3.133 \(\int \frac{\tanh ^{-1}(a x)^3}{x^3 (c+a c x)} \, dx\)

Optimal. Leaf size=305 \[ -\frac{3 a^2 \text{PolyLog}\left (2,\frac{2}{a x+1}-1\right )}{2 c}+\frac{3 a^2 \text{PolyLog}\left (3,\frac{2}{a x+1}-1\right )}{2 c}-\frac{3 a^2 \text{PolyLog}\left (4,\frac{2}{a x+1}-1\right )}{4 c}-\frac{3 a^2 \tanh ^{-1}(a x)^2 \text{PolyLog}\left (2,\frac{2}{a x+1}-1\right )}{2 c}+\frac{3 a^2 \tanh ^{-1}(a x) \text{PolyLog}\left (2,\frac{2}{a x+1}-1\right )}{c}-\frac{3 a^2 \tanh ^{-1}(a x) \text{PolyLog}\left (3,\frac{2}{a x+1}-1\right )}{2 c}-\frac{a^2 \tanh ^{-1}(a x)^3}{2 c}+\frac{3 a^2 \tanh ^{-1}(a x)^2}{2 c}+\frac{a^2 \log \left (2-\frac{2}{a x+1}\right ) \tanh ^{-1}(a x)^3}{c}-\frac{3 a^2 \log \left (2-\frac{2}{a x+1}\right ) \tanh ^{-1}(a x)^2}{c}+\frac{3 a^2 \log \left (2-\frac{2}{a x+1}\right ) \tanh ^{-1}(a x)}{c}-\frac{\tanh ^{-1}(a x)^3}{2 c x^2}+\frac{a \tanh ^{-1}(a x)^3}{c x}-\frac{3 a \tanh ^{-1}(a x)^2}{2 c x} \]

[Out]

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

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Rubi [A]  time = 0.748083, antiderivative size = 305, normalized size of antiderivative = 1., number of steps used = 18, number of rules used = 10, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.556, Rules used = {5934, 5916, 5982, 5988, 5932, 2447, 5948, 6056, 6610, 6060} \[ -\frac{3 a^2 \text{PolyLog}\left (2,\frac{2}{a x+1}-1\right )}{2 c}+\frac{3 a^2 \text{PolyLog}\left (3,\frac{2}{a x+1}-1\right )}{2 c}-\frac{3 a^2 \text{PolyLog}\left (4,\frac{2}{a x+1}-1\right )}{4 c}-\frac{3 a^2 \tanh ^{-1}(a x)^2 \text{PolyLog}\left (2,\frac{2}{a x+1}-1\right )}{2 c}+\frac{3 a^2 \tanh ^{-1}(a x) \text{PolyLog}\left (2,\frac{2}{a x+1}-1\right )}{c}-\frac{3 a^2 \tanh ^{-1}(a x) \text{PolyLog}\left (3,\frac{2}{a x+1}-1\right )}{2 c}-\frac{a^2 \tanh ^{-1}(a x)^3}{2 c}+\frac{3 a^2 \tanh ^{-1}(a x)^2}{2 c}+\frac{a^2 \log \left (2-\frac{2}{a x+1}\right ) \tanh ^{-1}(a x)^3}{c}-\frac{3 a^2 \log \left (2-\frac{2}{a x+1}\right ) \tanh ^{-1}(a x)^2}{c}+\frac{3 a^2 \log \left (2-\frac{2}{a x+1}\right ) \tanh ^{-1}(a x)}{c}-\frac{\tanh ^{-1}(a x)^3}{2 c x^2}+\frac{a \tanh ^{-1}(a x)^3}{c x}-\frac{3 a \tanh ^{-1}(a x)^2}{2 c x} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[a*x]^3/(x^3*(c + a*c*x)),x]

[Out]

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

Rule 5934

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

Rule 5916

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcT
anh[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTanh[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 5982

