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

Optimal. Leaf size=34 \[ -\frac{1}{2 b^2 \coth ^{-1}(\tanh (a+b x))}-\frac{x}{2 b \coth ^{-1}(\tanh (a+b x))^2} \]

[Out]

-x/(2*b*ArcCoth[Tanh[a + b*x]]^2) - 1/(2*b^2*ArcCoth[Tanh[a + b*x]])

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Rubi [A]  time = 0.0141126, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {2168, 2157, 30} \[ -\frac{1}{2 b^2 \coth ^{-1}(\tanh (a+b x))}-\frac{x}{2 b \coth ^{-1}(\tanh (a+b x))^2} \]

Antiderivative was successfully verified.

[In]

Int[x/ArcCoth[Tanh[a + b*x]]^3,x]

[Out]

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

Mathematica [A]  time = 0.0493936, size = 27, normalized size = 0.79 \[ -\frac{\coth ^{-1}(\tanh (a+b x))+b x}{2 b^2 \coth ^{-1}(\tanh (a+b x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.171, size = 634, normalized size = 18.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*I*(-2*Pi*csgn(I/(exp(2*b*x+2*a)+1))^2+Pi*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x
+2*a)/(exp(2*b*x+2*a)+1))-Pi*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+2*Pi*csgn(
I/(exp(2*b*x+2*a)+1))^3+Pi*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))-2*Pi*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*
x+2*a))^2+Pi*csgn(I*exp(2*b*x+2*a))^3-Pi*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+Pi
*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3+2*Pi+4*I*b*x+4*I*ln(exp(b*x+a)))/b^2/(-2*Pi*csgn(I/(exp(2*b*x+2*a
)+1))^2+Pi*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))-Pi*csgn
(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+2*Pi*csgn(I/(exp(2*b*x+2*a)+1))^3+Pi*csgn(I
*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))-2*Pi*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^2+Pi*csgn(I*exp(2*b*x+2*a
))^3-Pi*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+Pi*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x
+2*a)+1))^3+2*Pi+4*I*ln(exp(b*x+a)))^2

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Maxima [C]  time = 3.53148, size = 82, normalized size = 2.41 \begin{align*} \frac{8 i \, \pi - 32 \, b x - 16 \, a}{32 \, b^{4} x^{2} - 8 \, \pi ^{2} b^{2} - 32 i \, \pi a b^{2} + 32 \, a^{2} b^{2} +{\left (-32 i \, \pi b^{3} + 64 \, a b^{3}\right )} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(8*I*pi - 32*b*x - 16*a)/(32*b^4*x^2 - 8*pi^2*b^2 - 32*I*pi*a*b^2 + 32*a^2*b^2 + (-32*I*pi*b^3 + 64*a*b^3)*x)

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Fricas [B]  time = 1.64833, size = 262, normalized size = 7.71 \begin{align*} -\frac{2 \,{\left (8 \, b^{3} x^{3} + 20 \, a b^{2} x^{2} + 16 \, a^{2} b x + \pi ^{2} a + 4 \, a^{3}\right )}}{16 \, b^{6} x^{4} + 64 \, a b^{5} x^{3} + \pi ^{4} b^{2} + 8 \, \pi ^{2} a^{2} b^{2} + 16 \, a^{4} b^{2} + 8 \,{\left (\pi ^{2} b^{4} + 12 \, a^{2} b^{4}\right )} x^{2} + 16 \,{\left (\pi ^{2} a b^{3} + 4 \, a^{3} b^{3}\right )} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(8*b^3*x^3 + 20*a*b^2*x^2 + 16*a^2*b*x + pi^2*a + 4*a^3)/(16*b^6*x^4 + 64*a*b^5*x^3 + pi^4*b^2 + 8*pi^2*a^2
*b^2 + 16*a^4*b^2 + 8*(pi^2*b^4 + 12*a^2*b^4)*x^2 + 16*(pi^2*a*b^3 + 4*a^3*b^3)*x)

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Sympy [A]  time = 24.1672, size = 42, normalized size = 1.24 \begin{align*} \begin{cases} - \frac{x}{2 b \operatorname{acoth}^{2}{\left (\tanh{\left (a + b x \right )} \right )}} - \frac{1}{2 b^{2} \operatorname{acoth}{\left (\tanh{\left (a + b x \right )} \right )}} & \text{for}\: b \neq 0 \\\frac{x^{2}}{2 \operatorname{acoth}^{3}{\left (\tanh{\left (a \right )} \right )}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-x/(2*b*acoth(tanh(a + b*x))**2) - 1/(2*b**2*acoth(tanh(a + b*x))), Ne(b, 0)), (x**2/(2*acoth(tanh(
a))**3), True))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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