3.170 \(\int \frac{x}{\coth ^{-1}(\tanh (a+b x))^2} \, dx\)

Optimal. Leaf size=28 \[ \frac{\log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b^2}-\frac{x}{b \coth ^{-1}(\tanh (a+b x))} \]

[Out]

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

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Rubi [A]  time = 0.013374, antiderivative size = 28, 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, 29} \[ \frac{\log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b^2}-\frac{x}{b \coth ^{-1}(\tanh (a+b x))} \]

Antiderivative was successfully verified.

[In]

Int[x/ArcCoth[Tanh[a + b*x]]^2,x]

[Out]

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

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 29

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

Rubi steps

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

Mathematica [A]  time = 0.0654313, size = 27, normalized size = 0.96 \[ \frac{-\frac{b x}{\coth ^{-1}(\tanh (a+b x))}+\log \left (\coth ^{-1}(\tanh (a+b x))\right )+1}{b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x/ArcCoth[Tanh[a + b*x]]^2,x]

[Out]

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

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Maple [C]  time = 0.187, size = 625, normalized size = 22.3 \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))^2,x)

[Out]

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

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Maxima [C]  time = 2.4401, size = 63, normalized size = 2.25 \begin{align*} \frac{4 \,{\left (-i \, \pi + 2 \, a\right )}}{8 \, b^{3} x - 4 i \, \pi b^{2} + 8 \, a b^{2}} + \frac{\log \left (-i \, \pi + 2 \, b x + 2 \, a\right )}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/arccoth(tanh(b*x+a))^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.68068, size = 213, normalized size = 7.61 \begin{align*} \frac{8 \, a b x + 2 \, \pi ^{2} + 8 \, a^{2} +{\left (4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}\right )} \log \left (4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}\right )}{2 \,{\left (4 \, b^{4} x^{2} + 8 \, a b^{3} x + \pi ^{2} b^{2} + 4 \, a^{2} b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/arccoth(tanh(b*x+a))^2,x, algorithm="fricas")

[Out]

1/2*(8*a*b*x + 2*pi^2 + 8*a^2 + (4*b^2*x^2 + 8*a*b*x + pi^2 + 4*a^2)*log(4*b^2*x^2 + 8*a*b*x + pi^2 + 4*a^2))/
(4*b^4*x^2 + 8*a*b^3*x + pi^2*b^2 + 4*a^2*b^2)

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Sympy [A]  time = 16.2979, size = 36, normalized size = 1.29 \begin{align*} \begin{cases} - \frac{x}{b \operatorname{acoth}{\left (\tanh{\left (a + b x \right )} \right )}} + \frac{\log{\left (\operatorname{acoth}{\left (\tanh{\left (a + b x \right )} \right )} \right )}}{b^{2}} & \text{for}\: b \neq 0 \\\frac{x^{2}}{2 \operatorname{acoth}^{2}{\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))**2,x)

[Out]

Piecewise((-x/(b*acoth(tanh(a + b*x))) + log(acoth(tanh(a + b*x)))/b**2, Ne(b, 0)), (x**2/(2*acoth(tanh(a))**2
), 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 )^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/arccoth(tanh(b*x+a))^2,x, algorithm="giac")

[Out]

integrate(x/arccoth(tanh(b*x + a))^2, x)