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

Rule 5988

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

Rule 5932

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Simp[((a + b*ArcTanh[c*
x])^p*Log[2 - 2/(1 + (e*x)/d)])/d, x] - Dist[(b*c*p)/d, Int[((a + b*ArcTanh[c*x])^(p - 1)*Log[2 - 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 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rule 5948

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[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 6056

Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[((a + b*ArcTa
nh[c*x])^p*PolyLog[2, 1 - u])/(2*c*d), x] - Dist[(b*p)/2, Int[((a + b*ArcTanh[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]

Rule 6060

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

Rubi steps

\begin{align*} \int \frac{\tanh ^{-1}(a x)^3}{x^3 (c+a c x)} \, dx &=-\left (a \int \frac{\tanh ^{-1}(a x)^3}{x^2 (c+a c x)} \, dx\right )+\frac{\int \frac{\tanh ^{-1}(a x)^3}{x^3} \, dx}{c}\\ &=-\frac{\tanh ^{-1}(a x)^3}{2 c x^2}+a^2 \int \frac{\tanh ^{-1}(a x)^3}{x (c+a c x)} \, dx-\frac{a \int \frac{\tanh ^{-1}(a x)^3}{x^2} \, dx}{c}+\frac{(3 a) \int \frac{\tanh ^{-1}(a x)^2}{x^2 \left (1-a^2 x^2\right )} \, dx}{2 c}\\ &=-\frac{\tanh ^{-1}(a x)^3}{2 c x^2}+\frac{a \tanh ^{-1}(a x)^3}{c x}+\frac{a^2 \tanh ^{-1}(a x)^3 \log \left (2-\frac{2}{1+a x}\right )}{c}+\frac{(3 a) \int \frac{\tanh ^{-1}(a x)^2}{x^2} \, dx}{2 c}-\frac{\left (3 a^2\right ) \int \frac{\tanh ^{-1}(a x)^2}{x \left (1-a^2 x^2\right )} \, dx}{c}+\frac{\left (3 a^3\right ) \int \frac{\tanh ^{-1}(a x)^2}{1-a^2 x^2} \, dx}{2 c}-\frac{\left (3 a^3\right ) \int \frac{\tanh ^{-1}(a x)^2 \log \left (2-\frac{2}{1+a x}\right )}{1-a^2 x^2} \, dx}{c}\\ &=-\frac{3 a \tanh ^{-1}(a x)^2}{2 c x}-\frac{a^2 \tanh ^{-1}(a x)^3}{2 c}-\frac{\tanh ^{-1}(a x)^3}{2 c x^2}+\frac{a \tanh ^{-1}(a x)^3}{c x}+\frac{a^2 \tanh ^{-1}(a x)^3 \log \left (2-\frac{2}{1+a x}\right )}{c}-\frac{3 a^2 \tanh ^{-1}(a x)^2 \text{Li}_2\left (-1+\frac{2}{1+a x}\right )}{2 c}+\frac{\left (3 a^2\right ) \int \frac{\tanh ^{-1}(a x)}{x \left (1-a^2 x^2\right )} \, dx}{c}-\frac{\left (3 a^2\right ) \int \frac{\tanh ^{-1}(a x)^2}{x (1+a x)} \, dx}{c}+\frac{\left (3 a^3\right ) \int \frac{\tanh ^{-1}(a x) \text{Li}_2\left (-1+\frac{2}{1+a x}\right )}{1-a^2 x^2} \, dx}{c}\\ &=\frac{3 a^2 \tanh ^{-1}(a x)^2}{2 c}-\frac{3 a \tanh ^{-1}(a x)^2}{2 c x}-\frac{a^2 \tanh ^{-1}(a x)^3}{2 c}-\frac{\tanh ^{-1}(a x)^3}{2 c x^2}+\frac{a \tanh ^{-1}(a x)^3}{c x}-\frac{3 a^2 \tanh ^{-1}(a x)^2 \log \left (2-\frac{2}{1+a x}\right )}{c}+\frac{a^2 \tanh ^{-1}(a x)^3 \log \left (2-\frac{2}{1+a x}\right )}{c}-\frac{3 a^2 \tanh ^{-1}(a x)^2 \text{Li}_2\left (-1+\frac{2}{1+a x}\right )}{2 c}-\frac{3 a^2 \tanh ^{-1}(a x) \text{Li}_3\left (-1+\frac{2}{1+a x}\right )}{2 c}+\frac{\left (3 a^2\right ) \int \frac{\tanh ^{-1}(a x)}{x (1+a x)} \, dx}{c}+\frac{\left (3 a^3\right ) \int \frac{\text{Li}_3\left (-1+\frac{2}{1+a x}\right )}{1-a^2 x^2} \, dx}{2 c}+\frac{\left (6 a^3\right ) \int \frac{\tanh ^{-1}(a x) \log \left (2-\frac{2}{1+a x}\right )}{1-a^2 x^2} \, dx}{c}\\ &=\frac{3 a^2 \tanh ^{-1}(a x)^2}{2 c}-\frac{3 a \tanh ^{-1}(a x)^2}{2 c x}-\frac{a^2 \tanh ^{-1}(a x)^3}{2 c}-\frac{\tanh ^{-1}(a x)^3}{2 c x^2}+\frac{a \tanh ^{-1}(a x)^3}{c x}+\frac{3 a^2 \tanh ^{-1}(a x) \log \left (2-\frac{2}{1+a x}\right )}{c}-\frac{3 a^2 \tanh ^{-1}(a x)^2 \log \left (2-\frac{2}{1+a x}\right )}{c}+\frac{a^2 \tanh ^{-1}(a x)^3 \log \left (2-\frac{2}{1+a x}\right )}{c}+\frac{3 a^2 \tanh ^{-1}(a x) \text{Li}_2\left (-1+\frac{2}{1+a x}\right )}{c}-\frac{3 a^2 \tanh ^{-1}(a x)^2 \text{Li}_2\left (-1+\frac{2}{1+a x}\right )}{2 c}-\frac{3 a^2 \tanh ^{-1}(a x) \text{Li}_3\left (-1+\frac{2}{1+a x}\right )}{2 c}-\frac{3 a^2 \text{Li}_4\left (-1+\frac{2}{1+a x}\right )}{4 c}-\frac{\left (3 a^3\right ) \int \frac{\log \left (2-\frac{2}{1+a x}\right )}{1-a^2 x^2} \, dx}{c}-\frac{\left (3 a^3\right ) \int \frac{\text{Li}_2\left (-1+\frac{2}{1+a x}\right )}{1-a^2 x^2} \, dx}{c}\\ &=\frac{3 a^2 \tanh ^{-1}(a x)^2}{2 c}-\frac{3 a \tanh ^{-1}(a x)^2}{2 c x}-\frac{a^2 \tanh ^{-1}(a x)^3}{2 c}-\frac{\tanh ^{-1}(a x)^3}{2 c x^2}+\frac{a \tanh ^{-1}(a x)^3}{c x}+\frac{3 a^2 \tanh ^{-1}(a x) \log \left (2-\frac{2}{1+a x}\right )}{c}-\frac{3 a^2 \tanh ^{-1}(a x)^2 \log \left (2-\frac{2}{1+a x}\right )}{c}+\frac{a^2 \tanh ^{-1}(a x)^3 \log \left (2-\frac{2}{1+a x}\right )}{c}-\frac{3 a^2 \text{Li}_2\left (-1+\frac{2}{1+a x}\right )}{2 c}+\frac{3 a^2 \tanh ^{-1}(a x) \text{Li}_2\left (-1+\frac{2}{1+a x}\right )}{c}-\frac{3 a^2 \tanh ^{-1}(a x)^2 \text{Li}_2\left (-1+\frac{2}{1+a x}\right )}{2 c}+\frac{3 a^2 \text{Li}_3\left (-1+\frac{2}{1+a x}\right )}{2 c}-\frac{3 a^2 \tanh ^{-1}(a x) \text{Li}_3\left (-1+\frac{2}{1+a x}\right )}{2 c}-\frac{3 a^2 \text{Li}_4\left (-1+\frac{2}{1+a x}\right )}{4 c}\\ \end{align*}

