3.2.86 \(\int x^3 \coth ^{-1}(\tanh (a+b x))^n \, dx\) [186]

Optimal. Leaf size=121 \[ \frac {x^3 \coth ^{-1}(\tanh (a+b x))^{1+n}}{b (1+n)}-\frac {3 x^2 \coth ^{-1}(\tanh (a+b x))^{2+n}}{b^2 (1+n) (2+n)}+\frac {6 x \coth ^{-1}(\tanh (a+b x))^{3+n}}{b^3 (1+n) (2+n) (3+n)}-\frac {6 \coth ^{-1}(\tanh (a+b x))^{4+n}}{b^4 (1+n) (2+n) (3+n) (4+n)} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2199, 2188, 30} \begin {gather*} -\frac {6 \coth ^{-1}(\tanh (a+b x))^{n+4}}{b^4 (n+1) (n+2) (n+3) (n+4)}+\frac {6 x \coth ^{-1}(\tanh (a+b x))^{n+3}}{b^3 (n+1) (n+2) (n+3)}-\frac {3 x^2 \coth ^{-1}(\tanh (a+b x))^{n+2}}{b^2 (n+1) (n+2)}+\frac {x^3 \coth ^{-1}(\tanh (a+b x))^{n+1}}{b (n+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*ArcCoth[Tanh[a + b*x]]^n,x]

[Out]

(x^3*ArcCoth[Tanh[a + b*x]]^(1 + n))/(b*(1 + n)) - (3*x^2*ArcCoth[Tanh[a + b*x]]^(2 + n))/(b^2*(1 + n)*(2 + n)
) + (6*x*ArcCoth[Tanh[a + b*x]]^(3 + n))/(b^3*(1 + n)*(2 + n)*(3 + n)) - (6*ArcCoth[Tanh[a + b*x]]^(4 + n))/(b
^4*(1 + n)*(2 + n)*(3 + n)*(4 + n))

Rule 30

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

Rule 2188

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 2199

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

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Mathematica [A]
time = 0.06, size = 106, normalized size = 0.88 \begin {gather*} \frac {\coth ^{-1}(\tanh (a+b x))^{1+n} \left (b^3 \left (24+26 n+9 n^2+n^3\right ) x^3-3 b^2 \left (12+7 n+n^2\right ) x^2 \coth ^{-1}(\tanh (a+b x))+6 b (4+n) x \coth ^{-1}(\tanh (a+b x))^2-6 \coth ^{-1}(\tanh (a+b x))^3\right )}{b^4 (1+n) (2+n) (3+n) (4+n)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(ArcCoth[Tanh[a + b*x]]^(1 + n)*(b^3*(24 + 26*n + 9*n^2 + n^3)*x^3 - 3*b^2*(12 + 7*n + n^2)*x^2*ArcCoth[Tanh[a
+ b*x]] + 6*b*(4 + n)*x*ArcCoth[Tanh[a + b*x]]^2 - 6*ArcCoth[Tanh[a + b*x]]^3))/(b^4*(1 + n)*(2 + n)*(3 + n)*
(4 + n))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 11.51, size = 129477, normalized size = 1070.06

method	result	size
risch	\(\text {Expression too large to display}\)	\(129477\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arccoth(tanh(b*x+a))^n,x,method=_RETURNVERBOSE)

[Out]

