\(\int \frac {(a+b \text {arctanh}(c x))^3}{x^3} \, dx\) [32]

Optimal result

Integrand size = 14, antiderivative size = 123 \[ \int \frac {(a+b \text {arctanh}(c x))^3}{x^3} \, dx=\frac {3}{2} b c^2 (a+b \text {arctanh}(c x))^2-\frac {3 b c (a+b \text {arctanh}(c x))^2}{2 x}+\frac {1}{2} c^2 (a+b \text {arctanh}(c x))^3-\frac {(a+b \text {arctanh}(c x))^3}{2 x^2}+3 b^2 c^2 (a+b \text {arctanh}(c x)) \log \left (2-\frac {2}{1+c x}\right )-\frac {3}{2} b^3 c^2 \operatorname {PolyLog}\left (2,-1+\frac {2}{1+c x}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.56 \[ \int \frac {(a+b \text {arctanh}(c x))^3}{x^3} \, dx=\frac {6 b^2 (-1+c x) (a+a c x+b c x) \text {arctanh}(c x)^2+2 b^3 \left (-1+c^2 x^2\right ) \text {arctanh}(c x)^3-6 b \text {arctanh}(c x) \left (a^2+2 a b c x-2 b^2 c^2 x^2 \log \left (1-e^{-2 \text {arctanh}(c x)}\right )\right )+a \left (-2 a^2-6 a b c x-3 a b c^2 x^2 \log (1-c x)+3 a b c^2 x^2 \log (1+c x)+12 b^2 c^2 x^2 \log \left (\frac {c x}{\sqrt {1-c^2 x^2}}\right )\right )-6 b^3 c^2 x^2 \operatorname {PolyLog}\left (2,e^{-2 \text {arctanh}(c x)}\right )}{4 x^2} \] Input:

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

Output:

(6*b^2*(-1 + c*x)*(a + a*c*x + b*c*x)*ArcTanh[c*x]^2 + 2*b^3*(-1 + c^2*x^2 
)*ArcTanh[c*x]^3 - 6*b*ArcTanh[c*x]*(a^2 + 2*a*b*c*x - 2*b^2*c^2*x^2*Log[1 
 - E^(-2*ArcTanh[c*x])]) + a*(-2*a^2 - 6*a*b*c*x - 3*a*b*c^2*x^2*Log[1 - c 
*x] + 3*a*b*c^2*x^2*Log[1 + c*x] + 12*b^2*c^2*x^2*Log[(c*x)/Sqrt[1 - c^2*x 
^2]]) - 6*b^3*c^2*x^2*PolyLog[2, E^(-2*ArcTanh[c*x])])/(4*x^2)
 

Rubi [A] (verified)

Time = 1.10 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.97, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6452, 6544, 6452, 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.

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

\(\Big \downarrow \) 6452

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

\(\Big \downarrow \) 6544

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

\(\Big \downarrow \) 6452

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

\(\Big \downarrow \) 6510

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

\(\Big \downarrow \) 6550

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

\(\Big \downarrow \) 6494

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

\(\Big \downarrow \) 2897

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

Input:

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

Output:

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

Defintions of rubi rules used

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]
 

Maple [C] (warning: unable to verify)

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

Time = 8.77 (sec) , antiderivative size = 4599, normalized size of antiderivative = 37.39

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(4599\)
default	\(\text {Expression too large to display}\)	\(4599\)
parts	\(\text {Expression too large to display}\)	\(4601\)

Input:

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

Output:

c^2*(-1/2*a^3/c^2/x^2+b^3*(3/8*I*Pi*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))^2*c 
sgn(I*(c*x+1)^2/(c^2*x^2-1))*arctanh(c*x)*ln(1-(c*x+1)/(-c^2*x^2+1)^(1/2)) 
-3/8*I*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)))*polylog(2,-(c*x+1) 
/(-c^2*x^2+1)^(1/2))-3/8*I*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)) 
)*polylog(2,(c*x+1)/(-c^2*x^2+1)^(1/2))+3/8*I*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)))*dilog(1+(c*x+1)/(-c^2*x^2+1)^(1/2))-3/8*I*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)))*dilog((c*x+1)/(-c^2*x^2+1)^(1/2))-3/ 
8*I*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)*ln(1-(c*x+1)/(-c^2*x^2+1)^(1/2))+3/8*I*P 
i*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)^2-3/2*arctanh 
(c*x)^2+3/8*I*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)*ln(1-(c*x+1)/(-c^2*x^2+1)^ 
(1/2))+3/2*polylog(2,-(c*x+1)/(-c^2*x^2+1)^(1/2))-3/8*I*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)*ln(1-(c*x+1)/(-c^2*x^2+1)^(...
 

Fricas [F]

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

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(a+b \text {arctanh}(c x))^3}{x^3} \, dx=\frac {\mathit {atanh} \left (c x \right )^{3} b^{3} c^{2} x^{2}-\mathit {atanh} \left (c x \right )^{3} b^{3}+3 \mathit {atanh} \left (c x \right )^{2} a \,b^{2} c^{2} x^{2}-3 \mathit {atanh} \left (c x \right )^{2} a \,b^{2}-3 \mathit {atanh} \left (c x \right )^{2} b^{3} c x +3 \mathit {atanh} \left (c x \right ) a^{2} b \,c^{2} x^{2}-3 \mathit {atanh} \left (c x \right ) a^{2} b -6 \mathit {atanh} \left (c x \right ) a \,b^{2} c^{2} x^{2}-6 \mathit {atanh} \left (c x \right ) a \,b^{2} c x -3 \mathit {atanh} \left (c x \right ) b^{3} c^{2} x^{2}+3 \mathit {atanh} \left (c x \right ) b^{3}-6 \left (\int \frac {\mathit {atanh} \left (c x \right )}{c^{2} x^{5}-x^{3}}d x \right ) b^{3} x^{2}-6 \,\mathrm {log}\left (c^{2} x -c \right ) a \,b^{2} c^{2} x^{2}+6 \,\mathrm {log}\left (x \right ) a \,b^{2} c^{2} x^{2}-a^{3}-3 a^{2} b c x +3 b^{3} c x}{2 x^{2}} \] Input:

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

Output:

(atanh(c*x)**3*b**3*c**2*x**2 - atanh(c*x)**3*b**3 + 3*atanh(c*x)**2*a*b** 
2*c**2*x**2 - 3*atanh(c*x)**2*a*b**2 - 3*atanh(c*x)**2*b**3*c*x + 3*atanh( 
c*x)*a**2*b*c**2*x**2 - 3*atanh(c*x)*a**2*b - 6*atanh(c*x)*a*b**2*c**2*x** 
2 - 6*atanh(c*x)*a*b**2*c*x - 3*atanh(c*x)*b**3*c**2*x**2 + 3*atanh(c*x)*b 
**3 - 6*int(atanh(c*x)/(c**2*x**5 - x**3),x)*b**3*x**2 - 6*log(c**2*x - c) 
*a*b**2*c**2*x**2 + 6*log(x)*a*b**2*c**2*x**2 - a**3 - 3*a**2*b*c*x + 3*b* 
*3*c*x)/(2*x**2)