\(\int \frac {a+b \text {arctanh}(c x)}{d+c d x} \, dx\) [46]

Optimal result

Integrand size = 17, antiderivative size = 51 \[ \int \frac {a+b \text {arctanh}(c x)}{d+c d x} \, dx=-\frac {(a+b \text {arctanh}(c x)) \log \left (\frac {2}{1+c x}\right )}{c d}+\frac {b \operatorname {PolyLog}\left (2,1-\frac {2}{1+c x}\right )}{2 c d} \] Output:

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.02 \[ \int \frac {a+b \text {arctanh}(c x)}{d+c d x} \, dx=\frac {-2 b \text {arctanh}(c x) \log \left (1+e^{-2 \text {arctanh}(c x)}\right )+2 a \log (1+c x)+b \operatorname {PolyLog}\left (2,-e^{-2 \text {arctanh}(c x)}\right )}{2 c d} \] Input:

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

Output:

(-2*b*ArcTanh[c*x]*Log[1 + E^(-2*ArcTanh[c*x])] + 2*a*Log[1 + c*x] + b*Pol 
yLog[2, -E^(-2*ArcTanh[c*x])])/(2*c*d)
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {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.

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

\(\Big \downarrow \) 6470

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

\(\Big \downarrow \) 2849

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

\(\Big \downarrow \) 2752

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

Input:

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

Output:

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

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 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]
 

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.53

method	result	size
derivativedivides	\(\frac {\frac {a \ln \left (c x +1\right )}{d}+\frac {b \left (\operatorname {arctanh}\left (c x \right ) \ln \left (c x +1\right )-\frac {\ln \left (c x +1\right )^{2}}{4}+\frac {\left (\ln \left (c x +1\right )-\ln \left (\frac {c x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {c x}{2}+\frac {1}{2}\right )}{2}\right )}{d}}{c}\)	\(78\)
default	\(\frac {\frac {a \ln \left (c x +1\right )}{d}+\frac {b \left (\operatorname {arctanh}\left (c x \right ) \ln \left (c x +1\right )-\frac {\ln \left (c x +1\right )^{2}}{4}+\frac {\left (\ln \left (c x +1\right )-\ln \left (\frac {c x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {c x}{2}+\frac {1}{2}\right )}{2}\right )}{d}}{c}\)	\(78\)
parts	\(\frac {a \ln \left (c x +1\right )}{d c}+\frac {b \left (\operatorname {arctanh}\left (c x \right ) \ln \left (c x +1\right )-\frac {\ln \left (c x +1\right )^{2}}{4}+\frac {\left (\ln \left (c x +1\right )-\ln \left (\frac {c x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {c x}{2}+\frac {1}{2}\right )}{2}\right )}{d c}\)	\(80\)
risch	\(\frac {b \ln \left (c x +1\right )^{2}}{4 c d}-\frac {\ln \left (-c x +1\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right ) b}{2 c d}+\frac {\ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right ) b}{2 c d}+\frac {a \ln \left (-c x -1\right )}{c d}+\frac {\operatorname {dilog}\left (-\frac {c x}{2}+\frac {1}{2}\right ) b}{2 c d}\)	\(96\)

Input:

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

Output:

1/c*(a/d*ln(c*x+1)+b/d*(arctanh(c*x)*ln(c*x+1)-1/4*ln(c*x+1)^2+1/2*(ln(c*x 
+1)-ln(1/2*c*x+1/2))*ln(-1/2*c*x+1/2)-1/2*dilog(1/2*c*x+1/2)))
 

Fricas [F]

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

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

Output:

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

Sympy [F]

\[ \int \frac {a+b \text {arctanh}(c x)}{d+c d x} \, dx=\frac {\int \frac {a}{c x + 1}\, dx + \int \frac {b \operatorname {atanh}{\left (c x \right )}}{c x + 1}\, dx}{d} \] Input:

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

Output:

(Integral(a/(c*x + 1), x) + Integral(b*atanh(c*x)/(c*x + 1), x))/d
 

Maxima [F]

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

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

Output:

1/2*(2*c*integrate(x*log(c*x + 1)/(c^2*d*x^2 - d), x) - log(c*x + 1)*log(- 
c*x + 1)/(c*d))*b + a*log(c*d*x + d)/(c*d)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(int(atanh(c*x)/(c*x + 1),x)*b*c + log(c*x + 1)*a)/(c*d)