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

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

[Out]

1/3*arctanh(tanh(b*x+a))^3/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} \[ \frac {\tanh ^{-1}(\tanh (a+b x))^3}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^2,x]

[Out]

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

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Mathematica [A]  time = 0.01, size = 16, normalized size = 1.00 \[ \frac {\tanh ^{-1}(\tanh (a+b x))^3}{3 b} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^2,x]

[Out]

ArcTanh[Tanh[a + b*x]]^3/(3*b)

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fricas [A]  time = 0.64, size = 20, normalized size = 1.25 \[ \frac {1}{3} \, b^{2} x^{3} + a b x^{2} + a^{2} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b^2*x^3 + a*b*x^2 + a^2*x

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giac [A]  time = 0.16, size = 20, normalized size = 1.25 \[ \frac {1}{3} \, b^{2} x^{3} + a b x^{2} + a^{2} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^2,x, algorithm="giac")

[Out]

1/3*b^2*x^3 + a*b*x^2 + a^2*x

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maple [A]  time = 0.03, size = 15, normalized size = 0.94 \[ \frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{3}}{3 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.45, size = 33, normalized size = 2.06 \[ \frac {1}{3} \, b^{2} x^{3} - b x^{2} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right ) + x \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.07, size = 33, normalized size = 2.06 \[ \frac {b^2\,x^3}{3}-b\,x^2\,\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )+x\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2 \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^2,x)

[Out]

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

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sympy [A]  time = 0.30, size = 20, normalized size = 1.25 \[ \begin {cases} \frac {\operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{3 b} & \text {for}\: b \neq 0 \\x \operatorname {atanh}^{2}{\left (\tanh {\relax (a )} \right )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**2,x)

[Out]

Piecewise((atanh(tanh(a + b*x))**3/(3*b), Ne(b, 0)), (x*atanh(tanh(a))**2, True))

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