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

Optimal. Leaf size=48 \[ \frac{x \coth ^{-1}(\tanh (a+b x))^{n+1}}{b (n+1)}-\frac{\coth ^{-1}(\tanh (a+b x))^{n+2}}{b^2 (n+1) (n+2)} \]

[Out]

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

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Rubi [A]  time = 0.0200901, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {2168, 2157, 30} \[ \frac{x \coth ^{-1}(\tanh (a+b x))^{n+1}}{b (n+1)}-\frac{\coth ^{-1}(\tanh (a+b x))^{n+2}}{b^2 (n+1) (n+2)} \]

Antiderivative was successfully verified.

[In]

Int[x*ArcCoth[Tanh[a + b*x]]^n,x]

[Out]

(x*ArcCoth[Tanh[a + b*x]]^(1 + n))/(b*(1 + n)) - ArcCoth[Tanh[a + b*x]]^(2 + n)/(b^2*(1 + n)*(2 + 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 \coth ^{-1}(\tanh (a+b x))^n \, dx &=\frac{x \coth ^{-1}(\tanh (a+b x))^{1+n}}{b (1+n)}-\frac{\int \coth ^{-1}(\tanh (a+b x))^{1+n} \, dx}{b (1+n)}\\ &=\frac{x \coth ^{-1}(\tanh (a+b x))^{1+n}}{b (1+n)}-\frac{\operatorname{Subst}\left (\int x^{1+n} \, dx,x,\coth ^{-1}(\tanh (a+b x))\right )}{b^2 (1+n)}\\ &=\frac{x \coth ^{-1}(\tanh (a+b x))^{1+n}}{b (1+n)}-\frac{\coth ^{-1}(\tanh (a+b x))^{2+n}}{b^2 (1+n) (2+n)}\\ \end{align*}

Mathematica [A]  time = 0.0428894, size = 41, normalized size = 0.85 \[ \frac{\left (b (n+2) x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))^{n+1}}{b^2 (n+1) (n+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 15.611, size = 71611, normalized size = 1491.9 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [C]  time = 1.79195, size = 138, normalized size = 2.88 \begin{align*} \frac{{\left (4 \, b^{2}{\left (n + 1\right )} x^{2} + \pi ^{2} + 4 i \, \pi a - 4 \, a^{2} - 2 \,{\left (i \, \pi b n - 2 \, a 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^{2} + 3 \cdot 2^{n + 2} n + 2^{n + 3}\right )} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*b^2*(n + 1)*x^2 + pi^2 + 4*I*pi*a - 4*a^2 - 2*(I*pi*b*n - 2*a*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^2 + 3*2^(n + 2)*n + 2^(n + 3))*b^2)

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Fricas [B]  time = 1.60396, size = 489, normalized size = 10.19 \begin{align*} \frac{{\left (4 \, a b n x + 4 \,{\left (b^{2} n + b^{2}\right )} x^{2} + \pi ^{2} - 4 \, a^{2}\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 (\pi b n x - 2 \, \pi a\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^{2} n^{2} + 3 \, b^{2} n + 2 \, b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((4*a*b*n*x + 4*(b^2*n + b^2)*x^2 + pi^2 - 4*a^2)*(b^2*x^2 + 2*a*b*x + 1/4*pi^2 + a^2)^(1/2*n)*cos(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)) - 2*(pi*b*n*x - 2*pi*a)*(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^2*n^2 + 3*b^2*n + 2*b^2)

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

[Out]

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