\(\int \frac {\coth ^{-1}(x)}{1-x^2} \, dx\) [57]

Optimal result

Integrand size = 12, antiderivative size = 8 \[ \int \frac {\coth ^{-1}(x)}{1-x^2} \, dx=\frac {1}{2} \coth ^{-1}(x)^2 \]

[Out]

1/2*arccoth(x)^2

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6096} \[ \int \frac {\coth ^{-1}(x)}{1-x^2} \, dx=\frac {1}{2} \coth ^{-1}(x)^2 \]

[In]

Int[ArcCoth[x]/(1 - x^2),x]

[Out]

ArcCoth[x]^2/2

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 {1}{2} \coth ^{-1}(x)^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00 \[ \int \frac {\coth ^{-1}(x)}{1-x^2} \, dx=\frac {1}{2} \coth ^{-1}(x)^2 \]

[In]

Integrate[ArcCoth[x]/(1 - x^2),x]

[Out]

ArcCoth[x]^2/2

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.62

method	result	size
default	\(\operatorname {arctanh}\left (x \right ) \operatorname {arccoth}\left (x \right )-\frac {\operatorname {arctanh}\left (x \right )^{2}}{2}\)	\(13\)
parts	\(\operatorname {arctanh}\left (x \right ) \operatorname {arccoth}\left (x \right )-\frac {\operatorname {arctanh}\left (x \right )^{2}}{2}\)	\(13\)
risch	\(\frac {\ln \left (x -1\right )^{2}}{8}-\frac {\ln \left (1+x \right ) \ln \left (x -1\right )}{4}+\frac {\ln \left (1+x \right )^{2}}{8}\)	\(28\)

[In]

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

[Out]

arctanh(x)*arccoth(x)-1/2*arctanh(x)^2

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 14 vs. \(2 (6) = 12\).

Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.75 \[ \int \frac {\coth ^{-1}(x)}{1-x^2} \, dx=\frac {1}{8} \, \log \left (\frac {x + 1}{x - 1}\right )^{2} \]

[In]

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

[Out]

1/8*log((x + 1)/(x - 1))^2

Sympy [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 5, normalized size of antiderivative = 0.62 \[ \int \frac {\coth ^{-1}(x)}{1-x^2} \, dx=\frac {\operatorname {acoth}^{2}{\left (x \right )}}{2} \]

[In]

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

[Out]

acoth(x)**2/2

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \frac {\coth ^{-1}(x)}{1-x^2} \, dx=\frac {1}{2} \, \operatorname {arcoth}\left (x\right )^{2} \]

[In]

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

[Out]

1/2*arccoth(x)^2

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 14 vs. \(2 (6) = 12\).

Time = 0.27 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.75 \[ \int \frac {\coth ^{-1}(x)}{1-x^2} \, dx=\frac {1}{8} \, \log \left (\frac {x + 1}{x - 1}\right )^{2} \]

[In]

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

[Out]

1/8*log((x + 1)/(x - 1))^2

Mupad [B] (verification not implemented)

Time = 4.09 (sec) , antiderivative size = 21, normalized size of antiderivative = 2.62 \[ \int \frac {\coth ^{-1}(x)}{1-x^2} \, dx=\frac {{\left (\ln \left (1-\frac {1}{x}\right )-\ln \left (\frac {1}{x}+1\right )\right )}^2}{8} \]

[In]

int(-acoth(x)/(x^2 - 1),x)

[Out]

(log(1 - 1/x) - log(1/x + 1))^2/8