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]

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

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Rubi [A]  time = 0.01, antiderivative size = 23, normalized size of antiderivative = 1.00, 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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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]

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*}

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Mathematica [A]  time = 0.02, 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]

-1/6*(x^2*(b*x - 3*ArcTanh[Coth[a + b*x]]))

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fricas [A]  time = 0.60, size = 13, normalized size = 0.57 \[ \frac {1}{3} \, b x^{3} + \frac {1}{2} \, a x^{2} \]

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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giac [B]  time = 0.15, size = 71, normalized size = 3.09 \[ -\frac {1}{6} \, b x^{3} + \frac {1}{4} \, x^{2} \log \left (-\frac {\frac {e^{\left (2 \, b x + 2 \, a\right )} + 1}{e^{\left (2 \, b x + 2 \, a\right )} - 1} + 1}{\frac {e^{\left (2 \, b x + 2 \, a\right )} + 1}{e^{\left (2 \, b x + 2 \, a\right )} - 1} - 1}\right ) \]

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))

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maple [A]  time = 0.14, size = 20, normalized size = 0.87 \[ -\frac {b \,x^{3}}{6}+\frac {x^{2} \arctanh \left (\coth \left (b x +a \right )\right )}{2} \]

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 = 0.39, size = 19, normalized size = 0.83 \[ -\frac {1}{6} \, b x^{3} + \frac {1}{2} \, x^{2} \operatorname {artanh}\left (\coth \left (b x + a\right )\right ) \]

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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mupad [B]  time = 0.06, size = 19, normalized size = 0.83 \[ \frac {x^2\,\mathrm {atanh}\left (\mathrm {coth}\left (a+b\,x\right )\right )}{2}-\frac {b\,x^3}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*atanh(coth(a + b*x)),x)

[Out]

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

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sympy [A]  time = 5.55, size = 49, normalized size = 2.13 \[ \begin {cases} \left \langle - \frac {\pi }{4}, \frac {\pi }{4}\right \rangle i x^{2} & \text {for}\: a = \log {\left (- e^{- b x} \right )} \vee a = \log {\left (e^{- b x} \right )} \\- \frac {b x^{3}}{6} + \frac {x^{2} \operatorname {atanh}{\left (\frac {1}{\tanh {\left (a + b x \right )}} \right )}}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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