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

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

[Out]

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

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Rubi [A]  time = 0.0077072, 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 = {6242, 30} \[ \frac{1}{2} x^2 \coth ^{-1}(\coth (a+b x))-\frac{b x^3}{6} \]

Antiderivative was successfully verified.

[In]

Int[x*ArcCoth[Coth[a + b*x]],x]

[Out]

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

Rule 6242

Int[ArcCoth[(c_.) + Coth[(a_.) + (b_.)*(x_)]*(d_.)]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^(
m + 1)*ArcCoth[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 \coth ^{-1}(\coth (a+b x)) \, dx &=\frac{1}{2} x^2 \coth ^{-1}(\coth (a+b x))-\frac{1}{2} b \int x^2 \, dx\\ &=-\frac{b x^3}{6}+\frac{1}{2} x^2 \coth ^{-1}(\coth (a+b x))\\ \end{align*}

Mathematica [A]  time = 0.015436, size = 20, normalized size = 0.87 \[ -\frac{1}{6} x^2 \left (b x-3 \coth ^{-1}(\coth (a+b x))\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcCoth[Coth[a + b*x]],x]

[Out]

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

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Maple [B]  time = 0.075, size = 48, normalized size = 2.1 \begin{align*}{\frac{{x}^{2}{\rm arccoth} \left ({\rm coth} \left (bx+a\right )\right )}{2}}+{\frac{1}{2\,{b}^{2}} \left ( -{\frac{ \left ( bx+a \right ) ^{3}}{3}}+ \left ( bx+a \right ) ^{2}a-{a}^{2} \left ( bx+a \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.933736, size = 18, normalized size = 0.78 \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*arccoth(coth(b*x+a)),x, algorithm="maxima")

[Out]

1/3*b*x^3 + 1/2*a*x^2

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Fricas [A]  time = 1.26847, size = 31, normalized size = 1.35 \begin{align*} \frac{1}{3} x^{3} b + \frac{1}{2} x^{2} a \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*b + 1/2*x^2*a

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Sympy [A]  time = 37.9831, size = 60, normalized size = 2.61 \begin{align*} \begin{cases} 0 & \text{for}\: a = \log{\left (- e^{- b x} \right )} \vee a = \log{\left (e^{- b x} \right )} \\\frac{x^{2} \operatorname{acoth}{\left (\coth{\left (a \right )} \right )}}{2} & \text{for}\: b = 0 \\\frac{x \operatorname{acoth}^{2}{\left (\frac{1}{\tanh{\left (a + b x \right )}} \right )}}{2 b} - \frac{\operatorname{acoth}^{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*acoth(coth(b*x+a)),x)

[Out]

Piecewise((0, Eq(a, log(exp(-b*x))) | Eq(a, log(-exp(-b*x)))), (x**2*acoth(coth(a))/2, Eq(b, 0)), (x*acoth(1/t
anh(a + b*x))**2/(2*b) - acoth(1/tanh(a + b*x))**3/(6*b**2), True))

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Giac [A]  time = 1.13859, size = 18, normalized size = 0.78 \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*arccoth(coth(b*x+a)),x, algorithm="giac")

[Out]

1/3*b*x^3 + 1/2*a*x^2