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

Optimal. Leaf size=16 \[ \frac{\coth ^{-1}(\tanh (a+b x))^3}{3 b} \]

[Out]

ArcCoth[Tanh[a + b*x]]^3/(3*b)

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Rubi [A]  time = 0.0053532, antiderivative size = 16, 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} \[ \frac{\coth ^{-1}(\tanh (a+b x))^3}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[Tanh[a + b*x]]^2,x]

[Out]

ArcCoth[Tanh[a + b*x]]^3/(3*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))^2 \, dx &=\frac{\operatorname{Subst}\left (\int x^2 \, dx,x,\coth ^{-1}(\tanh (a+b x))\right )}{b}\\ &=\frac{\coth ^{-1}(\tanh (a+b x))^3}{3 b}\\ \end{align*}

Mathematica [A]  time = 0.00554, size = 16, normalized size = 1. \[ \frac{\coth ^{-1}(\tanh (a+b x))^3}{3 b} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[Tanh[a + b*x]]^2,x]

[Out]

ArcCoth[Tanh[a + b*x]]^3/(3*b)

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Maple [A]  time = 0.098, size = 15, normalized size = 0.9 \begin{align*}{\frac{ \left ({\rm arccoth} \left (\tanh \left ( bx+a \right ) \right ) \right ) ^{3}}{3\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*arccoth(tanh(b*x+a))^3/b

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Maxima [B]  time = 1.37255, size = 45, normalized size = 2.81 \begin{align*} \frac{1}{3} \, b^{2} x^{3} - b x^{2} \operatorname{arcoth}\left (\tanh \left (b x + a\right )\right ) + x \operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b^2*x^3 - b*x^2*arccoth(tanh(b*x + a)) + x*arccoth(tanh(b*x + a))^2

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Fricas [A]  time = 1.62762, size = 62, normalized size = 3.88 \begin{align*} \frac{1}{3} \, b^{2} x^{3} + a b x^{2} - \frac{1}{4} \,{\left (\pi ^{2} - 4 \, a^{2}\right )} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b^2*x^3 + a*b*x^2 - 1/4*(pi^2 - 4*a^2)*x

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Sympy [A]  time = 0.414211, size = 20, normalized size = 1.25 \begin{align*} \begin{cases} \frac{\operatorname{acoth}^{3}{\left (\tanh{\left (a + b x \right )} \right )}}{3 b} & \text{for}\: b \neq 0 \\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(acoth(tanh(b*x+a))**2,x)

[Out]

Piecewise((acoth(tanh(a + b*x))**3/(3*b), 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 \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(arccoth(tanh(b*x+a))^2,x, algorithm="giac")

[Out]

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