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

Optimal. Leaf size=165 \[ -\frac{4 x^3 \coth ^{-1}(\tanh (a+b x))^{n+2}}{b^2 (n+1) (n+2)}+\frac{12 x^2 \coth ^{-1}(\tanh (a+b x))^{n+3}}{b^3 (n+1) (n+2) (n+3)}-\frac{24 x \coth ^{-1}(\tanh (a+b x))^{n+4}}{b^4 (n+1) (n+2) (n+3) (n+4)}+\frac{24 \coth ^{-1}(\tanh (a+b x))^{n+5}}{b^5 (n+1) (n+2) (n+3) (n+4) (n+5)}+\frac{x^4 \coth ^{-1}(\tanh (a+b x))^{n+1}}{b (n+1)} \]

[Out]

(x^4*ArcCoth[Tanh[a + b*x]]^(1 + n))/(b*(1 + n)) - (4*x^3*ArcCoth[Tanh[a + b*x]]^(2 + n))/(b^2*(1 + n)*(2 + n)
) + (12*x^2*ArcCoth[Tanh[a + b*x]]^(3 + n))/(b^3*(1 + n)*(2 + n)*(3 + n)) - (24*x*ArcCoth[Tanh[a + b*x]]^(4 +
n))/(b^4*(1 + n)*(2 + n)*(3 + n)*(4 + n)) + (24*ArcCoth[Tanh[a + b*x]]^(5 + n))/(b^5*(1 + n)*(2 + n)*(3 + n)*(
4 + n)*(5 + n))

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Rubi [A]  time = 0.133038, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {2168, 2157, 30} \[ -\frac{4 x^3 \coth ^{-1}(\tanh (a+b x))^{n+2}}{b^2 (n+1) (n+2)}+\frac{12 x^2 \coth ^{-1}(\tanh (a+b x))^{n+3}}{b^3 (n+1) (n+2) (n+3)}-\frac{24 x \coth ^{-1}(\tanh (a+b x))^{n+4}}{b^4 (n+1) (n+2) (n+3) (n+4)}+\frac{24 \coth ^{-1}(\tanh (a+b x))^{n+5}}{b^5 (n+1) (n+2) (n+3) (n+4) (n+5)}+\frac{x^4 \coth ^{-1}(\tanh (a+b x))^{n+1}}{b (n+1)} \]

Antiderivative was successfully verified.

[In]

Int[x^4*ArcCoth[Tanh[a + b*x]]^n,x]

[Out]

(x^4*ArcCoth[Tanh[a + b*x]]^(1 + n))/(b*(1 + n)) - (4*x^3*ArcCoth[Tanh[a + b*x]]^(2 + n))/(b^2*(1 + n)*(2 + n)
) + (12*x^2*ArcCoth[Tanh[a + b*x]]^(3 + n))/(b^3*(1 + n)*(2 + n)*(3 + n)) - (24*x*ArcCoth[Tanh[a + b*x]]^(4 +
n))/(b^4*(1 + n)*(2 + n)*(3 + n)*(4 + n)) + (24*ArcCoth[Tanh[a + b*x]]^(5 + n))/(b^5*(1 + n)*(2 + n)*(3 + n)*(
4 + n)*(5 + n))

