3.1.34 \(\int \frac {(a+b \text {arctanh}(c x))^3}{x^5} \, dx\) [34]

3.1.34.1 Optimal result

Integrand size = 14, antiderivative size = 187 \[ \int \frac {(a+b \text {arctanh}(c x))^3}{x^5} \, dx=-\frac {b^3 c^3}{4 x}+\frac {1}{4} b^3 c^4 \text {arctanh}(c x)-\frac {b^2 c^2 (a+b \text {arctanh}(c x))}{4 x^2}+b c^4 (a+b \text {arctanh}(c x))^2-\frac {b c (a+b \text {arctanh}(c x))^2}{4 x^3}-\frac {3 b c^3 (a+b \text {arctanh}(c x))^2}{4 x}+\frac {1}{4} c^4 (a+b \text {arctanh}(c x))^3-\frac {(a+b \text {arctanh}(c x))^3}{4 x^4}+2 b^2 c^4 (a+b \text {arctanh}(c x)) \log \left (2-\frac {2}{1+c x}\right )-b^3 c^4 \operatorname {PolyLog}\left (2,-1+\frac {2}{1+c x}\right ) \]

output

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

3.1.34.2 Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.58 \[ \int \frac {(a+b \text {arctanh}(c x))^3}{x^5} \, dx=-\frac {2 a^3+2 a^2 b c x+2 a b^2 c^2 x^2+6 a^2 b c^3 x^3+2 b^3 c^3 x^3-2 a b^2 c^4 x^4+2 b^2 \left (b c x \left (1+3 c^2 x^2-4 c^3 x^3\right )+a \left (3-3 c^4 x^4\right )\right ) \text {arctanh}(c x)^2-2 b^3 \left (-1+c^4 x^4\right ) \text {arctanh}(c x)^3+2 b \text {arctanh}(c x) \left (3 a^2+b^2 c^2 x^2 \left (1-c^2 x^2\right )+2 a b c x \left (1+3 c^2 x^2\right )-8 b^2 c^4 x^4 \log \left (1-e^{-2 \text {arctanh}(c x)}\right )\right )+3 a^2 b c^4 x^4 \log (1-c x)-3 a^2 b c^4 x^4 \log (1+c x)-16 a b^2 c^4 x^4 \log \left (\frac {c x}{\sqrt {1-c^2 x^2}}\right )+8 b^3 c^4 x^4 \operatorname {PolyLog}\left (2,e^{-2 \text {arctanh}(c x)}\right )}{8 x^4} \]

input

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

output

-1/8*(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*Arc 
Tanh[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])])/x^4
 

3.1.34.3 Rubi [A] (verified)

Time = 1.82 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.29, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.786, Rules used = {6452, 6544, 6452, 6544, 6452, 264, 219, 6510, 6550, 6494, 2897}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {(a+b \text {arctanh}(c x))^3}{x^5} \, dx\)

\(\Big \downarrow \) 6452

\(\displaystyle \frac {3}{4} b c \int \frac {(a+b \text {arctanh}(c x))^2}{x^4 \left (1-c^2 x^2\right )}dx-\frac {(a+b \text {arctanh}(c x))^3}{4 x^4}\)

\(\Big \downarrow \) 6544

\(\displaystyle \frac {3}{4} b c \left (c^2 \int \frac {(a+b \text {arctanh}(c x))^2}{x^2 \left (1-c^2 x^2\right )}dx+\int \frac {(a+b \text {arctanh}(c x))^2}{x^4}dx\right )-\frac {(a+b \text {arctanh}(c x))^3}{4 x^4}\)

\(\Big \downarrow \) 6452

\(\displaystyle \frac {3}{4} b c \left (c^2 \int \frac {(a+b \text {arctanh}(c x))^2}{x^2 \left (1-c^2 x^2\right )}dx+\frac {2}{3} b c \int \frac {a+b \text {arctanh}(c x)}{x^3 \left (1-c^2 x^2\right )}dx-\frac {(a+b \text {arctanh}(c x))^2}{3 x^3}\right )-\frac {(a+b \text {arctanh}(c x))^3}{4 x^4}\)

