3.16 \(\int x \coth ^{-1}(a x)^2 \, dx\)

Optimal. Leaf size=54 \[ \frac {\log \left (1-a^2 x^2\right )}{2 a^2}-\frac {\coth ^{-1}(a x)^2}{2 a^2}+\frac {1}{2} x^2 \coth ^{-1}(a x)^2+\frac {x \coth ^{-1}(a x)}{a} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {5917, 5981, 5911, 260, 5949} \[ \frac {\log \left (1-a^2 x^2\right )}{2 a^2}-\frac {\coth ^{-1}(a x)^2}{2 a^2}+\frac {1}{2} x^2 \coth ^{-1}(a x)^2+\frac {x \coth ^{-1}(a x)}{a} \]

Antiderivative was successfully verified.

[In]

Int[x*ArcCoth[a*x]^2,x]

[Out]

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

Rule 260

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 5911

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

Rule 5917

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcC
oth[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcCoth[c*x])^(p - 1))/(1 -
c^2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 5949

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]

Rule 5981

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

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 43, normalized size = 0.80 \[ \frac {\log \left (1-a^2 x^2\right )+\left (a^2 x^2-1\right ) \coth ^{-1}(a x)^2+2 a x \coth ^{-1}(a x)}{2 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcCoth[a*x]^2,x]

[Out]

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

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fricas [A]  time = 0.53, size = 62, normalized size = 1.15 \[ \frac {4 \, a x \log \left (\frac {a x + 1}{a x - 1}\right ) + {\left (a^{2} x^{2} - 1\right )} \log \left (\frac {a x + 1}{a x - 1}\right )^{2} + 4 \, \log \left (a^{2} x^{2} - 1\right )}{8 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccoth(a*x)^2,x, algorithm="fricas")

[Out]

1/8*(4*a*x*log((a*x + 1)/(a*x - 1)) + (a^2*x^2 - 1)*log((a*x + 1)/(a*x - 1))^2 + 4*log(a^2*x^2 - 1))/a^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccoth(a*x)^2,x, algorithm="giac")

[Out]

integrate(x*arccoth(a*x)^2, x)

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maple [B]  time = 0.06, size = 155, normalized size = 2.87 \[ \frac {x^{2} \mathrm {arccoth}\left (a x \right )^{2}}{2}+\frac {x \,\mathrm {arccoth}\left (a x \right )}{a}+\frac {\mathrm {arccoth}\left (a x \right ) \ln \left (a x -1\right )}{2 a^{2}}-\frac {\mathrm {arccoth}\left (a x \right ) \ln \left (a x +1\right )}{2 a^{2}}+\frac {\ln \left (a x -1\right )^{2}}{8 a^{2}}-\frac {\ln \left (a x -1\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{4 a^{2}}+\frac {\ln \left (a x -1\right )}{2 a^{2}}+\frac {\ln \left (a x +1\right )}{2 a^{2}}+\frac {\ln \left (a x +1\right )^{2}}{8 a^{2}}+\frac {\ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{4 a^{2}}-\frac {\ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (a x +1\right )}{4 a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arccoth(a*x)^2,x)

[Out]

1/2*x^2*arccoth(a*x)^2+x*arccoth(a*x)/a+1/2/a^2*arccoth(a*x)*ln(a*x-1)-1/2/a^2*arccoth(a*x)*ln(a*x+1)+1/8/a^2*
ln(a*x-1)^2-1/4/a^2*ln(a*x-1)*ln(1/2+1/2*a*x)+1/2/a^2*ln(a*x-1)+1/2/a^2*ln(a*x+1)+1/8/a^2*ln(a*x+1)^2+1/4/a^2*
ln(-1/2*a*x+1/2)*ln(1/2+1/2*a*x)-1/4/a^2*ln(-1/2*a*x+1/2)*ln(a*x+1)

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maxima [B]  time = 0.32, size = 97, normalized size = 1.80 \[ \frac {1}{2} \, x^{2} \operatorname {arcoth}\left (a x\right )^{2} + \frac {1}{2} \, a {\left (\frac {2 \, x}{a^{2}} - \frac {\log \left (a x + 1\right )}{a^{3}} + \frac {\log \left (a x - 1\right )}{a^{3}}\right )} \operatorname {arcoth}\left (a x\right ) - \frac {2 \, {\left (\log \left (a x - 1\right ) - 2\right )} \log \left (a x + 1\right ) - \log \left (a x + 1\right )^{2} - \log \left (a x - 1\right )^{2} - 4 \, \log \left (a x - 1\right )}{8 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccoth(a*x)^2,x, algorithm="maxima")

[Out]

1/2*x^2*arccoth(a*x)^2 + 1/2*a*(2*x/a^2 - log(a*x + 1)/a^3 + log(a*x - 1)/a^3)*arccoth(a*x) - 1/8*(2*(log(a*x
- 1) - 2)*log(a*x + 1) - log(a*x + 1)^2 - log(a*x - 1)^2 - 4*log(a*x - 1))/a^2

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mupad [B]  time = 1.21, size = 44, normalized size = 0.81 \[ \frac {x^2\,{\mathrm {acoth}\left (a\,x\right )}^2}{2}+\frac {-\frac {{\mathrm {acoth}\left (a\,x\right )}^2}{2}+a\,x\,\mathrm {acoth}\left (a\,x\right )+\frac {\ln \left (a^2\,x^2-1\right )}{2}}{a^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*acoth(a*x)^2,x)

[Out]

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

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sympy [A]  time = 0.78, size = 60, normalized size = 1.11 \[ \begin {cases} \frac {x^{2} \operatorname {acoth}^{2}{\left (a x \right )}}{2} + \frac {x \operatorname {acoth}{\left (a x \right )}}{a} + \frac {\log {\left (a x + 1 \right )}}{a^{2}} - \frac {\operatorname {acoth}^{2}{\left (a x \right )}}{2 a^{2}} - \frac {\operatorname {acoth}{\left (a x \right )}}{a^{2}} & \text {for}\: a \neq 0 \\- \frac {\pi ^{2} x^{2}}{8} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*acoth(a*x)**2,x)

[Out]

Piecewise((x**2*acoth(a*x)**2/2 + x*acoth(a*x)/a + log(a*x + 1)/a**2 - acoth(a*x)**2/(2*a**2) - acoth(a*x)/a**
2, Ne(a, 0)), (-pi**2*x**2/8, True))

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