3.149 \(\int x^2 \coth ^{-1}(\tanh (a+b x))^3 \, dx\)

Optimal. Leaf size=53 \[ \frac{\coth ^{-1}(\tanh (a+b x))^6}{60 b^3}-\frac{x \coth ^{-1}(\tanh (a+b x))^5}{10 b^2}+\frac{x^2 \coth ^{-1}(\tanh (a+b x))^4}{4 b} \]

[Out]

(x^2*ArcCoth[Tanh[a + b*x]]^4)/(4*b) - (x*ArcCoth[Tanh[a + b*x]]^5)/(10*b^2) + ArcCoth[Tanh[a + b*x]]^6/(60*b^
3)

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Rubi [A]  time = 0.0312969, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {2168, 2157, 30} \[ \frac{\coth ^{-1}(\tanh (a+b x))^6}{60 b^3}-\frac{x \coth ^{-1}(\tanh (a+b x))^5}{10 b^2}+\frac{x^2 \coth ^{-1}(\tanh (a+b x))^4}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[x^2*ArcCoth[Tanh[a + b*x]]^3,x]

[Out]

(x^2*ArcCoth[Tanh[a + b*x]]^4)/(4*b) - (x*ArcCoth[Tanh[a + b*x]]^5)/(10*b^2) + ArcCoth[Tanh[a + b*x]]^6/(60*b^
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 2157

Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Dist[1/c, Subst[Int[x^m, x], x, u], x]] /; FreeQ[m,
x] && PiecewiseLinearQ[u, x]

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}(\tanh (a+b x))^3 \, dx &=\frac{x^2 \coth ^{-1}(\tanh (a+b x))^4}{4 b}-\frac{\int x \coth ^{-1}(\tanh (a+b x))^4 \, dx}{2 b}\\ &=\frac{x^2 \coth ^{-1}(\tanh (a+b x))^4}{4 b}-\frac{x \coth ^{-1}(\tanh (a+b x))^5}{10 b^2}+\frac{\int \coth ^{-1}(\tanh (a+b x))^5 \, dx}{10 b^2}\\ &=\frac{x^2 \coth ^{-1}(\tanh (a+b x))^4}{4 b}-\frac{x \coth ^{-1}(\tanh (a+b x))^5}{10 b^2}+\frac{\operatorname{Subst}\left (\int x^5 \, dx,x,\coth ^{-1}(\tanh (a+b x))\right )}{10 b^3}\\ &=\frac{x^2 \coth ^{-1}(\tanh (a+b x))^4}{4 b}-\frac{x \coth ^{-1}(\tanh (a+b x))^5}{10 b^2}+\frac{\coth ^{-1}(\tanh (a+b x))^6}{60 b^3}\\ \end{align*}

Mathematica [A]  time = 0.0222424, size = 54, normalized size = 1.02 \[ -\frac{1}{60} x^3 \left (-6 b^2 x^2 \coth ^{-1}(\tanh (a+b x))+15 b x \coth ^{-1}(\tanh (a+b x))^2-20 \coth ^{-1}(\tanh (a+b x))^3+b^3 x^3\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcCoth[Tanh[a + b*x]]^3,x]

[Out]

-(x^3*(b^3*x^3 - 6*b^2*x^2*ArcCoth[Tanh[a + b*x]] + 15*b*x*ArcCoth[Tanh[a + b*x]]^2 - 20*ArcCoth[Tanh[a + b*x]
]^3))/60

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Maple [C]  time = 1.081, size = 18111, normalized size = 341.7 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arccoth(tanh(b*x+a))^3,x)

[Out]

result too large to display

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Maxima [A]  time = 1.54597, size = 73, normalized size = 1.38 \begin{align*} -\frac{1}{4} \, b x^{4} \operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )^{2} + \frac{1}{3} \, x^{3} \operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )^{3} - \frac{1}{60} \,{\left (b^{2} x^{6} - 6 \, b x^{5} \operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arccoth(tanh(b*x+a))^3,x, algorithm="maxima")

[Out]

-1/4*b*x^4*arccoth(tanh(b*x + a))^2 + 1/3*x^3*arccoth(tanh(b*x + a))^3 - 1/60*(b^2*x^6 - 6*b*x^5*arccoth(tanh(
b*x + a)))*b

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Fricas [A]  time = 1.55735, size = 120, normalized size = 2.26 \begin{align*} \frac{1}{6} \, b^{3} x^{6} + \frac{3}{5} \, a b^{2} x^{5} - \frac{3}{16} \,{\left (\pi ^{2} b - 4 \, a^{2} b\right )} x^{4} - \frac{1}{12} \,{\left (3 \, \pi ^{2} a - 4 \, a^{3}\right )} x^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arccoth(tanh(b*x+a))^3,x, algorithm="fricas")

[Out]

1/6*b^3*x^6 + 3/5*a*b^2*x^5 - 3/16*(pi^2*b - 4*a^2*b)*x^4 - 1/12*(3*pi^2*a - 4*a^3)*x^3

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Sympy [A]  time = 3.3897, size = 60, normalized size = 1.13 \begin{align*} \begin{cases} \frac{x^{2} \operatorname{acoth}^{4}{\left (\tanh{\left (a + b x \right )} \right )}}{4 b} - \frac{x \operatorname{acoth}^{5}{\left (\tanh{\left (a + b x \right )} \right )}}{10 b^{2}} + \frac{\operatorname{acoth}^{6}{\left (\tanh{\left (a + b x \right )} \right )}}{60 b^{3}} & \text{for}\: b \neq 0 \\\frac{x^{3} \operatorname{acoth}^{3}{\left (\tanh{\left (a \right )} \right )}}{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*acoth(tanh(b*x+a))**3,x)

[Out]

Piecewise((x**2*acoth(tanh(a + b*x))**4/(4*b) - x*acoth(tanh(a + b*x))**5/(10*b**2) + acoth(tanh(a + b*x))**6/
(60*b**3), Ne(b, 0)), (x**3*acoth(tanh(a))**3/3, True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arccoth(tanh(b*x+a))^3,x, algorithm="giac")

[Out]

integrate(x^2*arccoth(tanh(b*x + a))^3, x)