result too large to display

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Maxima [C] Result contains complex when optimal does not.
time = 0.56, size = 257, normalized size = 2.12 \begin {gather*} \frac {{\left (8 \, {\left (n^{3} + 6 \, n^{2} + 11 \, n + 6\right )} b^{4} x^{4} - 3 \, \pi ^{4} - 24 i \, \pi ^{3} a + 72 \, \pi ^{2} a^{2} + 96 i \, \pi a^{3} - 48 \, a^{4} - 4 \, {\left (i \, \pi {\left (n^{3} + 3 \, n^{2} + 2 \, n\right )} b^{3} - 2 \, {\left (n^{3} + 3 \, n^{2} + 2 \, n\right )} a b^{3}\right )} x^{3} + 6 \, {\left (\pi ^{2} {\left (n^{2} + n\right )} b^{2} + 4 i \, \pi {\left (n^{2} + n\right )} a b^{2} - 4 \, {\left (n^{2} + n\right )} a^{2} b^{2}\right )} x^{2} - 6 \, {\left (-i \, \pi ^{3} b n + 6 \, \pi ^{2} a b n + 12 i \, \pi a^{2} b n - 8 \, a^{3} 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 + 3} n^{4} + 5 \cdot 2^{n + 4} n^{3} + 35 \cdot 2^{n + 3} n^{2} + 25 \cdot 2^{n + 4} n + 3 \cdot 2^{n + 6}\right )} b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(8*(n^3 + 6*n^2 + 11*n + 6)*b^4*x^4 - 3*pi^4 - 24*I*pi^3*a + 72*pi^2*a^2 + 96*I*pi*a^3 - 48*a^4 - 4*(I*pi*(n^3
+ 3*n^2 + 2*n)*b^3 - 2*(n^3 + 3*n^2 + 2*n)*a*b^3)*x^3 + 6*(pi^2*(n^2 + n)*b^2 + 4*I*pi*(n^2 + n)*a*b^2 - 4*(n
^2 + n)*a^2*b^2)*x^2 - 6*(-I*pi^3*b*n + 6*pi^2*a*b*n + 12*I*pi*a^2*b*n - 8*a^3*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 + 3)*n^4 + 5*2^(n + 4)*n^3 + 35*2^(n + 3)*n^2 + 25*2^(
n + 4)*n + 3*2^(n + 6))*b^4)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 411 vs. \(2 (121) = 242\).
time = 0.40, size = 411, normalized size = 3.40 \begin {gather*} \frac {{\left (8 \, {\left (b^{4} n^{3} + 6 \, b^{4} n^{2} + 11 \, b^{4} n + 6 \, b^{4}\right )} x^{4} - 3 \, \pi ^{4} + 72 \, \pi ^{2} a^{2} - 48 \, a^{4} + 8 \, {\left (a b^{3} n^{3} + 3 \, a b^{3} n^{2} + 2 \, a b^{3} n\right )} x^{3} - 6 \, {\left (4 \, a^{2} b^{2} n^{2} + 4 \, a^{2} b^{2} n - \pi ^{2} {\left (b^{2} n^{2} + b^{2} n\right )}\right )} x^{2} - 12 \, {\left (3 \, \pi ^{2} a b n - 4 \, 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} \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 ) - 2 \, {\left (2 \, \pi {\left (b^{3} n^{3} + 3 \, b^{3} n^{2} + 2 \, b^{3} n\right )} x^{3} + 12 \, \pi ^{3} a - 48 \, \pi a^{3} - 12 \, \pi {\left (a b^{2} n^{2} + a b^{2} n\right )} x^{2} - 3 \, {\left (\pi ^{3} b n - 12 \, \pi a^{2} 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 )}{8 \, {\left (b^{4} n^{4} + 10 \, b^{4} n^{3} + 35 \, b^{4} n^{2} + 50 \, b^{4} n + 24 \, b^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((8*(b^4*n^3 + 6*b^4*n^2 + 11*b^4*n + 6*b^4)*x^4 - 3*pi^4 + 72*pi^2*a^2 - 48*a^4 + 8*(a*b^3*n^3 + 3*a*b^3*
n^2 + 2*a*b^3*n)*x^3 - 6*(4*a^2*b^2*n^2 + 4*a^2*b^2*n - pi^2*(b^2*n^2 + b^2*n))*x^2 - 12*(3*pi^2*a*b*n - 4*a^3
*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*(2*pi*(b^3*n^3 + 3*b^3*n^2 + 2*b^3*n)*x^3 + 12*pi^3*a - 48*pi*a^3 - 12*pi*(a*b^2
*n^2 + a*b^2*n)*x^2 - 3*(pi^3*b*n - 12*pi*a^2*b*n)*x)*(b^2*x^2 + 2*a*b*x + 1/4*pi^2 + a^2)^(1/2*n)*sin(2*n*arc
tan(-2*b*x/pi - 2*a/pi + sqrt(4*b^2*x^2 + 8*a*b*x + pi^2 + 4*a^2)/pi)))/(b^4*n^4 + 10*b^4*n^3 + 35*b^4*n^2 + 5
