3.133 \(\int \tanh ^{-1}(\tanh (a+b x))^{5/2} \, dx\)

Optimal. Leaf size=18 \[ \frac{2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b} \]

[Out]

(2*ArcTanh[Tanh[a + b*x]]^(7/2))/(7*b)

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Rubi [A]  time = 0.0046375, 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} \[ \frac{2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^(5/2),x]

[Out]

(2*ArcTanh[Tanh[a + b*x]]^(7/2))/(7*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))^{5/2} \, dx &=\frac{\operatorname{Subst}\left (\int x^{5/2} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{b}\\ &=\frac{2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b}\\ \end{align*}

Mathematica [A]  time = 0.0065517, size = 18, normalized size = 1. \[ \frac{2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*ArcTanh[Tanh[a + b*x]]^(7/2))/(7*b)

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Maple [A]  time = 0.029, size = 15, normalized size = 0.8 \begin{align*}{\frac{2}{7\,b} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*arctanh(tanh(b*x+a))^(7/2)/b

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Maxima [A]  time = 1.71599, size = 16, normalized size = 0.89 \begin{align*} \frac{2 \,{\left (b x + a\right )}^{\frac{7}{2}}}{7 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

2/7*(b*x + a)^(7/2)/b

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Fricas [B]  time = 1.97339, size = 85, normalized size = 4.72 \begin{align*} \frac{2 \,{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right )} \sqrt{b x + a}}{7 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

2/7*(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)*sqrt(b*x + a)/b

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 1.12041, size = 116, normalized size = 6.44 \begin{align*} \frac{\sqrt{2}{\left (35 \, \sqrt{2}{\left (b x + a\right )}^{\frac{3}{2}} a^{2} + 14 \, \sqrt{2}{\left (3 \,{\left (b x + a\right )}^{\frac{5}{2}} - 5 \,{\left (b x + a\right )}^{\frac{3}{2}} a\right )} a + \sqrt{2}{\left (15 \,{\left (b x + a\right )}^{\frac{7}{2}} - 42 \,{\left (b x + a\right )}^{\frac{5}{2}} a + 35 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{2}\right )}\right )}}{105 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2),x, algorithm="giac")

[Out]

1/105*sqrt(2)*(35*sqrt(2)*(b*x + a)^(3/2)*a^2 + 14*sqrt(2)*(3*(b*x + a)^(5/2) - 5*(b*x + a)^(3/2)*a)*a + sqrt(
2)*(15*(b*x + a)^(7/2) - 42*(b*x + a)^(5/2)*a + 35*(b*x + a)^(3/2)*a^2))/b