\(\int \text {arctanh}(c+d \tan (a+b x)) \, dx\) [319]

Optimal result

Integrand size = 11, antiderivative size = 194 \[ \int \text {arctanh}(c+d \tan (a+b x)) \, dx=x \text {arctanh}(c+d \tan (a+b x))+\frac {1}{2} x \log \left (1+\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right )-\frac {1}{2} x \log \left (1+\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right )-\frac {i \operatorname {PolyLog}\left (2,-\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right )}{4 b}+\frac {i \operatorname {PolyLog}\left (2,-\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right )}{4 b} \] Output:

x*arctanh(c+d*tan(b*x+a))+1/2*x*ln(1+(1-c+I*d)*exp(2*I*a+2*I*b*x)/(1-c-I*d 
))-1/2*x*ln(1+(1+c-I*d)*exp(2*I*a+2*I*b*x)/(1+c+I*d))-1/4*I*polylog(2,-(1- 
c+I*d)*exp(2*I*a+2*I*b*x)/(1-c-I*d))/b+1/4*I*polylog(2,-(1+c-I*d)*exp(2*I* 
a+2*I*b*x)/(1+c+I*d))/b
 

Mathematica [A] (warning: unable to verify)

Time = 3.73 (sec) , antiderivative size = 365, normalized size of antiderivative = 1.88 \[ \int \text {arctanh}(c+d \tan (a+b x)) \, dx=x \left (\text {arctanh}(c+d \tan (a+b x))+\frac {2 a \log (1-c-d \tan (a+b x))-i \log (1-i \tan (a+b x)) \log \left (\frac {-1+c+d \tan (a+b x)}{-1+c-i d}\right )+i \log (1+i \tan (a+b x)) \log \left (\frac {-1+c+d \tan (a+b x)}{-1+c+i d}\right )-2 a \log (1+c+d \tan (a+b x))+i \log (1-i \tan (a+b x)) \log \left (\frac {1+c+d \tan (a+b x)}{1+c-i d}\right )-i \log (1+i \tan (a+b x)) \log \left (\frac {1+c+d \tan (a+b x)}{1+c+i d}\right )+i \operatorname {PolyLog}\left (2,-\frac {d (-i+\tan (a+b x))}{-1+c+i d}\right )-i \operatorname {PolyLog}\left (2,-\frac {d (-i+\tan (a+b x))}{1+c+i d}\right )-i \operatorname {PolyLog}\left (2,-\frac {d (i+\tan (a+b x))}{-1+c-i d}\right )+i \operatorname {PolyLog}\left (2,-\frac {d (i+\tan (a+b x))}{1+c-i d}\right )}{4 a-2 i \log (1-i \tan (a+b x))+2 i \log (1+i \tan (a+b x))}\right ) \] Input:

Integrate[ArcTanh[c + d*Tan[a + b*x]],x]
 

Output:

x*(ArcTanh[c + d*Tan[a + b*x]] + (2*a*Log[1 - c - d*Tan[a + b*x]] - I*Log[ 
1 - I*Tan[a + b*x]]*Log[(-1 + c + d*Tan[a + b*x])/(-1 + c - I*d)] + I*Log[ 
1 + I*Tan[a + b*x]]*Log[(-1 + c + d*Tan[a + b*x])/(-1 + c + I*d)] - 2*a*Lo 
g[1 + c + d*Tan[a + b*x]] + I*Log[1 - I*Tan[a + b*x]]*Log[(1 + c + d*Tan[a 
 + b*x])/(1 + c - I*d)] - I*Log[1 + I*Tan[a + b*x]]*Log[(1 + c + d*Tan[a + 
 b*x])/(1 + c + I*d)] + I*PolyLog[2, -((d*(-I + Tan[a + b*x]))/(-1 + c + I 
*d))] - I*PolyLog[2, -((d*(-I + Tan[a + b*x]))/(1 + c + I*d))] - I*PolyLog 
[2, -((d*(I + Tan[a + b*x]))/(-1 + c - I*d))] + I*PolyLog[2, -((d*(I + Tan 
[a + b*x]))/(1 + c - I*d))])/(4*a - (2*I)*Log[1 - I*Tan[a + b*x]] + (2*I)* 
Log[1 + I*Tan[a + b*x]]))
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.48, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {6813, 2620, 2715, 2838}

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.

