\(\int \text {arctanh}(a+b x)^2 \, dx\) [4]

Optimal result

Integrand size = 8, antiderivative size = 81 \[ \int \text {arctanh}(a+b x)^2 \, dx=\frac {\text {arctanh}(a+b x)^2}{b}+\frac {(a+b x) \text {arctanh}(a+b x)^2}{b}-\frac {2 \text {arctanh}(a+b x) \log \left (\frac {2}{1-a-b x}\right )}{b}-\frac {\operatorname {PolyLog}\left (2,-\frac {1+a+b x}{1-a-b x}\right )}{b} \] Output:

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.68 \[ \int \text {arctanh}(a+b x)^2 \, dx=\frac {\text {arctanh}(a+b x) \left ((-1+a+b x) \text {arctanh}(a+b x)-2 \log \left (1+e^{-2 \text {arctanh}(a+b x)}\right )\right )+\operatorname {PolyLog}\left (2,-e^{-2 \text {arctanh}(a+b x)}\right )}{b} \] Input:

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

Output:

(ArcTanh[a + b*x]*((-1 + a + b*x)*ArcTanh[a + b*x] - 2*Log[1 + E^(-2*ArcTa 
nh[a + b*x])]) + PolyLog[2, -E^(-2*ArcTanh[a + b*x])])/b
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.94, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {6653, 6436, 6546, 6470, 2849, 2752}

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.

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

\(\Big \downarrow \) 6653

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

\(\Big \downarrow \) 6436

\(\displaystyle \frac {(a+b x) \text {arctanh}(a+b x)^2-2 \int \frac {(a+b x) \text {arctanh}(a+b x)}{1-(a+b x)^2}d(a+b x)}{b}\)

\(\Big \downarrow \) 6546

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

\(\Big \downarrow \) 6470

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

\(\Big \downarrow \) 2849

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

\(\Big \downarrow \) 2752

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2752

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLo 
g[2, 1 - c*x], x] /; FreeQ[{c, d, e}, x] && EqQ[e + c*d, 0]
 

rule 2849

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Simp 
[-e/g   Subst[Int[Log[2*d*x]/(1 - 2*d*x), x], x, 1/(d + e*x)], x] /; FreeQ[ 
{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]
 

rule 6436

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

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 6546

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), 
 x_Symbol] :> Simp[(a + b*ArcTanh[c*x])^(p + 1)/(b*e*(p + 1)), x] + Simp[1/ 
(c*d)   Int[(a + b*ArcTanh[c*x])^p/(1 - c*x), x], x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[c^2*d + e, 0] && IGtQ[p, 0]
 

rule 6653

Int[((a_.) + ArcTanh[(c_) + (d_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[1/d 
   Subst[Int[(a + b*ArcTanh[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d 
}, x] && IGtQ[p, 0]
 

Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.12

method	result	size
derivativedivides	\(\frac {\operatorname {arctanh}\left (b x +a \right )^{2} \left (b x +a -1\right )+2 \operatorname {arctanh}\left (b x +a \right )^{2}-2 \,\operatorname {arctanh}\left (b x +a \right ) \ln \left (1+\frac {\left (b x +a +1\right )^{2}}{1-\left (b x +a \right )^{2}}\right )-\operatorname {polylog}\left (2, -\frac {\left (b x +a +1\right )^{2}}{1-\left (b x +a \right )^{2}}\right )}{b}\)	\(91\)
default	\(\frac {\operatorname {arctanh}\left (b x +a \right )^{2} \left (b x +a -1\right )+2 \operatorname {arctanh}\left (b x +a \right )^{2}-2 \,\operatorname {arctanh}\left (b x +a \right ) \ln \left (1+\frac {\left (b x +a +1\right )^{2}}{1-\left (b x +a \right )^{2}}\right )-\operatorname {polylog}\left (2, -\frac {\left (b x +a +1\right )^{2}}{1-\left (b x +a \right )^{2}}\right )}{b}\)	\(91\)
risch	\(\frac {\left (b x +a +1\right ) \ln \left (b x +a +1\right )^{2}}{4 b}+\left (-\frac {\ln \left (-b x -a +1\right ) x}{2}+\frac {-\ln \left (-b x -a +1\right ) a +\ln \left (-b x -a +1\right )}{2 b}\right ) \ln \left (b x +a +1\right )+\frac {x \ln \left (-b x -a +1\right )^{2}}{4}+\frac {\ln \left (-b x -a +1\right )^{2} a}{4 b}-\frac {\ln \left (-b x -a +1\right )^{2}}{4 b}+\frac {\ln \left (\frac {b x}{2}+\frac {a}{2}+\frac {1}{2}\right ) \ln \left (-\frac {b x}{2}-\frac {a}{2}+\frac {1}{2}\right )}{b}-\frac {\ln \left (\frac {b x}{2}+\frac {a}{2}+\frac {1}{2}\right ) \ln \left (-b x -a +1\right )}{b}+\frac {\operatorname {dilog}\left (-\frac {b x}{2}-\frac {a}{2}+\frac {1}{2}\right )}{b}\)	\(186\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.72 \[ \int \text {arctanh}(a+b x)^2 \, dx=-\frac {1}{4} \, b^{2} {\left (\frac {{\left (a + 1\right )} \log \left (b x + a + 1\right )^{2} - 2 \, {\left (a + 1\right )} \log \left (b x + a + 1\right ) \log \left (b x + a - 1\right ) + {\left (a - 1\right )} \log \left (b x + a - 1\right )^{2}}{b^{3}} + \frac {4 \, {\left (\log \left (b x + a - 1\right ) \log \left (\frac {1}{2} \, b x + \frac {1}{2} \, a + \frac {1}{2}\right ) + {\rm Li}_2\left (-\frac {1}{2} \, b x - \frac {1}{2} \, a + \frac {1}{2}\right )\right )}}{b^{3}}\right )} + b {\left (\frac {{\left (a + 1\right )} \log \left (b x + a + 1\right )}{b^{2}} - \frac {{\left (a - 1\right )} \log \left (b x + a - 1\right )}{b^{2}}\right )} \operatorname {artanh}\left (b x + a\right ) + x \operatorname {artanh}\left (b x + a\right )^{2} \] Input:

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

Output:

-1/4*b^2*(((a + 1)*log(b*x + a + 1)^2 - 2*(a + 1)*log(b*x + a + 1)*log(b*x 
 + a - 1) + (a - 1)*log(b*x + a - 1)^2)/b^3 + 4*(log(b*x + a - 1)*log(1/2* 
b*x + 1/2*a + 1/2) + dilog(-1/2*b*x - 1/2*a + 1/2))/b^3) + b*((a + 1)*log( 
b*x + a + 1)/b^2 - (a - 1)*log(b*x + a - 1)/b^2)*arctanh(b*x + a) + x*arct 
anh(b*x + a)^2
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(atanh(a + b*x)**2*a**2 + 2*atanh(a + b*x)**2*a*b*x - atanh(a + b*x)**2 + 
2*atanh(a + b*x)*a + 2*atanh(a + b*x)*b*x + 2*atanh(a + b*x) - 2*int((atan 
h(a + b*x)*x**2)/(a**2 + 2*a*b*x + b**2*x**2 - 1),x)*b**3 + 2*log(a + b*x 
- 1))/(2*a*b)