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^4 \coth ^{-1}(\tanh (a+b x))^n \, dx &=\frac{x^4 \coth ^{-1}(\tanh (a+b x))^{1+n}}{b (1+n)}-\frac{4 \int x^3 \coth ^{-1}(\tanh (a+b x))^{1+n} \, dx}{b (1+n)}\\ &=\frac{x^4 \coth ^{-1}(\tanh (a+b x))^{1+n}}{b (1+n)}-\frac{4 x^3 \coth ^{-1}(\tanh (a+b x))^{2+n}}{b^2 (1+n) (2+n)}+\frac{12 \int x^2 \coth ^{-1}(\tanh (a+b x))^{2+n} \, dx}{b^2 (1+n) (2+n)}\\ &=\frac{x^4 \coth ^{-1}(\tanh (a+b x))^{1+n}}{b (1+n)}-\frac{4 x^3 \coth ^{-1}(\tanh (a+b x))^{2+n}}{b^2 (1+n) (2+n)}+\frac{12 x^2 \coth ^{-1}(\tanh (a+b x))^{3+n}}{b^3 (1+n) (2+n) (3+n)}-\frac{24 \int x \coth ^{-1}(\tanh (a+b x))^{3+n} \, dx}{b^3 (1+n) (2+n) (3+n)}\\ &=\frac{x^4 \coth ^{-1}(\tanh (a+b x))^{1+n}}{b (1+n)}-\frac{4 x^3 \coth ^{-1}(\tanh (a+b x))^{2+n}}{b^2 (1+n) (2+n)}+\frac{12 x^2 \coth ^{-1}(\tanh (a+b x))^{3+n}}{b^3 (1+n) (2+n) (3+n)}-\frac{24 x \coth ^{-1}(\tanh (a+b x))^{4+n}}{b^4 (1+n) (2+n) (3+n) (4+n)}+\frac{24 \int \coth ^{-1}(\tanh (a+b x))^{4+n} \, dx}{b^4 (1+n) (2+n) (3+n) (4+n)}\\ &=\frac{x^4 \coth ^{-1}(\tanh (a+b x))^{1+n}}{b (1+n)}-\frac{4 x^3 \coth ^{-1}(\tanh (a+b x))^{2+n}}{b^2 (1+n) (2+n)}+\frac{12 x^2 \coth ^{-1}(\tanh (a+b x))^{3+n}}{b^3 (1+n) (2+n) (3+n)}-\frac{24 x \coth ^{-1}(\tanh (a+b x))^{4+n}}{b^4 (1+n) (2+n) (3+n) (4+n)}+\frac{24 \operatorname{Subst}\left (\int x^{4+n} \, dx,x,\coth ^{-1}(\tanh (a+b x))\right )}{b^5 (1+n) (2+n) (3+n) (4+n)}\\ &=\frac{x^4 \coth ^{-1}(\tanh (a+b x))^{1+n}}{b (1+n)}-\frac{4 x^3 \coth ^{-1}(\tanh (a+b x))^{2+n}}{b^2 (1+n) (2+n)}+\frac{12 x^2 \coth ^{-1}(\tanh (a+b x))^{3+n}}{b^3 (1+n) (2+n) (3+n)}-\frac{24 x \coth ^{-1}(\tanh (a+b x))^{4+n}}{b^4 (1+n) (2+n) (3+n) (4+n)}+\frac{24 \coth ^{-1}(\tanh (a+b x))^{5+n}}{b^5 (1+n) (2+n) (3+n) (4+n) (5+n)}\\ \end{align*}

Mathematica [A]  time = 0.103393, size = 146, normalized size = 0.88 \[ \frac{\coth ^{-1}(\tanh (a+b x))^{n+1} \left (-4 b^3 \left (n^3+12 n^2+47 n+60\right ) x^3 \coth ^{-1}(\tanh (a+b x))+12 b^2 \left (n^2+9 n+20\right ) x^2 \coth ^{-1}(\tanh (a+b x))^2-24 b (n+5) x \coth ^{-1}(\tanh (a+b x))^3+24 \coth ^{-1}(\tanh (a+b x))^4+b^4 \left (n^4+14 n^3+71 n^2+154 n+120\right ) x^4\right )}{b^5 (n+1) (n+2) (n+3) (n+4) (n+5)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(ArcCoth[Tanh[a + b*x]]^(1 + n)*(b^4*(120 + 154*n + 71*n^2 + 14*n^3 + n^4)*x^4 - 4*b^3*(60 + 47*n + 12*n^2 + n
^3)*x^3*ArcCoth[Tanh[a + b*x]] + 12*b^2*(20 + 9*n + n^2)*x^2*ArcCoth[Tanh[a + b*x]]^2 - 24*b*(5 + n)*x*ArcCoth
[Tanh[a + b*x]]^3 + 24*ArcCoth[Tanh[a + b*x]]^4))/(b^5*(1 + n)*(2 + n)*(3 + n)*(4 + n)*(5 + n))

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Maple [B]  time = 87.445, size = 2984280, normalized size = 18086.6 \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))^n,x)

