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

Optimal. Leaf size=139 \[ -\frac{\text{PolyLog}\left (2,1-\frac{2}{1-a x}\right )}{a^4}+\frac{x^2 \coth ^{-1}(a x)}{4 a^2}+\frac{x}{4 a^3}-\frac{\tanh ^{-1}(a x)}{4 a^4}+\frac{3 x \coth ^{-1}(a x)^2}{4 a^3}-\frac{\coth ^{-1}(a x)^3}{4 a^4}+\frac{\coth ^{-1}(a x)^2}{a^4}-\frac{2 \log \left (\frac{2}{1-a x}\right ) \coth ^{-1}(a x)}{a^4}+\frac{1}{4} x^4 \coth ^{-1}(a x)^3+\frac{x^3 \coth ^{-1}(a x)^2}{4 a} \]

[Out]

x/(4*a^3) + (x^2*ArcCoth[a*x])/(4*a^2) + ArcCoth[a*x]^2/a^4 + (3*x*ArcCoth[a*x]^2)/(4*a^3) + (x^3*ArcCoth[a*x]
^2)/(4*a) - ArcCoth[a*x]^3/(4*a^4) + (x^4*ArcCoth[a*x]^3)/4 - ArcTanh[a*x]/(4*a^4) - (2*ArcCoth[a*x]*Log[2/(1
- a*x)])/a^4 - PolyLog[2, 1 - 2/(1 - a*x)]/a^4

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Rubi [A]  time = 0.41668, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 18, number of rules used = 10, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1., Rules used = {5917, 5981, 321, 206, 5985, 5919, 2402, 2315, 5911, 5949} \[ -\frac{\text{PolyLog}\left (2,1-\frac{2}{1-a x}\right )}{a^4}+\frac{x^2 \coth ^{-1}(a x)}{4 a^2}+\frac{x}{4 a^3}-\frac{\tanh ^{-1}(a x)}{4 a^4}+\frac{3 x \coth ^{-1}(a x)^2}{4 a^3}-\frac{\coth ^{-1}(a x)^3}{4 a^4}+\frac{\coth ^{-1}(a x)^2}{a^4}-\frac{2 \log \left (\frac{2}{1-a x}\right ) \coth ^{-1}(a x)}{a^4}+\frac{1}{4} x^4 \coth ^{-1}(a x)^3+\frac{x^3 \coth ^{-1}(a x)^2}{4 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/(4*a^3) + (x^2*ArcCoth[a*x])/(4*a^2) + ArcCoth[a*x]^2/a^4 + (3*x*ArcCoth[a*x]^2)/(4*a^3) + (x^3*ArcCoth[a*x]
^2)/(4*a) - ArcCoth[a*x]^3/(4*a^4) + (x^4*ArcCoth[a*x]^3)/4 - ArcTanh[a*x]/(4*a^4) - (2*ArcCoth[a*x]*Log[2/(1
- a*x)])/a^4 - PolyLog[2, 1 - 2/(1 - a*x)]/a^4

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 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]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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 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 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 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 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 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]

