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

Optimal. Leaf size=58 \[ -\frac {\text {Li}_2\left (1-\frac {2}{1-a x}\right )}{a}+x \coth ^{-1}(a x)^2+\frac {\coth ^{-1}(a x)^2}{a}-\frac {2 \log \left (\frac {2}{1-a x}\right ) \coth ^{-1}(a x)}{a} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.833, Rules used = {5911, 5985, 5919, 2402, 2315} \[ -\frac {\text {PolyLog}\left (2,1-\frac {2}{1-a x}\right )}{a}+x \coth ^{-1}(a x)^2+\frac {\coth ^{-1}(a x)^2}{a}-\frac {2 \log \left (\frac {2}{1-a x}\right ) \coth ^{-1}(a x)}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

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 5919

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

Rule 5985

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

Rubi steps

\begin {align*} \int \coth ^{-1}(a x)^2 \, dx &=x \coth ^{-1}(a x)^2-(2 a) \int \frac {x \coth ^{-1}(a x)}{1-a^2 x^2} \, dx\\ &=\frac {\coth ^{-1}(a x)^2}{a}+x \coth ^{-1}(a x)^2-2 \int \frac {\coth ^{-1}(a x)}{1-a x} \, dx\\ &=\frac {\coth ^{-1}(a x)^2}{a}+x \coth ^{-1}(a x)^2-\frac {2 \coth ^{-1}(a x) \log \left (\frac {2}{1-a x}\right )}{a}+2 \int \frac {\log \left (\frac {2}{1-a x}\right )}{1-a^2 x^2} \, dx\\ &=\frac {\coth ^{-1}(a x)^2}{a}+x \coth ^{-1}(a x)^2-\frac {2 \coth ^{-1}(a x) \log \left (\frac {2}{1-a x}\right )}{a}-\frac {2 \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-a x}\right )}{a}\\ &=\frac {\coth ^{-1}(a x)^2}{a}+x \coth ^{-1}(a x)^2-\frac {2 \coth ^{-1}(a x) \log \left (\frac {2}{1-a x}\right )}{a}-\frac {\text {Li}_2\left (1-\frac {2}{1-a x}\right )}{a}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 46, normalized size = 0.79 \[ \frac {\text {Li}_2\left (e^{-2 \coth ^{-1}(a x)}\right )+\coth ^{-1}(a x) \left ((a x-1) \coth ^{-1}(a x)-2 \log \left (1-e^{-2 \coth ^{-1}(a x)}\right )\right )}{a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\operatorname {arcoth}\left (a x\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arccoth(a*x)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccoth(a*x)^2, x)

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maple [B]  time = 0.30, size = 122, normalized size = 2.10 \[ x \mathrm {arccoth}\left (a x \right )^{2}+\frac {\mathrm {arccoth}\left (a x \right )^{2}}{a}-\frac {2 \,\mathrm {arccoth}\left (a x \right ) \ln \left (1-\frac {1}{\sqrt {\frac {a x -1}{a x +1}}}\right )}{a}-\frac {2 \,\mathrm {arccoth}\left (a x \right ) \ln \left (1+\frac {1}{\sqrt {\frac {a x -1}{a x +1}}}\right )}{a}-\frac {2 \polylog \left (2, \frac {1}{\sqrt {\frac {a x -1}{a x +1}}}\right )}{a}-\frac {2 \polylog \left (2, -\frac {1}{\sqrt {\frac {a x -1}{a x +1}}}\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.32, size = 135, normalized size = 2.33 \[ x \operatorname {arcoth}\left (a x\right )^{2} + \frac {1}{4} \, {\left (a {\left (\frac {\log \left (a x + 1\right )^{2} + 2 \, \log \left (a x + 1\right ) \log \left (a x - 1\right ) - \log \left (a x - 1\right )^{2}}{a^{3}} - \frac {4 \, {\left (\log \left (a x - 1\right ) \log \left (\frac {1}{2} \, a x + \frac {1}{2}\right ) + {\rm Li}_2\left (-\frac {1}{2} \, a x + \frac {1}{2}\right )\right )}}{a^{3}}\right )} - \frac {2 \, {\left (\frac {\log \left (a x + 1\right )}{a} - \frac {\log \left (a x - 1\right )}{a}\right )} \log \left (a^{2} x^{2} - 1\right )}{a}\right )} a + \frac {\operatorname {arcoth}\left (a x\right ) \log \left (a^{2} x^{2} - 1\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int {\mathrm {acoth}\left (a\,x\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(acoth(a*x)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(acoth(a*x)**2, x)

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