Integrand size = 16, antiderivative size = 155 \[ \int x^3 \coth ^{-1}(1+d+d \tanh (a+b x)) \, dx=\frac {b x^5}{20}+\frac {1}{4} x^4 \coth ^{-1}(1+d+d \tanh (a+b x))-\frac {1}{8} x^4 \log \left (1+(1+d) e^{2 a+2 b x}\right )-\frac {x^3 \operatorname {PolyLog}\left (2,-\left ((1+d) e^{2 a+2 b x}\right )\right )}{4 b}+\frac {3 x^2 \operatorname {PolyLog}\left (3,-\left ((1+d) e^{2 a+2 b x}\right )\right )}{8 b^2}-\frac {3 x \operatorname {PolyLog}\left (4,-\left ((1+d) e^{2 a+2 b x}\right )\right )}{8 b^3}+\frac {3 \operatorname {PolyLog}\left (5,-\left ((1+d) e^{2 a+2 b x}\right )\right )}{16 b^4} \]
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Time = 0.22 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {6375, 2215, 2221, 2611, 6744, 2320, 6724} \[ \int x^3 \coth ^{-1}(1+d+d \tanh (a+b x)) \, dx=\frac {3 \operatorname {PolyLog}\left (5,-\left ((d+1) e^{2 a+2 b x}\right )\right )}{16 b^4}-\frac {3 x \operatorname {PolyLog}\left (4,-\left ((d+1) e^{2 a+2 b x}\right )\right )}{8 b^3}+\frac {3 x^2 \operatorname {PolyLog}\left (3,-\left ((d+1) e^{2 a+2 b x}\right )\right )}{8 b^2}-\frac {x^3 \operatorname {PolyLog}\left (2,-\left ((d+1) e^{2 a+2 b x}\right )\right )}{4 b}-\frac {1}{8} x^4 \log \left ((d+1) e^{2 a+2 b x}+1\right )+\frac {1}{4} x^4 \coth ^{-1}(d \tanh (a+b x)+d+1)+\frac {b x^5}{20} \]
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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 = 0.12 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.95 \[ \int x^3 \coth ^{-1}(1+d+d \tanh (a+b x)) \, dx=\frac {4 b^4 x^4 \coth ^{-1}(1+d+d \tanh (a+b x))-2 b^4 x^4 \log \left (1+\frac {e^{-2 (a+b x)}}{1+d}\right )+4 b^3 x^3 \operatorname {PolyLog}\left (2,-\frac {e^{-2 (a+b x)}}{1+d}\right )+6 b^2 x^2 \operatorname {PolyLog}\left (3,-\frac {e^{-2 (a+b x)}}{1+d}\right )+6 b x \operatorname {PolyLog}\left (4,-\frac {e^{-2 (a+b x)}}{1+d}\right )+3 \operatorname {PolyLog}\left (5,-\frac {e^{-2 (a+b x)}}{1+d}\right )}{16 b^4} \]
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 2.09 (sec) , antiderivative size = 1684, normalized size of antiderivative = 10.86
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Leaf count of result is larger than twice the leaf count of optimal. 450 vs. \(2 (135) = 270\).
Time = 0.27 (sec) , antiderivative size = 450, normalized size of antiderivative = 2.90 \[ \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 {{\left (d + 2\right )} \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}{d \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}\right ) - 20 \, b^{3} x^{3} {\rm Li}_2\left (\frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 20 \, b^{3} x^{3} {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {-4 \, d - 4} {\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 ) + \sqrt {-4 \, d - 4}\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 ) - \sqrt {-4 \, d - 4}\right ) + 60 \, b^{2} x^{2} {\rm polylog}\left (3, \frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + 60 \, b^{2} x^{2} {\rm polylog}\left (3, -\frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 120 \, b x {\rm polylog}\left (4, \frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 120 \, b x {\rm polylog}\left (4, -\frac {1}{2} \, \sqrt {-4 \, d - 4} {\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 (\frac {1}{2} \, \sqrt {-4 \, d - 4} {\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 (-\frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) + 120 \, {\rm polylog}\left (5, \frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + 120 \, {\rm polylog}\left (5, -\frac {1}{2} \, \sqrt {-4 \, d - 4} {\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 = 149, normalized size of antiderivative = 0.96 \[ \int x^3 \coth ^{-1}(1+d+d \tanh (a+b x)) \, dx=\frac {1}{4} \, x^{4} \operatorname {arcoth}\left (d \tanh \left (b x + a\right ) + d + 1\right ) + \frac {1}{40} \, {\left (\frac {2 \, x^{5}}{d} - \frac {5 \, {\left (2 \, b^{4} x^{4} \log \left ({\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 4 \, b^{3} x^{3} {\rm Li}_2\left (-{\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}\right ) - 6 \, b^{2} x^{2} {\rm Li}_{3}(-{\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}) + 6 \, b x {\rm Li}_{4}(-{\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}) - 3 \, {\rm Li}_{5}(-{\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )})\right )}}{b^{5} d}\right )} b d \]
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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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