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]

-1/12*b*x^4+1/3*x^3*arccoth(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 = 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 30

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

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]))

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*}

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Mathematica [A]  time = 0.04, 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]

-1/12*(x^3*(b*x - 4*ArcCoth[Coth[a + b*x]]))

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fricas [A]  time = 0.39, size = 13, normalized size = 0.57 \[ \frac {1}{4} x^{4} b + \frac {1}{3} x^{3} a \]

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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giac [A]  time = 0.12, size = 13, normalized size = 0.57 \[ \frac {1}{4} \, b x^{4} + \frac {1}{3} \, a x^{3} \]

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

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maple [B]  time = 0.44, size = 59, normalized size = 2.57 \[ \frac {x^{3} \mathrm {arccoth}\left (\coth \left (b x +a \right )\right )}{3}+\frac {-\frac {\left (b x +a \right )^{4}}{4}+\left (b x +a \right )^{3} a -\frac {3 a^{2} \left (b x +a \right )^{2}}{2}+\left (b x +a \right ) a^{3}}{3 b^{3}} \]

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 = 0.31, size = 13, normalized size = 0.57 \[ \frac {1}{4} \, b x^{4} + \frac {1}{3} \, a x^{3} \]

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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mupad [B]  time = 0.08, size = 19, normalized size = 0.83 \[ \frac {x^3\,\mathrm {acoth}\left (\mathrm {coth}\left (a+b\,x\right )\right )}{3}-\frac {b\,x^4}{12} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*acoth(coth(a + b*x)),x)

[Out]

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

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sympy [A]  time = 11.34, size = 39, normalized size = 1.70 \[ \begin {cases} 0 & \text {for}\: a = \log {\left (- e^{- b x} \right )} \vee a = \log {\left (e^{- b x} \right )} \\- \frac {b x^{4}}{12} + \frac {x^{3} \operatorname {acoth}{\left (\frac {1}{\tanh {\left (a + b x \right )}} \right )}}{3} & \text {otherwise} \end {cases} \]

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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