[Out]

result too large to display

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Maxima [C]  time = 1.81928, size = 513, normalized size = 3.11 \begin{align*} \frac{{\left (4 \,{\left (n^{4} + 10 \, n^{3} + 35 \, n^{2} + 50 \, n + 24\right )} b^{5} x^{5} - 3 i \, \pi ^{5} + 30 \, \pi ^{4} a + 120 i \, \pi ^{3} a^{2} - 240 \, \pi ^{2} a^{3} - 240 i \, \pi a^{4} + 96 \, a^{5} - 2 \,{\left (i \, \pi{\left (n^{4} + 6 \, n^{3} + 11 \, n^{2} + 6 \, n\right )} b^{4} - 2 \,{\left (n^{4} + 6 \, n^{3} + 11 \, n^{2} + 6 \, n\right )} a b^{4}\right )} x^{4} +{\left (4 \, \pi ^{2}{\left (n^{3} + 3 \, n^{2} + 2 \, n\right )} b^{3} + 16 i \, \pi{\left (n^{3} + 3 \, n^{2} + 2 \, n\right )} a b^{3} - 16 \,{\left (n^{3} + 3 \, n^{2} + 2 \, n\right )} a^{2} b^{3}\right )} x^{3} +{\left (6 i \, \pi ^{3}{\left (n^{2} + n\right )} b^{2} - 36 \, \pi ^{2}{\left (n^{2} + n\right )} a b^{2} - 72 i \, \pi{\left (n^{2} + n\right )} a^{2} b^{2} + 48 \,{\left (n^{2} + n\right )} a^{3} b^{2}\right )} x^{2} -{\left (6 \, \pi ^{4} b n + 48 i \, \pi ^{3} a b n - 144 \, \pi ^{2} a^{2} b n - 192 i \, \pi a^{3} b n + 96 \, a^{4} b n\right )} x\right )}{\left (\cosh \left (-n \log \left (-i \, \pi + 2 \, b x + 2 \, a\right )\right ) - \sinh \left (-n \log \left (-i \, \pi + 2 \, b x + 2 \, a\right )\right )\right )}}{{\left (2^{n + 2} n^{5} + 15 \cdot 2^{n + 2} n^{4} + 85 \cdot 2^{n + 2} n^{3} + 225 \cdot 2^{n + 2} n^{2} + 137 \cdot 2^{n + 3} n + 15 \cdot 2^{n + 5}\right )} b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*(n^4 + 10*n^3 + 35*n^2 + 50*n + 24)*b^5*x^5 - 3*I*pi^5 + 30*pi^4*a + 120*I*pi^3*a^2 - 240*pi^2*a^3 - 240*I*
pi*a^4 + 96*a^5 - 2*(I*pi*(n^4 + 6*n^3 + 11*n^2 + 6*n)*b^4 - 2*(n^4 + 6*n^3 + 11*n^2 + 6*n)*a*b^4)*x^4 + (4*pi
^2*(n^3 + 3*n^2 + 2*n)*b^3 + 16*I*pi*(n^3 + 3*n^2 + 2*n)*a*b^3 - 16*(n^3 + 3*n^2 + 2*n)*a^2*b^3)*x^3 + (6*I*pi
^3*(n^2 + n)*b^2 - 36*pi^2*(n^2 + n)*a*b^2 - 72*I*pi*(n^2 + n)*a^2*b^2 + 48*(n^2 + n)*a^3*b^2)*x^2 - (6*pi^4*b
*n + 48*I*pi^3*a*b*n - 144*pi^2*a^2*b*n - 192*I*pi*a^3*b*n + 96*a^4*b*n)*x)*(cosh(-n*log(-I*pi + 2*b*x + 2*a))
- sinh(-n*log(-I*pi + 2*b*x + 2*a)))/((2^(n + 2)*n^5 + 15*2^(n + 2)*n^4 + 85*2^(n + 2)*n^3 + 225*2^(n + 2)*n^
2 + 137*2^(n + 3)*n + 15*2^(n + 5))*b^5)

