\(\int \frac {\text {arctanh}(a+b x)^2}{x} \, dx\) [5]

Optimal result

Integrand size = 12, antiderivative size = 148 \[ \int \frac {\text {arctanh}(a+b x)^2}{x} \, dx=-\text {arctanh}(a+b x)^2 \log \left (\frac {2}{1+a+b x}\right )+\text {arctanh}(a+b x)^2 \log \left (\frac {2 b x}{(1-a) (1+a+b x)}\right )+\text {arctanh}(a+b x) \operatorname {PolyLog}\left (2,1-\frac {2}{1+a+b x}\right )-\text {arctanh}(a+b x) \operatorname {PolyLog}\left (2,1-\frac {2 b x}{(1-a) (1+a+b x)}\right )+\frac {1}{2} \operatorname {PolyLog}\left (3,1-\frac {2}{1+a+b x}\right )-\frac {1}{2} \operatorname {PolyLog}\left (3,1-\frac {2 b x}{(1-a) (1+a+b x)}\right ) \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 2.26 (sec) , antiderivative size = 753, normalized size of antiderivative = 5.09 \[ \int \frac {\text {arctanh}(a+b x)^2}{x} \, dx =\text {Too large to display} \] Input:

Integrate[ArcTanh[a + b*x]^2/x,x]
 

Output:

(-4*ArcTanh[a + b*x]^3)/3 - (2*ArcTanh[a + b*x]^3)/(3*a) + (2*Sqrt[1 - a^2 
]*E^ArcTanh[a]*ArcTanh[a + b*x]^3)/(3*a) - ArcTanh[a + b*x]^2*Log[1 + E^(- 
2*ArcTanh[a + b*x])] - I*Pi*ArcTanh[a + b*x]*Log[(E^(-ArcTanh[a + b*x]) + 
E^ArcTanh[a + b*x])/2] - ArcTanh[a + b*x]^2*Log[1 - (Sqrt[-1 + a]*E^ArcTan 
h[a + b*x])/Sqrt[-1 - a]] - ArcTanh[a + b*x]^2*Log[1 + (Sqrt[-1 + a]*E^Arc 
Tanh[a + b*x])/Sqrt[-1 - a]] + ArcTanh[a + b*x]^2*Log[(1 + a - E^(2*ArcTan 
h[a + b*x]) + a*E^(2*ArcTanh[a + b*x]))/(2*E^ArcTanh[a + b*x])] + ArcTanh[ 
a + b*x]^2*Log[1 - E^(-ArcTanh[a] + ArcTanh[a + b*x])] + ArcTanh[a + b*x]^ 
2*Log[1 + E^(-ArcTanh[a] + ArcTanh[a + b*x])] - 2*ArcTanh[a]*ArcTanh[a + b 
*x]*Log[(I/2)*(-E^(ArcTanh[a] - ArcTanh[a + b*x]) + E^(-ArcTanh[a] + ArcTa 
nh[a + b*x]))] + ArcTanh[a + b*x]^2*Log[1 - E^(-2*ArcTanh[a] + 2*ArcTanh[a 
 + b*x])] + I*Pi*ArcTanh[a + b*x]*Log[1/Sqrt[1 - (a + b*x)^2]] - ArcTanh[a 
 + b*x]^2*Log[-((b*x)/Sqrt[1 - (a + b*x)^2])] + 2*ArcTanh[a]*ArcTanh[a + b 
*x]*Log[(-I)*Sinh[ArcTanh[a] - ArcTanh[a + b*x]]] + ArcTanh[a + b*x]*PolyL 
og[2, -E^(-2*ArcTanh[a + b*x])] - 2*ArcTanh[a + b*x]*PolyLog[2, -((Sqrt[-1 
 + a]*E^ArcTanh[a + b*x])/Sqrt[-1 - a])] - 2*ArcTanh[a + b*x]*PolyLog[2, ( 
Sqrt[-1 + a]*E^ArcTanh[a + b*x])/Sqrt[-1 - a]] + 2*ArcTanh[a + b*x]*PolyLo 
g[2, -E^(-ArcTanh[a] + ArcTanh[a + b*x])] + 2*ArcTanh[a + b*x]*PolyLog[2, 
E^(-ArcTanh[a] + ArcTanh[a + b*x])] + ArcTanh[a + b*x]*PolyLog[2, E^(-2*Ar 
cTanh[a] + 2*ArcTanh[a + b*x])] + PolyLog[3, -E^(-2*ArcTanh[a + b*x])]/...
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6661, 25, 27, 6474}

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.

