3.55 \(\int \frac {\coth ^{-1}(x)^2}{(1-x^2)^2} \, dx\)

Optimal. Leaf size=62 \[ \frac {x}{4 \left (1-x^2\right )}+\frac {x \coth ^{-1}(x)^2}{2 \left (1-x^2\right )}-\frac {\coth ^{-1}(x)}{2 \left (1-x^2\right )}+\frac {1}{4} \tanh ^{-1}(x)+\frac {1}{6} \coth ^{-1}(x)^3 \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5957, 5995, 199, 206} \[ \frac {x}{4 \left (1-x^2\right )}+\frac {x \coth ^{-1}(x)^2}{2 \left (1-x^2\right )}-\frac {\coth ^{-1}(x)}{2 \left (1-x^2\right )}+\frac {1}{4} \tanh ^{-1}(x)+\frac {1}{6} \coth ^{-1}(x)^3 \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[x]^2/(1 - x^2)^2,x]

[Out]

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

Rule 199

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] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 206

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

Rule 5957

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

Rule 5995

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*} \int \frac {\coth ^{-1}(x)^2}{\left (1-x^2\right )^2} \, dx &=\frac {x \coth ^{-1}(x)^2}{2 \left (1-x^2\right )}+\frac {1}{6} \coth ^{-1}(x)^3-\int \frac {x \coth ^{-1}(x)}{\left (1-x^2\right )^2} \, dx\\ &=-\frac {\coth ^{-1}(x)}{2 \left (1-x^2\right )}+\frac {x \coth ^{-1}(x)^2}{2 \left (1-x^2\right )}+\frac {1}{6} \coth ^{-1}(x)^3+\frac {1}{2} \int \frac {1}{\left (1-x^2\right )^2} \, dx\\ &=\frac {x}{4 \left (1-x^2\right )}-\frac {\coth ^{-1}(x)}{2 \left (1-x^2\right )}+\frac {x \coth ^{-1}(x)^2}{2 \left (1-x^2\right )}+\frac {1}{6} \coth ^{-1}(x)^3+\frac {1}{4} \int \frac {1}{1-x^2} \, dx\\ &=\frac {x}{4 \left (1-x^2\right )}-\frac {\coth ^{-1}(x)}{2 \left (1-x^2\right )}+\frac {x \coth ^{-1}(x)^2}{2 \left (1-x^2\right )}+\frac {1}{6} \coth ^{-1}(x)^3+\frac {1}{4} \tanh ^{-1}(x)\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.07, size = 61, normalized size = 0.98 \[ \frac {-3 \left (x^2-1\right ) \log (1-x)+3 \left (x^2-1\right ) \log (x+1)+4 \left (x^2-1\right ) \coth ^{-1}(x)^3-6 x-12 x \coth ^{-1}(x)^2+12 \coth ^{-1}(x)}{24 \left (x^2-1\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[x]^2/(1 - x^2)^2,x]

[Out]

(-6*x + 12*ArcCoth[x] - 12*x*ArcCoth[x]^2 + 4*(-1 + x^2)*ArcCoth[x]^3 - 3*(-1 + x^2)*Log[1 - x] + 3*(-1 + x^2)
*Log[1 + x])/(24*(-1 + x^2))

________________________________________________________________________________________

fricas [A]  time = 0.53, size = 63, normalized size = 1.02 \[ \frac {{\left (x^{2} - 1\right )} \log \left (\frac {x + 1}{x - 1}\right )^{3} - 6 \, x \log \left (\frac {x + 1}{x - 1}\right )^{2} + 6 \, {\left (x^{2} + 1\right )} \log \left (\frac {x + 1}{x - 1}\right ) - 12 \, x}{48 \, {\left (x^{2} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*((x^2 - 1)*log((x + 1)/(x - 1))^3 - 6*x*log((x + 1)/(x - 1))^2 + 6*(x^2 + 1)*log((x + 1)/(x - 1)) - 12*x)
/(x^2 - 1)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arcoth}\relax (x)^{2}}{{\left (x^{2} - 1\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccoth(x)^2/(x^2 - 1)^2, x)