\(\Big \downarrow \) 6544

\(\displaystyle \frac {3}{4} b c \left (c^2 \left (c^2 \int \frac {(a+b \text {arctanh}(c x))^2}{1-c^2 x^2}dx+\int \frac {(a+b \text {arctanh}(c x))^2}{x^2}dx\right )+\frac {2}{3} b c \left (c^2 \int \frac {a+b \text {arctanh}(c x)}{x \left (1-c^2 x^2\right )}dx+\int \frac {a+b \text {arctanh}(c x)}{x^3}dx\right )-\frac {(a+b \text {arctanh}(c x))^2}{3 x^3}\right )-\frac {(a+b \text {arctanh}(c x))^3}{4 x^4}\)

\(\Big \downarrow \) 6452

\(\displaystyle \frac {3}{4} b c \left (c^2 \left (c^2 \int \frac {(a+b \text {arctanh}(c x))^2}{1-c^2 x^2}dx+2 b c \int \frac {a+b \text {arctanh}(c x)}{x \left (1-c^2 x^2\right )}dx-\frac {(a+b \text {arctanh}(c x))^2}{x}\right )+\frac {2}{3} b c \left (c^2 \int \frac {a+b \text {arctanh}(c x)}{x \left (1-c^2 x^2\right )}dx+\frac {1}{2} b c \int \frac {1}{x^2 \left (1-c^2 x^2\right )}dx-\frac {a+b \text {arctanh}(c x)}{2 x^2}\right )-\frac {(a+b \text {arctanh}(c x))^2}{3 x^3}\right )-\frac {(a+b \text {arctanh}(c x))^3}{4 x^4}\)

\(\Big \downarrow \) 264

\(\displaystyle \frac {3}{4} b c \left (c^2 \left (c^2 \int \frac {(a+b \text {arctanh}(c x))^2}{1-c^2 x^2}dx+2 b c \int \frac {a+b \text {arctanh}(c x)}{x \left (1-c^2 x^2\right )}dx-\frac {(a+b \text {arctanh}(c x))^2}{x}\right )+\frac {2}{3} b c \left (c^2 \int \frac {a+b \text {arctanh}(c x)}{x \left (1-c^2 x^2\right )}dx+\frac {1}{2} b c \left (c^2 \int \frac {1}{1-c^2 x^2}dx-\frac {1}{x}\right )-\frac {a+b \text {arctanh}(c x)}{2 x^2}\right )-\frac {(a+b \text {arctanh}(c x))^2}{3 x^3}\right )-\frac {(a+b \text {arctanh}(c x))^3}{4 x^4}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {3}{4} b c \left (c^2 \left (c^2 \int \frac {(a+b \text {arctanh}(c x))^2}{1-c^2 x^2}dx+2 b c \int \frac {a+b \text {arctanh}(c x)}{x \left (1-c^2 x^2\right )}dx-\frac {(a+b \text {arctanh}(c x))^2}{x}\right )+\frac {2}{3} b c \left (c^2 \int \frac {a+b \text {arctanh}(c x)}{x \left (1-c^2 x^2\right )}dx-\frac {a+b \text {arctanh}(c x)}{2 x^2}+\frac {1}{2} b c \left (c \text {arctanh}(c x)-\frac {1}{x}\right )\right )-\frac {(a+b \text {arctanh}(c x))^2}{3 x^3}\right )-\frac {(a+b \text {arctanh}(c x))^3}{4 x^4}\)

\(\Big \downarrow \) 6510

\(\displaystyle \frac {3}{4} b c \left (c^2 \left (2 b c \int \frac {a+b \text {arctanh}(c x)}{x \left (1-c^2 x^2\right )}dx+\frac {c (a+b \text {arctanh}(c x))^3}{3 b}-\frac {(a+b \text {arctanh}(c x))^2}{x}\right )+\frac {2}{3} b c \left (c^2 \int \frac {a+b \text {arctanh}(c x)}{x \left (1-c^2 x^2\right )}dx-\frac {a+b \text {arctanh}(c x)}{2 x^2}+\frac {1}{2} b c \left (c \text {arctanh}(c x)-\frac {1}{x}\right )\right )-\frac {(a+b \text {arctanh}(c x))^2}{3 x^3}\right )-\frac {(a+b \text {arctanh}(c x))^3}{4 x^4}\)