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Fricas [B]  time = 1.72095, size = 1288, normalized size = 7.81 \begin{align*} \frac{2 \,{\left (2 \,{\left (b^{5} n^{4} + 10 \, b^{5} n^{3} + 35 \, b^{5} n^{2} + 50 \, b^{5} n + 24 \, b^{5}\right )} x^{5} + 15 \, \pi ^{4} a - 120 \, \pi ^{2} a^{3} + 48 \, a^{5} + 2 \,{\left (a b^{4} n^{4} + 6 \, a b^{4} n^{3} + 11 \, a b^{4} n^{2} + 6 \, a b^{4} n\right )} x^{4} - 2 \,{\left (4 \, a^{2} b^{3} n^{3} + 12 \, a^{2} b^{3} n^{2} + 8 \, a^{2} b^{3} n - \pi ^{2}{\left (b^{3} n^{3} + 3 \, b^{3} n^{2} + 2 \, b^{3} n\right )}\right )} x^{3} + 6 \,{\left (4 \, a^{3} b^{2} n^{2} + 4 \, a^{3} b^{2} n - 3 \, \pi ^{2}{\left (a b^{2} n^{2} + a b^{2} n\right )}\right )} x^{2} - 3 \,{\left (\pi ^{4} b n - 24 \, \pi ^{2} a^{2} b n + 16 \, a^{4} b n\right )} x\right )}{\left (b^{2} x^{2} + 2 \, a b x + \frac{1}{4} \, \pi ^{2} + a^{2}\right )}^{\frac{1}{2} \, n} \cos \left (2 \, n \arctan \left (-\frac{2 \, b x}{\pi } - \frac{2 \, a}{\pi } + \frac{\sqrt{4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}}}{\pi }\right )\right ) -{\left (2 \, \pi{\left (b^{4} n^{4} + 6 \, b^{4} n^{3} + 11 \, b^{4} n^{2} + 6 \, b^{4} n\right )} x^{4} + 3 \, \pi ^{5} - 120 \, \pi ^{3} a^{2} + 240 \, \pi a^{4} - 16 \, \pi{\left (a b^{3} n^{3} + 3 \, a b^{3} n^{2} + 2 \, a b^{3} n\right )} x^{3} - 6 \,{\left (\pi ^{3}{\left (b^{2} n^{2} + b^{2} n\right )} - 12 \, \pi{\left (a^{2} b^{2} n^{2} + a^{2} b^{2} n\right )}\right )} x^{2} + 48 \,{\left (\pi ^{3} a b n - 4 \, \pi a^{3} b n\right )} x\right )}{\left (b^{2} x^{2} + 2 \, a b x + \frac{1}{4} \, \pi ^{2} + a^{2}\right )}^{\frac{1}{2} \, n} \sin \left (2 \, n \arctan \left (-\frac{2 \, b x}{\pi } - \frac{2 \, a}{\pi } + \frac{\sqrt{4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}}}{\pi }\right )\right )}{4 \,{\left (b^{5} n^{5} + 15 \, b^{5} n^{4} + 85 \, b^{5} n^{3} + 225 \, b^{5} n^{2} + 274 \, b^{5} n + 120 \, b^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*(2*(b^5*n^4 + 10*b^5*n^3 + 35*b^5*n^2 + 50*b^5*n + 24*b^5)*x^5 + 15*pi^4*a - 120*pi^2*a^3 + 48*a^5 + 2*
(a*b^4*n^4 + 6*a*b^4*n^3 + 11*a*b^4*n^2 + 6*a*b^4*n)*x^4 - 2*(4*a^2*b^3*n^3 + 12*a^2*b^3*n^2 + 8*a^2*b^3*n - p
i^2*(b^3*n^3 + 3*b^3*n^2 + 2*b^3*n))*x^3 + 6*(4*a^3*b^2*n^2 + 4*a^3*b^2*n - 3*pi^2*(a*b^2*n^2 + a*b^2*n))*x^2
- 3*(pi^4*b*n - 24*pi^2*a^2*b*n + 16*a^4*b*n)*x)*(b^2*x^2 + 2*a*b*x + 1/4*pi^2 + a^2)^(1/2*n)*cos(2*n*arctan(-
2*b*x/pi - 2*a/pi + sqrt(4*b^2*x^2 + 8*a*b*x + pi^2 + 4*a^2)/pi)) - (2*pi*(b^4*n^4 + 6*b^4*n^3 + 11*b^4*n^2 +
6*b^4*n)*x^4 + 3*pi^5 - 120*pi^3*a^2 + 240*pi*a^4 - 16*pi*(a*b^3*n^3 + 3*a*b^3*n^2 + 2*a*b^3*n)*x^3 - 6*(pi^3*
(b^2*n^2 + b^2*n) - 12*pi*(a^2*b^2*n^2 + a^2*b^2*n))*x^2 + 48*(pi^3*a*b*n - 4*pi*a^3*b*n)*x)*(b^2*x^2 + 2*a*b*
x + 1/4*pi^2 + a^2)^(1/2*n)*sin(2*n*arctan(-2*b*x/pi - 2*a/pi + sqrt(4*b^2*x^2 + 8*a*b*x + pi^2 + 4*a^2)/pi)))
/(b^5*n^5 + 15*b^5*n^4 + 85*b^5*n^3 + 225*b^5*n^2 + 274*b^5*n + 120*b^5)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError

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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 )^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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