________________________________________________________________________________________

maple [C]  time = 2.00, size = 707, normalized size = 11.40 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*arccoth(x)^2/(-1+x)-1/4*arccoth(x)^2*ln(-1+x)-1/4*arccoth(x)^2/(1+x)+1/4*arccoth(x)^2*ln(1+x)+1/4*arccoth
(x)^2*ln((-1+x)/(1+x))+1/24*(-3*I*arccoth(x)^2*Pi*csgn(I/((1+x)/(-1+x)-1))*csgn(I*(1+x)/(-1+x)/((1+x)/(-1+x)-1
))^2*x^2+3*I*arccoth(x)^2*Pi*csgn(I*(1+x)/(-1+x))*csgn(I/((-1+x)/(1+x))^(1/2))^2*x^2+3*I*csgn(I*(1+x)/(-1+x)/(
(1+x)/(-1+x)-1))^2*csgn(I/((1+x)/(-1+x)-1))*arccoth(x)^2*Pi+3*I*arccoth(x)^2*Pi*csgn(I*(1+x)/(-1+x))*csgn(I/((
1+x)/(-1+x)-1))*csgn(I*(1+x)/(-1+x)/((1+x)/(-1+x)-1))*x^2+3*I*arccoth(x)^2*Pi*csgn(I*(1+x)/(-1+x))^3*x^2-3*I*c
sgn(I*(1+x)/(-1+x)/((1+x)/(-1+x)-1))*csgn(I/((1+x)/(-1+x)-1))*csgn(I*(1+x)/(-1+x))*arccoth(x)^2*Pi+6*I*csgn(I*
(1+x)/(-1+x))^2*csgn(I/((-1+x)/(1+x))^(1/2))*arccoth(x)^2*Pi+3*I*csgn(I*(1+x)/(-1+x)/((1+x)/(-1+x)-1))^2*csgn(
I*(1+x)/(-1+x))*arccoth(x)^2*Pi+3*I*arccoth(x)^2*Pi*csgn(I*(1+x)/(-1+x)/((1+x)/(-1+x)-1))^3*x^2-3*I*csgn(I*(1+
x)/(-1+x))*csgn(I/((-1+x)/(1+x))^(1/2))^2*arccoth(x)^2*Pi-6*I*arccoth(x)^2*Pi*csgn(I*(1+x)/(-1+x))^2*csgn(I/((
-1+x)/(1+x))^(1/2))*x^2-3*I*csgn(I*(1+x)/(-1+x)/((1+x)/(-1+x)-1))^3*arccoth(x)^2*Pi-3*I*arccoth(x)^2*Pi*csgn(I
*(1+x)/(-1+x))*csgn(I*(1+x)/(-1+x)/((1+x)/(-1+x)-1))^2*x^2-3*I*csgn(I*(1+x)/(-1+x))^3*arccoth(x)^2*Pi+4*arccot
h(x)^3*x^2-4*arccoth(x)^3+6*arccoth(x)*x^2+6*arccoth(x)-6*x)/(-1+x)/(1+x)