Rubi steps

\begin{align*} \int x^3 \coth ^{-1}(a x)^3 \, dx &=\frac{1}{4} x^4 \coth ^{-1}(a x)^3-\frac{1}{4} (3 a) \int \frac{x^4 \coth ^{-1}(a x)^2}{1-a^2 x^2} \, dx\\ &=\frac{1}{4} x^4 \coth ^{-1}(a x)^3+\frac{3 \int x^2 \coth ^{-1}(a x)^2 \, dx}{4 a}-\frac{3 \int \frac{x^2 \coth ^{-1}(a x)^2}{1-a^2 x^2} \, dx}{4 a}\\ &=\frac{x^3 \coth ^{-1}(a x)^2}{4 a}+\frac{1}{4} x^4 \coth ^{-1}(a x)^3-\frac{1}{2} \int \frac{x^3 \coth ^{-1}(a x)}{1-a^2 x^2} \, dx+\frac{3 \int \coth ^{-1}(a x)^2 \, dx}{4 a^3}-\frac{3 \int \frac{\coth ^{-1}(a x)^2}{1-a^2 x^2} \, dx}{4 a^3}\\ &=\frac{3 x \coth ^{-1}(a x)^2}{4 a^3}+\frac{x^3 \coth ^{-1}(a x)^2}{4 a}-\frac{\coth ^{-1}(a x)^3}{4 a^4}+\frac{1}{4} x^4 \coth ^{-1}(a x)^3+\frac{\int x \coth ^{-1}(a x) \, dx}{2 a^2}-\frac{\int \frac{x \coth ^{-1}(a x)}{1-a^2 x^2} \, dx}{2 a^2}-\frac{3 \int \frac{x \coth ^{-1}(a x)}{1-a^2 x^2} \, dx}{2 a^2}\\ &=\frac{x^2 \coth ^{-1}(a x)}{4 a^2}+\frac{\coth ^{-1}(a x)^2}{a^4}+\frac{3 x \coth ^{-1}(a x)^2}{4 a^3}+\frac{x^3 \coth ^{-1}(a x)^2}{4 a}-\frac{\coth ^{-1}(a x)^3}{4 a^4}+\frac{1}{4} x^4 \coth ^{-1}(a x)^3-\frac{\int \frac{\coth ^{-1}(a x)}{1-a x} \, dx}{2 a^3}-\frac{3 \int \frac{\coth ^{-1}(a x)}{1-a x} \, dx}{2 a^3}-\frac{\int \frac{x^2}{1-a^2 x^2} \, dx}{4 a}\\ &=\frac{x}{4 a^3}+\frac{x^2 \coth ^{-1}(a x)}{4 a^2}+\frac{\coth ^{-1}(a x)^2}{a^4}+\frac{3 x \coth ^{-1}(a x)^2}{4 a^3}+\frac{x^3 \coth ^{-1}(a x)^2}{4 a}-\frac{\coth ^{-1}(a x)^3}{4 a^4}+\frac{1}{4} x^4 \coth ^{-1}(a x)^3-\frac{2 \coth ^{-1}(a x) \log \left (\frac{2}{1-a x}\right )}{a^4}-\frac{\int \frac{1}{1-a^2 x^2} \, dx}{4 a^3}+\frac{\int \frac{\log \left (\frac{2}{1-a x}\right )}{1-a^2 x^2} \, dx}{2 a^3}+\frac{3 \int \frac{\log \left (\frac{2}{1-a x}\right )}{1-a^2 x^2} \, dx}{2 a^3}\\ &=\frac{x}{4 a^3}+\frac{x^2 \coth ^{-1}(a x)}{4 a^2}+\frac{\coth ^{-1}(a x)^2}{a^4}+\frac{3 x \coth ^{-1}(a x)^2}{4 a^3}+\frac{x^3 \coth ^{-1}(a x)^2}{4 a}-\frac{\coth ^{-1}(a x)^3}{4 a^4}+\frac{1}{4} x^4 \coth ^{-1}(a x)^3-\frac{\tanh ^{-1}(a x)}{4 a^4}-\frac{2 \coth ^{-1}(a x) \log \left (\frac{2}{1-a x}\right )}{a^4}-\frac{\operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-a x}\right )}{2 a^4}-\frac{3 \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-a x}\right )}{2 a^4}\\ &=\frac{x}{4 a^3}+\frac{x^2 \coth ^{-1}(a x)}{4 a^2}+\frac{\coth ^{-1}(a x)^2}{a^4}+\frac{3 x \coth ^{-1}(a x)^2}{4 a^3}+\frac{x^3 \coth ^{-1}(a x)^2}{4 a}-\frac{\coth ^{-1}(a x)^3}{4 a^4}+\frac{1}{4} x^4 \coth ^{-1}(a x)^3-\frac{\tanh ^{-1}(a x)}{4 a^4}-\frac{2 \coth ^{-1}(a x) \log \left (\frac{2}{1-a x}\right )}{a^4}-\frac{\text{Li}_2\left (1-\frac{2}{1-a x}\right )}{a^4}\\ \end{align*}

Mathematica [A]  time = 0.299733, size = 88, normalized size = 0.63 \[ \frac{4 \text{PolyLog}\left (2,e^{-2 \coth ^{-1}(a x)}\right )+\left (a^4 x^4-1\right ) \coth ^{-1}(a x)^3+\left (a^3 x^3+3 a x-4\right ) \coth ^{-1}(a x)^2+\coth ^{-1}(a x) \left (a^2 x^2-8 \log \left (1-e^{-2 \coth ^{-1}(a x)}\right )-1\right )+a x}{4 a^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a*x + (-4 + 3*a*x + a^3*x^3)*ArcCoth[a*x]^2 + (-1 + a^4*x^4)*ArcCoth[a*x]^3 + ArcCoth[a*x]*(-1 + a^2*x^2 - 8*
Log[1 - E^(-2*ArcCoth[a*x])]) + 4*PolyLog[2, E^(-2*ArcCoth[a*x])])/(4*a^4)

