\(\int \frac {\coth ^{-1}(x)^n}{1-x^2} \, dx\) [54]

Optimal result

Integrand size = 14, antiderivative size = 12 \[ \int \frac {\coth ^{-1}(x)^n}{1-x^2} \, dx=\frac {\coth ^{-1}(x)^{1+n}}{1+n} \]

[Out]

arccoth(x)^(1+n)/(1+n)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {6096} \[ \int \frac {\coth ^{-1}(x)^n}{1-x^2} \, dx=\frac {\coth ^{-1}(x)^{n+1}}{n+1} \]

[In]

Int[ArcCoth[x]^n/(1 - x^2),x]

[Out]

ArcCoth[x]^(1 + n)/(1 + n)

Rule 6096

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcCoth[c*x])^(p
+ 1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*d + e, 0] && NeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {\coth ^{-1}(x)^{1+n}}{1+n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {\coth ^{-1}(x)^n}{1-x^2} \, dx=\frac {\coth ^{-1}(x)^{1+n}}{1+n} \]

[In]

Integrate[ArcCoth[x]^n/(1 - x^2),x]

[Out]

ArcCoth[x]^(1 + n)/(1 + n)

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.08

method	result	size
default	\(\frac {\operatorname {arccoth}\left (x \right )^{1+n}}{1+n}\)	\(13\)

[In]

int(arccoth(x)^n/(-x^2+1),x,method=_RETURNVERBOSE)

[Out]

arccoth(x)^(1+n)/(1+n)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 62 vs. \(2 (12) = 24\).

Time = 0.25 (sec) , antiderivative size = 62, normalized size of antiderivative = 5.17 \[ \int \frac {\coth ^{-1}(x)^n}{1-x^2} \, dx=\frac {\cosh \left (n \log \left (\frac {1}{2} \, \log \left (\frac {x + 1}{x - 1}\right )\right )\right ) \log \left (\frac {x + 1}{x - 1}\right ) + \log \left (\frac {x + 1}{x - 1}\right ) \sinh \left (n \log \left (\frac {1}{2} \, \log \left (\frac {x + 1}{x - 1}\right )\right )\right )}{2 \, {\left (n + 1\right )}} \]

[In]

integrate(arccoth(x)^n/(-x^2+1),x, algorithm="fricas")

[Out]

1/2*(cosh(n*log(1/2*log((x + 1)/(x - 1))))*log((x + 1)/(x - 1)) + log((x + 1)/(x - 1))*sinh(n*log(1/2*log((x +
1)/(x - 1)))))/(n + 1)

Sympy [A] (verification not implemented)

Time = 0.81 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.25 \[ \int \frac {\coth ^{-1}(x)^n}{1-x^2} \, dx=\begin {cases} \frac {\operatorname {acoth}^{n + 1}{\left (x \right )}}{n + 1} & \text {for}\: n \neq -1 \\\log {\left (\operatorname {acoth}{\left (x \right )} \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(acoth(x)**n/(-x**2+1),x)

[Out]

Piecewise((acoth(x)**(n + 1)/(n + 1), Ne(n, -1)), (log(acoth(x)), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {\coth ^{-1}(x)^n}{1-x^2} \, dx=\frac {\operatorname {arcoth}\left (x\right )^{n + 1}}{n + 1} \]

[In]

integrate(arccoth(x)^n/(-x^2+1),x, algorithm="maxima")

[Out]

arccoth(x)^(n + 1)/(n + 1)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.83 \[ \int \frac {\coth ^{-1}(x)^n}{1-x^2} \, dx=\frac {\left (\frac {1}{2} \, \log \left (\frac {x + 1}{x - 1}\right )\right )^{n + 1}}{n + 1} \]

[In]

integrate(arccoth(x)^n/(-x^2+1),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.34 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.83 \[ \int \frac {\coth ^{-1}(x)^n}{1-x^2} \, dx=\left \{\begin {array}{cl} \ln \left (\mathrm {acoth}\left (x\right )\right ) & \text {\ if\ \ }n=-1\\ \frac {{\mathrm {acoth}\left (x\right )}^{n+1}}{n+1} & \text {\ if\ \ }n\neq -1 \end {array}\right . \]

[In]

int(-acoth(x)^n/(x^2 - 1),x)

[Out]

piecewise(n == -1, log(acoth(x)), n ~= -1, acoth(x)^(n + 1)/(n + 1))