3.28 \(\int \coth ^{-1}(a x)^3 \, dx\)

Optimal. Leaf size=85 \[ \frac{3 \text{PolyLog}\left (3,1-\frac{2}{1-a x}\right )}{2 a}-\frac{3 \coth ^{-1}(a x) \text{PolyLog}\left (2,1-\frac{2}{1-a x}\right )}{a}+x \coth ^{-1}(a x)^3+\frac{\coth ^{-1}(a x)^3}{a}-\frac{3 \log \left (\frac{2}{1-a x}\right ) \coth ^{-1}(a x)^2}{a} \]

[Out]

ArcCoth[a*x]^3/a + x*ArcCoth[a*x]^3 - (3*ArcCoth[a*x]^2*Log[2/(1 - a*x)])/a - (3*ArcCoth[a*x]*PolyLog[2, 1 - 2
/(1 - a*x)])/a + (3*PolyLog[3, 1 - 2/(1 - a*x)])/(2*a)

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Rubi [A]  time = 0.165484, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 6, integrand size = 6, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1., Rules used = {5911, 5985, 5919, 5949, 6059, 6610} \[ \frac{3 \text{PolyLog}\left (3,1-\frac{2}{1-a x}\right )}{2 a}-\frac{3 \coth ^{-1}(a x) \text{PolyLog}\left (2,1-\frac{2}{1-a x}\right )}{a}+x \coth ^{-1}(a x)^3+\frac{\coth ^{-1}(a x)^3}{a}-\frac{3 \log \left (\frac{2}{1-a x}\right ) \coth ^{-1}(a x)^2}{a} \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[a*x]^3,x]

[Out]

ArcCoth[a*x]^3/a + x*ArcCoth[a*x]^3 - (3*ArcCoth[a*x]^2*Log[2/(1 - a*x)])/a - (3*ArcCoth[a*x]*PolyLog[2, 1 - 2
/(1 - a*x)])/a + (3*PolyLog[3, 1 - 2/(1 - a*x)])/(2*a)

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 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]

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 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 6059

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

Rule 6610

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
;  !FalseQ[w]] /; FreeQ[n, x]

Rubi steps

\begin{align*} \int \coth ^{-1}(a x)^3 \, dx &=x \coth ^{-1}(a x)^3-(3 a) \int \frac{x \coth ^{-1}(a x)^2}{1-a^2 x^2} \, dx\\ &=\frac{\coth ^{-1}(a x)^3}{a}+x \coth ^{-1}(a x)^3-3 \int \frac{\coth ^{-1}(a x)^2}{1-a x} \, dx\\ &=\frac{\coth ^{-1}(a x)^3}{a}+x \coth ^{-1}(a x)^3-\frac{3 \coth ^{-1}(a x)^2 \log \left (\frac{2}{1-a x}\right )}{a}+6 \int \frac{\coth ^{-1}(a x) \log \left (\frac{2}{1-a x}\right )}{1-a^2 x^2} \, dx\\ &=\frac{\coth ^{-1}(a x)^3}{a}+x \coth ^{-1}(a x)^3-\frac{3 \coth ^{-1}(a x)^2 \log \left (\frac{2}{1-a x}\right )}{a}-\frac{3 \coth ^{-1}(a x) \text{Li}_2\left (1-\frac{2}{1-a x}\right )}{a}+3 \int \frac{\text{Li}_2\left (1-\frac{2}{1-a x}\right )}{1-a^2 x^2} \, dx\\ &=\frac{\coth ^{-1}(a x)^3}{a}+x \coth ^{-1}(a x)^3-\frac{3 \coth ^{-1}(a x)^2 \log \left (\frac{2}{1-a x}\right )}{a}-\frac{3 \coth ^{-1}(a x) \text{Li}_2\left (1-\frac{2}{1-a x}\right )}{a}+\frac{3 \text{Li}_3\left (1-\frac{2}{1-a x}\right )}{2 a}\\ \end{align*}

Mathematica [A]  time = 0.0976196, size = 79, normalized size = 0.93 \[ -\frac{3 \coth ^{-1}(a x) \text{PolyLog}\left (2,e^{2 \coth ^{-1}(a x)}\right )}{a}+\frac{3 \text{PolyLog}\left (3,e^{2 \coth ^{-1}(a x)}\right )}{2 a}+x \coth ^{-1}(a x)^3+\frac{\coth ^{-1}(a x)^3}{a}-\frac{3 \coth ^{-1}(a x)^2 \log \left (1-e^{2 \coth ^{-1}(a x)}\right )}{a} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcCoth[a*x]^3,x]

