\(\int \frac {(a+b \text {arctanh}(c x))^3}{1+c x} \, dx\) [123]

Optimal result

Integrand size = 18, antiderivative size = 111 \[ \int \frac {(a+b \text {arctanh}(c x))^3}{1+c x} \, dx=-\frac {(a+b \text {arctanh}(c x))^3 \log \left (\frac {2}{1+c x}\right )}{c}+\frac {3 b (a+b \text {arctanh}(c x))^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1+c x}\right )}{2 c}+\frac {3 b^2 (a+b \text {arctanh}(c x)) \operatorname {PolyLog}\left (3,1-\frac {2}{1+c x}\right )}{2 c}+\frac {3 b^3 \operatorname {PolyLog}\left (4,1-\frac {2}{1+c x}\right )}{4 c} \] Output:

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

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.37 \[ \int \frac {(a+b \text {arctanh}(c x))^3}{1+c x} \, dx=\frac {-12 a^2 b \text {arctanh}(c x) \log \left (1+e^{-2 \text {arctanh}(c x)}\right )-12 a b^2 \text {arctanh}(c x)^2 \log \left (1+e^{-2 \text {arctanh}(c x)}\right )-4 b^3 \text {arctanh}(c x)^3 \log \left (1+e^{-2 \text {arctanh}(c x)}\right )+4 a^3 \log (1+c x)+6 b (a+b \text {arctanh}(c x))^2 \operatorname {PolyLog}\left (2,-e^{-2 \text {arctanh}(c x)}\right )+6 b^2 (a+b \text {arctanh}(c x)) \operatorname {PolyLog}\left (3,-e^{-2 \text {arctanh}(c x)}\right )+3 b^3 \operatorname {PolyLog}\left (4,-e^{-2 \text {arctanh}(c x)}\right )}{4 c} \] Input:

Integrate[(a + b*ArcTanh[c*x])^3/(1 + c*x),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6470, 6618, 6622, 7164}

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.

\(\displaystyle \int \frac {(a+b \text {arctanh}(c x))^3}{c x+1} \, dx\)

\(\Big \downarrow \) 6470

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

\(\Big \downarrow \) 6618

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

\(\Big \downarrow \) 6622

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

\(\Big \downarrow \) 7164

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

Input:

Int[(a + b*ArcTanh[c*x])^3/(1 + c*x),x]
 

Output:

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

Defintions of rubi rules used

rule 6470

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol 
] :> Simp[(-(a + b*ArcTanh[c*x])^p)*(Log[2/(1 + e*(x/d))]/e), x] + Simp[b*c 
*(p/e)   Int[(a + b*ArcTanh[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 6618

Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^ 
2), x_Symbol] :> Simp[(a + b*ArcTanh[c*x])^p*(PolyLog[2, 1 - u]/(2*c*d)), x 
] - Simp[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 6622

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] + Simp[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]
 

rule 7164

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]
 

Maple [C] (warning: unable to verify)

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

Time = 1.54 (sec) , antiderivative size = 1215, normalized size of antiderivative = 10.95

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1215\)
default	\(\text {Expression too large to display}\)	\(1215\)
parts	\(\text {Expression too large to display}\)	\(1223\)

Input:

int((a+b*arctanh(c*x))^3/(c*x+1),x,method=_RETURNVERBOSE)
 

Output:

1/c*(a^3*ln(c*x+1)+b^3*(ln(c*x+1)*arctanh(c*x)^3-2*arctanh(c*x)^3*ln((c*x+ 
1)/(-c^2*x^2+1)^(1/2))+1/2*arctanh(c*x)^4-3/2*arctanh(c*x)^2*polylog(2,-(c 
*x+1)^2/(-c^2*x^2+1))+3/2*arctanh(c*x)*polylog(3,-(c*x+1)^2/(-c^2*x^2+1))- 
3/4*polylog(4,-(c*x+1)^2/(-c^2*x^2+1))-1/2*(I*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))+2*I*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+I*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1 
))^3-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-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)))*csgn(I/(1-(c*x+1)^2/(c^2*x^2-1))) 
+I*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1-(c*x+1)^2/(c^2*x^2-1)))^2*csgn(I/(1- 
(c*x+1)^2/(c^2*x^2-1)))+I*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1-(c*x+1)^2/(c^ 
2*x^2-1)))^3+2*ln(2))*arctanh(c*x)^3)+3*a*b^2*(arctanh(c*x)^2*ln(c*x+1)-2* 
arctanh(c*x)^2*ln((c*x+1)/(-c^2*x^2+1)^(1/2))+2/3*arctanh(c*x)^3-1/2*(I*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))+2*I*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+I*Pi*c 
sgn(I*(c*x+1)^2/(c^2*x^2-1))^3-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-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)))*csgn(I/ 
(1-(c*x+1)^2/(c^2*x^2-1)))+I*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1-(c*x+1)^2/ 
(c^2*x^2-1)))^2*csgn(I/(1-(c*x+1)^2/(c^2*x^2-1)))+I*Pi*csgn(I*(c*x+1)^2...
 