\(\displaystyle \int \frac {\text {arctanh}(a+b x)^2}{x} \, dx\)

\(\Big \downarrow \) 6661

\(\displaystyle \frac {\int \frac {\text {arctanh}(a+b x)^2}{x}d(a+b x)}{b}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\int -\frac {\text {arctanh}(a+b x)^2}{x}d(a+b x)}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle -\int -\frac {\text {arctanh}(a+b x)^2}{b x}d(a+b x)\)

\(\Big \downarrow \) 6474

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

Input:

Int[ArcTanh[a + b*x]^2/x,x]
 

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 6474

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

rule 6661

Int[((a_.) + ArcTanh[(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* 
ArcTanh[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 = 32.32 (sec) , antiderivative size = 901, normalized size of antiderivative = 6.09

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(901\)
default	\(\text {Expression too large to display}\)	\(901\)
parts	\(\text {Expression too large to display}\)	\(1544\)

Input:

int(arctanh(b*x+a)^2/x,x,method=_RETURNVERBOSE)
 

Output:

ln(-b*x)*arctanh(b*x+a)^2-arctanh(b*x+a)^2*ln(-(b*x+a+1)^2/(1-(b*x+a)^2)+1 
+a*(1+(b*x+a+1)^2/(1-(b*x+a)^2)))+1/2*I*Pi*csgn(I*((b*x+a+1)^2/((b*x+a)^2- 
1)+1+a*(1-(b*x+a+1)^2/((b*x+a)^2-1)))/(1-(b*x+a+1)^2/((b*x+a)^2-1)))*(csgn 
(I*((b*x+a+1)^2/((b*x+a)^2-1)+1+a*(1-(b*x+a+1)^2/((b*x+a)^2-1))))*csgn(I/( 
1-(b*x+a+1)^2/((b*x+a)^2-1)))-csgn(I*((b*x+a+1)^2/((b*x+a)^2-1)+1+a*(1-(b* 
x+a+1)^2/((b*x+a)^2-1)))/(1-(b*x+a+1)^2/((b*x+a)^2-1)))*csgn(I/(1-(b*x+a+1 
)^2/((b*x+a)^2-1)))-csgn(I*((b*x+a+1)^2/((b*x+a)^2-1)+1+a*(1-(b*x+a+1)^2/( 
(b*x+a)^2-1))))*csgn(I*((b*x+a+1)^2/((b*x+a)^2-1)+1+a*(1-(b*x+a+1)^2/((b*x 
+a)^2-1)))/(1-(b*x+a+1)^2/((b*x+a)^2-1)))+csgn(I*((b*x+a+1)^2/((b*x+a)^2-1 
)+1+a*(1-(b*x+a+1)^2/((b*x+a)^2-1)))/(1-(b*x+a+1)^2/((b*x+a)^2-1)))^2)*arc 
tanh(b*x+a)^2-arctanh(b*x+a)*polylog(2,-(b*x+a+1)^2/(1-(b*x+a)^2))+1/2*pol 
ylog(3,-(b*x+a+1)^2/(1-(b*x+a)^2))+a/(a-1)*arctanh(b*x+a)^2*ln(1-(a-1)*(b* 
x+a+1)^2/(1-(b*x+a)^2)/(-1-a))+a/(a-1)*arctanh(b*x+a)*polylog(2,(a-1)*(b*x 
+a+1)^2/(1-(b*x+a)^2)/(-1-a))-1/2*a/(a-1)*polylog(3,(a-1)*(b*x+a+1)^2/(1-( 
b*x+a)^2)/(-1-a))-1/(a-1)*arctanh(b*x+a)^2*ln(1-(a-1)*(b*x+a+1)^2/(1-(b*x+ 
a)^2)/(-1-a))-1/(a-1)*arctanh(b*x+a)*polylog(2,(a-1)*(b*x+a+1)^2/(1-(b*x+a 
)^2)/(-1-a))+1/2/(a-1)*polylog(3,(a-1)*(b*x+a+1)^2/(1-(b*x+a)^2)/(-1-a))
 

Fricas [F]

\[ \int \frac {\text {arctanh}(a+b x)^2}{x} \, dx=\int { \frac {\operatorname {artanh}\left (b x + a\right )^{2}}{x} \,d x } \] Input:

integrate(arctanh(b*x+a)^2/x,x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {\text {arctanh}(a+b x)^2}{x} \, dx=\int \frac {\operatorname {atanh}^{2}{\left (a + b x \right )}}{x}\, dx \] Input:

integrate(atanh(b*x+a)**2/x,x)
 

Output:

Integral(atanh(a + b*x)**2/x, x)
 

Maxima [F]

\[ \int \frac {\text {arctanh}(a+b x)^2}{x} \, dx=\int { \frac {\operatorname {artanh}\left (b x + a\right )^{2}}{x} \,d x } \] Input:

integrate(arctanh(b*x+a)^2/x,x, algorithm="maxima")
 

Output:

integrate(arctanh(b*x + a)^2/x, x)
 

Giac [F]

\[ \int \frac {\text {arctanh}(a+b x)^2}{x} \, dx=\int { \frac {\operatorname {artanh}\left (b x + a\right )^{2}}{x} \,d x } \] Input:

integrate(arctanh(b*x+a)^2/x,x, algorithm="giac")
 

Output:

integrate(arctanh(b*x + a)^2/x, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\text {arctanh}(a+b x)^2}{x} \, dx=\int \frac {{\mathrm {atanh}\left (a+b\,x\right )}^2}{x} \,d x \] Input:

int(atanh(a + b*x)^2/x,x)
 

Output:

int(atanh(a + b*x)^2/x, x)
 

Reduce [F]

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

int(atanh(b*x+a)^2/x,x)
                                                                                    
                                                                                    
 

Output:

int(atanh(a + b*x)**2/x,x)