3.124 \(\int \tanh ^{-1}(\tanh (a+b x))^{3/2} \, dx\)

Optimal. Leaf size=18 \[ \frac{2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b} \]

[Out]

(2*ArcTanh[Tanh[a + b*x]]^(5/2))/(5*b)

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Rubi [A]  time = 0.0047478, antiderivative size = 18, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {2157, 30} \[ \frac{2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^(3/2),x]

[Out]

(2*ArcTanh[Tanh[a + b*x]]^(5/2))/(5*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))^{3/2} \, dx &=\frac{\operatorname{Subst}\left (\int x^{3/2} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{b}\\ &=\frac{2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b}\\ \end{align*}

Mathematica [A]  time = 0.0061363, size = 18, normalized size = 1. \[ \frac{2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^(3/2),x]

[Out]

(2*ArcTanh[Tanh[a + b*x]]^(5/2))/(5*b)

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Maple [A]  time = 0.027, size = 15, normalized size = 0.8 \begin{align*}{\frac{2}{5\,b} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^(3/2),x)

[Out]

2/5*arctanh(tanh(b*x+a))^(5/2)/b

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Maxima [A]  time = 1.69263, size = 16, normalized size = 0.89 \begin{align*} \frac{2 \,{\left (b x + a\right )}^{\frac{5}{2}}}{5 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

2/5*(b*x + a)^(5/2)/b

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Fricas [A]  time = 2.02259, size = 63, normalized size = 3.5 \begin{align*} \frac{2 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt{b x + a}}{5 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

2/5*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(b*x + a)/b

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Sympy [A]  time = 19.8266, size = 26, normalized size = 1.44 \begin{align*} \begin{cases} \frac{2 \operatorname{atanh}^{\frac{5}{2}}{\left (\tanh{\left (a + b x \right )} \right )}}{5 b} & \text{for}\: b \neq 0 \\x \operatorname{atanh}^{\frac{3}{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(atanh(tanh(b*x+a))**(3/2),x)

[Out]

Piecewise((2*atanh(tanh(a + b*x))**(5/2)/(5*b), Ne(b, 0)), (x*atanh(tanh(a))**(3/2), True))

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Giac [B]  time = 1.13809, size = 62, normalized size = 3.44 \begin{align*} \frac{\sqrt{2}{\left (5 \, \sqrt{2}{\left (b x + a\right )}^{\frac{3}{2}} a + \sqrt{2}{\left (3 \,{\left (b x + a\right )}^{\frac{5}{2}} - 5 \,{\left (b x + a\right )}^{\frac{3}{2}} a\right )}\right )}}{15 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(3/2),x, algorithm="giac")

[Out]

1/15*sqrt(2)*(5*sqrt(2)*(b*x + a)^(3/2)*a + sqrt(2)*(3*(b*x + a)^(5/2) - 5*(b*x + a)^(3/2)*a))/b