3.34 \(\int \frac{(a+b \tanh ^{-1}(c x))^3}{x^5} \, dx\)

Optimal. Leaf size=187 \[ -b^3 c^4 \text{PolyLog}\left (2,\frac{2}{c x+1}-1\right )-\frac{b^2 c^2 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^2}+2 b^2 c^4 \log \left (2-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{4} c^4 \left (a+b \tanh ^{-1}(c x)\right )^3+b c^4 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{3 b c^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 x}-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )^2}{4 x^3}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{4 x^4}-\frac{b^3 c^3}{4 x}+\frac{1}{4} b^3 c^4 \tanh ^{-1}(c x) \]

[Out]

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

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Rubi [A]  time = 0.62579, antiderivative size = 187, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 8, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.571, Rules used = {5916, 5982, 325, 206, 5988, 5932, 2447, 5948} \[ -b^3 c^4 \text{PolyLog}\left (2,\frac{2}{c x+1}-1\right )-\frac{b^2 c^2 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^2}+2 b^2 c^4 \log \left (2-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{4} c^4 \left (a+b \tanh ^{-1}(c x)\right )^3+b c^4 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{3 b c^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 x}-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )^2}{4 x^3}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{4 x^4}-\frac{b^3 c^3}{4 x}+\frac{1}{4} b^3 c^4 \tanh ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])^3/x^5,x]

[Out]

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

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 325

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

Rubi steps

\begin{align*} \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{x^5} \, dx &=-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{4 x^4}+\frac{1}{4} (3 b c) \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{x^4 \left (1-c^2 x^2\right )} \, dx\\ &=-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{4 x^4}+\frac{1}{4} (3 b c) \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{x^4} \, dx+\frac{1}{4} \left (3 b c^3\right ) \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{x^2 \left (1-c^2 x^2\right )} \, dx\\ &=-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )^2}{4 x^3}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{4 x^4}+\frac{1}{2} \left (b^2 c^2\right ) \int \frac{a+b \tanh ^{-1}(c x)}{x^3 \left (1-c^2 x^2\right )} \, dx+\frac{1}{4} \left (3 b c^3\right ) \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{x^2} \, dx+\frac{1}{4} \left (3 b c^5\right ) \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2} \, dx\\ &=-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )^2}{4 x^3}-\frac{3 b c^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 x}+\frac{1}{4} c^4 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{4 x^4}+\frac{1}{2} \left (b^2 c^2\right ) \int \frac{a+b \tanh ^{-1}(c x)}{x^3} \, dx+\frac{1}{2} \left (b^2 c^4\right ) \int \frac{a+b \tanh ^{-1}(c x)}{x \left (1-c^2 x^2\right )} \, dx+\frac{1}{2} \left (3 b^2 c^4\right ) \int \frac{a+b \tanh ^{-1}(c x)}{x \left (1-c^2 x^2\right )} \, dx\\ &=-\frac{b^2 c^2 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^2}+b c^4 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )^2}{4 x^3}-\frac{3 b c^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 x}+\frac{1}{4} c^4 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{4 x^4}+\frac{1}{4} \left (b^3 c^3\right ) \int \frac{1}{x^2 \left (1-c^2 x^2\right )} \, dx+\frac{1}{2} \left (b^2 c^4\right ) \int \frac{a+b \tanh ^{-1}(c x)}{x (1+c x)} \, dx+\frac{1}{2} \left (3 b^2 c^4\right ) \int \frac{a+b \tanh ^{-1}(c x)}{x (1+c x)} \, dx\\ &=-\frac{b^3 c^3}{4 x}-\frac{b^2 c^2 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^2}+b c^4 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )^2}{4 x^3}-\frac{3 b c^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 x}+\frac{1}{4} c^4 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{4 x^4}+2 b^2 c^4 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac{2}{1+c x}\right )+\frac{1}{4} \left (b^3 c^5\right ) \int \frac{1}{1-c^2 x^2} \, dx-\frac{1}{2} \left (b^3 c^5\right ) \int \frac{\log \left (2-\frac{2}{1+c x}\right )}{1-c^2 x^2} \, dx-\frac{1}{2} \left (3 b^3 c^5\right ) \int \frac{\log \left (2-\frac{2}{1+c x}\right )}{1-c^2 x^2} \, dx\\ &=-\frac{b^3 c^3}{4 x}+\frac{1}{4} b^3 c^4 \tanh ^{-1}(c x)-\frac{b^2 c^2 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^2}+b c^4 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )^2}{4 x^3}-\frac{3 b c^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 x}+\frac{1}{4} c^4 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{4 x^4}+2 b^2 c^4 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac{2}{1+c x}\right )-b^3 c^4 \text{Li}_2\left (-1+\frac{2}{1+c x}\right )\\ \end{align*}

