3.147 \(\int x^4 \coth ^{-1}(\tanh (a+b x))^3 \, dx\)

Optimal. Leaf size=61 \[ \frac{1}{35} b^2 x^7 \coth ^{-1}(\tanh (a+b x))-\frac{1}{10} b x^6 \coth ^{-1}(\tanh (a+b x))^2+\frac{1}{5} x^5 \coth ^{-1}(\tanh (a+b x))^3-\frac{1}{280} b^3 x^8 \]

[Out]

-(b^3*x^8)/280 + (b^2*x^7*ArcCoth[Tanh[a + b*x]])/35 - (b*x^6*ArcCoth[Tanh[a + b*x]]^2)/10 + (x^5*ArcCoth[Tanh
[a + b*x]]^3)/5

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Rubi [A]  time = 0.0411641, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2168, 30} \[ \frac{1}{35} b^2 x^7 \coth ^{-1}(\tanh (a+b x))-\frac{1}{10} b x^6 \coth ^{-1}(\tanh (a+b x))^2+\frac{1}{5} x^5 \coth ^{-1}(\tanh (a+b x))^3-\frac{1}{280} b^3 x^8 \]

Antiderivative was successfully verified.

[In]

Int[x^4*ArcCoth[Tanh[a + b*x]]^3,x]

[Out]

-(b^3*x^8)/280 + (b^2*x^7*ArcCoth[Tanh[a + b*x]])/35 - (b*x^6*ArcCoth[Tanh[a + b*x]]^2)/10 + (x^5*ArcCoth[Tanh
[a + b*x]]^3)/5

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^4 \coth ^{-1}(\tanh (a+b x))^3 \, dx &=\frac{1}{5} x^5 \coth ^{-1}(\tanh (a+b x))^3-\frac{1}{5} (3 b) \int x^5 \coth ^{-1}(\tanh (a+b x))^2 \, dx\\ &=-\frac{1}{10} b x^6 \coth ^{-1}(\tanh (a+b x))^2+\frac{1}{5} x^5 \coth ^{-1}(\tanh (a+b x))^3+\frac{1}{5} b^2 \int x^6 \coth ^{-1}(\tanh (a+b x)) \, dx\\ &=\frac{1}{35} b^2 x^7 \coth ^{-1}(\tanh (a+b x))-\frac{1}{10} b x^6 \coth ^{-1}(\tanh (a+b x))^2+\frac{1}{5} x^5 \coth ^{-1}(\tanh (a+b x))^3-\frac{1}{35} b^3 \int x^7 \, dx\\ &=-\frac{1}{280} b^3 x^8+\frac{1}{35} b^2 x^7 \coth ^{-1}(\tanh (a+b x))-\frac{1}{10} b x^6 \coth ^{-1}(\tanh (a+b x))^2+\frac{1}{5} x^5 \coth ^{-1}(\tanh (a+b x))^3\\ \end{align*}

Mathematica [A]  time = 0.030511, size = 54, normalized size = 0.89 \[ -\frac{1}{280} x^5 \left (-8 b^2 x^2 \coth ^{-1}(\tanh (a+b x))+28 b x \coth ^{-1}(\tanh (a+b x))^2-56 \coth ^{-1}(\tanh (a+b x))^3+b^3 x^3\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*ArcCoth[Tanh[a + b*x]]^3,x]

[Out]

-(x^5*(b^3*x^3 - 8*b^2*x^2*ArcCoth[Tanh[a + b*x]] + 28*b*x*ArcCoth[Tanh[a + b*x]]^2 - 56*ArcCoth[Tanh[a + b*x]
]^3))/280

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Maple [C]  time = 1.102, size = 18111, normalized size = 296.9 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*arccoth(tanh(b*x+a))^3,x)

[Out]

result too large to display

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Maxima [A]  time = 1.54533, size = 73, normalized size = 1.2 \begin{align*} -\frac{1}{10} \, b x^{6} \operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )^{2} + \frac{1}{5} \, x^{5} \operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )^{3} - \frac{1}{280} \,{\left (b^{2} x^{8} - 8 \, b x^{7} \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^4*arccoth(tanh(b*x+a))^3,x, algorithm="maxima")

[Out]

-1/10*b*x^6*arccoth(tanh(b*x + a))^2 + 1/5*x^5*arccoth(tanh(b*x + a))^3 - 1/280*(b^2*x^8 - 8*b*x^7*arccoth(tan
h(b*x + a)))*b

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Fricas [A]  time = 1.59236, size = 119, normalized size = 1.95 \begin{align*} \frac{1}{8} \, b^{3} x^{8} + \frac{3}{7} \, a b^{2} x^{7} - \frac{1}{8} \,{\left (\pi ^{2} b - 4 \, a^{2} b\right )} x^{6} - \frac{1}{20} \,{\left (3 \, \pi ^{2} a - 4 \, a^{3}\right )} x^{5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arccoth(tanh(b*x+a))^3,x, algorithm="fricas")

[Out]

1/8*b^3*x^8 + 3/7*a*b^2*x^7 - 1/8*(pi^2*b - 4*a^2*b)*x^6 - 1/20*(3*pi^2*a - 4*a^3)*x^5

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Sympy [A]  time = 6.05556, size = 56, normalized size = 0.92 \begin{align*} - \frac{b^{3} x^{8}}{280} + \frac{b^{2} x^{7} \operatorname{acoth}{\left (\tanh{\left (a + b x \right )} \right )}}{35} - \frac{b x^{6} \operatorname{acoth}^{2}{\left (\tanh{\left (a + b x \right )} \right )}}{10} + \frac{x^{5} \operatorname{acoth}^{3}{\left (\tanh{\left (a + b x \right )} \right )}}{5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*acoth(tanh(b*x+a))**3,x)

[Out]

-b**3*x**8/280 + b**2*x**7*acoth(tanh(a + b*x))/35 - b*x**6*acoth(tanh(a + b*x))**2/10 + x**5*acoth(tanh(a + b
*x))**3/5

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{4} \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^4*arccoth(tanh(b*x+a))^3,x, algorithm="giac")

[Out]

integrate(x^4*arccoth(tanh(b*x + a))^3, x)