\(\int \frac {\text {arctanh}(x)}{a+b x} \, dx\) [23]

Optimal result

Integrand size = 10, antiderivative size = 86 \[ \int \frac {\text {arctanh}(x)}{a+b x} \, dx=-\frac {\text {arctanh}(x) \log \left (\frac {2}{1+x}\right )}{b}+\frac {\text {arctanh}(x) \log \left (\frac {2 (a+b x)}{(a+b) (1+x)}\right )}{b}+\frac {\operatorname {PolyLog}\left (2,1-\frac {2}{1+x}\right )}{2 b}-\frac {\operatorname {PolyLog}\left (2,1-\frac {2 (a+b x)}{(a+b) (1+x)}\right )}{2 b} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 0.06 (sec) , antiderivative size = 260, normalized size of antiderivative = 3.02 \[ \int \frac {\text {arctanh}(x)}{a+b x} \, dx=\frac {-\pi ^2+4 \text {arctanh}\left (\frac {a}{b}\right )^2+4 i \pi \text {arctanh}(x)+8 \text {arctanh}\left (\frac {a}{b}\right ) \text {arctanh}(x)+8 \text {arctanh}(x)^2-4 i \pi \log \left (1+e^{2 \text {arctanh}(x)}\right )-8 \text {arctanh}(x) \log \left (1+e^{2 \text {arctanh}(x)}\right )+8 \text {arctanh}\left (\frac {a}{b}\right ) \log \left (1-e^{-2 \left (\text {arctanh}\left (\frac {a}{b}\right )+\text {arctanh}(x)\right )}\right )+8 \text {arctanh}(x) \log \left (1-e^{-2 \left (\text {arctanh}\left (\frac {a}{b}\right )+\text {arctanh}(x)\right )}\right )+4 i \pi \log \left (\frac {2}{\sqrt {1-x^2}}\right )+8 \text {arctanh}(x) \log \left (\frac {2}{\sqrt {1-x^2}}\right )+4 \text {arctanh}(x) \log \left (1-x^2\right )+8 \text {arctanh}(x) \log \left (i \sinh \left (\text {arctanh}\left (\frac {a}{b}\right )+\text {arctanh}(x)\right )\right )-8 \text {arctanh}\left (\frac {a}{b}\right ) \log \left (2 i \sinh \left (\text {arctanh}\left (\frac {a}{b}\right )+\text {arctanh}(x)\right )\right )-8 \text {arctanh}(x) \log \left (2 i \sinh \left (\text {arctanh}\left (\frac {a}{b}\right )+\text {arctanh}(x)\right )\right )-4 \operatorname {PolyLog}\left (2,-e^{2 \text {arctanh}(x)}\right )-4 \operatorname {PolyLog}\left (2,e^{-2 \left (\text {arctanh}\left (\frac {a}{b}\right )+\text {arctanh}(x)\right )}\right )}{8 b} \] Input:

Integrate[ArcTanh[x]/(a + b*x),x]
 

Output:

(-Pi^2 + 4*ArcTanh[a/b]^2 + (4*I)*Pi*ArcTanh[x] + 8*ArcTanh[a/b]*ArcTanh[x 
] + 8*ArcTanh[x]^2 - (4*I)*Pi*Log[1 + E^(2*ArcTanh[x])] - 8*ArcTanh[x]*Log 
[1 + E^(2*ArcTanh[x])] + 8*ArcTanh[a/b]*Log[1 - E^(-2*(ArcTanh[a/b] + ArcT 
anh[x]))] + 8*ArcTanh[x]*Log[1 - E^(-2*(ArcTanh[a/b] + ArcTanh[x]))] + (4* 
I)*Pi*Log[2/Sqrt[1 - x^2]] + 8*ArcTanh[x]*Log[2/Sqrt[1 - x^2]] + 4*ArcTanh 
[x]*Log[1 - x^2] + 8*ArcTanh[x]*Log[I*Sinh[ArcTanh[a/b] + ArcTanh[x]]] - 8 
*ArcTanh[a/b]*Log[(2*I)*Sinh[ArcTanh[a/b] + ArcTanh[x]]] - 8*ArcTanh[x]*Lo 
g[(2*I)*Sinh[ArcTanh[a/b] + ArcTanh[x]]] - 4*PolyLog[2, -E^(2*ArcTanh[x])] 
 - 4*PolyLog[2, E^(-2*(ArcTanh[a/b] + ArcTanh[x]))])/(8*b)
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6472, 2849, 2752, 2897}

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.