________________________________________________________________________________________

maxima [B]  time = 0.31, size = 171, normalized size = 2.76 \[ -\frac {1}{4} \, {\left (\frac {2 \, x}{x^{2} - 1} - \log \left (x + 1\right ) + \log \left (x - 1\right )\right )} \operatorname {arcoth}\relax (x)^{2} - \frac {{\left ({\left (x^{2} - 1\right )} \log \left (x + 1\right )^{2} - 2 \, {\left (x^{2} - 1\right )} \log \left (x + 1\right ) \log \left (x - 1\right ) + {\left (x^{2} - 1\right )} \log \left (x - 1\right )^{2} - 4\right )} \operatorname {arcoth}\relax (x)}{8 \, {\left (x^{2} - 1\right )}} + \frac {{\left (x^{2} - 1\right )} \log \left (x + 1\right )^{3} - 3 \, {\left (x^{2} - 1\right )} \log \left (x + 1\right )^{2} \log \left (x - 1\right ) - {\left (x^{2} - 1\right )} \log \left (x - 1\right )^{3} + 3 \, {\left ({\left (x^{2} - 1\right )} \log \left (x - 1\right )^{2} + 2 \, x^{2} - 2\right )} \log \left (x + 1\right ) - 6 \, {\left (x^{2} - 1\right )} \log \left (x - 1\right ) - 12 \, x}{48 \, {\left (x^{2} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*x/(x^2 - 1) - log(x + 1) + log(x - 1))*arccoth(x)^2 - 1/8*((x^2 - 1)*log(x + 1)^2 - 2*(x^2 - 1)*log(x
+ 1)*log(x - 1) + (x^2 - 1)*log(x - 1)^2 - 4)*arccoth(x)/(x^2 - 1) + 1/48*((x^2 - 1)*log(x + 1)^3 - 3*(x^2 - 1
)*log(x + 1)^2*log(x - 1) - (x^2 - 1)*log(x - 1)^3 + 3*((x^2 - 1)*log(x - 1)^2 + 2*x^2 - 2)*log(x + 1) - 6*(x^
2 - 1)*log(x - 1) - 12*x)/(x^2 - 1)

________________________________________________________________________________________

mupad [B]  time = 2.68, size = 201, normalized size = 3.24 \[ \frac {{\ln \left (\frac {1}{x}+1\right )}^3}{48}-\frac {{\ln \left (1-\frac {1}{x}\right )}^3}{48}-\frac {x}{4\,\left (x^2-1\right )}+\ln \left (1-\frac {1}{x}\right )\,\left (\frac {\frac {3\,x}{32}-\frac {1}{8}}{x^2-1}-\frac {\frac {x}{8}+\frac {1}{8}}{x^2-1}-\frac {{\ln \left (\frac {1}{x}+1\right )}^2}{16}+\frac {x}{32\,\left (x^2-1\right )}+\ln \left (\frac {1}{x}+1\right )\,\left (\frac {\frac {x}{4}+\frac {1}{16}}{x^2-1}-\frac {1}{16\,\left (x^2-1\right )}\right )\right )+{\ln \left (1-\frac {1}{x}\right )}^2\,\left (\frac {\ln \left (\frac {1}{x}+1\right )}{16}-\frac {x}{8\,\left (x^2-1\right )}\right )+\frac {\ln \left (\frac {1}{x}+1\right )}{4\,\left (x^2-1\right )}-\frac {x\,{\ln \left (\frac {1}{x}+1\right )}^2}{8\,\left (x^2-1\right )}-\frac {\mathrm {atan}\left (x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(x)^2/(x^2 - 1)^2,x)

[Out]

log(1/x + 1)^3/48 - (atan(x*1i)*1i)/4 - log(1 - 1/x)^3/48 - x/(4*(x^2 - 1)) + log(1 - 1/x)*(((3*x)/32 - 1/8)/(
x^2 - 1) - (x/8 + 1/8)/(x^2 - 1) - log(1/x + 1)^2/16 + x/(32*(x^2 - 1)) + log(1/x + 1)*((x/4 + 1/16)/(x^2 - 1)
- 1/(16*(x^2 - 1)))) + log(1 - 1/x)^2*(log(1/x + 1)/16 - x/(8*(x^2 - 1))) + log(1/x + 1)/(4*(x^2 - 1)) - (x*l
og(1/x + 1)^2)/(8*(x^2 - 1))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {acoth}^{2}{\relax (x )}}{\left (x - 1\right )^{2} \left (x + 1\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________