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

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

[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))

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Rubi [A]  time = 0.0796348, antiderivative size = 121, normalized size of antiderivative = 1., 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 = {2168, 2157, 30} \[ -\frac{3 x^2 \coth ^{-1}(\tanh (a+b x))^{n+2}}{b^2 (n+1) (n+2)}+\frac{6 x \coth ^{-1}(\tanh (a+b x))^{n+3}}{b^3 (n+1) (n+2) (n+3)}-\frac{6 \coth ^{-1}(\tanh (a+b x))^{n+4}}{b^4 (n+1) (n+2) (n+3) (n+4)}+\frac{x^3 \coth ^{-1}(\tanh (a+b x))^{n+1}}{b (n+1)} \]

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 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^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 \operatorname{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*}

Mathematica [A]  time = 0.0773392, size = 106, normalized size = 0.88 \[ \frac{\coth ^{-1}(\tanh (a+b x))^{n+1} \left (-3 b^2 \left (n^2+7 n+12\right ) x^2 \coth ^{-1}(\tanh (a+b x))+6 b (n+4) x \coth ^{-1}(\tanh (a+b x))^2-6 \coth ^{-1}(\tanh (a+b x))^3+b^3 \left (n^3+9 n^2+26 n+24\right ) x^3\right )}{b^4 (n+1) (n+2) (n+3) (n+4)} \]

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 [B]  time = 36.951, size = 953037, normalized size = 7876.3 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [C]  time = 1.79852, size = 344, normalized size = 2.84 \begin{align*} \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} +{\left (-4 i \, \pi{\left (n^{3} + 3 \, n^{2} + 2 \, n\right )} b^{3} + 8 \,{\left (n^{3} + 3 \, n^{2} + 2 \, n\right )} a b^{3}\right )} x^{3} +{\left (6 \, \pi ^{2}{\left (n^{2} + n\right )} b^{2} + 24 i \, \pi{\left (n^{2} + n\right )} a b^{2} - 24 \,{\left (n^{2} + n\right )} a^{2} b^{2}\right )} x^{2} +{\left (6 i \, \pi ^{3} b n - 36 \, \pi ^{2} a b n - 72 i \, \pi a^{2} b n + 48 \, 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{align*}

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 + 8*(n^3 + 3*n^2 + 2*n)*a*b^3)*x^3 + (6*pi^2*(n^2 + n)*b^2 + 24*I*pi*(n^2 + n)*a*b^2 - 24
*(n^2 + n)*a^2*b^2)*x^2 + (6*I*pi^3*b*n - 36*pi^2*a*b*n - 72*I*pi*a^2*b*n + 48*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]  time = 1.76948, size = 917, normalized size = 7.58 \begin{align*} \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{align*}

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(-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**3*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^{3} \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^3*arccoth(tanh(b*x+a))^n,x, algorithm="giac")

[Out]

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