3.58 \(\int \tanh ^{-1}(\tanh (a+b x))^3 \, dx\)

Optimal. Leaf size=16 \[ \frac{\tanh ^{-1}(\tanh (a+b x))^4}{4 b} \]

[Out]

ArcTanh[Tanh[a + b*x]]^4/(4*b)

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Rubi [A]  time = 0.0044888, 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} \[ \frac{\tanh ^{-1}(\tanh (a+b x))^4}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^3,x]

[Out]

ArcTanh[Tanh[a + b*x]]^4/(4*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}(\tanh (a+b x))^3 \, dx &=\frac{\operatorname{Subst}\left (\int x^3 \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{b}\\ &=\frac{\tanh ^{-1}(\tanh (a+b x))^4}{4 b}\\ \end{align*}

Mathematica [A]  time = 0.0057057, size = 16, normalized size = 1. \[ \frac{\tanh ^{-1}(\tanh (a+b x))^4}{4 b} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^3,x]

[Out]

ArcTanh[Tanh[a + b*x]]^4/(4*b)

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Maple [A]  time = 0.028, size = 15, normalized size = 0.9 \begin{align*}{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{4}}{4\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^3,x)

[Out]

1/4*arctanh(tanh(b*x+a))^4/b

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Maxima [B]  time = 1.53072, size = 69, normalized size = 4.31 \begin{align*} -\frac{3}{2} \, b x^{2} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{2} + x \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{3} - \frac{1}{4} \,{\left (b^{2} x^{4} - 4 \, b x^{3} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^3,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.41287, size = 66, normalized size = 4.12 \begin{align*} \frac{1}{4} \, b^{3} x^{4} + a b^{2} x^{3} + \frac{3}{2} \, a^{2} b x^{2} + a^{3} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^3,x, algorithm="fricas")

[Out]

1/4*b^3*x^4 + a*b^2*x^3 + 3/2*a^2*b*x^2 + a^3*x

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Sympy [A]  time = 3.0068, size = 20, normalized size = 1.25 \begin{align*} \begin{cases} \frac{\operatorname{atanh}^{4}{\left (\tanh{\left (a + b x \right )} \right )}}{4 b} & \text{for}\: b \neq 0 \\x \operatorname{atanh}^{3}{\left (\tanh{\left (a \right )} \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**3,x)

[Out]

Piecewise((atanh(tanh(a + b*x))**4/(4*b), Ne(b, 0)), (x*atanh(tanh(a))**3, True))

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Giac [B]  time = 1.11, size = 42, normalized size = 2.62 \begin{align*} \frac{1}{4} \, b^{3} x^{4} + a b^{2} x^{3} + \frac{3}{2} \, a^{2} b x^{2} + a^{3} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^3,x, algorithm="giac")

[Out]

1/4*b^3*x^4 + a*b^2*x^3 + 3/2*a^2*b*x^2 + a^3*x