3.171 \(\int \frac{1}{\coth ^{-1}(\tanh (a+b x))^2} \, dx\)

Optimal. Leaf size=14 \[ -\frac{1}{b \coth ^{-1}(\tanh (a+b x))} \]

[Out]

-(1/(b*ArcCoth[Tanh[a + b*x]]))

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Rubi [A]  time = 0.0048201, antiderivative size = 14, 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} \[ -\frac{1}{b \coth ^{-1}(\tanh (a+b x))} \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[Tanh[a + b*x]]^(-2),x]

[Out]

-(1/(b*ArcCoth[Tanh[a + b*x]]))

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

Mathematica [A]  time = 0.0058844, size = 14, normalized size = 1. \[ -\frac{1}{b \coth ^{-1}(\tanh (a+b x))} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[Tanh[a + b*x]]^(-2),x]

[Out]

-(1/(b*ArcCoth[Tanh[a + b*x]]))

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Maple [A]  time = 0.078, size = 15, normalized size = 1.1 \begin{align*} -{\frac{1}{b{\rm arccoth} \left (\tanh \left ( bx+a \right ) \right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C]  time = 1.47959, size = 24, normalized size = 1.71 \begin{align*} \frac{4}{{\left (-2 i \, \pi - 4 \, b x - 4 \, a\right )} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arccoth(tanh(b*x+a))^2,x, algorithm="maxima")

[Out]

4/((-2*I*pi - 4*b*x - 4*a)*b)

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Fricas [B]  time = 1.61425, size = 77, normalized size = 5.5 \begin{align*} -\frac{4 \,{\left (b x + a\right )}}{4 \, b^{3} x^{2} + 8 \, a b^{2} x + \pi ^{2} b + 4 \, a^{2} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arccoth(tanh(b*x+a))^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 16.1339, size = 20, normalized size = 1.43 \begin{align*} \begin{cases} - \frac{1}{b \operatorname{acoth}{\left (\tanh{\left (a + b x \right )} \right )}} & \text{for}\: b \neq 0 \\\frac{x}{\operatorname{acoth}^{2}{\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))**2,x)

[Out]

Piecewise((-1/(b*acoth(tanh(a + b*x))), Ne(b, 0)), (x/acoth(tanh(a))**2, 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 )^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arccoth(tanh(b*x+a))^2,x, algorithm="giac")

[Out]

integrate(arccoth(tanh(b*x + a))^(-2), x)