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

Optimal. Leaf size=17 \[ b \log (x)-\frac{\tanh ^{-1}(\coth (a+b x))}{x} \]

[Out]

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

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Rubi [A]  time = 0.0082163, antiderivative size = 17, 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, 29} \[ b \log (x)-\frac{\tanh ^{-1}(\coth (a+b x))}{x} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Coth[a + b*x]]/x^2,x]

[Out]

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

Mathematica [A]  time = 0.0154534, size = 18, normalized size = 1.06 \[ -\frac{\tanh ^{-1}(\coth (a+b x))}{x}+b \log (x)+b \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Coth[a + b*x]]/x^2,x]

[Out]

b - ArcTanh[Coth[a + b*x]]/x + b*Log[x]

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Maple [A]  time = 0.033, size = 18, normalized size = 1.1 \begin{align*} -{\frac{{\it Artanh} \left ({\rm coth} \left (bx+a\right ) \right ) }{x}}+b\ln \left ( x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(coth(b*x+a))/x^2,x)

[Out]

-arctanh(coth(b*x+a))/x+b*ln(x)

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Maxima [A]  time = 1.13869, size = 23, normalized size = 1.35 \begin{align*} b \log \left (x\right ) - \frac{\operatorname{artanh}\left (\coth \left (b x + a\right )\right )}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*log(x) - arctanh(coth(b*x + a))/x

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Fricas [A]  time = 2.08442, size = 27, normalized size = 1.59 \begin{align*} \frac{b x \log \left (x\right ) - a}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*x*log(x) - a)/x

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Sympy [A]  time = 17.0947, size = 42, normalized size = 2.47 \begin{align*} \begin{cases} \frac{\left \langle - \frac{\pi }{2}, \frac{\pi }{2}\right \rangle i}{x} & \text{for}\: a = \log{\left (- e^{- b x} \right )} \vee a = \log{\left (e^{- b x} \right )} \\b \log{\left (x \right )} - \frac{\operatorname{atanh}{\left (\frac{1}{\tanh{\left (a + b x \right )}} \right )}}{x} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(coth(b*x+a))/x**2,x)

[Out]

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

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Giac [B]  time = 1.18725, size = 95, normalized size = 5.59 \begin{align*} b \log \left ({\left | x \right |}\right ) - \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 )}{2 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*log(abs(x)) - 1/2*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