Fricas [F]

\[ \int \frac {(a+b \text {arctanh}(c x))^3}{1+c x} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{3}}{c x + 1} \,d x } \] Input:

integrate((a+b*arctanh(c*x))^3/(c*x+1),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)/(c*x + 1), x)
 

Sympy [F]

\[ \int \frac {(a+b \text {arctanh}(c x))^3}{1+c x} \, dx=\int \frac {\left (a + b \operatorname {atanh}{\left (c x \right )}\right )^{3}}{c x + 1}\, dx \] Input:

integrate((a+b*atanh(c*x))**3/(c*x+1),x)
 

Output:

Integral((a + b*atanh(c*x))**3/(c*x + 1), x)
 

Maxima [F]

\[ \int \frac {(a+b \text {arctanh}(c x))^3}{1+c x} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{3}}{c x + 1} \,d x } \] Input:

integrate((a+b*arctanh(c*x))^3/(c*x+1),x, algorithm="maxima")
 

Output:

-1/8*b^3*log(c*x + 1)*log(-c*x + 1)^3/c + a^3*log(c*x + 1)/c + integrate(1 
/8*((b^3*c*x - b^3)*log(c*x + 1)^3 + 6*(a*b^2*c*x - a*b^2)*log(c*x + 1)^2 
+ 6*(b^3*c*x*log(c*x + 1) + a*b^2*c*x - a*b^2)*log(-c*x + 1)^2 + 12*(a^2*b 
*c*x - a^2*b)*log(c*x + 1) - 3*(4*a^2*b*c*x - 4*a^2*b + (b^3*c*x - b^3)*lo 
g(c*x + 1)^2 + 4*(a*b^2*c*x - a*b^2)*log(c*x + 1))*log(-c*x + 1))/(c^2*x^2 
 - 1), x)
 

Giac [F]

\[ \int \frac {(a+b \text {arctanh}(c x))^3}{1+c x} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{3}}{c x + 1} \,d x } \] Input:

integrate((a+b*arctanh(c*x))^3/(c*x+1),x, algorithm="giac")
 

Output:

integrate((b*arctanh(c*x) + a)^3/(c*x + 1), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \text {arctanh}(c x))^3}{1+c x} \, dx=\int \frac {{\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}^3}{c\,x+1} \,d x \] Input:

int((a + b*atanh(c*x))^3/(c*x + 1),x)
 

Output:

int((a + b*atanh(c*x))^3/(c*x + 1), x)
 

Reduce [F]

\[ \int \frac {(a+b \text {arctanh}(c x))^3}{1+c x} \, dx=\frac {3 \left (\int \frac {\mathit {atanh} \left (c x \right )}{c x +1}d x \right ) a^{2} b c +\left (\int \frac {\mathit {atanh} \left (c x \right )^{3}}{c x +1}d x \right ) b^{3} c +3 \left (\int \frac {\mathit {atanh} \left (c x \right )^{2}}{c x +1}d x \right ) a \,b^{2} c +\mathrm {log}\left (c x +1\right ) a^{3}}{c} \] Input:

int((a+b*atanh(c*x))^3/(c*x+1),x)
                                                                                    
                                                                                    
 

Output:

(3*int(atanh(c*x)/(c*x + 1),x)*a**2*b*c + int(atanh(c*x)**3/(c*x + 1),x)*b 
**3*c + 3*int(atanh(c*x)**2/(c*x + 1),x)*a*b**2*c + log(c*x + 1)*a**3)/c