[Out]

ArcCoth[a*x]^3/a + x*ArcCoth[a*x]^3 - (3*ArcCoth[a*x]^2*Log[1 - E^(2*ArcCoth[a*x])])/a - (3*ArcCoth[a*x]*PolyL
og[2, E^(2*ArcCoth[a*x])])/a + (3*PolyLog[3, E^(2*ArcCoth[a*x])])/(2*a)

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Maple [B]  time = 0.115, size = 180, normalized size = 2.1 \begin{align*} x \left ({\rm arccoth} \left (ax\right ) \right ) ^{3}-3\,{\frac{ \left ({\rm arccoth} \left (ax\right ) \right ) ^{2}}{a}\ln \left ( 1+{\frac{1}{\sqrt{{\frac{ax-1}{ax+1}}}}} \right ) }-3\,{\frac{ \left ({\rm arccoth} \left (ax\right ) \right ) ^{2}}{a}\ln \left ( 1-{\frac{1}{\sqrt{{\frac{ax-1}{ax+1}}}}} \right ) }+{\frac{ \left ({\rm arccoth} \left (ax\right ) \right ) ^{3}}{a}}-6\,{\frac{{\rm arccoth} \left (ax\right )}{a}{\it polylog} \left ( 2,-{\frac{1}{\sqrt{{\frac{ax-1}{ax+1}}}}} \right ) }-6\,{\frac{{\rm arccoth} \left (ax\right )}{a}{\it polylog} \left ( 2,{\frac{1}{\sqrt{{\frac{ax-1}{ax+1}}}}} \right ) }+6\,{\frac{1}{a}{\it polylog} \left ( 3,-{\frac{1}{\sqrt{{\frac{ax-1}{ax+1}}}}} \right ) }+6\,{\frac{1}{a}{\it polylog} \left ( 3,{\frac{1}{\sqrt{{\frac{ax-1}{ax+1}}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccoth(a*x)^3,x)

[Out]

x*arccoth(a*x)^3-3/a*arccoth(a*x)^2*ln(1+1/((a*x-1)/(a*x+1))^(1/2))-3/a*arccoth(a*x)^2*ln(1-1/((a*x-1)/(a*x+1)
)^(1/2))+arccoth(a*x)^3/a-6/a*arccoth(a*x)*polylog(2,-1/((a*x-1)/(a*x+1))^(1/2))-6/a*arccoth(a*x)*polylog(2,1/
((a*x-1)/(a*x+1))^(1/2))+6/a*polylog(3,-1/((a*x-1)/(a*x+1))^(1/2))+6/a*polylog(3,1/((a*x-1)/(a*x+1))^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (a x + 1\right )} \log \left (a x + 1\right )^{3} - 3 \,{\left (a x - 1\right )} \log \left (a x + 1\right )^{2} \log \left (a x - 1\right )}{8 \, a} + \frac{1}{8} \, \int -\frac{{\left (a x + 1\right )} \log \left (a x - 1\right )^{3} - 3 \,{\left ({\left (a x + 1\right )} \log \left (a x - 1\right )^{2} + 2 \,{\left (a x - 1\right )} \log \left (a x - 1\right )\right )} \log \left (a x + 1\right )}{a x + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(a*x)^3,x, algorithm="maxima")

[Out]

1/8*((a*x + 1)*log(a*x + 1)^3 - 3*(a*x - 1)*log(a*x + 1)^2*log(a*x - 1))/a + 1/8*integrate(-((a*x + 1)*log(a*x
- 1)^3 - 3*((a*x + 1)*log(a*x - 1)^2 + 2*(a*x - 1)*log(a*x - 1))*log(a*x + 1))/(a*x + 1), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\operatorname{arcoth}\left (a x\right )^{3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(a*x)^3,x, algorithm="fricas")

[Out]

integral(arccoth(a*x)^3, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \operatorname{acoth}^{3}{\left (a x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acoth(a*x)**3,x)

[Out]

Integral(acoth(a*x)**3, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \operatorname{arcoth}\left (a x\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(a*x)^3,x, algorithm="giac")

[Out]

integrate(arccoth(a*x)^3, x)