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

\(\Big \downarrow \) 6813

\(\displaystyle -b (i c+d+i) \int \frac {e^{2 i a+2 i b x} x}{c+(c-i d+1) e^{2 i a+2 i b x}+i d+1}dx+b (-d+i (1-c)) \int \frac {e^{2 i a+2 i b x} x}{-c+(-c+i d+1) e^{2 i a+2 i b x}-i d+1}dx+x \text {arctanh}(d \tan (a+b x)+c)\)

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

\(\displaystyle b (-d+i (1-c)) \left (\frac {i \int e^{-2 i a-2 i b x} \log \left (\frac {e^{2 i a+2 i b x} (-c+i d+1)}{-c-i d+1}+1\right )de^{2 i a+2 i b x}}{4 b^2 (-d+i (1-c))}+\frac {x \log \left (1+\frac {(-c+i d+1) e^{2 i a+2 i b x}}{-c-i d+1}\right )}{2 b (-d+i (1-c))}\right )-b (i c+d+i) \left (\frac {i \int e^{-2 i a-2 i b x} \log \left (\frac {e^{2 i a+2 i b x} (c-i d+1)}{c+i d+1}+1\right )de^{2 i a+2 i b x}}{4 b (b d+i (b c+b))}+\frac {x \log \left (1+\frac {(c-i d+1) e^{2 i a+2 i b x}}{c+i d+1}\right )}{2 (b d+i (b c+b))}\right )+x \text {arctanh}(d \tan (a+b x)+c)\)

\(\Big \downarrow \) 2838

\(\displaystyle x \text {arctanh}(d \tan (a+b x)+c)+b (-d+i (1-c)) \left (\frac {x \log \left (1+\frac {(-c+i d+1) e^{2 i a+2 i b x}}{-c-i d+1}\right )}{2 b (-d+i (1-c))}-\frac {i \operatorname {PolyLog}\left (2,-\frac {(-c+i d+1) e^{2 i a+2 i b x}}{-c-i d+1}\right )}{4 b^2 (-d+i (1-c))}\right )-b (i c+d+i) \left (\frac {x \log \left (1+\frac {(c-i d+1) e^{2 i a+2 i b x}}{c+i d+1}\right )}{2 (b d+i (b c+b))}-\frac {i \operatorname {PolyLog}\left (2,-\frac {(c-i d+1) e^{2 i a+2 i b x}}{c+i d+1}\right )}{4 b (b d+i (b c+b))}\right )\)

Input:

Int[ArcTanh[c + d*Tan[a + b*x]],x]
 

Output:

x*ArcTanh[c + d*Tan[a + b*x]] + b*(I*(1 - c) - d)*((x*Log[1 + ((1 - c + I* 
d)*E^((2*I)*a + (2*I)*b*x))/(1 - c - I*d)])/(2*b*(I*(1 - c) - d)) - ((I/4) 
*PolyLog[2, -(((1 - c + I*d)*E^((2*I)*a + (2*I)*b*x))/(1 - c - I*d))])/(b^ 
2*(I*(1 - c) - d))) - b*(I + I*c + d)*((x*Log[1 + ((1 + c - I*d)*E^((2*I)* 
a + (2*I)*b*x))/(1 + c + I*d)])/(2*(I*(b + b*c) + b*d)) - ((I/4)*PolyLog[2 
, -(((1 + c - I*d)*E^((2*I)*a + (2*I)*b*x))/(1 + c + I*d))])/(b*(I*(b + b* 
c) + b*d)))
 