Mathematica [C]  time = 0.629177, size = 222, normalized size = 0.73 \[ \frac{a^2 \left (96 \left (\tanh ^{-1}(a x)-2\right ) \tanh ^{-1}(a x) \text{PolyLog}\left (2,e^{2 \tanh ^{-1}(a x)}\right )-96 \tanh ^{-1}(a x) \text{PolyLog}\left (3,e^{2 \tanh ^{-1}(a x)}\right )-96 \text{PolyLog}\left (2,e^{-2 \tanh ^{-1}(a x)}\right )+96 \text{PolyLog}\left (3,e^{2 \tanh ^{-1}(a x)}\right )+48 \text{PolyLog}\left (4,e^{2 \tanh ^{-1}(a x)}\right )-\frac{32 \tanh ^{-1}(a x)^3}{a^2 x^2}-32 \tanh ^{-1}(a x)^4+\frac{64 \tanh ^{-1}(a x)^3}{a x}+96 \tanh ^{-1}(a x)^3-\frac{96 \tanh ^{-1}(a x)^2}{a x}+96 \tanh ^{-1}(a x)^2+64 \tanh ^{-1}(a x)^3 \log \left (1-e^{2 \tanh ^{-1}(a x)}\right )-192 \tanh ^{-1}(a x)^2 \log \left (1-e^{2 \tanh ^{-1}(a x)}\right )+192 \tanh ^{-1}(a x) \log \left (1-e^{-2 \tanh ^{-1}(a x)}\right )+\pi ^4-8 i \pi ^3\right )}{64 c} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcTanh[a*x]^3/(x^3*(c + a*c*x)),x]

[Out]

(a^2*((-8*I)*Pi^3 + Pi^4 + 96*ArcTanh[a*x]^2 - (96*ArcTanh[a*x]^2)/(a*x) + 96*ArcTanh[a*x]^3 - (32*ArcTanh[a*x
]^3)/(a^2*x^2) + (64*ArcTanh[a*x]^3)/(a*x) - 32*ArcTanh[a*x]^4 + 192*ArcTanh[a*x]*Log[1 - E^(-2*ArcTanh[a*x])]
- 192*ArcTanh[a*x]^2*Log[1 - E^(2*ArcTanh[a*x])] + 64*ArcTanh[a*x]^3*Log[1 - E^(2*ArcTanh[a*x])] - 96*PolyLog
[2, E^(-2*ArcTanh[a*x])] + 96*(-2 + ArcTanh[a*x])*ArcTanh[a*x]*PolyLog[2, E^(2*ArcTanh[a*x])] + 96*PolyLog[3,
E^(2*ArcTanh[a*x])] - 96*ArcTanh[a*x]*PolyLog[3, E^(2*ArcTanh[a*x])] + 48*PolyLog[4, E^(2*ArcTanh[a*x])]))/(64
*c)

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Maple [B]  time = 0.984, size = 664, normalized size = 2.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(a*x)^3/x^3/(a*c*x+c),x)

[Out]

