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

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

[Out]

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

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Rubi [A]  time = 0.014254, 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} \[ \frac{1}{3} x^3 \coth ^{-1}(\coth (a+b x))-\frac{b x^4}{12} \]

Antiderivative was successfully verified.

[In]

Int[x^2*ArcCoth[Coth[a + b*x]],x]

[Out]

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

Mathematica [A]  time = 0.0213134, size = 20, normalized size = 0.87 \[ -\frac{1}{12} x^3 \left (b x-4 \coth ^{-1}(\coth (a+b x))\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.092, size = 59, normalized size = 2.6 \begin{align*}{\frac{{x}^{3}{\rm arccoth} \left ({\rm coth} \left (bx+a\right )\right )}{3}}+{\frac{1}{3\,{b}^{3}} \left ( -{\frac{ \left ( bx+a \right ) ^{4}}{4}}+ \left ( bx+a \right ) ^{3}a-{\frac{3\,{a}^{2} \left ( bx+a \right ) ^{2}}{2}}+ \left ( bx+a \right ){a}^{3} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*arccoth(coth(b*x+a))+1/3/b^3*(-1/4*(b*x+a)^4+(b*x+a)^3*a-3/2*a^2*(b*x+a)^2+(b*x+a)*a^3)

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Maxima [A]  time = 0.945767, size = 18, normalized size = 0.78 \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*arccoth(coth(b*x+a)),x, algorithm="maxima")

[Out]

1/4*b*x^4 + 1/3*a*x^3

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Fricas [A]  time = 1.34393, size = 31, normalized size = 1.35 \begin{align*} \frac{1}{4} x^{4} b + \frac{1}{3} x^{3} a \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4*b + 1/3*x^3*a

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Sympy [A]  time = 91.8494, size = 39, normalized size = 1.7 \begin{align*} \begin{cases} 0 & \text{for}\: a = \log{\left (- e^{- b x} \right )} \vee a = \log{\left (e^{- b x} \right )} \\- \frac{b x^{4}}{12} + \frac{x^{3} \operatorname{acoth}{\left (\frac{1}{\tanh{\left (a + b x \right )}} \right )}}{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((0, Eq(a, log(exp(-b*x))) | Eq(a, log(-exp(-b*x)))), (-b*x**4/12 + x**3*acoth(1/tanh(a + b*x))/3, Tr
ue))

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Giac [A]  time = 1.17349, size = 18, normalized size = 0.78 \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*arccoth(coth(b*x+a)),x, algorithm="giac")

[Out]

1/4*b*x^4 + 1/3*a*x^3