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

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

[Out]

ArcTanh[Coth[a + b*x]]^2/(2*b)

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Rubi [A]  time = 0.0031824, antiderivative size = 16, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2157, 30} \[ \frac{\tanh ^{-1}(\coth (a+b x))^2}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Coth[a + b*x]],x]

[Out]

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

Mathematica [A]  time = 0.0065838, size = 18, normalized size = 1.12 \[ x \tanh ^{-1}(\coth (a+b x))-\frac{b x^2}{2} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Coth[a + b*x]],x]

[Out]

-(b*x^2)/2 + x*ArcTanh[Coth[a + b*x]]

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Maple [A]  time = 0.026, size = 15, normalized size = 0.9 \begin{align*}{\frac{ \left ({\it Artanh} \left ({\rm coth} \left (bx+a\right ) \right ) \right ) ^{2}}{2\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(coth(b*x+a)),x)

[Out]

1/2*arctanh(coth(b*x+a))^2/b

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Maxima [A]  time = 1.12481, size = 22, normalized size = 1.38 \begin{align*} -\frac{1}{2} \, b x^{2} + x \operatorname{artanh}\left (\coth \left (b x + a\right )\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(coth(b*x+a)),x, algorithm="maxima")

[Out]

-1/2*b*x^2 + x*arctanh(coth(b*x + a))

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Fricas [A]  time = 2.10572, size = 23, normalized size = 1.44 \begin{align*} \frac{1}{2} \, b x^{2} + a x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(coth(b*x+a)),x, algorithm="fricas")

[Out]

1/2*b*x^2 + a*x

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Sympy [A]  time = 9.43627, size = 46, normalized size = 2.88 \begin{align*} \begin{cases} x \operatorname{atanh}{\left (\coth{\left (a \right )} \right )} & \text{for}\: b = 0 \\\left \langle - \frac{\pi }{2}, \frac{\pi }{2}\right \rangle i x & \text{for}\: a = \log{\left (- e^{- b x} \right )} \vee a = \log{\left (e^{- b x} \right )} \\\frac{\operatorname{atanh}^{2}{\left (\frac{1}{\tanh{\left (a + b x \right )}} \right )}}{2 b} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(coth(b*x+a)),x)

[Out]

Piecewise((x*atanh(coth(a)), Eq(b, 0)), (AccumBounds(-pi/2, pi/2)*I*x, Eq(a, log(exp(-b*x))) | Eq(a, log(-exp(
-b*x)))), (atanh(1/tanh(a + b*x))**2/(2*b), True))

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Giac [B]  time = 1.1999, size = 89, normalized size = 5.56 \begin{align*} \frac{\log \left (-\frac{\frac{e^{\left (2 \, b x + 2 \, a\right )} + 1}{e^{\left (2 \, b x + 2 \, a\right )} - 1} + 1}{\frac{e^{\left (2 \, b x + 2 \, a\right )} + 1}{e^{\left (2 \, b x + 2 \, a\right )} - 1} - 1}\right )^{2}}{8 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(coth(b*x+a)),x, algorithm="giac")

[Out]

1/8*log(-((e^(2*b*x + 2*a) + 1)/(e^(2*b*x + 2*a) - 1) + 1)/((e^(2*b*x + 2*a) + 1)/(e^(2*b*x + 2*a) - 1) - 1))^
2/b