\(\Big \downarrow \) 6550

\(\displaystyle \frac {3}{4} b c \left (\frac {2}{3} b c \left (c^2 \left (\int \frac {a+b \text {arctanh}(c x)}{x (c x+1)}dx+\frac {(a+b \text {arctanh}(c x))^2}{2 b}\right )-\frac {a+b \text {arctanh}(c x)}{2 x^2}+\frac {1}{2} b c \left (c \text {arctanh}(c x)-\frac {1}{x}\right )\right )+c^2 \left (2 b c \left (\int \frac {a+b \text {arctanh}(c x)}{x (c x+1)}dx+\frac {(a+b \text {arctanh}(c x))^2}{2 b}\right )+\frac {c (a+b \text {arctanh}(c x))^3}{3 b}-\frac {(a+b \text {arctanh}(c x))^2}{x}\right )-\frac {(a+b \text {arctanh}(c x))^2}{3 x^3}\right )-\frac {(a+b \text {arctanh}(c x))^3}{4 x^4}\)

\(\Big \downarrow \) 6494

\(\displaystyle \frac {3}{4} b c \left (c^2 \left (2 b c \left (-b c \int \frac {\log \left (2-\frac {2}{c x+1}\right )}{1-c^2 x^2}dx+\frac {(a+b \text {arctanh}(c x))^2}{2 b}+\log \left (2-\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))\right )+\frac {c (a+b \text {arctanh}(c x))^3}{3 b}-\frac {(a+b \text {arctanh}(c x))^2}{x}\right )+\frac {2}{3} b c \left (c^2 \left (-b c \int \frac {\log \left (2-\frac {2}{c x+1}\right )}{1-c^2 x^2}dx+\frac {(a+b \text {arctanh}(c x))^2}{2 b}+\log \left (2-\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))\right )-\frac {a+b \text {arctanh}(c x)}{2 x^2}+\frac {1}{2} b c \left (c \text {arctanh}(c x)-\frac {1}{x}\right )\right )-\frac {(a+b \text {arctanh}(c x))^2}{3 x^3}\right )-\frac {(a+b \text {arctanh}(c x))^3}{4 x^4}\)

\(\Big \downarrow \) 2897

\(\displaystyle \frac {3}{4} b c \left (\frac {2}{3} b c \left (c^2 \left (\frac {(a+b \text {arctanh}(c x))^2}{2 b}+\log \left (2-\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))-\frac {1}{2} b \operatorname {PolyLog}\left (2,\frac {2}{c x+1}-1\right )\right )-\frac {a+b \text {arctanh}(c x)}{2 x^2}+\frac {1}{2} b c \left (c \text {arctanh}(c x)-\frac {1}{x}\right )\right )+c^2 \left (2 b c \left (\frac {(a+b \text {arctanh}(c x))^2}{2 b}+\log \left (2-\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))-\frac {1}{2} b \operatorname {PolyLog}\left (2,\frac {2}{c x+1}-1\right )\right )+\frac {c (a+b \text {arctanh}(c x))^3}{3 b}-\frac {(a+b \text {arctanh}(c x))^2}{x}\right )-\frac {(a+b \text {arctanh}(c x))^2}{3 x^3}\right )-\frac {(a+b \text {arctanh}(c x))^3}{4 x^4}\)

input

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

output

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

3.1.34.3.1 Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 264

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( 
m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c 
^2*(m + 1)))   Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p 
}, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
 

rule 2897

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[RationalFunctionExponents[u, 
 x][[2]], Expon[Pq, x]]
 

rule 6452

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] : 
> Simp[x^(m + 1)*((a + b*ArcTanh[c*x^n])^p/(m + 1)), x] - Simp[b*c*n*(p/(m 
+ 1))   Int[x^(m + n)*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))), x 
], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1 
] && IntegerQ[m])) && NeQ[m, -1]
 

rule 6494

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] - 
Simp[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 6510

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symb 
ol] :> 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 6544

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + ( 
e_.)*(x_)^2), x_Symbol] :> Simp[1/d   Int[(f*x)^m*(a + b*ArcTanh[c*x])^p, x 
], x] - Simp[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 6550

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] + Simp[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]
 