a*arctanh(a*x)^3/c/x-3/2*a*arctanh(a*x)^2/c/x-1/2*arctanh(a*x)^3/c/x^2-1/2*a^2*arctanh(a*x)^4/c+6*a^2/c*polylo
g(4,-(a*x+1)/(-a^2*x^2+1)^(1/2))+6*a^2/c*polylog(4,(a*x+1)/(-a^2*x^2+1)^(1/2))-3/2*a^2*arctanh(a*x)^2/c+3*a^2/
c*polylog(2,-(a*x+1)/(-a^2*x^2+1)^(1/2))+3*a^2/c*polylog(2,(a*x+1)/(-a^2*x^2+1)^(1/2))+3/2*a^2*arctanh(a*x)^3/
c+6*a^2/c*polylog(3,-(a*x+1)/(-a^2*x^2+1)^(1/2))+6*a^2/c*polylog(3,(a*x+1)/(-a^2*x^2+1)^(1/2))+3*a^2/c*arctanh
(a*x)^2*polylog(2,(a*x+1)/(-a^2*x^2+1)^(1/2))-6*a^2/c*arctanh(a*x)*polylog(3,(a*x+1)/(-a^2*x^2+1)^(1/2))+3*a^2
/c*arctanh(a*x)*ln(1+(a*x+1)/(-a^2*x^2+1)^(1/2))+3*a^2/c*arctanh(a*x)*ln(1-(a*x+1)/(-a^2*x^2+1)^(1/2))-3*a^2/c
*arctanh(a*x)^2*ln(1+(a*x+1)/(-a^2*x^2+1)^(1/2))-6*a^2/c*arctanh(a*x)*polylog(2,-(a*x+1)/(-a^2*x^2+1)^(1/2))-3
*a^2/c*arctanh(a*x)^2*ln(1-(a*x+1)/(-a^2*x^2+1)^(1/2))-6*a^2/c*arctanh(a*x)*polylog(2,(a*x+1)/(-a^2*x^2+1)^(1/
2))+a^2/c*arctanh(a*x)^3*ln(1+(a*x+1)/(-a^2*x^2+1)^(1/2))+3*a^2/c*arctanh(a*x)^2*polylog(2,-(a*x+1)/(-a^2*x^2+
1)^(1/2))-6*a^2/c*arctanh(a*x)*polylog(3,-(a*x+1)/(-a^2*x^2+1)^(1/2))+a^2/c*arctanh(a*x)^3*ln(1-(a*x+1)/(-a^2*
x^2+1)^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (2 \, a^{2} x^{2} \log \left (a x + 1\right ) - 2 \, a x + 1\right )} \log \left (-a x + 1\right )^{3}}{16 \, c x^{2}} - \frac{1}{8} \, \int -\frac{2 \,{\left (a x - 1\right )} \log \left (a x + 1\right )^{3} - 6 \,{\left (a x - 1\right )} \log \left (a x + 1\right )^{2} \log \left (-a x + 1\right ) + 3 \,{\left (2 \, a^{3} x^{3} + a^{2} x^{2} - a x - 2 \,{\left (a^{4} x^{4} + a^{3} x^{3} - a x + 1\right )} \log \left (a x + 1\right )\right )} \log \left (-a x + 1\right )^{2}}{2 \,{\left (a^{2} c x^{5} - c x^{3}\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)^3/x^3/(a*c*x+c),x, algorithm="maxima")

[Out]

1/16*(2*a^2*x^2*log(a*x + 1) - 2*a*x + 1)*log(-a*x + 1)^3/(c*x^2) - 1/8*integrate(-1/2*(2*(a*x - 1)*log(a*x +
1)^3 - 6*(a*x - 1)*log(a*x + 1)^2*log(-a*x + 1) + 3*(2*a^3*x^3 + a^2*x^2 - a*x - 2*(a^4*x^4 + a^3*x^3 - a*x +
1)*log(a*x + 1))*log(-a*x + 1)^2)/(a^2*c*x^5 - c*x^3), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\operatorname{artanh}\left (a x\right )^{3}}{a c x^{4} + c x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)^3/x^3/(a*c*x+c),x, algorithm="fricas")

[Out]

integral(arctanh(a*x)^3/(a*c*x^4 + c*x^3), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\operatorname{atanh}^{3}{\left (a x \right )}}{a x^{4} + x^{3}}\, dx}{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(a*x)**3/x**3/(a*c*x+c),x)

[Out]

Integral(atanh(a*x)**3/(a*x**4 + x**3), x)/c

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)^3/x^3/(a*c*x+c),x, algorithm="giac")

[Out]

integrate(arctanh(a*x)^3/((a*c*x + c)*x^3), x)