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

3.1.4.1 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
 

3.1.4.2 Mathematica [A] (verified)

Time = 0.07 (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
 

3.1.4.3 Rubi [A] (verified)

Time = 0.45 (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
 

3.1.4.3.1 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]
 

3.1.4.4 Maple [A] (verified)

Time = 0.12 (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	\(\frac {\left (b x +a +1\right ) \ln \left (b x +a +1\right )^{2}}{4 b}+\left (-\frac {x \ln \left (-b x -a +1\right )}{2}+\frac {-\ln \left (-b x -a +1\right ) a +\ln \left (-b x -a +1\right )}{2 b}\right ) \ln \left (b x +a +1\right )+\frac {x \ln \left (-b x -a +1\right )^{2}}{4}+\frac {\ln \left (-b x -a +1\right )^{2} a}{4 b}-\frac {\ln \left (-b x -a +1\right )^{2}}{4 b}+\frac {\ln \left (\frac {b x}{2}+\frac {a}{2}+\frac {1}{2}\right ) \ln \left (-\frac {b x}{2}-\frac {a}{2}+\frac {1}{2}\right )}{b}-\frac {\ln \left (\frac {b x}{2}+\frac {a}{2}+\frac {1}{2}\right ) \ln \left (-b x -a +1\right )}{b}+\frac {\operatorname {dilog}\left (-\frac {b x}{2}-\frac {a}{2}+\frac {1}{2}\right )}{b}\)	\(186\)

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)))
 

3.1.4.5 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)
 

3.1.4.6 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)
 

3.1.4.7 Maxima [A] (verification not implemented)

Time = 0.18 (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
 

3.1.4.8 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)
 

3.1.4.9 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)