Optimal. Leaf size=110 \[ \frac{6 b^2 x^{m+3} \coth ^{-1}(\tanh (a+b x))}{m^3+6 m^2+11 m+6}-\frac{3 b x^{m+2} \coth ^{-1}(\tanh (a+b x))^2}{m^2+3 m+2}+\frac{x^{m+1} \coth ^{-1}(\tanh (a+b x))^3}{m+1}-\frac{6 b^3 x^{m+4}}{(m+1) \left (m^3+9 m^2+26 m+24\right )} \]
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Rubi [A] time = 0.0593837, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2168, 30} \[ \frac{6 b^2 x^{m+3} \coth ^{-1}(\tanh (a+b x))}{m^3+6 m^2+11 m+6}-\frac{3 b x^{m+2} \coth ^{-1}(\tanh (a+b x))^2}{m^2+3 m+2}+\frac{x^{m+1} \coth ^{-1}(\tanh (a+b x))^3}{m+1}-\frac{6 b^3 x^{m+4}}{(m+1) \left (m^3+9 m^2+26 m+24\right )} \]
Antiderivative was successfully verified.
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Rule 2168
Rule 30
Rubi steps
\begin{align*} \int x^m \coth ^{-1}(\tanh (a+b x))^3 \, dx &=\frac{x^{1+m} \coth ^{-1}(\tanh (a+b x))^3}{1+m}-\frac{(3 b) \int x^{1+m} \coth ^{-1}(\tanh (a+b x))^2 \, dx}{1+m}\\ &=-\frac{3 b x^{2+m} \coth ^{-1}(\tanh (a+b x))^2}{2+3 m+m^2}+\frac{x^{1+m} \coth ^{-1}(\tanh (a+b x))^3}{1+m}+\frac{\left (6 b^2\right ) \int x^{2+m} \coth ^{-1}(\tanh (a+b x)) \, dx}{2+3 m+m^2}\\ &=\frac{6 b^2 x^{3+m} \coth ^{-1}(\tanh (a+b x))}{6+11 m+6 m^2+m^3}-\frac{3 b x^{2+m} \coth ^{-1}(\tanh (a+b x))^2}{2+3 m+m^2}+\frac{x^{1+m} \coth ^{-1}(\tanh (a+b x))^3}{1+m}-\frac{\left (6 b^3\right ) \int x^{3+m} \, dx}{6+11 m+6 m^2+m^3}\\ &=-\frac{6 b^3 x^{4+m}}{(4+m) \left (6+11 m+6 m^2+m^3\right )}+\frac{6 b^2 x^{3+m} \coth ^{-1}(\tanh (a+b x))}{6+11 m+6 m^2+m^3}-\frac{3 b x^{2+m} \coth ^{-1}(\tanh (a+b x))^2}{2+3 m+m^2}+\frac{x^{1+m} \coth ^{-1}(\tanh (a+b x))^3}{1+m}\\ \end{align*}
Mathematica [A] time = 0.113509, size = 97, normalized size = 0.88 \[ \frac{x^{m+1} \left (6 b^2 (m+4) x^2 \coth ^{-1}(\tanh (a+b x))-3 b \left (m^2+7 m+12\right ) x \coth ^{-1}(\tanh (a+b x))^2+\left (m^3+9 m^2+26 m+24\right ) \coth ^{-1}(\tanh (a+b x))^3-6 b^3 x^3\right )}{(m+1) (m+2) (m+3) (m+4)} \]
Antiderivative was successfully verified.
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Maple [C] time = 6.384, size = 63382, normalized size = 576.2 \begin{align*} \text{output too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.71674, size = 468, normalized size = 4.25 \begin{align*} \frac{{\left (4 \,{\left (b^{3} m^{3} + 6 \, b^{3} m^{2} + 11 \, b^{3} m + 6 \, b^{3}\right )} x^{4} + 12 \,{\left (a b^{2} m^{3} + 7 \, a b^{2} m^{2} + 14 \, a b^{2} m + 8 \, a b^{2}\right )} x^{3} + 3 \,{\left (4 \, a^{2} b m^{3} + 32 \, a^{2} b m^{2} + 76 \, a^{2} b m - \pi ^{2}{\left (b m^{3} + 8 \, b m^{2} + 19 \, b m + 12 \, b\right )} + 48 \, a^{2} b\right )} x^{2} +{\left (4 \, a^{3} m^{3} + 36 \, a^{3} m^{2} + 104 \, a^{3} m - 3 \, \pi ^{2}{\left (a m^{3} + 9 \, a m^{2} + 26 \, a m + 24 \, a\right )} + 96 \, a^{3}\right )} x\right )} x^{m}}{4 \,{\left (m^{4} + 10 \, m^{3} + 35 \, m^{2} + 50 \, m + 24\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{m} \operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )^{3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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