Defintions of rubi rules used

rule 2620

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[1/(d*e*n*Log[F])   Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) 
))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
 

rule 2838

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

rule 6813

Int[ArcTanh[(c_.) + (d_.)*Tan[(a_.) + (b_.)*(x_)]], x_Symbol] :> Simp[x*Arc 
Tanh[c + d*Tan[a + b*x]], x] + (-Simp[I*b*(1 + c - I*d)   Int[x*(E^(2*I*a + 
 2*I*b*x)/(1 + c + I*d + (1 + c - I*d)*E^(2*I*a + 2*I*b*x))), x], x] + Simp 
[I*b*(1 - c + I*d)   Int[x*(E^(2*I*a + 2*I*b*x)/(1 - c - I*d + (1 - c + I*d 
)*E^(2*I*a + 2*I*b*x))), x], x]) /; FreeQ[{a, b, c, d}, x] && NeQ[(c + I*d) 
^2, 1]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 556 vs. \(2 (164 ) = 328\).

Time = 3.42 (sec) , antiderivative size = 557, normalized size of antiderivative = 2.87

method	result	size
derivativedivides	\(\frac {d \arctan \left (\tan \left (b x +a \right )\right ) \operatorname {arctanh}\left (c +d \tan \left (b x +a \right )\right )-d^{2} \left (\frac {\arctan \left (-\frac {c +d \tan \left (b x +a \right )}{d}+\frac {c}{d}\right ) \ln \left (d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )+c -1\right )}{2 d}-\frac {\arctan \left (-\frac {c +d \tan \left (b x +a \right )}{d}+\frac {c}{d}\right ) \ln \left (d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )+c +1\right )}{2 d}+\frac {i \ln \left (d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )+c +1\right ) \left (\ln \left (\frac {i d -d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )}{i d +c +1}\right )-\ln \left (\frac {i d +d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )}{i d -c -1}\right )\right )}{4 d}+\frac {i \left (\operatorname {dilog}\left (\frac {i d -d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )}{i d +c +1}\right )-\operatorname {dilog}\left (\frac {i d +d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )}{i d -c -1}\right )\right )}{4 d}-\frac {i \ln \left (d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )+c -1\right ) \left (\ln \left (\frac {i d -d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )}{i d +c -1}\right )-\ln \left (\frac {i d +d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )}{i d -c +1}\right )\right )}{4 d}-\frac {i \left (\operatorname {dilog}\left (\frac {i d -d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )}{i d +c -1}\right )-\operatorname {dilog}\left (\frac {i d +d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )}{i d -c +1}\right )\right )}{4 d}\right )}{b d}\)	\(557\)
default	\(\frac {d \arctan \left (\tan \left (b x +a \right )\right ) \operatorname {arctanh}\left (c +d \tan \left (b x +a \right )\right )-d^{2} \left (\frac {\arctan \left (-\frac {c +d \tan \left (b x +a \right )}{d}+\frac {c}{d}\right ) \ln \left (d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )+c -1\right )}{2 d}-\frac {\arctan \left (-\frac {c +d \tan \left (b x +a \right )}{d}+\frac {c}{d}\right ) \ln \left (d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )+c +1\right )}{2 d}+\frac {i \ln \left (d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )+c +1\right ) \left (\ln \left (\frac {i d -d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )}{i d +c +1}\right )-\ln \left (\frac {i d +d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )}{i d -c -1}\right )\right )}{4 d}+\frac {i \left (\operatorname {dilog}\left (\frac {i d -d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )}{i d +c +1}\right )-\operatorname {dilog}\left (\frac {i d +d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )}{i d -c -1}\right )\right )}{4 d}-\frac {i \ln \left (d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )+c -1\right ) \left (\ln \left (\frac {i d -d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )}{i d +c -1}\right )-\ln \left (\frac {i d +d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )}{i d -c +1}\right )\right )}{4 d}-\frac {i \left (\operatorname {dilog}\left (\frac {i d -d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )}{i d +c -1}\right )-\operatorname {dilog}\left (\frac {i d +d \left (\frac {c +d \tan \left (b x +a \right )}{d}-\frac {c}{d}\right )}{i d -c +1}\right )\right )}{4 d}\right )}{b d}\)	\(557\)
risch	\(\text {Expression too large to display}\)	\(4027\)

