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

Optimal. Leaf size=18 \[ -\frac{2}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

[Out]

-2/(3*b*ArcTanh[Tanh[a + b*x]]^(3/2))

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Rubi [A]  time = 0.0045551, 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} \[ -\frac{2}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^(-5/2),x]

[Out]

-2/(3*b*ArcTanh[Tanh[a + b*x]]^(3/2))

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

Mathematica [A]  time = 0.006595, size = 18, normalized size = 1. \[ -\frac{2}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-2/(3*b*ArcTanh[Tanh[a + b*x]]^(3/2))

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Maple [A]  time = 0.029, size = 15, normalized size = 0.8 \begin{align*} -{\frac{2}{3\,b} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/arctanh(tanh(b*x+a))^(5/2),x)

[Out]

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

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Maxima [A]  time = 1.7228, size = 16, normalized size = 0.89 \begin{align*} -\frac{2}{3 \,{\left (b x + a\right )}^{\frac{3}{2}} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arctanh(tanh(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

-2/3/((b*x + a)^(3/2)*b)

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Fricas [B]  time = 2.10169, size = 68, normalized size = 3.78 \begin{align*} -\frac{2 \, \sqrt{b x + a}}{3 \,{\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arctanh(tanh(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 70.418, size = 27, normalized size = 1.5 \begin{align*} \begin{cases} - \frac{2}{3 b \operatorname{atanh}^{\frac{3}{2}}{\left (\tanh{\left (a + b x \right )} \right )}} & \text{for}\: b \neq 0 \\\frac{x}{\operatorname{atanh}^{\frac{5}{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(1/atanh(tanh(b*x+a))**(5/2),x)

[Out]

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

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Giac [A]  time = 1.14559, size = 16, normalized size = 0.89 \begin{align*} -\frac{2}{3 \,{\left (b x + a\right )}^{\frac{3}{2}} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arctanh(tanh(b*x+a))^(5/2),x, algorithm="giac")

[Out]

-2/3/((b*x + a)^(3/2)*b)