0*b^4*n + 24*b^4)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \begin {cases} \frac {x^{4} \operatorname {acoth}^{n}{\left (\tanh {\left (a \right )} \right )}}{4} & \text {for}\: b = 0 \\- \frac {x^{3}}{3 b \operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}} - \frac {x^{2}}{2 b^{2} \operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}} - \frac {x}{b^{3} \operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )}} + \frac {\log {\left (\operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )} \right )}}{b^{4}} & \text {for}\: n = -4 \\\int \frac {x^{3}}{\operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx & \text {for}\: n = -3 \\\int \frac {x^{3}}{\operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx & \text {for}\: n = -2 \\\int \frac {x^{3}}{\operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx & \text {for}\: n = -1 \\\frac {b^{3} n^{3} x^{3} \operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )} \operatorname {acoth}^{n}{\left (\tanh {\left (a + b x \right )} \right )}}{b^{4} n^{4} + 10 b^{4} n^{3} + 35 b^{4} n^{2} + 50 b^{4} n + 24 b^{4}} + \frac {9 b^{3} n^{2} x^{3} \operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )} \operatorname {acoth}^{n}{\left (\tanh {\left (a + b x \right )} \right )}}{b^{4} n^{4} + 10 b^{4} n^{3} + 35 b^{4} n^{2} + 50 b^{4} n + 24 b^{4}} + \frac {26 b^{3} n x^{3} \operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )} \operatorname {acoth}^{n}{\left (\tanh {\left (a + b x \right )} \right )}}{b^{4} n^{4} + 10 b^{4} n^{3} + 35 b^{4} n^{2} + 50 b^{4} n + 24 b^{4}} + \frac {24 b^{3} x^{3} \operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )} \operatorname {acoth}^{n}{\left (\tanh {\left (a + b x \right )} \right )}}{b^{4} n^{4} + 10 b^{4} n^{3} + 35 b^{4} n^{2} + 50 b^{4} n + 24 b^{4}} - \frac {3 b^{2} n^{2} x^{2} \operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )} \operatorname {acoth}^{n}{\left (\tanh {\left (a + b x \right )} \right )}}{b^{4} n^{4} + 10 b^{4} n^{3} + 35 b^{4} n^{2} + 50 b^{4} n + 24 b^{4}} - \frac {21 b^{2} n x^{2} \operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )} \operatorname {acoth}^{n}{\left (\tanh {\left (a + b x \right )} \right )}}{b^{4} n^{4} + 10 b^{4} n^{3} + 35 b^{4} n^{2} + 50 b^{4} n + 24 b^{4}} - \frac {36 b^{2} x^{2} \operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )} \operatorname {acoth}^{n}{\left (\tanh {\left (a + b x \right )} \right )}}{b^{4} n^{4} + 10 b^{4} n^{3} + 35 b^{4} n^{2} + 50 b^{4} n + 24 b^{4}} + \frac {6 b n x \operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )} \operatorname {acoth}^{n}{\left (\tanh {\left (a + b x \right )} \right )}}{b^{4} n^{4} + 10 b^{4} n^{3} + 35 b^{4} n^{2} + 50 b^{4} n + 24 b^{4}} + \frac {24 b x \operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )} \operatorname {acoth}^{n}{\left (\tanh {\left (a + b x \right )} \right )}}{b^{4} n^{4} + 10 b^{4} n^{3} + 35 b^{4} n^{2} + 50 b^{4} n + 24 b^{4}} - \frac {6 \operatorname {acoth}^{4}{\left (\tanh {\left (a + b x \right )} \right )} \operatorname {acoth}^{n}{\left (\tanh {\left (a + b x \right )} \right )}}{b^{4} n^{4} + 10 b^{4} n^{3} + 35 b^{4} n^{2} + 50 b^{4} n + 24 b^{4}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x**4*acoth(tanh(a))**n/4, Eq(b, 0)), (-x**3/(3*b*acoth(tanh(a + b*x))**3) - x**2/(2*b**2*acoth(tanh
(a + b*x))**2) - x/(b**3*acoth(tanh(a + b*x))) + log(acoth(tanh(a + b*x)))/b**4, Eq(n, -4)), (Integral(x**3/ac