Input:

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

Output:

1/b/d*(d*arctan(tan(b*x+a))*arctanh(c+d*tan(b*x+a))-d^2*(1/2*arctan(-(c+d* 
tan(b*x+a))/d+c/d)/d*ln(d*((c+d*tan(b*x+a))/d-c/d)+c-1)-1/2*arctan(-(c+d*t 
an(b*x+a))/d+c/d)/d*ln(d*((c+d*tan(b*x+a))/d-c/d)+c+1)+1/4*I*ln(d*((c+d*ta 
n(b*x+a))/d-c/d)+c+1)*(ln((I*d-d*((c+d*tan(b*x+a))/d-c/d))/(1+c+I*d))-ln(( 
I*d+d*((c+d*tan(b*x+a))/d-c/d))/(I*d-c-1)))/d+1/4*I*(dilog((I*d-d*((c+d*ta 
n(b*x+a))/d-c/d))/(1+c+I*d))-dilog((I*d+d*((c+d*tan(b*x+a))/d-c/d))/(I*d-c 
-1)))/d-1/4*I*ln(d*((c+d*tan(b*x+a))/d-c/d)+c-1)*(ln((I*d-d*((c+d*tan(b*x+ 
a))/d-c/d))/(I*d+c-1))-ln((I*d+d*((c+d*tan(b*x+a))/d-c/d))/(1-c+I*d)))/d-1 
/4*I*(dilog((I*d-d*((c+d*tan(b*x+a))/d-c/d))/(I*d+c-1))-dilog((I*d+d*((c+d 
*tan(b*x+a))/d-c/d))/(1-c+I*d)))/d))
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1185 vs. \(2 (136) = 272\).

Time = 0.15 (sec) , antiderivative size = 1185, normalized size of antiderivative = 6.11 \[ \int \text {arctanh}(c+d \tan (a+b x)) \, dx=\text {Too large to display} \] Input:

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

Output:

1/8*(4*b*x*log(-(d*tan(b*x + a) + c + 1)/(d*tan(b*x + a) + c - 1)) - 2*(b* 
x + a)*log(-2*((I*(c + 1)*d - d^2)*tan(b*x + a)^2 - c^2 - I*(c + 1)*d + (I 
*c^2 - 2*(c + 1)*d - I*d^2 + 2*I*c + I)*tan(b*x + a) - 2*c - 1)/((c^2 + d^ 
2 + 2*c + 1)*tan(b*x + a)^2 + c^2 + d^2 + 2*c + 1)) - 2*(b*x + a)*log(-2*( 
(-I*(c + 1)*d - d^2)*tan(b*x + a)^2 - c^2 + I*(c + 1)*d + (-I*c^2 - 2*(c + 
 1)*d + I*d^2 - 2*I*c - I)*tan(b*x + a) - 2*c - 1)/((c^2 + d^2 + 2*c + 1)* 
tan(b*x + a)^2 + c^2 + d^2 + 2*c + 1)) + 2*(b*x + a)*log(-2*((I*(c - 1)*d 
- d^2)*tan(b*x + a)^2 - c^2 - I*(c - 1)*d + (I*c^2 - 2*(c - 1)*d - I*d^2 - 
 2*I*c + I)*tan(b*x + a) + 2*c - 1)/((c^2 + d^2 - 2*c + 1)*tan(b*x + a)^2 
+ c^2 + d^2 - 2*c + 1)) + 2*(b*x + a)*log(-2*((-I*(c - 1)*d - d^2)*tan(b*x 
 + a)^2 - c^2 + I*(c - 1)*d + (-I*c^2 - 2*(c - 1)*d + I*d^2 + 2*I*c - I)*t 
an(b*x + a) + 2*c - 1)/((c^2 + d^2 - 2*c + 1)*tan(b*x + a)^2 + c^2 + d^2 - 
 2*c + 1)) + 2*a*log(((I*(c + 1)*d + d^2)*tan(b*x + a)^2 - c^2 + I*(c + 1) 
*d + (I*c^2 + I*d^2 + 2*I*c + I)*tan(b*x + a) - 2*c - 1)/(tan(b*x + a)^2 + 
 1)) + 2*a*log(((I*(c + 1)*d - d^2)*tan(b*x + a)^2 + c^2 + I*(c + 1)*d + ( 
I*c^2 + I*d^2 + 2*I*c + I)*tan(b*x + a) + 2*c + 1)/(tan(b*x + a)^2 + 1)) - 
 2*a*log(((I*(c - 1)*d + d^2)*tan(b*x + a)^2 - c^2 + I*(c - 1)*d + (I*c^2 
+ I*d^2 - 2*I*c + I)*tan(b*x + a) + 2*c - 1)/(tan(b*x + a)^2 + 1)) - 2*a*l 
og(((I*(c - 1)*d - d^2)*tan(b*x + a)^2 + c^2 + I*(c - 1)*d + (I*c^2 + I*d^ 
2 - 2*I*c + I)*tan(b*x + a) - 2*c + 1)/(tan(b*x + a)^2 + 1)) - I*dilog(...
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 372 vs. \(2 (136) = 272\).

