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

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

[Out]

ArcCoth[Tanh[a + b*x]]^2/(2*b)

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Rubi [A]  time = 0.003068, 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} \[ \frac{\coth ^{-1}(\tanh (a+b x))^2}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[Tanh[a + b*x]],x]

[Out]

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

Mathematica [A]  time = 0.0083259, size = 18, normalized size = 1.12 \[ x \coth ^{-1}(\tanh (a+b x))-\frac{b x^2}{2} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[Tanh[a + b*x]],x]

[Out]

-(b*x^2)/2 + x*ArcCoth[Tanh[a + b*x]]

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Maple [B]  time = 0.06, size = 32, normalized size = 2. \begin{align*}{\frac{1}{b} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ){\rm arccoth} \left (\tanh \left ( bx+a \right ) \right )-{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.15234, size = 22, normalized size = 1.38 \begin{align*} -\frac{1}{2} \, b x^{2} + x \operatorname{arcoth}\left (\tanh \left (b x + a\right )\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b*x^2 + x*arccoth(tanh(b*x + a))

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Fricas [A]  time = 1.57382, 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(arccoth(tanh(b*x+a)),x, algorithm="fricas")

[Out]

1/2*b*x^2 + a*x

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Sympy [A]  time = 0.194614, size = 19, normalized size = 1.19 \begin{align*} \begin{cases} \frac{\operatorname{acoth}^{2}{\left (\tanh{\left (a + b x \right )} \right )}}{2 b} & \text{for}\: b \neq 0 \\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(acoth(tanh(b*x+a)),x)

[Out]

Piecewise((acoth(tanh(a + b*x))**2/(2*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 \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(arccoth(tanh(b*x+a)),x, algorithm="giac")

[Out]

integrate(arccoth(tanh(b*x + a)), x)