3.1.34.4 Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 18.94 (sec) , antiderivative size = 1136, normalized size of antiderivative = 6.07

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1136\)
default	\(\text {Expression too large to display}\)	\(1136\)
parts	\(\text {Expression too large to display}\)	\(1193\)

input

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

output

c^4*(-3/16*I*b^3*Pi*csgn(I/(1-(c*x+1)^2/(c^2*x^2-1)))*csgn(I*(c*x+1)^2/(c^ 
2*x^2-1)/(1-(c*x+1)^2/(c^2*x^2-1)))^2*arctanh(c*x)^2-3/8*I*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*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*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)/(1-(c*x+1)^2/(c^2*x^2-1)))^2*arctanh(c*x)^2+2*b^3*dil 
og(1+(c*x+1)/(-c^2*x^2+1)^(1/2))-3/4*b^3*arctanh(c*x)^2*ln((c*x+1)/(-c^2*x 
^2+1)^(1/2))-3/8*b^3*arctanh(c*x)^2*ln(c*x-1)+3/8*b^3*arctanh(c*x)^2*ln(c* 
x+1)+3/16*I*b^3*Pi*csgn(I/(1-(c*x+1)^2/(c^2*x^2-1)))*csgn(I*(c*x+1)^2/(c^2 
*x^2-1))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1-(c*x+1)^2/(c^2*x^2-1)))*arctanh(c 
*x)^2-3/4*b^3/c/x*arctanh(c*x)^2+1/4*b^3*arctanh(c*x)^3-b^3*arctanh(c*x)^2 
+1/4*b^3*arctanh(c*x)-2*b^3*dilog((c*x+1)/(-c^2*x^2+1)^(1/2))+3*a*b^2*(-1/ 
4/c^4/x^4*arctanh(c*x)^2-1/6/c^3/x^3*arctanh(c*x)-1/2/c/x*arctanh(c*x)+1/4 
*arctanh(c*x)*ln(c*x+1)-1/4*arctanh(c*x)*ln(c*x-1)+1/8*ln(c*x-1)*ln(1/2*c* 
x+1/2)-1/16*ln(c*x-1)^2+1/8*(ln(c*x+1)-ln(1/2*c*x+1/2))*ln(-1/2*c*x+1/2)-1 
/16*ln(c*x+1)^2-1/12/c^2/x^2+2/3*ln(c*x)-1/3*ln(c*x+1)-1/3*ln(c*x-1))-3/16 
*I*b^3*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1-(c*x+1)^2/(c^2*x^2-1)))^3*arctan 
h(c*x)^2+3/8*I*b^3*Pi*csgn(I/(1-(c*x+1)^2/(c^2*x^2-1)))^3*arctanh(c*x)^2-3 
/8*I*b^3*Pi*csgn(I/(1-(c*x+1)^2/(c^2*x^2-1)))^2*arctanh(c*x)^2-3/16*I*b^3* 
Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))^3*arctanh(c*x)^2-1/4*a^3/c^4/x^4+2*b^3...
 

3.1.34.5 Fricas [F]

\[ \int \frac {(a+b \text {arctanh}(c x))^3}{x^5} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{3}}{x^{5}} \,d x } \]

input

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

output

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)
 

3.1.34.6 Sympy [F]

\[ \int \frac {(a+b \text {arctanh}(c x))^3}{x^5} \, dx=\int \frac {\left (a + b \operatorname {atanh}{\left (c x \right )}\right )^{3}}{x^{5}}\, dx \]

input

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

output

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

3.1.34.7 Maxima [F]

\[ \int \frac {(a+b \text {arctanh}(c x))^3}{x^5} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{3}}{x^{5}} \,d x } \]

input

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

output

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*lo 
g(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 - 1/3 
2*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*arc 
tanh(c*x)^2/x^4 - 1/4*a^3/x^4
 

3.1.34.8 Giac [F]

\[ \int \frac {(a+b \text {arctanh}(c x))^3}{x^5} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{3}}{x^{5}} \,d x } \]

input

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

output

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

3.1.34.9 Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \text {arctanh}(c x))^3}{x^5} \, dx=\int \frac {{\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}^3}{x^5} \,d x \]

input

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

output

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