3.155 \(\int \frac{\coth ^{-1}(\tanh (a+b x))^3}{x^4} \, dx\)

Optimal. Leaf size=55 \[ -\frac{b^2 \coth ^{-1}(\tanh (a+b x))}{x}-\frac{b \coth ^{-1}(\tanh (a+b x))^2}{2 x^2}-\frac{\coth ^{-1}(\tanh (a+b x))^3}{3 x^3}+b^3 \log (x) \]

[Out]

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

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Rubi [A]  time = 0.0381152, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2168, 29} \[ -\frac{b^2 \coth ^{-1}(\tanh (a+b x))}{x}-\frac{b \coth ^{-1}(\tanh (a+b x))^2}{2 x^2}-\frac{\coth ^{-1}(\tanh (a+b x))^3}{3 x^3}+b^3 \log (x) \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[Tanh[a + b*x]]^3/x^4,x]

[Out]

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

Rule 2168

Int[(u_)^(m_)*(v_)^(n_.), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(u^(m + 1)*v^
n)/(a*(m + 1)), x] - Dist[(b*n)/(a*(m + 1)), Int[u^(m + 1)*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m
, n}, x] && PiecewiseLinearQ[u, v, x] && NeQ[m, -1] && ((LtQ[m, -1] && GtQ[n, 0] &&  !(ILtQ[m + n, -2] && (Fra
ctionQ[m] || GeQ[2*n + m + 1, 0]))) || (IGtQ[n, 0] && IGtQ[m, 0] && LeQ[n, m]) || (IGtQ[n, 0] &&  !IntegerQ[m]
) || (ILtQ[m, 0] &&  !IntegerQ[n]))

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rubi steps

\begin{align*} \int \frac{\coth ^{-1}(\tanh (a+b x))^3}{x^4} \, dx &=-\frac{\coth ^{-1}(\tanh (a+b x))^3}{3 x^3}+b \int \frac{\coth ^{-1}(\tanh (a+b x))^2}{x^3} \, dx\\ &=-\frac{b \coth ^{-1}(\tanh (a+b x))^2}{2 x^2}-\frac{\coth ^{-1}(\tanh (a+b x))^3}{3 x^3}+b^2 \int \frac{\coth ^{-1}(\tanh (a+b x))}{x^2} \, dx\\ &=-\frac{b^2 \coth ^{-1}(\tanh (a+b x))}{x}-\frac{b \coth ^{-1}(\tanh (a+b x))^2}{2 x^2}-\frac{\coth ^{-1}(\tanh (a+b x))^3}{3 x^3}+b^3 \int \frac{1}{x} \, dx\\ &=-\frac{b^2 \coth ^{-1}(\tanh (a+b x))}{x}-\frac{b \coth ^{-1}(\tanh (a+b x))^2}{2 x^2}-\frac{\coth ^{-1}(\tanh (a+b x))^3}{3 x^3}+b^3 \log (x)\\ \end{align*}

Mathematica [A]  time = 0.0243607, size = 60, normalized size = 1.09 \[ \frac{-6 b^2 x^2 \coth ^{-1}(\tanh (a+b x))-3 b x \coth ^{-1}(\tanh (a+b x))^2-2 \coth ^{-1}(\tanh (a+b x))^3+b^3 x^3 (6 \log (x)+11)}{6 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[Tanh[a + b*x]]^3/x^4,x]

[Out]

(-6*b^2*x^2*ArcCoth[Tanh[a + b*x]] - 3*b*x*ArcCoth[Tanh[a + b*x]]^2 - 2*ArcCoth[Tanh[a + b*x]]^3 + b^3*x^3*(11
+ 6*Log[x]))/(6*x^3)

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Maple [C]  time = 1.385, size = 17237, normalized size = 313.4 \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^4,x)

[Out]

result too large to display

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Maxima [A]  time = 1.59677, size = 70, normalized size = 1.27 \begin{align*}{\left (b^{2} \log \left (x\right ) - \frac{b \operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )}{x}\right )} b - \frac{b \operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )^{2}}{2 \, x^{2}} - \frac{\operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )^{3}}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(tanh(b*x+a))^3/x^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.65861, size = 120, normalized size = 2.18 \begin{align*} \frac{24 \, b^{3} x^{3} \log \left (x\right ) - 72 \, a b^{2} x^{2} + 6 \, \pi ^{2} a - 8 \, a^{3} + 9 \,{\left (\pi ^{2} b - 4 \, a^{2} b\right )} x}{24 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(tanh(b*x+a))^3/x^4,x, algorithm="fricas")

[Out]

1/24*(24*b^3*x^3*log(x) - 72*a*b^2*x^2 + 6*pi^2*a - 8*a^3 + 9*(pi^2*b - 4*a^2*b)*x)/x^3

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Sympy [A]  time = 1.21213, size = 51, normalized size = 0.93 \begin{align*} b^{3} \log{\left (x \right )} - \frac{b^{2} \operatorname{acoth}{\left (\tanh{\left (a + b x \right )} \right )}}{x} - \frac{b \operatorname{acoth}^{2}{\left (\tanh{\left (a + b x \right )} \right )}}{2 x^{2}} - \frac{\operatorname{acoth}^{3}{\left (\tanh{\left (a + b x \right )} \right )}}{3 x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acoth(tanh(b*x+a))**3/x**4,x)

[Out]

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

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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^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(tanh(b*x+a))^3/x^4,x, algorithm="giac")

[Out]

integrate(arccoth(tanh(b*x + a))^3/x^4, x)