Mathematica [A]  time = 0.638678, size = 295, normalized size = 1.58 \[ -\frac{8 b^3 c^4 x^4 \text{PolyLog}\left (2,e^{-2 \tanh ^{-1}(c x)}\right )+2 b \tanh ^{-1}(c x) \left (3 a^2+2 a b c x \left (3 c^2 x^2+1\right )+b^2 c^2 x^2 \left (1-c^2 x^2\right )-8 b^2 c^4 x^4 \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )\right )+6 a^2 b c^3 x^3+3 a^2 b c^4 x^4 \log (1-c x)-3 a^2 b c^4 x^4 \log (c x+1)+2 a^2 b c x+2 a^3-2 a b^2 c^4 x^4+2 a b^2 c^2 x^2-16 a b^2 c^4 x^4 \log \left (\frac{c x}{\sqrt{1-c^2 x^2}}\right )+2 b^2 \tanh ^{-1}(c x)^2 \left (a \left (3-3 c^4 x^4\right )+b c x \left (-4 c^3 x^3+3 c^2 x^2+1\right )\right )+2 b^3 c^3 x^3-2 b^3 \left (c^4 x^4-1\right ) \tanh ^{-1}(c x)^3}{8 x^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcTanh[c*x])^3/x^5,x]

[Out]

-(2*a^3 + 2*a^2*b*c*x + 2*a*b^2*c^2*x^2 + 6*a^2*b*c^3*x^3 + 2*b^3*c^3*x^3 - 2*a*b^2*c^4*x^4 + 2*b^2*(b*c*x*(1
+ 3*c^2*x^2 - 4*c^3*x^3) + a*(3 - 3*c^4*x^4))*ArcTanh[c*x]^2 - 2*b^3*(-1 + c^4*x^4)*ArcTanh[c*x]^3 + 2*b*ArcTa
nh[c*x]*(3*a^2 + b^2*c^2*x^2*(1 - c^2*x^2) + 2*a*b*c*x*(1 + 3*c^2*x^2) - 8*b^2*c^4*x^4*Log[1 - E^(-2*ArcTanh[c
*x])]) + 3*a^2*b*c^4*x^4*Log[1 - c*x] - 3*a^2*b*c^4*x^4*Log[1 + c*x] - 16*a*b^2*c^4*x^4*Log[(c*x)/Sqrt[1 - c^2
*x^2]] + 8*b^3*c^4*x^4*PolyLog[2, E^(-2*ArcTanh[c*x])])/(8*x^4)

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Maple [C]  time = 0.873, size = 1281, normalized size = 6.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x))^3/x^5,x)

[Out]

