\(\int \frac {(a+b \coth ^{-1}(c+d x))^3}{e+f x} \, dx\) [36]

Optimal result

Integrand size = 20, antiderivative size = 308 \[ \int \frac {\left (a+b \coth ^{-1}(c+d x)\right )^3}{e+f x} \, dx=-\frac {\left (a+b \coth ^{-1}(c+d x)\right )^3 \log \left (\frac {2}{1+c+d x}\right )}{f}+\frac {\left (a+b \coth ^{-1}(c+d x)\right )^3 \log \left (\frac {2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{f}+\frac {3 b \left (a+b \coth ^{-1}(c+d x)\right )^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1+c+d x}\right )}{2 f}-\frac {3 b \left (a+b \coth ^{-1}(c+d x)\right )^2 \operatorname {PolyLog}\left (2,1-\frac {2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{2 f}+\frac {3 b^2 \left (a+b \coth ^{-1}(c+d x)\right ) \operatorname {PolyLog}\left (3,1-\frac {2}{1+c+d x}\right )}{2 f}-\frac {3 b^2 \left (a+b \coth ^{-1}(c+d x)\right ) \operatorname {PolyLog}\left (3,1-\frac {2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{2 f}+\frac {3 b^3 \operatorname {PolyLog}\left (4,1-\frac {2}{1+c+d x}\right )}{4 f}-\frac {3 b^3 \operatorname {PolyLog}\left (4,1-\frac {2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{4 f} \] Output:

-(a+b*arccoth(d*x+c))^3*ln(2/(d*x+c+1))/f+(a+b*arccoth(d*x+c))^3*ln(2*d*(f 
*x+e)/(-c*f+d*e+f)/(d*x+c+1))/f+3/2*b*(a+b*arccoth(d*x+c))^2*polylog(2,1-2 
/(d*x+c+1))/f-3/2*b*(a+b*arccoth(d*x+c))^2*polylog(2,1-2*d*(f*x+e)/(-c*f+d 
*e+f)/(d*x+c+1))/f+3/2*b^2*(a+b*arccoth(d*x+c))*polylog(3,1-2/(d*x+c+1))/f 
-3/2*b^2*(a+b*arccoth(d*x+c))*polylog(3,1-2*d*(f*x+e)/(-c*f+d*e+f)/(d*x+c+ 
1))/f+3/4*b^3*polylog(4,1-2/(d*x+c+1))/f-3/4*b^3*polylog(4,1-2*d*(f*x+e)/( 
-c*f+d*e+f)/(d*x+c+1))/f
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 22.97 (sec) , antiderivative size = 3317, normalized size of antiderivative = 10.77 \[ \int \frac {\left (a+b \coth ^{-1}(c+d x)\right )^3}{e+f x} \, dx=\text {Result too large to show} \] Input:

Integrate[(a + b*ArcCoth[c + d*x])^3/(e + f*x),x]
 

Output:

(a^3*Log[e + f*x])/f + 3*a^2*b*(((ArcCoth[c + d*x] - ArcTanh[c + d*x])*Log 
[e + f*x])/f - (I*(I*ArcTanh[c + d*x]*(-Log[1/Sqrt[1 - (c + d*x)^2]] + Log 
[I*Sinh[ArcTanh[(d*e - c*f)/f] + ArcTanh[c + d*x]]]) + ((-I)*(I*ArcTanh[(d 
*e - c*f)/f] + I*ArcTanh[c + d*x])^2 - (I/4)*(Pi - (2*I)*ArcTanh[c + d*x]) 
^2 + 2*(I*ArcTanh[(d*e - c*f)/f] + I*ArcTanh[c + d*x])*Log[1 - E^((2*I)*(I 
*ArcTanh[(d*e - c*f)/f] + I*ArcTanh[c + d*x]))] + (Pi - (2*I)*ArcTanh[c + 
d*x])*Log[1 - E^(I*(Pi - (2*I)*ArcTanh[c + d*x]))] - (Pi - (2*I)*ArcTanh[c 
 + d*x])*Log[2*Sin[(Pi - (2*I)*ArcTanh[c + d*x])/2]] - 2*(I*ArcTanh[(d*e - 
 c*f)/f] + I*ArcTanh[c + d*x])*Log[(2*I)*Sinh[ArcTanh[(d*e - c*f)/f] + Arc 
Tanh[c + d*x]]] - I*PolyLog[2, E^((2*I)*(I*ArcTanh[(d*e - c*f)/f] + I*ArcT 
anh[c + d*x]))] - I*PolyLog[2, E^(I*(Pi - (2*I)*ArcTanh[c + d*x]))])/2))/f 
) - (3*a*b^2*(d*e - c*f + f*(c + d*x))*(1 - (c + d*x)^2)*(-1/24*(I*f*Pi^3 
- 8*d*e*ArcCoth[c + d*x]^3 - 8*f*ArcCoth[c + d*x]^3 + 8*c*f*ArcCoth[c + d* 
x]^3 + 24*f*ArcCoth[c + d*x]^2*Log[1 - E^(2*ArcCoth[c + d*x])] + 24*f*ArcC 
oth[c + d*x]*PolyLog[2, E^(2*ArcCoth[c + d*x])] - 12*f*PolyLog[3, E^(2*Arc 
Coth[c + d*x])])/f^2 + ((-(d*e) - f + c*f)*(-(d*e) + f + c*f)*(2*d*e*ArcCo 
th[c + d*x]^3 - 6*f*ArcCoth[c + d*x]^3 - 2*c*f*ArcCoth[c + d*x]^3 - (4*d*e 
*Sqrt[(d^2*e^2 - 2*c*d*e*f + (-1 + c^2)*f^2)/(d*e - c*f)^2]*ArcCoth[c + d* 
x]^3)/E^ArcTanh[f/(d*e - c*f)] + (4*c*f*Sqrt[(d^2*e^2 - 2*c*d*e*f + (-1 + 
c^2)*f^2)/(d*e - c*f)^2]*ArcCoth[c + d*x]^3)/E^ArcTanh[f/(d*e - c*f)] +...
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 344, normalized size of antiderivative = 1.12, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {6662, 27, 6477}

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.

\(\displaystyle \int \frac {\left (a+b \coth ^{-1}(c+d x)\right )^3}{e+f x} \, dx\)

\(\Big \downarrow \) 6662

\(\displaystyle \frac {\int \frac {d \left (a+b \coth ^{-1}(c+d x)\right )^3}{d \left (e-\frac {c f}{d}\right )+f (c+d x)}d(c+d x)}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \int \frac {\left (a+b \coth ^{-1}(c+d x)\right )^3}{f (c+d x)-c f+d e}d(c+d x)\)

\(\Big \downarrow \) 6477

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

Input:

Int[(a + b*ArcCoth[c + d*x])^3/(e + f*x),x]
 

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 6477

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^3/((d_) + (e_.)*(x_)), x_Symbol] :> 
 Simp[(-(a + b*ArcCoth[c*x])^3)*(Log[2/(1 + c*x)]/e), x] + (Simp[(a + b*Arc 
Coth[c*x])^3*(Log[2*c*((d + e*x)/((c*d + e)*(1 + c*x)))]/e), x] + Simp[3*b* 
(a + b*ArcCoth[c*x])^2*(PolyLog[2, 1 - 2/(1 + c*x)]/(2*e)), x] - Simp[3*b*( 
a + b*ArcCoth[c*x])^2*(PolyLog[2, 1 - 2*c*((d + e*x)/((c*d + e)*(1 + c*x))) 
]/(2*e)), x] + Simp[3*b^2*(a + b*ArcCoth[c*x])*(PolyLog[3, 1 - 2/(1 + c*x)] 
/(2*e)), x] - Simp[3*b^2*(a + b*ArcCoth[c*x])*(PolyLog[3, 1 - 2*c*((d + e*x 
)/((c*d + e)*(1 + c*x)))]/(2*e)), x] + Simp[3*b^3*(PolyLog[4, 1 - 2/(1 + c* 
x)]/(4*e)), x] - Simp[3*b^3*(PolyLog[4, 1 - 2*c*((d + e*x)/((c*d + e)*(1 + 
c*x)))]/(4*e)), x]) /; FreeQ[{a, b, c, d, e}, x] && NeQ[c^2*d^2 - e^2, 0]
 

rule 6662

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

Maple [C] (warning: unable to verify)

