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

Optimal. Leaf size=16 \[ -\frac{1}{2 b \coth ^{-1}(\tanh (a+b x))^2} \]

[Out]

-1/(2*b*ArcCoth[Tanh[a + b*x]]^2)

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Rubi [A]  time = 0.0053586, 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} \[ -\frac{1}{2 b \coth ^{-1}(\tanh (a+b x))^2} \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[Tanh[a + b*x]]^(-3),x]

[Out]

-1/(2*b*ArcCoth[Tanh[a + b*x]]^2)

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 \frac{1}{\coth ^{-1}(\tanh (a+b x))^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^3} \, dx,x,\coth ^{-1}(\tanh (a+b x))\right )}{b}\\ &=-\frac{1}{2 b \coth ^{-1}(\tanh (a+b x))^2}\\ \end{align*}

Mathematica [A]  time = 0.0055339, size = 16, normalized size = 1. \[ -\frac{1}{2 b \coth ^{-1}(\tanh (a+b x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[Tanh[a + b*x]]^(-3),x]

[Out]

-1/(2*b*ArcCoth[Tanh[a + b*x]]^2)

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Maple [A]  time = 0.074, size = 15, normalized size = 0.9 \begin{align*} -{\frac{1}{2\,b \left ({\rm arccoth} \left (\tanh \left ( bx+a \right ) \right ) \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/arccoth(tanh(b*x+a))^3,x)

[Out]

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

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Maxima [C]  time = 1.47068, size = 41, normalized size = 2.56 \begin{align*} \frac{8}{{\left (4 \, \pi ^{2} - 16 i \, \pi{\left (b x + a\right )} - 16 \,{\left (b x + a\right )}^{2}\right )} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arccoth(tanh(b*x+a))^3,x, algorithm="maxima")

[Out]

8/((4*pi^2 - 16*I*pi*(b*x + a) - 16*(b*x + a)^2)*b)

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Fricas [B]  time = 1.60991, size = 227, normalized size = 14.19 \begin{align*} -\frac{2 \,{\left (4 \, b^{2} x^{2} + 8 \, a b x - \pi ^{2} + 4 \, a^{2}\right )}}{16 \, b^{5} x^{4} + 64 \, a b^{4} x^{3} + \pi ^{4} b + 8 \, \pi ^{2} a^{2} b + 16 \, a^{4} b + 8 \,{\left (\pi ^{2} b^{3} + 12 \, a^{2} b^{3}\right )} x^{2} + 16 \,{\left (\pi ^{2} a b^{2} + 4 \, a^{3} b^{2}\right )} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arccoth(tanh(b*x+a))^3,x, algorithm="fricas")

[Out]

-2*(4*b^2*x^2 + 8*a*b*x - pi^2 + 4*a^2)/(16*b^5*x^4 + 64*a*b^4*x^3 + pi^4*b + 8*pi^2*a^2*b + 16*a^4*b + 8*(pi^
2*b^3 + 12*a^2*b^3)*x^2 + 16*(pi^2*a*b^2 + 4*a^3*b^2)*x)

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Sympy [A]  time = 16.4599, size = 24, normalized size = 1.5 \begin{align*} \begin{cases} - \frac{1}{2 b \operatorname{acoth}^{2}{\left (\tanh{\left (a + b x \right )} \right )}} & \text{for}\: b \neq 0 \\\frac{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(1/acoth(tanh(b*x+a))**3,x)

[Out]

Piecewise((-1/(2*b*acoth(tanh(a + b*x))**2), 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 \frac{1}{\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(1/arccoth(tanh(b*x+a))^3,x, algorithm="giac")

[Out]

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