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

\(\Big \downarrow \) 6472

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

\(\Big \downarrow \) 2849

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

\(\Big \downarrow \) 2752

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

\(\Big \downarrow \) 2897

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

Input:

Int[ArcTanh[x]/(a + b*x),x]
 

Output:

-((ArcTanh[x]*Log[2/(1 + x)])/b) + (ArcTanh[x]*Log[(2*(a + b*x))/((a + b)* 
(1 + x))])/b + PolyLog[2, 1 - 2/(1 + x)]/(2*b) - PolyLog[2, 1 - (2*(a + b* 
x))/((a + b)*(1 + 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 2897

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[Pq^m*((1 - u)/ 
D[u, x])]}, Simp[C*PolyLog[2, 1 - u], x] /; FreeQ[C, x]] /; IntegerQ[m] && 
PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponents[u, 
 x][[2]], Expon[Pq, x]]
 

rule 6472

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

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.23

method	result	size
default	\(\frac {\ln \left (b x +a \right ) \operatorname {arctanh}\left (x \right )}{b}-\frac {\frac {b \left (\operatorname {dilog}\left (\frac {b x +b}{-a +b}\right )+\ln \left (b x +a \right ) \ln \left (\frac {b x +b}{-a +b}\right )\right )}{2}-\frac {b \left (\operatorname {dilog}\left (\frac {b x -b}{-a -b}\right )+\ln \left (b x +a \right ) \ln \left (\frac {b x -b}{-a -b}\right )\right )}{2}}{b^{2}}\)	\(106\)
parts	\(\frac {\ln \left (b x +a \right ) \operatorname {arctanh}\left (x \right )}{b}-\frac {\frac {b \left (\operatorname {dilog}\left (\frac {b x +b}{-a +b}\right )+\ln \left (b x +a \right ) \ln \left (\frac {b x +b}{-a +b}\right )\right )}{2}-\frac {b \left (\operatorname {dilog}\left (\frac {b x -b}{-a -b}\right )+\ln \left (b x +a \right ) \ln \left (\frac {b x -b}{-a -b}\right )\right )}{2}}{b^{2}}\)	\(106\)
risch	\(-\frac {\operatorname {dilog}\left (\frac {\left (1-x \right ) b -a -b}{-a -b}\right )}{2 b}-\frac {\ln \left (1-x \right ) \ln \left (\frac {\left (1-x \right ) b -a -b}{-a -b}\right )}{2 b}+\frac {\operatorname {dilog}\left (\frac {\left (1+x \right ) b +a -b}{a -b}\right )}{2 b}+\frac {\ln \left (1+x \right ) \ln \left (\frac {\left (1+x \right ) b +a -b}{a -b}\right )}{2 b}\)	\(120\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.38 \[ \int \frac {\text {arctanh}(x)}{a+b x} \, dx=-\frac {{\left (\log \left (x + 1\right ) - \log \left (x - 1\right )\right )} \log \left (b x + a\right )}{2 \, b} + \frac {\operatorname {artanh}\left (x\right ) \log \left (b x + a\right )}{b} - \frac {\log \left (x - 1\right ) \log \left (\frac {b x - b}{a + b} + 1\right ) + {\rm Li}_2\left (-\frac {b x - b}{a + b}\right )}{2 \, b} + \frac {\log \left (x + 1\right ) \log \left (\frac {b x + b}{a - b} + 1\right ) + {\rm Li}_2\left (-\frac {b x + b}{a - b}\right )}{2 \, b} \] Input:

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

Output:

-1/2*(log(x + 1) - log(x - 1))*log(b*x + a)/b + arctanh(x)*log(b*x + a)/b 
- 1/2*(log(x - 1)*log((b*x - b)/(a + b) + 1) + dilog(-(b*x - b)/(a + b)))/ 
b + 1/2*(log(x + 1)*log((b*x + b)/(a - b) + 1) + dilog(-(b*x + b)/(a - b)) 
)/b
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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