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

Optimal. Leaf size=23 \[ \frac{1}{2} x^2 \tanh ^{-1}(\coth (a+b x))-\frac{b x^3}{6} \]

[Out]

-(b*x^3)/6 + (x^2*ArcTanh[Coth[a + b*x]])/2

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Rubi [A]  time = 0.0070311, antiderivative size = 23, normalized size of antiderivative = 1., 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 = {6241, 30} \[ \frac{1}{2} x^2 \tanh ^{-1}(\coth (a+b x))-\frac{b x^3}{6} \]

Antiderivative was successfully verified.

[In]

Int[x*ArcTanh[Coth[a + b*x]],x]

[Out]

-(b*x^3)/6 + (x^2*ArcTanh[Coth[a + b*x]])/2

Rule 6241

Int[ArcTanh[(c_.) + Coth[(a_.) + (b_.)*(x_)]*(d_.)]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^(
m + 1)*ArcTanh[c + d*Coth[a + b*x]])/(f*(m + 1)), x] + Dist[b/(f*(m + 1)), Int[(e + f*x)^(m + 1)/(c - d - c*E^
(2*a + 2*b*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && EqQ[(c - d)^2, 1]

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 x \tanh ^{-1}(\coth (a+b x)) \, dx &=\frac{1}{2} x^2 \tanh ^{-1}(\coth (a+b x))-\frac{1}{2} b \int x^2 \, dx\\ &=-\frac{b x^3}{6}+\frac{1}{2} x^2 \tanh ^{-1}(\coth (a+b x))\\ \end{align*}

Mathematica [A]  time = 0.0141067, size = 20, normalized size = 0.87 \[ -\frac{1}{6} x^2 \left (b x-3 \tanh ^{-1}(\coth (a+b x))\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcTanh[Coth[a + b*x]],x]

[Out]

-(x^2*(b*x - 3*ArcTanh[Coth[a + b*x]]))/6

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Maple [A]  time = 0.032, size = 20, normalized size = 0.9 \begin{align*} -{\frac{b{x}^{3}}{6}}+{\frac{{x}^{2}{\it Artanh} \left ({\rm coth} \left (bx+a\right ) \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arctanh(coth(b*x+a)),x)

[Out]

-1/6*b*x^3+1/2*x^2*arctanh(coth(b*x+a))

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Maxima [A]  time = 1.14285, size = 26, normalized size = 1.13 \begin{align*} -\frac{1}{6} \, b x^{3} + \frac{1}{2} \, x^{2} \operatorname{artanh}\left (\coth \left (b x + a\right )\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*b*x^3 + 1/2*x^2*arctanh(coth(b*x + a))

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Fricas [A]  time = 2.12269, size = 31, normalized size = 1.35 \begin{align*} \frac{1}{3} \, b x^{3} + \frac{1}{2} \, a x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b*x^3 + 1/2*a*x^2

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Sympy [A]  time = 124.543, size = 70, normalized size = 3.04 \begin{align*} \begin{cases} \frac{x^{2} \operatorname{atanh}{\left (\coth{\left (a \right )} \right )}}{2} & \text{for}\: b = 0 \\\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{x \operatorname{atanh}^{2}{\left (\frac{1}{\tanh{\left (a + b x \right )}} \right )}}{2 b} - \frac{\operatorname{atanh}^{3}{\left (\frac{1}{\tanh{\left (a + b x \right )}} \right )}}{6 b^{2}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x**2*atanh(coth(a))/2, Eq(b, 0)), (AccumBounds(-pi/4, pi/4)*I*x**2, Eq(a, log(exp(-b*x))) | Eq(a, l
og(-exp(-b*x)))), (x*atanh(1/tanh(a + b*x))**2/(2*b) - atanh(1/tanh(a + b*x))**3/(6*b**2), True))

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Giac [B]  time = 1.20196, size = 96, normalized size = 4.17 \begin{align*} -\frac{1}{6} \, b x^{3} + \frac{1}{4} \, x^{2} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*b*x^3 + 1/4*x^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))