3.157 \(\int \frac{\coth ^{-1}(\tanh (a+b x))^3}{x^6} \, dx\)

Optimal. Leaf size=64 \[ \frac{\coth ^{-1}(\tanh (a+b x))^4}{5 x^5 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}+\frac{b \coth ^{-1}(\tanh (a+b x))^4}{20 x^4 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2} \]

[Out]

(b*ArcCoth[Tanh[a + b*x]]^4)/(20*x^4*(b*x - ArcCoth[Tanh[a + b*x]])^2) + ArcCoth[Tanh[a + b*x]]^4/(5*x^5*(b*x
- ArcCoth[Tanh[a + b*x]]))

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Rubi [A]  time = 0.0335485, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2171, 2167} \[ \frac{\coth ^{-1}(\tanh (a+b x))^4}{5 x^5 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}+\frac{b \coth ^{-1}(\tanh (a+b x))^4}{20 x^4 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[Tanh[a + b*x]]^3/x^6,x]

[Out]

(b*ArcCoth[Tanh[a + b*x]]^4)/(20*x^4*(b*x - ArcCoth[Tanh[a + b*x]])^2) + ArcCoth[Tanh[a + b*x]]^4/(5*x^5*(b*x
- ArcCoth[Tanh[a + b*x]]))

Rule 2171

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] + Dist[(b*(m + n + 2))/((m + 1)*(b*u - a*v)), Int[u^(m + 1)*v^n, x], x] /;
NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && NeQ[m + n + 2, 0] && LtQ[m, -1]

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^6} \, dx &=\frac{\coth ^{-1}(\tanh (a+b x))^4}{5 x^5 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}+\frac{b \int \frac{\coth ^{-1}(\tanh (a+b x))^3}{x^5} \, dx}{5 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}\\ &=\frac{b \coth ^{-1}(\tanh (a+b x))^4}{20 x^4 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}+\frac{\coth ^{-1}(\tanh (a+b x))^4}{5 x^5 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}\\ \end{align*}

Mathematica [A]  time = 0.0339501, size = 54, normalized size = 0.84 \[ -\frac{2 b^2 x^2 \coth ^{-1}(\tanh (a+b x))+3 b x \coth ^{-1}(\tanh (a+b x))^2+4 \coth ^{-1}(\tanh (a+b x))^3+b^3 x^3}{20 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[Tanh[a + b*x]]^3/x^6,x]

[Out]

-(b^3*x^3 + 2*b^2*x^2*ArcCoth[Tanh[a + b*x]] + 3*b*x*ArcCoth[Tanh[a + b*x]]^2 + 4*ArcCoth[Tanh[a + b*x]]^3)/(2
0*x^5)

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Maple [C]  time = 1.273, size = 17234, normalized size = 269.3 \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^6,x)

[Out]

result too large to display

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Maxima [A]  time = 1.59881, size = 73, normalized size = 1.14 \begin{align*} -\frac{1}{20} \, b{\left (\frac{b^{2}}{x^{2}} + \frac{2 \, b \operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )}{x^{3}}\right )} - \frac{3 \, b \operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )^{2}}{20 \, x^{4}} - \frac{\operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )^{3}}{5 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(tanh(b*x+a))^3/x^6,x, algorithm="maxima")

[Out]

-1/20*b*(b^2/x^2 + 2*b*arccoth(tanh(b*x + a))/x^3) - 3/20*b*arccoth(tanh(b*x + a))^2/x^4 - 1/5*arccoth(tanh(b*
x + a))^3/x^5

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Fricas [A]  time = 1.69803, size = 116, normalized size = 1.81 \begin{align*} -\frac{40 \, b^{3} x^{3} + 80 \, a b^{2} x^{2} - 12 \, \pi ^{2} a + 16 \, a^{3} - 15 \,{\left (\pi ^{2} b - 4 \, a^{2} b\right )} x}{80 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(tanh(b*x+a))^3/x^6,x, algorithm="fricas")

[Out]

-1/80*(40*b^3*x^3 + 80*a*b^2*x^2 - 12*pi^2*a + 16*a^3 - 15*(pi^2*b - 4*a^2*b)*x)/x^5

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Sympy [A]  time = 3.61206, size = 60, normalized size = 0.94 \begin{align*} - \frac{b^{3}}{20 x^{2}} - \frac{b^{2} \operatorname{acoth}{\left (\tanh{\left (a + b x \right )} \right )}}{10 x^{3}} - \frac{3 b \operatorname{acoth}^{2}{\left (\tanh{\left (a + b x \right )} \right )}}{20 x^{4}} - \frac{\operatorname{acoth}^{3}{\left (\tanh{\left (a + b x \right )} \right )}}{5 x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acoth(tanh(b*x+a))**3/x**6,x)

[Out]

-b**3/(20*x**2) - b**2*acoth(tanh(a + b*x))/(10*x**3) - 3*b*acoth(tanh(a + b*x))**2/(20*x**4) - acoth(tanh(a +
b*x))**3/(5*x**5)

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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^{6}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(tanh(b*x+a))^3/x^6,x, algorithm="giac")

[Out]

integrate(arccoth(tanh(b*x + a))^3/x^6, x)