oth(tanh(a + b*x))**3, x), Eq(n, -3)), (Integral(x**3/acoth(tanh(a + b*x))**2, x), Eq(n, -2)), (Integral(x**3/
acoth(tanh(a + b*x)), x), Eq(n, -1)), (b**3*n**3*x**3*acoth(tanh(a + b*x))*acoth(tanh(a + b*x))**n/(b**4*n**4
+ 10*b**4*n**3 + 35*b**4*n**2 + 50*b**4*n + 24*b**4) + 9*b**3*n**2*x**3*acoth(tanh(a + b*x))*acoth(tanh(a + b*
x))**n/(b**4*n**4 + 10*b**4*n**3 + 35*b**4*n**2 + 50*b**4*n + 24*b**4) + 26*b**3*n*x**3*acoth(tanh(a + b*x))*a
coth(tanh(a + b*x))**n/(b**4*n**4 + 10*b**4*n**3 + 35*b**4*n**2 + 50*b**4*n + 24*b**4) + 24*b**3*x**3*acoth(ta
nh(a + b*x))*acoth(tanh(a + b*x))**n/(b**4*n**4 + 10*b**4*n**3 + 35*b**4*n**2 + 50*b**4*n + 24*b**4) - 3*b**2*
n**2*x**2*acoth(tanh(a + b*x))**2*acoth(tanh(a + b*x))**n/(b**4*n**4 + 10*b**4*n**3 + 35*b**4*n**2 + 50*b**4*n
+ 24*b**4) - 21*b**2*n*x**2*acoth(tanh(a + b*x))**2*acoth(tanh(a + b*x))**n/(b**4*n**4 + 10*b**4*n**3 + 35*b*
*4*n**2 + 50*b**4*n + 24*b**4) - 36*b**2*x**2*acoth(tanh(a + b*x))**2*acoth(tanh(a + b*x))**n/(b**4*n**4 + 10*
b**4*n**3 + 35*b**4*n**2 + 50*b**4*n + 24*b**4) + 6*b*n*x*acoth(tanh(a + b*x))**3*acoth(tanh(a + b*x))**n/(b**
4*n**4 + 10*b**4*n**3 + 35*b**4*n**2 + 50*b**4*n + 24*b**4) + 24*b*x*acoth(tanh(a + b*x))**3*acoth(tanh(a + b*
x))**n/(b**4*n**4 + 10*b**4*n**3 + 35*b**4*n**2 + 50*b**4*n + 24*b**4) - 6*acoth(tanh(a + b*x))**4*acoth(tanh(
a + b*x))**n/(b**4*n**4 + 10*b**4*n**3 + 35*b**4*n**2 + 50*b**4*n + 24*b**4), True))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.50, size = 418, normalized size = 3.45 \begin {gather*} -{\left (\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}{2}-\frac {\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}{2}\right )}^n\,\left (\frac {3\,{\left (\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^4}{8\,b^4\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )}-\frac {x^4\,\left (n^3+6\,n^2+11\,n+6\right )}{n^4+10\,n^3+35\,n^2+50\,n+24}+\frac {3\,n\,x\,{\left (\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^3}{4\,b^3\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )}+\frac {n\,x^3\,\left (\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )\,\left (n^2+3\,n+2\right )}{2\,b\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )}+\frac {3\,n\,x^2\,\left (n+1\right )\,{\left (\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^2}{4\,b^2\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*acoth(tanh(a + b*x))^n,x)

[Out]

-(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1))/2 - log(-2/(exp(2*a)*exp(2*b*x) - 1))/2)^n*((3*(log(-
2/(exp(2*a)*exp(2*b*x) - 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x)^4)/(8*b^4*(50*n
+ 35*n^2 + 10*n^3 + n^4 + 24)) - (x^4*(11*n + 6*n^2 + n^3 + 6))/(50*n + 35*n^2 + 10*n^3 + n^4 + 24) + (3*n*x*
(log(-2/(exp(2*a)*exp(2*b*x) - 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x)^3)/(4*b^3
*(50*n + 35*n^2 + 10*n^3 + n^4 + 24)) + (n*x^3*(log(-2/(exp(2*a)*exp(2*b*x) - 1)) - log((2*exp(2*a)*exp(2*b*x)
)/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x)*(3*n + n^2 + 2))/(2*b*(50*n + 35*n^2 + 10*n^3 + n^4 + 24)) + (3*n*x^2*(n
+ 1)*(log(-2/(exp(2*a)*exp(2*b*x) - 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x)^2)/
(4*b^2*(50*n + 35*n^2 + 10*n^3 + n^4 + 24)))

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