2}-\frac {3 x \operatorname {PolyLog}\left (4,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{8 b^3}+\frac {3 \int \operatorname {PolyLog}\left (4,(-1+d) e^{2 a+2 b x}\right ) \, dx}{8 b^3} \\ & = \frac {b x^5}{20}+\frac {1}{4} x^4 \coth ^{-1}(1-d-d \tanh (a+b x))-\frac {1}{8} x^4 \log \left (1+(1-d) e^{2 a+2 b x}\right )-\frac {x^3 \operatorname {PolyLog}\left (2,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{4 b}+\frac {3 x^2 \operatorname {PolyLog}\left (3,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{8 b^2}-\frac {3 x \operatorname {PolyLog}\left (4,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{8 b^3}+\frac {3 \text {Subst}\left (\int \frac {\operatorname {PolyLog}(4,(-1+d) x)}{x} \, dx,x,e^{2 a+2 b x}\right )}{16 b^4} \\ & = \frac {b x^5}{20}+\frac {1}{4} x^4 \coth ^{-1}(1-d-d \tanh (a+b x))-\frac {1}{8} x^4 \log \left (1+(1-d) e^{2 a+2 b x}\right )-\frac {x^3 \operatorname {PolyLog}\left (2,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{4 b}+\frac {3 x^2 \operatorname {PolyLog}\left (3,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{8 b^2}-\frac {3 x \operatorname {PolyLog}\left (4,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{8 b^3}+\frac {3 \operatorname {PolyLog}\left (5,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{16 b^4} \\ \end{align*}
Time = 0.13 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.88 \[ \int x^3 \coth ^{-1}(1-d-d \tanh (a+b x)) \, dx=\frac {4 b^4 x^4 \coth ^{-1}(1-d-d \tanh (a+b x))-2 b^4 x^4 \log \left (1-\frac {e^{-2 (a+b x)}}{-1+d}\right )+4 b^3 x^3 \operatorname {PolyLog}\left (2,\frac {e^{-2 (a+b x)}}{-1+d}\right )+6 b^2 x^2 \operatorname {PolyLog}\left (3,\frac {e^{-2 (a+b x)}}{-1+d}\right )+6 b x \operatorname {PolyLog}\left (4,\frac {e^{-2 (a+b x)}}{-1+d}\right )+3 \operatorname {PolyLog}\left (5,\frac {e^{-2 (a+b x)}}{-1+d}\right )}{16 b^4} \]
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 2.02 (sec) , antiderivative size = 1754, normalized size of antiderivative = 10.44
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Leaf count of result is larger than twice the leaf count of optimal. 423 vs. \(2 (135) = 270\).
Time = 0.27 (sec) , antiderivative size = 423, normalized size of antiderivative = 2.52 \[ \int x^3 \coth ^{-1}(1-d-d \tanh (a+b x)) \, dx=\frac {2 \, b^{5} x^{5} - 5 \, b^{4} x^{4} \log \left (\frac {d \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}{{\left (d - 2\right )} \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}\right ) - 20 \, b^{3} x^{3} {\rm Li}_2\left (\sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 20 \, b^{3} x^{3} {\rm Li}_2\left (-\sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 5 \, a^{4} \log \left (2 \, {\left (d - 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (d - 1\right )} \sinh \left (b x + a\right ) + 2 \, \sqrt {d - 1}\right ) - 5 \, a^{4} \log \left (2 \, {\left (d - 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (d - 1\right )} \sinh \left (b x + a\right ) - 2 \, \sqrt {d - 1}\right ) + 60 \, b^{2} x^{2} {\rm polylog}\left (3, \sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + 60 \, b^{2} x^{2} {\rm polylog}\left (3, -\sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 120 \, b x {\rm polylog}\left (4, \sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 120 \, b x {\rm polylog}\left (4, -\sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 5 \, {\left (b^{4} x^{4} - a^{4}\right )} \log \left (\sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) - 5 \, {\left (b^{4} x^{4} - a^{4}\right )} \log \left (-\sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) + 120 \, {\rm polylog}\left (5, \sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + 120 \, {\rm polylog}\left (5, -\sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right )}{40 \, b^{4}} \]
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\[ \int x^3 \coth ^{-1}(1-d-d \tanh (a+b x)) \, dx=- \int x^{3} \operatorname {acoth}{\left (d \tanh {\left (a + b x \right )} + d - 1 \right )}\, dx \]
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Time = 0.79 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.87 \[ \int x^3 \coth ^{-1}(1-d-d \tanh (a+b x)) \, dx=-\frac {1}{4} \, x^{4} \operatorname {arcoth}\left (d \tanh \left (b x + a\right ) + d - 1\right ) + \frac {1}{40} \, {\left (\frac {2 \, x^{5}}{d} - \frac {5 \, {\left (2 \, b^{4} x^{4} \log \left (-{\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 4 \, b^{3} x^{3} {\rm Li}_2\left ({\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}\right ) - 6 \, b^{2} x^{2} {\rm Li}_{3}({\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}) + 6 \, b x {\rm Li}_{4}({\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}) - 3 \, {\rm Li}_{5}({\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )})\right )}}{b^{5} d}\right )} b d \]
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\[ \int x^3 \coth ^{-1}(1-d-d \tanh (a+b x)) \, dx=\int { x^{3} \operatorname {arcoth}\left (-d \tanh \left (b x + a\right ) - d + 1\right ) \,d x } \]
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Timed out. \[ \int x^3 \coth ^{-1}(1-d-d \tanh (a+b x)) \, dx=\int -x^3\,\mathrm {acoth}\left (d+d\,\mathrm {tanh}\left (a+b\,x\right )-1\right ) \,d x \]
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