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

Time = 13.08 (sec) , antiderivative size = 3250, normalized size of antiderivative = 10.55

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(3250\)
default	\(\text {Expression too large to display}\)	\(3250\)
parts	\(\text {Expression too large to display}\)	\(3419\)

Input:

int((a+b*arccoth(d*x+c))^3/(f*x+e),x,method=_RETURNVERBOSE)
 

Output:

1/d*(a^3*d*ln(c*f-d*e-f*(d*x+c))/f-b^3*d*(-ln(c*f-d*e-f*(d*x+c))/f*arccoth 
(d*x+c)^3-3/f*(-1/3*arccoth(d*x+c)^3*ln(f*c*((d*x+c+1)/(d*x+c-1)-1)+(-(d*x 
+c+1)/(d*x+c-1)+1)*e*d+(-(d*x+c+1)/(d*x+c-1)-1)*f)+1/6*I*Pi*csgn(I*(f*c*(( 
d*x+c+1)/(d*x+c-1)-1)+(-(d*x+c+1)/(d*x+c-1)+1)*e*d+(-(d*x+c+1)/(d*x+c-1)-1 
)*f)/((d*x+c+1)/(d*x+c-1)-1))*(csgn(I*(f*c*((d*x+c+1)/(d*x+c-1)-1)+(-(d*x+ 
c+1)/(d*x+c-1)+1)*e*d+(-(d*x+c+1)/(d*x+c-1)-1)*f))*csgn(I/((d*x+c+1)/(d*x+ 
c-1)-1))-csgn(I*(f*c*((d*x+c+1)/(d*x+c-1)-1)+(-(d*x+c+1)/(d*x+c-1)+1)*e*d+ 
(-(d*x+c+1)/(d*x+c-1)-1)*f)/((d*x+c+1)/(d*x+c-1)-1))*csgn(I/((d*x+c+1)/(d* 
x+c-1)-1))-csgn(I*(f*c*((d*x+c+1)/(d*x+c-1)-1)+(-(d*x+c+1)/(d*x+c-1)+1)*e* 
d+(-(d*x+c+1)/(d*x+c-1)-1)*f))*csgn(I*(f*c*((d*x+c+1)/(d*x+c-1)-1)+(-(d*x+ 
c+1)/(d*x+c-1)+1)*e*d+(-(d*x+c+1)/(d*x+c-1)-1)*f)/((d*x+c+1)/(d*x+c-1)-1)) 
+csgn(I*(f*c*((d*x+c+1)/(d*x+c-1)-1)+(-(d*x+c+1)/(d*x+c-1)+1)*e*d+(-(d*x+c 
+1)/(d*x+c-1)-1)*f)/((d*x+c+1)/(d*x+c-1)-1))^2)*arccoth(d*x+c)^3+1/3*arcco 
th(d*x+c)^3*ln((d*x+c+1)/(d*x+c-1)-1)-1/3*arccoth(d*x+c)^3*ln(1+1/((d*x+c- 
1)/(d*x+c+1))^(1/2))-arccoth(d*x+c)^2*polylog(2,-1/((d*x+c-1)/(d*x+c+1))^( 
1/2))+2*arccoth(d*x+c)*polylog(3,-1/((d*x+c-1)/(d*x+c+1))^(1/2))-2*polylog 
(4,-1/((d*x+c-1)/(d*x+c+1))^(1/2))-1/3*arccoth(d*x+c)^3*ln(1-1/((d*x+c-1)/ 
(d*x+c+1))^(1/2))-arccoth(d*x+c)^2*polylog(2,1/((d*x+c-1)/(d*x+c+1))^(1/2) 
)+2*arccoth(d*x+c)*polylog(3,1/((d*x+c-1)/(d*x+c+1))^(1/2))-2*polylog(4,1/ 
((d*x+c-1)/(d*x+c+1))^(1/2))+1/3*f*c/(c*f-d*e-f)*arccoth(d*x+c)^3*ln(1-...
 

