\(\int \frac {x \coth ^{-1}(x)}{(1-x^2)^3} \, dx\) [60]

Optimal result

Integrand size = 13, antiderivative size = 50 \[ \int \frac {x \coth ^{-1}(x)}{\left (1-x^2\right )^3} \, dx=-\frac {x}{16 \left (1-x^2\right )^2}-\frac {3 x}{32 \left (1-x^2\right )}+\frac {\coth ^{-1}(x)}{4 \left (1-x^2\right )^2}-\frac {3 \text {arctanh}(x)}{32} \]

[Out]

-1/16*x/(-x^2+1)^2-3/32*x/(-x^2+1)+1/4*arccoth(x)/(-x^2+1)^2-3/32*arctanh(x)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {6142, 205, 212} \[ \int \frac {x \coth ^{-1}(x)}{\left (1-x^2\right )^3} \, dx=-\frac {3 \text {arctanh}(x)}{32}-\frac {3 x}{32 \left (1-x^2\right )}-\frac {x}{16 \left (1-x^2\right )^2}+\frac {\coth ^{-1}(x)}{4 \left (1-x^2\right )^2} \]

[In]

Int[(x*ArcCoth[x])/(1 - x^2)^3,x]

[Out]

-1/16*x/(1 - x^2)^2 - (3*x)/(32*(1 - x^2)) + ArcCoth[x]/(4*(1 - x^2)^2) - (3*ArcTanh[x])/32

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 6142

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

Rubi steps \begin{align*} \text {integral}& = \frac {\coth ^{-1}(x)}{4 \left (1-x^2\right )^2}-\frac {1}{4} \int \frac {1}{\left (1-x^2\right )^3} \, dx \\ & = -\frac {x}{16 \left (1-x^2\right )^2}+\frac {\coth ^{-1}(x)}{4 \left (1-x^2\right )^2}-\frac {3}{16} \int \frac {1}{\left (1-x^2\right )^2} \, dx \\ & = -\frac {x}{16 \left (1-x^2\right )^2}-\frac {3 x}{32 \left (1-x^2\right )}+\frac {\coth ^{-1}(x)}{4 \left (1-x^2\right )^2}-\frac {3}{32} \int \frac {1}{1-x^2} \, dx \\ & = -\frac {x}{16 \left (1-x^2\right )^2}-\frac {3 x}{32 \left (1-x^2\right )}+\frac {\coth ^{-1}(x)}{4 \left (1-x^2\right )^2}-\frac {3 \text {arctanh}(x)}{32} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00 \[ \int \frac {x \coth ^{-1}(x)}{\left (1-x^2\right )^3} \, dx=\frac {1}{64} \left (-\frac {4 x}{\left (-1+x^2\right )^2}+\frac {6 x}{-1+x^2}+\frac {16 \coth ^{-1}(x)}{\left (-1+x^2\right )^2}+3 \log (1-x)-3 \log (1+x)\right ) \]

[In]

Integrate[(x*ArcCoth[x])/(1 - x^2)^3,x]

[Out]

((-4*x)/(-1 + x^2)^2 + (6*x)/(-1 + x^2) + (16*ArcCoth[x])/(-1 + x^2)^2 + 3*Log[1 - x] - 3*Log[1 + x])/64

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.74

method	result	size
parallelrisch	\(-\frac {3 \,\operatorname {arccoth}\left (x \right ) x^{4}-3 x^{3}-6 x^{2} \operatorname {arccoth}\left (x \right )+5 x -5 \,\operatorname {arccoth}\left (x \right )}{32 \left (x^{2}-1\right )^{2}}\)	\(37\)
default	\(\frac {\operatorname {arccoth}\left (x \right )}{4 \left (x^{2}-1\right )^{2}}+\frac {1}{64 \left (1+x \right )^{2}}+\frac {3}{64 \left (1+x \right )}-\frac {3 \ln \left (1+x \right )}{64}-\frac {1}{64 \left (x -1\right )^{2}}+\frac {3}{64 \left (x -1\right )}+\frac {3 \ln \left (x -1\right )}{64}\)	\(53\)
parts	\(\frac {\operatorname {arccoth}\left (x \right )}{4 \left (x^{2}-1\right )^{2}}+\frac {1}{64 \left (1+x \right )^{2}}+\frac {3}{64 \left (1+x \right )}-\frac {3 \ln \left (1+x \right )}{64}-\frac {1}{64 \left (x -1\right )^{2}}+\frac {3}{64 \left (x -1\right )}+\frac {3 \ln \left (x -1\right )}{64}\)	\(53\)
risch	\(-\frac {\ln \left (x -1\right )}{8 \left (x^{2}-1\right )^{2}}-\frac {3 x^{4} \ln \left (1+x \right )-3 \ln \left (x -1\right ) x^{4}-6 x^{2} \ln \left (1+x \right )+6 \ln \left (x -1\right ) x^{2}-6 x^{3}-5 \ln \left (1+x \right )-3 \ln \left (x -1\right )+10 x}{64 \left (x -1\right ) \left (1+x \right ) \left (x^{2}-1\right )}\)	\(91\)

[In]

int(x*arccoth(x)/(-x^2+1)^3,x,method=_RETURNVERBOSE)

[Out]

-1/32*(3*arccoth(x)*x^4-3*x^3-6*x^2*arccoth(x)+5*x-5*arccoth(x))/(x^2-1)^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.94 \[ \int \frac {x \coth ^{-1}(x)}{\left (1-x^2\right )^3} \, dx=\frac {6 \, x^{3} - {\left (3 \, x^{4} - 6 \, x^{2} - 5\right )} \log \left (\frac {x + 1}{x - 1}\right ) - 10 \, x}{64 \, {\left (x^{4} - 2 \, x^{2} + 1\right )}} \]

[In]

integrate(x*arccoth(x)/(-x^2+1)^3,x, algorithm="fricas")

[Out]

1/64*(6*x^3 - (3*x^4 - 6*x^2 - 5)*log((x + 1)/(x - 1)) - 10*x)/(x^4 - 2*x^2 + 1)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 88 vs. \(2 (37) = 74\).