1/4*c^4*b^3*arctanh(c*x)^3-c^4*b^3*arctanh(c*x)^2+2*c^4*b^3*dilog(1+(c*x+1)/(-c^2*x^2+1)^(1/2))-2*c^4*b^3*dilo
g((c*x+1)/(-c^2*x^2+1)^(1/2))-1/4*b^3/x^4*arctanh(c*x)^3-1/2*c*a*b^2*arctanh(c*x)/x^3-3/8*I*c^4*b^3*Pi*csgn(I*
(c*x+1)/(-c^2*x^2+1)^(1/2))*csgn(I*(c*x+1)^2/(c^2*x^2-1))^2*arctanh(c*x)^2-3/16*I*c^4*b^3*Pi*csgn(I/((c*x+1)^2
/(-c^2*x^2+1)+1))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^2*arctanh(c*x)^2-3/16*I*c^4*b^3*Pi*
csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))^2*csgn(I*(c*x+1)^2/(c^2*x^2-1))*arctanh(c*x)^2+3/16*I*c^4*b^3*Pi*csgn(I*(c*
x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^2*arctanh(c*x)^2+1/4*b^3*c^4*arct
anh(c*x)-1/4*a^3/x^4+3/16*I*c^4*b^3*Pi*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I
*(c*x+1)^2/(c^2*x^2-1)/((c*x+1)^2/(-c^2*x^2+1)+1))*arctanh(c*x)^2-3/8*c^4*a^2*b*ln(c*x-1)+3/8*c^4*a^2*b*ln(c*x
+1)-3/16*c^4*a*b^2*ln(c*x-1)^2-3/16*c^4*a*b^2*ln(c*x+1)^2-3/8*c^4*b^3*arctanh(c*x)^2*ln(c*x-1)+3/8*c^4*b^3*arc
tanh(c*x)^2*ln(c*x+1)-3/4*c^4*b^3*arctanh(c*x)^2*ln((c*x+1)/(-c^2*x^2+1)^(1/2))+2*c^4*b^3*arctanh(c*x)*ln(1+(c
*x+1)/(-c^2*x^2+1)^(1/2))-1/4*c^4*b^3/(c*x+1-(-c^2*x^2+1)^(1/2))*(-c^2*x^2+1)^(1/2)+1/4*c^4*b^3/((-c^2*x^2+1)^
(1/2)+c*x+1)*(-c^2*x^2+1)^(1/2)-1/4*c*b^3*arctanh(c*x)^2/x^3-1/4*c^2*b^3*arctanh(c*x)/x^2-3/4*c^3*b^3*arctanh(
c*x)^2/x+3/8*I*c^4*b^3*Pi*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))^3*arctanh(c*x)^2-3/4*c^3*a^2*b/x-1/4*c^2*a*b^2/x^
2-1/4*c*a^2*b/x^3-3/4*a*b^2/x^4*arctanh(c*x)^2-3/4*a^2*b/x^4*arctanh(c*x)+2*c^4*a*b^2*ln(c*x)-c^4*a*b^2*ln(c*x
-1)-c^4*a*b^2*ln(c*x+1)-3/16*I*c^4*b^3*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))^3*arctanh(c*x)^2-3/8*I*c^4*b^3*Pi*csgn
(I/((c*x+1)^2/(-c^2*x^2+1)+1))^2*arctanh(c*x)^2-3/16*I*c^4*b^3*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1)/((c*x+1)^2/(-c^
2*x^2+1)+1))^3*arctanh(c*x)^2-3/2*c^3*a*b^2/x*arctanh(c*x)-3/4*c^4*a*b^2*arctanh(c*x)*ln(c*x-1)+3/4*c^4*a*b^2*
arctanh(c*x)*ln(c*x+1)+3/8*c^4*a*b^2*ln(c*x-1)*ln(1/2+1/2*c*x)+3/8*c^4*a*b^2*ln(-1/2*c*x+1/2)*ln(c*x+1)-3/8*c^
4*a*b^2*ln(-1/2*c*x+1/2)*ln(1/2+1/2*c*x)+3/8*I*c^4*b^3*Pi*arctanh(c*x)^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{8} \,{\left ({\left (3 \, c^{3} \log \left (c x + 1\right ) - 3 \, c^{3} \log \left (c x - 1\right ) - \frac{2 \,{\left (3 \, c^{2} x^{2} + 1\right )}}{x^{3}}\right )} c - \frac{6 \, \operatorname{artanh}\left (c x\right )}{x^{4}}\right )} a^{2} b + \frac{1}{16} \,{\left ({\left (32 \, c^{2} \log \left (x\right ) - \frac{3 \, c^{2} x^{2} \log \left (c x + 1\right )^{2} + 3 \, c^{2} x^{2} \log \left (c x - 1\right )^{2} + 16 \, c^{2} x^{2} \log \left (c x - 1\right ) - 2 \,{\left (3 \, c^{2} x^{2} \log \left (c x - 1\right ) - 8 \, c^{2} x^{2}\right )} \log \left (c x + 1\right ) + 4}{x^{2}}\right )} c^{2} + 4 \,{\left (3 \, c^{3} \log \left (c x + 1\right ) - 3 \, c^{3} \log \left (c x - 1\right ) - \frac{2 \,{\left (3 \, c^{2} x^{2} + 1\right )}}{x^{3}}\right )} c \operatorname{artanh}\left (c x\right )\right )} a b^{2} - \frac{1}{32} \, b^{3}{\left (\frac{{\left (c^{4} x^{4} - 1\right )} \log \left (-c x + 1\right )^{3} +{\left (6 \, c^{3} x^{3} + 2 \, c x - 3 \,{\left (c^{4} x^{4} - 1\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )^{2}}{x^{4}} + 4 \, \int -\frac{2 \,{\left (c x - 1\right )} \log \left (c x + 1\right )^{3} +{\left (6 \, c^{4} x^{4} + 2 \, c^{2} x^{2} - 6 \,{\left (c x - 1\right )} \log \left (c x + 1\right )^{2} - 3 \,{\left (c^{5} x^{5} - c x\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{2 \,{\left (c x^{6} - x^{5}\right )}}\,{d x}\right )} - \frac{3 \, a b^{2} \operatorname{artanh}\left (c x\right )^{2}}{4 \, x^{4}} - \frac{a^{3}}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))^3/x^5,x, algorithm="maxima")

