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]

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

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Mathematica [A]  time = 0.01, size = 16, normalized size = 1.00 \[ \frac {\tanh ^{-1}(\tanh (a+b x))^4}{4 b} \]

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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fricas [B]  time = 0.73, size = 31, normalized size = 1.94 \[ \frac {1}{4} \, b^{3} x^{4} + a b^{2} x^{3} + \frac {3}{2} \, a^{2} b x^{2} + a^{3} x \]

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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giac [B]  time = 0.17, size = 31, normalized size = 1.94 \[ \frac {1}{2} \, {\left (b x^{2} + 2 \, a x\right )} a^{2} + \frac {1}{4} \, {\left (b x^{2} + 2 \, a x\right )}^{2} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 15, normalized size = 0.94 \[ \frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{4}}{4 b} \]

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 = 0.52, size = 51, normalized size = 3.19 \[ -\frac {3}{2} \, b x^{2} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} + x \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3} - \frac {1}{4} \, {\left (b^{2} x^{4} - 4 \, b x^{3} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )\right )} b \]

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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mupad [B]  time = 0.10, size = 47, normalized size = 2.94 \[ \frac {x\,\left (2\,\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )-b\,x\right )\,\left (b^2\,x^2-2\,b\,x\,\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )+2\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^3,x)

[Out]

(x*(2*atanh(tanh(a + b*x)) - b*x)*(2*atanh(tanh(a + b*x))^2 + b^2*x^2 - 2*b*x*atanh(tanh(a + b*x))))/4

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sympy [A]  time = 0.61, size = 20, normalized size = 1.25 \[ \begin {cases} \frac {\operatorname {atanh}^{4}{\left (\tanh {\left (a + b x \right )} \right )}}{4 b} & \text {for}\: b \neq 0 \\x \operatorname {atanh}^{3}{\left (\tanh {\relax (a )} \right )} & \text {otherwise} \end {cases} \]

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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