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

Optimal. Leaf size=20 \[ \frac{\coth ^{-1}(\tanh (a+b x))^{n+1}}{b (n+1)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0065514, antiderivative size = 20, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 9, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2157, 30} \[ \frac{\coth ^{-1}(\tanh (a+b x))^{n+1}}{b (n+1)} \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[Tanh[a + b*x]]^n,x]

[Out]

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

Mathematica [A]  time = 0.0142894, size = 20, normalized size = 1. \[ \frac{\coth ^{-1}(\tanh (a+b x))^{n+1}}{b (n+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[Tanh[a + b*x]]^n,x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.087, size = 21, normalized size = 1.1 \begin{align*}{\frac{ \left ({\rm arccoth} \left (\tanh \left ( bx+a \right ) \right ) \right ) ^{1+n}}{b \left ( 1+n \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arccoth(tanh(b*x+a))^(1+n)/b/(1+n)

________________________________________________________________________________________

Maxima [C]  time = 1.73917, size = 88, normalized size = 4.4 \begin{align*} \frac{{\left (-i \, \pi + 2 \, b x + 2 \, a\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 + 1} n + 2^{n + 1}\right )} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-I*pi + 2*b*x + 2*a)*(cosh(-n*log(-I*pi + 2*b*x + 2*a)) - sinh(-n*log(-I*pi + 2*b*x + 2*a)))/((2^(n + 1)*n +
2^(n + 1))*b)

________________________________________________________________________________________

Fricas [B]  time = 1.7029, size = 389, normalized size = 19.45 \begin{align*} \frac{2 \,{\left (b x + a\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 ) - \pi{\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 )}{2 \,{\left (b n + b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(b*x + a)*(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)) - pi*(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*n + b)

________________________________________________________________________________________

Sympy [A]  time = 1.04688, size = 51, normalized size = 2.55 \begin{align*} \begin{cases} \frac{x}{\operatorname{acoth}{\left (\tanh{\left (a \right )} \right )}} & \text{for}\: b = 0 \wedge n = -1 \\x \operatorname{acoth}^{n}{\left (\tanh{\left (a \right )} \right )} & \text{for}\: b = 0 \\\frac{\log{\left (\operatorname{acoth}{\left (\tanh{\left (a + b x \right )} \right )} \right )}}{b} & \text{for}\: n = -1 \\\frac{\operatorname{acoth}{\left (\tanh{\left (a + b x \right )} \right )} \operatorname{acoth}^{n}{\left (\tanh{\left (a + b x \right )} \right )}}{b n + b} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x/acoth(tanh(a)), Eq(b, 0) & Eq(n, -1)), (x*acoth(tanh(a))**n, Eq(b, 0)), (log(acoth(tanh(a + b*x))
)/b, Eq(n, -1)), (acoth(tanh(a + b*x))*acoth(tanh(a + b*x))**n/(b*n + b), True))

________________________________________________________________________________________

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

[Out]

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