[Out]

1/8*((3*c^3*log(c*x + 1) - 3*c^3*log(c*x - 1) - 2*(3*c^2*x^2 + 1)/x^3)*c - 6*arctanh(c*x)/x^4)*a^2*b + 1/16*((
32*c^2*log(x) - (3*c^2*x^2*log(c*x + 1)^2 + 3*c^2*x^2*log(c*x - 1)^2 + 16*c^2*x^2*log(c*x - 1) - 2*(3*c^2*x^2*
log(c*x - 1) - 8*c^2*x^2)*log(c*x + 1) + 4)/x^2)*c^2 + 4*(3*c^3*log(c*x + 1) - 3*c^3*log(c*x - 1) - 2*(3*c^2*x
^2 + 1)/x^3)*c*arctanh(c*x))*a*b^2 - 1/32*b^3*(((c^4*x^4 - 1)*log(-c*x + 1)^3 + (6*c^3*x^3 + 2*c*x - 3*(c^4*x^
4 - 1)*log(c*x + 1))*log(-c*x + 1)^2)/x^4 + 4*integrate(-1/2*(2*(c*x - 1)*log(c*x + 1)^3 + (6*c^4*x^4 + 2*c^2*
x^2 - 6*(c*x - 1)*log(c*x + 1)^2 - 3*(c^5*x^5 - c*x)*log(c*x + 1))*log(-c*x + 1))/(c*x^6 - x^5), x)) - 3/4*a*b
^2*arctanh(c*x)^2/x^4 - 1/4*a^3/x^4

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{3} \operatorname{artanh}\left (c x\right )^{3} + 3 \, a b^{2} \operatorname{artanh}\left (c x\right )^{2} + 3 \, a^{2} b \operatorname{artanh}\left (c x\right ) + a^{3}}{x^{5}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))^3/x^5,x, algorithm="fricas")

[Out]

integral((b^3*arctanh(c*x)^3 + 3*a*b^2*arctanh(c*x)^2 + 3*a^2*b*arctanh(c*x) + a^3)/x^5, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{atanh}{\left (c x \right )}\right )^{3}}{x^{5}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x))**3/x**5,x)

[Out]

Integral((a + b*atanh(c*x))**3/x**5, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{artanh}\left (c x\right ) + a\right )}^{3}}{x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))^3/x^5,x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x) + a)^3/x^5, x)