Fricas [F]

\[ \int \frac {\left (a+b \coth ^{-1}(c+d x)\right )^3}{e+f x} \, dx=\int { \frac {{\left (b \operatorname {arcoth}\left (d x + c\right ) + a\right )}^{3}}{f x + e} \,d x } \] Input:

integrate((a+b*arccoth(d*x+c))^3/(f*x+e),x, algorithm="fricas")
 

Output:

integral((b^3*arccoth(d*x + c)^3 + 3*a*b^2*arccoth(d*x + c)^2 + 3*a^2*b*ar 
ccoth(d*x + c) + a^3)/(f*x + e), x)
 

Sympy [F]

\[ \int \frac {\left (a+b \coth ^{-1}(c+d x)\right )^3}{e+f x} \, dx=\int \frac {\left (a + b \operatorname {acoth}{\left (c + d x \right )}\right )^{3}}{e + f x}\, dx \] Input:

integrate((a+b*acoth(d*x+c))**3/(f*x+e),x)
 

Output:

Integral((a + b*acoth(c + d*x))**3/(e + f*x), x)
 

Maxima [F]

\[ \int \frac {\left (a+b \coth ^{-1}(c+d x)\right )^3}{e+f x} \, dx=\int { \frac {{\left (b \operatorname {arcoth}\left (d x + c\right ) + a\right )}^{3}}{f x + e} \,d x } \] Input:

integrate((a+b*arccoth(d*x+c))^3/(f*x+e),x, algorithm="maxima")
 

Output:

a^3*log(f*x + e)/f + integrate(1/8*b^3*(log(1/(d*x + c) + 1) - log(-1/(d*x 
 + c) + 1))^3/(f*x + e) + 3/4*a*b^2*(log(1/(d*x + c) + 1) - log(-1/(d*x + 
c) + 1))^2/(f*x + e) + 3/2*a^2*b*(log(1/(d*x + c) + 1) - log(-1/(d*x + c) 
+ 1))/(f*x + e), x)
 

Giac [F]

\[ \int \frac {\left (a+b \coth ^{-1}(c+d x)\right )^3}{e+f x} \, dx=\int { \frac {{\left (b \operatorname {arcoth}\left (d x + c\right ) + a\right )}^{3}}{f x + e} \,d x } \] Input:

integrate((a+b*arccoth(d*x+c))^3/(f*x+e),x, algorithm="giac")
 

Output:

integrate((b*arccoth(d*x + c) + a)^3/(f*x + e), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b \coth ^{-1}(c+d x)\right )^3}{e+f x} \, dx=\int \frac {{\left (a+b\,\mathrm {acoth}\left (c+d\,x\right )\right )}^3}{e+f\,x} \,d x \] Input:

int((a + b*acoth(c + d*x))^3/(e + f*x),x)
 

Output:

int((a + b*acoth(c + d*x))^3/(e + f*x), x)
 

Reduce [F]

\[ \int \frac {\left (a+b \coth ^{-1}(c+d x)\right )^3}{e+f x} \, dx=\frac {3 \left (\int \frac {\mathit {acoth} \left (d x +c \right )}{f x +e}d x \right ) a^{2} b f +\left (\int \frac {\mathit {acoth} \left (d x +c \right )^{3}}{f x +e}d x \right ) b^{3} f +3 \left (\int \frac {\mathit {acoth} \left (d x +c \right )^{2}}{f x +e}d x \right ) a \,b^{2} f +\mathrm {log}\left (f x +e \right ) a^{3}}{f} \] Input:

int((a+b*acoth(d*x+c))^3/(f*x+e),x)
                                                                                    
                                                                                    
 

Output:

(3*int(acoth(c + d*x)/(e + f*x),x)*a**2*b*f + int(acoth(c + d*x)**3/(e + f 
*x),x)*b**3*f + 3*int(acoth(c + d*x)**2/(e + f*x),x)*a*b**2*f + log(e + f* 
x)*a**3)/f