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Maple [C]  time = 0.619, size = 684, normalized size = 4.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arccoth(a*x)^2/a^4+3/8/a^4*arccoth(a*x)^2*ln(a*x-1)-3/8/a^4*arccoth(a*x)^2*ln(a*x+1)-2/a^4*arccoth(a*x)*ln(1+1
/((a*x-1)/(a*x+1))^(1/2))-3/8/a^4*arccoth(a*x)^2*ln((a*x-1)/(a*x+1))-1/4/a^4/(-1+((a*x-1)/(a*x+1))^(1/2))*((a*
x-1)/(a*x+1))^(1/2)-1/4/a^4/(((a*x-1)/(a*x+1))^(1/2)+1)*((a*x-1)/(a*x+1))^(1/2)-1/4/a^4*arccoth(a*x)+1/4*x^4*a
rccoth(a*x)^3+3/4*x*arccoth(a*x)^2/a^3+1/4*x^3*arccoth(a*x)^2/a+1/4*x^2*arccoth(a*x)/a^2+2/a^4*dilog(1/((a*x-1
)/(a*x+1))^(1/2))-2/a^4*dilog(1+1/((a*x-1)/(a*x+1))^(1/2))-3/16*I/a^4*Pi*csgn(I*(a*x+1)/(a*x-1))^3*arccoth(a*x
)^2-3/16*I/a^4*Pi*csgn(I*(a*x+1)/(a*x-1)/((a*x+1)/(a*x-1)-1))^3*arccoth(a*x)^2+3/16*I/a^4*Pi*csgn(I*(a*x+1)/(a
*x-1)/((a*x+1)/(a*x-1)-1))^2*csgn(I/((a*x+1)/(a*x-1)-1))*arccoth(a*x)^2-1/4*arccoth(a*x)^3/a^4-3/16*I/a^4*Pi*c
sgn(I/((a*x-1)/(a*x+1))^(1/2))^2*csgn(I*(a*x+1)/(a*x-1))*arccoth(a*x)^2+3/8*I/a^4*Pi*csgn(I/((a*x-1)/(a*x+1))^
(1/2))*csgn(I*(a*x+1)/(a*x-1))^2*arccoth(a*x)^2+3/16*I/a^4*Pi*csgn(I*(a*x+1)/(a*x-1)/((a*x+1)/(a*x-1)-1))^2*cs
gn(I*(a*x+1)/(a*x-1))*arccoth(a*x)^2-3/16*I/a^4*Pi*csgn(I*(a*x+1)/(a*x-1)/((a*x+1)/(a*x-1)-1))*csgn(I*(a*x+1)/
(a*x-1))*csgn(I/((a*x+1)/(a*x-1)-1))*arccoth(a*x)^2

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Maxima [B]  time = 0.994781, size = 354, normalized size = 2.55 \begin{align*} \frac{1}{4} \, x^{4} \operatorname{arcoth}\left (a x\right )^{3} + \frac{1}{8} \, a{\left (\frac{2 \,{\left (a^{2} x^{3} + 3 \, x\right )}}{a^{4}} - \frac{3 \, \log \left (a x + 1\right )}{a^{5}} + \frac{3 \, \log \left (a x - 1\right )}{a^{5}}\right )} \operatorname{arcoth}\left (a x\right )^{2} + \frac{1}{32} \, a{\left (\frac{\frac{{\left (3 \, \log \left (a x - 1\right ) - 8\right )} \log \left (a x + 1\right )^{2} - \log \left (a x + 1\right )^{3} + \log \left (a x - 1\right )^{3} + 8 \, a x -{\left (3 \, \log \left (a x - 1\right )^{2} - 16 \, \log \left (a x - 1\right )\right )} \log \left (a x + 1\right ) + 8 \, \log \left (a x - 1\right )^{2} + 4 \, \log \left (a x - 1\right )}{a} - \frac{32 \,{\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} - \frac{4 \, \log \left (a x + 1\right )}{a}}{a^{4}} + \frac{2 \,{\left (4 \, a^{2} x^{2} - 2 \,{\left (3 \, \log \left (a x - 1\right ) - 8\right )} \log \left (a x + 1\right ) + 3 \, \log \left (a x + 1\right )^{2} + 3 \, \log \left (a x - 1\right )^{2} + 16 \, \log \left (a x - 1\right )\right )} \operatorname{arcoth}\left (a x\right )}{a^{5}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4*arccoth(a*x)^3 + 1/8*a*(2*(a^2*x^3 + 3*x)/a^4 - 3*log(a*x + 1)/a^5 + 3*log(a*x - 1)/a^5)*arccoth(a*x)^
2 + 1/32*a*((((3*log(a*x - 1) - 8)*log(a*x + 1)^2 - log(a*x + 1)^3 + log(a*x - 1)^3 + 8*a*x - (3*log(a*x - 1)^
2 - 16*log(a*x - 1))*log(a*x + 1) + 8*log(a*x - 1)^2 + 4*log(a*x - 1))/a - 32*(log(a*x - 1)*log(1/2*a*x + 1/2)
+ dilog(-1/2*a*x + 1/2))/a - 4*log(a*x + 1)/a)/a^4 + 2*(4*a^2*x^2 - 2*(3*log(a*x - 1) - 8)*log(a*x + 1) + 3*l
og(a*x + 1)^2 + 3*log(a*x - 1)^2 + 16*log(a*x - 1))*arccoth(a*x)/a^5)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^3*arccoth(a*x)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3*acoth(a*x)**3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3*arccoth(a*x)^3, x)