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]

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

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Rubi [A]  time = 0.00, antiderivative size = 16, normalized size of antiderivative = 1.00, 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 30

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

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]

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*}

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Mathematica [A]  time = 0.01, 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]

-1/2*(b*x^2) + x*ArcTanh[Coth[a + b*x]]

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fricas [A]  time = 0.49, size = 10, normalized size = 0.62 \[ \frac {1}{2} \, b x^{2} + a x \]

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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giac [B]  time = 0.14, size = 69, normalized size = 4.31 \[ -\frac {1}{2} \, b x^{2} + \frac {1}{2} \, x \log \left (-\frac {\frac {e^{\left (2 \, b x + 2 \, a\right )} + 1}{e^{\left (2 \, b x + 2 \, a\right )} - 1} + 1}{\frac {e^{\left (2 \, b x + 2 \, a\right )} + 1}{e^{\left (2 \, b x + 2 \, a\right )} - 1} - 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b*x^2 + 1/2*x*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))

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maple [A]  time = 0.03, size = 15, normalized size = 0.94 \[ \frac {\arctanh \left (\coth \left (b x +a \right )\right )^{2}}{2 b} \]

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 = 0.38, size = 16, normalized size = 1.00 \[ -\frac {1}{2} \, b x^{2} + x \operatorname {artanh}\left (\coth \left (b x + a\right )\right ) \]

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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mupad [B]  time = 0.02, size = 16, normalized size = 1.00 \[ x\,\mathrm {atanh}\left (\mathrm {coth}\left (a+b\,x\right )\right )-\frac {b\,x^2}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(coth(a + b*x)),x)

[Out]

x*atanh(coth(a + b*x)) - (b*x^2)/2

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sympy [A]  time = 2.82, size = 46, normalized size = 2.88 \[ \begin {cases} x \operatorname {atanh}{\left (\coth {\relax (a )} \right )} & \text {for}\: b = 0 \\\left \langle - \frac {\pi }{2}, \frac {\pi }{2}\right \rangle i x & \text {for}\: a = \log {\left (- e^{- b x} \right )} \vee a = \log {\left (e^{- b x} \right )} \\\frac {\operatorname {atanh}^{2}{\left (\frac {1}{\tanh {\left (a + b x \right )}} \right )}}{2 b} & \text {otherwise} \end {cases} \]

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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