\(\int \coth ^{-1}(\tanh (a+b x))^n \, dx\) [189]

Optimal result

Integrand size = 9, antiderivative size = 20 \[ \int \coth ^{-1}(\tanh (a+b x))^n \, dx=\frac {\coth ^{-1}(\tanh (a+b x))^{1+n}}{b (1+n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2188, 30} \[ \int \coth ^{-1}(\tanh (a+b x))^n \, dx=\frac {\coth ^{-1}(\tanh (a+b x))^{n+1}}{b (n+1)} \]

[In]

Int[ArcCoth[Tanh[a + b*x]]^n,x]

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \frac {\text {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] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \coth ^{-1}(\tanh (a+b x))^n \, dx=\frac {\coth ^{-1}(\tanh (a+b x))^{1+n}}{b (1+n)} \]

[In]

Integrate[ArcCoth[Tanh[a + b*x]]^n,x]

[Out]

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

Maple [A] (verified)

Time = 9.14 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05

method	result	size
derivativedivides	\(\frac {\operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{1+n}}{b \left (1+n \right )}\)	\(21\)
default	\(\frac {\operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{1+n}}{b \left (1+n \right )}\)	\(21\)
parallelrisch	\(\frac {\operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{n} \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )}{b \left (1+n \right )}\)	\(26\)
risch	\(\frac {\left (\frac {1}{2}\right )^{n} {\left (2 \ln \left ({\mathrm e}^{b x +a}\right )-\frac {i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) {\left (-\operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )+\operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right )\right )}^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right ) \left (-\operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )+\operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )\right ) \left (-\operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )+\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )\right )}{2}-i \pi -i \pi \operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2} \left (\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )-1\right )\right )}^{1+n}}{2 b \left (1+n \right )}\)	\(237\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.90 \[ \int \coth ^{-1}(\tanh (a+b x))^n \, dx=\frac {{\left (i \, \pi + 2 \, b x + 2 \, a\right )} \cosh \left (n \log \left (\frac {1}{2} i \, \pi + b x + a\right )\right ) + {\left (i \, \pi + 2 \, b x + 2 \, a\right )} \sinh \left (n \log \left (\frac {1}{2} i \, \pi + b x + a\right )\right )}{2 \, {\left (b n + b\right )}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (15) = 30\).

Time = 0.27 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.55 \[ \int \coth ^{-1}(\tanh (a+b x))^n \, dx=\begin {cases} \frac {x}{\operatorname {acoth}{\left (\tanh {\left (a \right )} \right )}} & \text {for}\: b = 0 \wedge n = -1 \\x \operatorname {acoth}^{n}{\left (\tanh {\left (a \right )} \right )} & \text {for}\: b = 0 \\\frac {\log {\left (\operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )} \right )}}{b} & \text {for}\: n = -1 \\\frac {\operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )} \operatorname {acoth}^{n}{\left (\tanh {\left (a + b x \right )} \right )}}{b n + b} & \text {otherwise} \end {cases} \]

[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))

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.35 (sec) , antiderivative size = 65, normalized size of antiderivative = 3.25 \[ \int \coth ^{-1}(\tanh (a+b x))^n \, dx=\frac {{\left (-i \, \pi + 2 \, b x + 2 \, a\right )} {\left (\cosh \left (-n \log \left (-i \, \pi + 2 \, b x + 2 \, a\right )\right ) - \sinh \left (-n \log \left (-i \, \pi + 2 \, b x + 2 \, a\right )\right )\right )}}{{\left (2^{n + 1} n + 2^{n + 1}\right )} b} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.35 \[ \int \coth ^{-1}(\tanh (a+b x))^n \, dx=\frac {\left (\frac {1}{2} \, \log \left (-e^{\left (2 \, b x + 2 \, a\right )}\right )\right )^{n + 1}}{b {\left (n + 1\right )}} \]

[In]

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

[Out]

(1/2*log(-e^(2*b*x + 2*a)))^(n + 1)/(b*(n + 1))

Mupad [B] (verification not implemented)

Time = 3.89 (sec) , antiderivative size = 121, normalized size of antiderivative = 6.05 \[ \int \coth ^{-1}(\tanh (a+b x))^n \, dx={\left (\frac {1}{2}\right )}^n\,\left (\frac {x}{n+1}-\frac {\frac {\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}{2}-\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}{2}+b\,x}{b\,\left (n+1\right )}\right )\,{\left (\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\right )}^n \]

[In]

int(acoth(tanh(a + b*x))^n,x)

[Out]

(1/2)^n*(x/(n + 1) - (log(-2/(exp(2*a)*exp(2*b*x) - 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) -
1))/2 + b*x)/(b*(n + 1)))*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) - log(-2/(exp(2*a)*exp(2*b*
x) - 1)))^n