\(\int \coth ^{-1}(a x) \, dx\) [6]

Optimal result

Integrand size = 4, antiderivative size = 25 \[ \int \coth ^{-1}(a x) \, dx=x \coth ^{-1}(a x)+\frac {\log \left (1-a^2 x^2\right )}{2 a} \]

[Out]

x*arccoth(a*x)+1/2*ln(-a^2*x^2+1)/a

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6022, 266} \[ \int \coth ^{-1}(a x) \, dx=\frac {\log \left (1-a^2 x^2\right )}{2 a}+x \coth ^{-1}(a x) \]

[In]

Int[ArcCoth[a*x],x]

[Out]

x*ArcCoth[a*x] + Log[1 - a^2*x^2]/(2*a)

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 6022

Int[((a_.) + ArcCoth[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcCoth[c*x^n])^p, x] - Dist[b
*c*n*p, Int[x^n*((a + b*ArcCoth[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p
, 0] && (EqQ[n, 1] || EqQ[p, 1])

Rubi steps \begin{align*} \text {integral}& = x \coth ^{-1}(a x)-a \int \frac {x}{1-a^2 x^2} \, dx \\ & = x \coth ^{-1}(a x)+\frac {\log \left (1-a^2 x^2\right )}{2 a} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \coth ^{-1}(a x) \, dx=x \coth ^{-1}(a x)+\frac {\log \left (1-a^2 x^2\right )}{2 a} \]

[In]

Integrate[ArcCoth[a*x],x]

[Out]

x*ArcCoth[a*x] + Log[1 - a^2*x^2]/(2*a)

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92

method	result	size
parts	\(x \,\operatorname {arccoth}\left (a x \right )+\frac {\ln \left (a^{2} x^{2}-1\right )}{2 a}\)	\(23\)
derivativedivides	\(\frac {a x \,\operatorname {arccoth}\left (a x \right )+\frac {\ln \left (a^{2} x^{2}-1\right )}{2}}{a}\)	\(25\)
default	\(\frac {a x \,\operatorname {arccoth}\left (a x \right )+\frac {\ln \left (a^{2} x^{2}-1\right )}{2}}{a}\)	\(25\)
parallelrisch	\(-\frac {-a x \,\operatorname {arccoth}\left (a x \right )-\ln \left (a x -1\right )-\operatorname {arccoth}\left (a x \right )}{a}\)	\(29\)
risch	\(\frac {x \ln \left (a x +1\right )}{2}-\frac {\ln \left (a x -1\right ) x}{2}+\frac {\ln \left (a^{2} x^{2}-1\right )}{2 a}\)	\(35\)

[In]

int(arccoth(a*x),x,method=_RETURNVERBOSE)

[Out]

x*arccoth(a*x)+1/2/a*ln(a^2*x^2-1)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32 \[ \int \coth ^{-1}(a x) \, dx=\frac {a x \log \left (\frac {a x + 1}{a x - 1}\right ) + \log \left (a^{2} x^{2} - 1\right )}{2 \, a} \]

[In]

integrate(arccoth(a*x),x, algorithm="fricas")

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \coth ^{-1}(a x) \, dx=\begin {cases} x \operatorname {acoth}{\left (a x \right )} + \frac {\log {\left (a x + 1 \right )}}{a} - \frac {\operatorname {acoth}{\left (a x \right )}}{a} & \text {for}\: a \neq 0 \\\frac {i \pi x}{2} & \text {otherwise} \end {cases} \]

[In]

integrate(acoth(a*x),x)

[Out]

Piecewise((x*acoth(a*x) + log(a*x + 1)/a - acoth(a*x)/a, Ne(a, 0)), (I*pi*x/2, True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \coth ^{-1}(a x) \, dx=\frac {2 \, a x \operatorname {arcoth}\left (a x\right ) + \log \left (-a^{2} x^{2} + 1\right )}{2 \, a} \]

[In]

integrate(arccoth(a*x),x, algorithm="maxima")

[Out]

1/2*(2*a*x*arccoth(a*x) + log(-a^2*x^2 + 1))/a

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 153 vs. \(2 (23) = 46\).

Time = 0.25 (sec) , antiderivative size = 153, normalized size of antiderivative = 6.12 \[ \int \coth ^{-1}(a x) \, dx=a {\left (\frac {\log \left (\frac {{\left | a x + 1 \right |}}{{\left | a x - 1 \right |}}\right )}{a^{2}} - \frac {\log \left ({\left | \frac {a x + 1}{a x - 1} - 1 \right |}\right )}{a^{2}} + \frac {\log \left (-\frac {\frac {\frac {{\left (a x + 1\right )} a}{a x - 1} - a}{a {\left (\frac {a x + 1}{a x - 1} + 1\right )}} + 1}{\frac {\frac {{\left (a x + 1\right )} a}{a x - 1} - a}{a {\left (\frac {a x + 1}{a x - 1} + 1\right )}} - 1}\right )}{a^{2} {\left (\frac {a x + 1}{a x - 1} - 1\right )}}\right )} \]

[In]

integrate(arccoth(a*x),x, algorithm="giac")

[Out]

a*(log(abs(a*x + 1)/abs(a*x - 1))/a^2 - log(abs((a*x + 1)/(a*x - 1) - 1))/a^2 + log(-(((a*x + 1)*a/(a*x - 1) -
a)/(a*((a*x + 1)/(a*x - 1) + 1)) + 1)/(((a*x + 1)*a/(a*x - 1) - a)/(a*((a*x + 1)/(a*x - 1) + 1)) - 1))/(a^2*(
(a*x + 1)/(a*x - 1) - 1)))

Mupad [B] (verification not implemented)

Time = 4.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \coth ^{-1}(a x) \, dx=x\,\mathrm {acoth}\left (a\,x\right )+\frac {\ln \left (a^2\,x^2-1\right )}{2\,a} \]

[In]

int(acoth(a*x),x)

[Out]

x*acoth(a*x) + log(a^2*x^2 - 1)/(2*a)