Time = 0.29 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.76 \[ \int \frac {x \coth ^{-1}(x)}{\left (1-x^2\right )^3} \, dx=- \frac {3 x^{4} \operatorname {acoth}{\left (x \right )}}{32 x^{4} - 64 x^{2} + 32} + \frac {3 x^{3}}{32 x^{4} - 64 x^{2} + 32} + \frac {6 x^{2} \operatorname {acoth}{\left (x \right )}}{32 x^{4} - 64 x^{2} + 32} - \frac {5 x}{32 x^{4} - 64 x^{2} + 32} + \frac {5 \operatorname {acoth}{\left (x \right )}}{32 x^{4} - 64 x^{2} + 32} \]

[In]

integrate(x*acoth(x)/(-x**2+1)**3,x)

[Out]

-3*x**4*acoth(x)/(32*x**4 - 64*x**2 + 32) + 3*x**3/(32*x**4 - 64*x**2 + 32) + 6*x**2*acoth(x)/(32*x**4 - 64*x*
*2 + 32) - 5*x/(32*x**4 - 64*x**2 + 32) + 5*acoth(x)/(32*x**4 - 64*x**2 + 32)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.94 \[ \int \frac {x \coth ^{-1}(x)}{\left (1-x^2\right )^3} \, dx=\frac {3 \, x^{3} - 5 \, x}{32 \, {\left (x^{4} - 2 \, x^{2} + 1\right )}} + \frac {\operatorname {arcoth}\left (x\right )}{4 \, {\left (x^{2} - 1\right )}^{2}} - \frac {3}{64} \, \log \left (x + 1\right ) + \frac {3}{64} \, \log \left (x - 1\right ) \]

[In]

integrate(x*arccoth(x)/(-x^2+1)^3,x, algorithm="maxima")

[Out]

1/32*(3*x^3 - 5*x)/(x^4 - 2*x^2 + 1) + 1/4*arccoth(x)/(x^2 - 1)^2 - 3/64*log(x + 1) + 3/64*log(x - 1)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 154 vs. \(2 (36) = 72\).

Time = 0.26 (sec) , antiderivative size = 154, normalized size of antiderivative = 3.08 \[ \int \frac {x \coth ^{-1}(x)}{\left (1-x^2\right )^3} \, dx=-\frac {1}{128} \, {\left (\frac {{\left (x - 1\right )}^{2} {\left (\frac {4 \, {\left (x + 1\right )}}{x - 1} - 1\right )}}{{\left (x + 1\right )}^{2}} - \frac {{\left (x + 1\right )}^{2}}{{\left (x - 1\right )}^{2}} + \frac {4 \, {\left (x + 1\right )}}{x - 1}\right )} \log \left (-\frac {\frac {\frac {x + 1}{x - 1} - 1}{\frac {x + 1}{x - 1} + 1} + 1}{\frac {\frac {x + 1}{x - 1} - 1}{\frac {x + 1}{x - 1} + 1} - 1}\right ) - \frac {{\left (x - 1\right )}^{2} {\left (\frac {8 \, {\left (x + 1\right )}}{x - 1} - 1\right )}}{256 \, {\left (x + 1\right )}^{2}} - \frac {{\left (x + 1\right )}^{2}}{256 \, {\left (x - 1\right )}^{2}} + \frac {x + 1}{32 \, {\left (x - 1\right )}} \]

[In]

integrate(x*arccoth(x)/(-x^2+1)^3,x, algorithm="giac")

[Out]

-1/128*((x - 1)^2*(4*(x + 1)/(x - 1) - 1)/(x + 1)^2 - (x + 1)^2/(x - 1)^2 + 4*(x + 1)/(x - 1))*log(-(((x + 1)/
(x - 1) - 1)/((x + 1)/(x - 1) + 1) + 1)/(((x + 1)/(x - 1) - 1)/((x + 1)/(x - 1) + 1) - 1)) - 1/256*(x - 1)^2*(
8*(x + 1)/(x - 1) - 1)/(x + 1)^2 - 1/256*(x + 1)^2/(x - 1)^2 + 1/32*(x + 1)/(x - 1)

Mupad [B] (verification not implemented)

Time = 4.50 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.68 \[ \int \frac {x \coth ^{-1}(x)}{\left (1-x^2\right )^3} \, dx=\frac {3\,\ln \left (x-1\right )}{64}-\frac {3\,\ln \left (x+1\right )}{64}+\frac {\frac {\mathrm {acoth}\left (x\right )}{4}-\frac {5\,x}{32}+\frac {3\,x^3}{32}}{{\left (x^2-1\right )}^2} \]

[In]

int(-(x*acoth(x))/(x^2 - 1)^3,x)

[Out]

(3*log(x - 1))/64 - (3*log(x + 1))/64 + (acoth(x)/4 - (5*x)/32 + (3*x^3)/32)/(x^2 - 1)^2