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

Optimal. Leaf size=23 \[ \frac{1}{3} x^3 \tanh ^{-1}(\coth (a+b x))-\frac{b x^4}{12} \]

[Out]

-(b*x^4)/12 + (x^3*ArcTanh[Coth[a + b*x]])/3

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Rubi [A]  time = 0.0196044, 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} \[ \frac{1}{3} x^3 \tanh ^{-1}(\coth (a+b x))-\frac{b x^4}{12} \]

Antiderivative was successfully verified.

[In]

Int[x^2*ArcTanh[Coth[a + b*x]],x]

[Out]

-(b*x^4)/12 + (x^3*ArcTanh[Coth[a + b*x]])/3

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 x^2 \tanh ^{-1}(\coth (a+b x)) \, dx &=\frac{1}{3} x^3 \tanh ^{-1}(\coth (a+b x))-\frac{1}{3} b \int x^3 \, dx\\ &=-\frac{b x^4}{12}+\frac{1}{3} x^3 \tanh ^{-1}(\coth (a+b x))\\ \end{align*}

Mathematica [A]  time = 0.0237691, size = 20, normalized size = 0.87 \[ -\frac{1}{12} x^3 \left (b x-4 \tanh ^{-1}(\coth (a+b x))\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x^3*(b*x - 4*ArcTanh[Coth[a + b*x]]))/12

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Maple [A]  time = 0.034, size = 20, normalized size = 0.9 \begin{align*} -{\frac{b{x}^{4}}{12}}+{\frac{{x}^{3}{\it Artanh} \left ({\rm coth} \left (bx+a\right ) \right ) }{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*b*x^4+1/3*x^3*arctanh(coth(b*x+a))

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Maxima [A]  time = 1.13334, size = 26, normalized size = 1.13 \begin{align*} -\frac{1}{12} \, b x^{4} + \frac{1}{3} \, x^{3} \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^2*arctanh(coth(b*x+a)),x, algorithm="maxima")

[Out]

-1/12*b*x^4 + 1/3*x^3*arctanh(coth(b*x + a))

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Fricas [A]  time = 2.08463, size = 31, normalized size = 1.35 \begin{align*} \frac{1}{4} \, b x^{4} + \frac{1}{3} \, a x^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b*x^4 + 1/3*a*x^3

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.17625, size = 96, normalized size = 4.17 \begin{align*} -\frac{1}{12} \, b x^{4} + \frac{1}{6} \, x^{3} \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^2*arctanh(coth(b*x+a)),x, algorithm="giac")

[Out]

-1/12*b*x^4 + 1/6*x^3*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))