3.151 \(\int \coth ^{-1}(\tanh (a+b x))^3 \, dx\)

Optimal. Leaf size=16 \[ \frac{\coth ^{-1}(\tanh (a+b x))^4}{4 b} \]

[Out]

ArcCoth[Tanh[a + b*x]]^4/(4*b)

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Rubi [A]  time = 0.0051923, antiderivative size = 16, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 9, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2157, 30} \[ \frac{\coth ^{-1}(\tanh (a+b x))^4}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[Tanh[a + b*x]]^3,x]

[Out]

ArcCoth[Tanh[a + b*x]]^4/(4*b)

Rule 2157

Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Dist[1/c, Subst[Int[x^m, x], x, u], x]] /; FreeQ[m,
x] && PiecewiseLinearQ[u, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \coth ^{-1}(\tanh (a+b x))^3 \, dx &=\frac{\operatorname{Subst}\left (\int x^3 \, dx,x,\coth ^{-1}(\tanh (a+b x))\right )}{b}\\ &=\frac{\coth ^{-1}(\tanh (a+b x))^4}{4 b}\\ \end{align*}

Mathematica [A]  time = 0.0056073, size = 16, normalized size = 1. \[ \frac{\coth ^{-1}(\tanh (a+b x))^4}{4 b} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[Tanh[a + b*x]]^3,x]

[Out]

ArcCoth[Tanh[a + b*x]]^4/(4*b)

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Maple [A]  time = 0.106, size = 15, normalized size = 0.9 \begin{align*}{\frac{ \left ({\rm arccoth} \left (\tanh \left ( bx+a \right ) \right ) \right ) ^{4}}{4\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccoth(tanh(b*x+a))^3,x)

[Out]

1/4*arccoth(tanh(b*x+a))^4/b

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Maxima [B]  time = 1.55637, size = 69, normalized size = 4.31 \begin{align*} -\frac{3}{2} \, b x^{2} \operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )^{2} + x \operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )^{3} - \frac{1}{4} \,{\left (b^{2} x^{4} - 4 \, b x^{3} \operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(tanh(b*x+a))^3,x, algorithm="maxima")

[Out]

-3/2*b*x^2*arccoth(tanh(b*x + a))^2 + x*arccoth(tanh(b*x + a))^3 - 1/4*(b^2*x^4 - 4*b*x^3*arccoth(tanh(b*x + a
)))*b

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Fricas [B]  time = 1.59564, size = 109, normalized size = 6.81 \begin{align*} \frac{1}{4} \, b^{3} x^{4} + a b^{2} x^{3} - \frac{3}{8} \,{\left (\pi ^{2} b - 4 \, a^{2} b\right )} x^{2} - \frac{1}{4} \,{\left (3 \, \pi ^{2} a - 4 \, a^{3}\right )} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(tanh(b*x+a))^3,x, algorithm="fricas")

[Out]

1/4*b^3*x^4 + a*b^2*x^3 - 3/8*(pi^2*b - 4*a^2*b)*x^2 - 1/4*(3*pi^2*a - 4*a^3)*x

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Sympy [A]  time = 0.803763, size = 20, normalized size = 1.25 \begin{align*} \begin{cases} \frac{\operatorname{acoth}^{4}{\left (\tanh{\left (a + b x \right )} \right )}}{4 b} & \text{for}\: b \neq 0 \\x \operatorname{acoth}^{3}{\left (\tanh{\left (a \right )} \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acoth(tanh(b*x+a))**3,x)

[Out]

Piecewise((acoth(tanh(a + b*x))**4/(4*b), Ne(b, 0)), (x*acoth(tanh(a))**3, True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \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.

[In]

integrate(arccoth(tanh(b*x+a))^3,x, algorithm="giac")

[Out]

integrate(arccoth(tanh(b*x + a))^3, x)