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

Optimal. Leaf size=23 \[ -\frac{\tanh ^{-1}(\coth (a+b x))}{2 x^2}-\frac{b}{2 x} \]

[Out]

-b/(2*x) - ArcTanh[Coth[a + b*x]]/(2*x^2)

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Rubi [A]  time = 0.0089536, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {2168, 30} \[ -\frac{\tanh ^{-1}(\coth (a+b x))}{2 x^2}-\frac{b}{2 x} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Coth[a + b*x]]/x^3,x]

[Out]

-b/(2*x) - ArcTanh[Coth[a + b*x]]/(2*x^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 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{\tanh ^{-1}(\coth (a+b x))}{x^3} \, dx &=-\frac{\tanh ^{-1}(\coth (a+b x))}{2 x^2}+\frac{1}{2} b \int \frac{1}{x^2} \, dx\\ &=-\frac{b}{2 x}-\frac{\tanh ^{-1}(\coth (a+b x))}{2 x^2}\\ \end{align*}

Mathematica [A]  time = 0.0135759, size = 18, normalized size = 0.78 \[ -\frac{\tanh ^{-1}(\coth (a+b x))+b x}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Coth[a + b*x]]/x^3,x]

[Out]

-(b*x + ArcTanh[Coth[a + b*x]])/(2*x^2)

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Maple [A]  time = 0.032, size = 20, normalized size = 0.9 \begin{align*} -{\frac{b}{2\,x}}-{\frac{{\it Artanh} \left ({\rm coth} \left (bx+a\right ) \right ) }{2\,{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(coth(b*x+a))/x^3,x)

[Out]

-1/2*b/x-1/2*arctanh(coth(b*x+a))/x^2

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Maxima [A]  time = 1.14687, size = 26, normalized size = 1.13 \begin{align*} -\frac{b}{2 \, x} - \frac{\operatorname{artanh}\left (\coth \left (b x + a\right )\right )}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b/x - 1/2*arctanh(coth(b*x + a))/x^2

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Fricas [A]  time = 2.12019, size = 30, normalized size = 1.3 \begin{align*} -\frac{2 \, b x + a}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*b*x + a)/x^2

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Sympy [A]  time = 142.164, size = 49, normalized size = 2.13 \begin{align*} \begin{cases} \frac{\left \langle - \frac{\pi }{4}, \frac{\pi }{4}\right \rangle i}{x^{2}} & \text{for}\: a = \log{\left (- e^{- b x} \right )} \vee a = \log{\left (e^{- b x} \right )} \\- \frac{b}{2 x} - \frac{\operatorname{atanh}{\left (\frac{1}{\tanh{\left (a + b x \right )}} \right )}}{2 x^{2}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(coth(b*x+a))/x**3,x)

[Out]

Piecewise((AccumBounds(-pi/4, pi/4)*I/x**2, Eq(a, log(exp(-b*x))) | Eq(a, log(-exp(-b*x)))), (-b/(2*x) - atanh
(1/tanh(a + b*x))/(2*x**2), True))

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Giac [B]  time = 1.19043, size = 96, normalized size = 4.17 \begin{align*} -\frac{b}{2 \, x} - \frac{\log \left (-\frac{\frac{e^{\left (2 \, b x + 2 \, a\right )} + 1}{e^{\left (2 \, b x + 2 \, a\right )} - 1} + 1}{\frac{e^{\left (2 \, b x + 2 \, a\right )} + 1}{e^{\left (2 \, b x + 2 \, a\right )} - 1} - 1}\right )}{4 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b/x - 1/4*log(-((e^(2*b*x + 2*a) + 1)/(e^(2*b*x + 2*a) - 1) + 1)/((e^(2*b*x + 2*a) + 1)/(e^(2*b*x + 2*a)
- 1) - 1))/x^2