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

Optimal. Leaf size=16 \[ -\frac {1}{2 b \coth ^{-1}(\tanh (a+b x))^2} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2157, 30} \[ -\frac {1}{2 b \coth ^{-1}(\tanh (a+b x))^2} \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[Tanh[a + b*x]]^(-3),x]

[Out]

-1/(2*b*ArcCoth[Tanh[a + b*x]]^2)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 16, normalized size = 1.00 \[ -\frac {1}{2 b \coth ^{-1}(\tanh (a+b x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[Tanh[a + b*x]]^(-3),x]

[Out]

-1/2*1/(b*ArcCoth[Tanh[a + b*x]]^2)

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fricas [B]  time = 0.57, size = 107, normalized size = 6.69 \[ -\frac {2 \, {\left (4 \, b^{2} x^{2} + 8 \, a b x - \pi ^{2} + 4 \, a^{2}\right )}}{16 \, b^{5} x^{4} + 64 \, a b^{4} x^{3} + \pi ^{4} b + 8 \, \pi ^{2} a^{2} b + 16 \, a^{4} b + 8 \, {\left (\pi ^{2} b^{3} + 12 \, a^{2} b^{3}\right )} x^{2} + 16 \, {\left (\pi ^{2} a b^{2} + 4 \, a^{3} b^{2}\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arccoth(tanh(b*x+a))^3,x, algorithm="fricas")

[Out]

-2*(4*b^2*x^2 + 8*a*b*x - pi^2 + 4*a^2)/(16*b^5*x^4 + 64*a*b^4*x^3 + pi^4*b + 8*pi^2*a^2*b + 16*a^4*b + 8*(pi^
2*b^3 + 12*a^2*b^3)*x^2 + 16*(pi^2*a*b^2 + 4*a^3*b^2)*x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arccoth(tanh(b*x+a))^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.08, size = 15, normalized size = 0.94 \[ -\frac {1}{2 b \mathrm {arccoth}\left (\tanh \left (b x +a \right )\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/arccoth(tanh(b*x+a))^3,x)

[Out]

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

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maxima [C]  time = 0.43, size = 30, normalized size = 1.88 \[ \frac {8}{{\left (4 \, \pi ^{2} - 16 i \, \pi {\left (b x + a\right )} - 16 \, {\left (b x + a\right )}^{2}\right )} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arccoth(tanh(b*x+a))^3,x, algorithm="maxima")

[Out]

8/((4*pi^2 - 16*I*pi*(b*x + a) - 16*(b*x + a)^2)*b)

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mupad [B]  time = 0.07, size = 14, normalized size = 0.88 \[ -\frac {1}{2\,b\,{\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/acoth(tanh(a + b*x))^3,x)

[Out]

-1/(2*b*acoth(tanh(a + b*x))^2)

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/acoth(tanh(b*x+a))**3,x)

[Out]

Exception raised: TypeError

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