3.156 \(\int \frac{\coth ^{-1}(\tanh (a+b x))^3}{x^5} \, dx\)

Optimal. Leaf size=31 \[ \frac{\coth ^{-1}(\tanh (a+b x))^4}{4 x^4 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )} \]

[Out]

ArcCoth[Tanh[a + b*x]]^4/(4*x^4*(b*x - ArcCoth[Tanh[a + b*x]]))

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Rubi [A]  time = 0.0137358, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {2167} \[ \frac{\coth ^{-1}(\tanh (a+b x))^4}{4 x^4 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )} \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[Tanh[a + b*x]]^3/x^5,x]

[Out]

ArcCoth[Tanh[a + b*x]]^4/(4*x^4*(b*x - ArcCoth[Tanh[a + b*x]]))

Rule 2167

Int[(u_)^(m_)*(v_)^(n_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, -Simp[(u^(m + 1)*v^
(n + 1))/((m + 1)*(b*u - a*v)), x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m, n}, x] && PiecewiseLinearQ[u, v, x] && E
qQ[m + n + 2, 0] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{\coth ^{-1}(\tanh (a+b x))^3}{x^5} \, dx &=\frac{\coth ^{-1}(\tanh (a+b x))^4}{4 x^4 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}\\ \end{align*}

Mathematica [A]  time = 0.0222898, size = 50, normalized size = 1.61 \[ -\frac{b^2 x^2 \coth ^{-1}(\tanh (a+b x))+b x \coth ^{-1}(\tanh (a+b x))^2+\coth ^{-1}(\tanh (a+b x))^3+b^3 x^3}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[Tanh[a + b*x]]^3/x^5,x]

[Out]

-(b^3*x^3 + b^2*x^2*ArcCoth[Tanh[a + b*x]] + b*x*ArcCoth[Tanh[a + b*x]]^2 + ArcCoth[Tanh[a + b*x]]^3)/(4*x^4)

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Maple [C]  time = 1.347, size = 17235, normalized size = 556. \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccoth(tanh(b*x+a))^3/x^5,x)

[Out]

result too large to display

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Maxima [A]  time = 1.59718, size = 72, normalized size = 2.32 \begin{align*} -\frac{1}{4} \, b{\left (\frac{b^{2}}{x} + \frac{b \operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )}{x^{2}}\right )} - \frac{b \operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )^{2}}{4 \, x^{3}} - \frac{\operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )^{3}}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(tanh(b*x+a))^3/x^5,x, algorithm="maxima")

[Out]

-1/4*b*(b^2/x + b*arccoth(tanh(b*x + a))/x^2) - 1/4*b*arccoth(tanh(b*x + a))^2/x^3 - 1/4*arccoth(tanh(b*x + a)
)^3/x^4

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Fricas [A]  time = 1.60225, size = 112, normalized size = 3.61 \begin{align*} -\frac{16 \, b^{3} x^{3} + 24 \, a b^{2} x^{2} - 3 \, \pi ^{2} a + 4 \, a^{3} - 4 \,{\left (\pi ^{2} b - 4 \, a^{2} b\right )} x}{16 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(tanh(b*x+a))^3/x^5,x, algorithm="fricas")

[Out]

-1/16*(16*b^3*x^3 + 24*a*b^2*x^2 - 3*pi^2*a + 4*a^3 - 4*(pi^2*b - 4*a^2*b)*x)/x^4

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Sympy [B]  time = 2.14484, size = 56, normalized size = 1.81 \begin{align*} - \frac{b^{3}}{4 x} - \frac{b^{2} \operatorname{acoth}{\left (\tanh{\left (a + b x \right )} \right )}}{4 x^{2}} - \frac{b \operatorname{acoth}^{2}{\left (\tanh{\left (a + b x \right )} \right )}}{4 x^{3}} - \frac{\operatorname{acoth}^{3}{\left (\tanh{\left (a + b x \right )} \right )}}{4 x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acoth(tanh(b*x+a))**3/x**5,x)

[Out]

-b**3/(4*x) - b**2*acoth(tanh(a + b*x))/(4*x**2) - b*acoth(tanh(a + b*x))**2/(4*x**3) - acoth(tanh(a + b*x))**
3/(4*x**4)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )^{3}}{x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(tanh(b*x+a))^3/x^5,x, algorithm="giac")

[Out]

integrate(arccoth(tanh(b*x + a))^3/x^5, x)