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

Optimal. Leaf size=12 \[ \frac{\log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b} \]

[Out]

Log[ArcCoth[Tanh[a + b*x]]]/b

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Rubi [A]  time = 0.0040205, antiderivative size = 12, 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, 29} \[ \frac{\log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[Tanh[a + b*x]]^(-1),x]

[Out]

Log[ArcCoth[Tanh[a + b*x]]]/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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rubi steps

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

Mathematica [A]  time = 0.04961, size = 12, normalized size = 1. \[ \frac{\log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[Tanh[a + b*x]]^(-1),x]

[Out]

Log[ArcCoth[Tanh[a + b*x]]]/b

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Maple [A]  time = 0.076, size = 13, normalized size = 1.1 \begin{align*}{\frac{\ln \left ({\rm arccoth} \left (\tanh \left ( bx+a \right ) \right ) \right ) }{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(arccoth(tanh(b*x+a)))/b

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Maxima [C]  time = 1.45703, size = 22, normalized size = 1.83 \begin{align*} \frac{\log \left (-\frac{1}{2} i \, \pi - b x - a\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(-1/2*I*pi - b*x - a)/b

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Fricas [B]  time = 1.63011, size = 63, normalized size = 5.25 \begin{align*} \frac{\log \left (4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 8.22448, size = 17, normalized size = 1.42 \begin{align*} \begin{cases} \frac{\log{\left (\operatorname{acoth}{\left (\tanh{\left (a + b x \right )} \right )} \right )}}{b} & \text{for}\: b \neq 0 \\\frac{x}{\operatorname{acoth}{\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)),x)

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/arccoth(tanh(b*x + a)), x)