Time = 0.20 (sec) , antiderivative size = 372, normalized size of antiderivative = 1.92 \[ \int \text {arctanh}(c+d \tan (a+b x)) \, dx=\frac {4 \, {\left (b x + a\right )} \operatorname {artanh}\left (d \tan \left (b x + a\right ) + c\right ) + {\left (\arctan \left (\frac {d^{2} \tan \left (b x + a\right ) + {\left (c + 1\right )} d}{c^{2} + d^{2} + 2 \, c + 1}, \frac {{\left (c + 1\right )} d \tan \left (b x + a\right ) + c^{2} + 2 \, c + 1}{c^{2} + d^{2} + 2 \, c + 1}\right ) - \arctan \left (\frac {d^{2} \tan \left (b x + a\right ) + {\left (c - 1\right )} d}{c^{2} + d^{2} - 2 \, c + 1}, \frac {{\left (c - 1\right )} d \tan \left (b x + a\right ) + c^{2} - 2 \, c + 1}{c^{2} + d^{2} - 2 \, c + 1}\right )\right )} \log \left (\tan \left (b x + a\right )^{2} + 1\right ) - {\left (b x + a\right )} \log \left (\frac {d^{2} \tan \left (b x + a\right )^{2} + 2 \, {\left (c + 1\right )} d \tan \left (b x + a\right ) + c^{2} + 2 \, c + 1}{c^{2} + d^{2} + 2 \, c + 1}\right ) + {\left (b x + a\right )} \log \left (\frac {d^{2} \tan \left (b x + a\right )^{2} + 2 \, {\left (c - 1\right )} d \tan \left (b x + a\right ) + c^{2} - 2 \, c + 1}{c^{2} + d^{2} - 2 \, c + 1}\right ) - i \, {\rm Li}_2\left (-\frac {i \, d \tan \left (b x + a\right ) - d}{i \, c + d + i}\right ) + i \, {\rm Li}_2\left (-\frac {i \, d \tan \left (b x + a\right ) - d}{i \, c + d - i}\right ) - i \, {\rm Li}_2\left (\frac {i \, d \tan \left (b x + a\right ) + d}{-i \, c + d + i}\right ) + i \, {\rm Li}_2\left (\frac {i \, d \tan \left (b x + a\right ) + d}{-i \, c + d - i}\right )}{4 \, b} \] Input:

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

Output:

1/4*(4*(b*x + a)*arctanh(d*tan(b*x + a) + c) + (arctan2((d^2*tan(b*x + a) 
+ (c + 1)*d)/(c^2 + d^2 + 2*c + 1), ((c + 1)*d*tan(b*x + a) + c^2 + 2*c + 
1)/(c^2 + d^2 + 2*c + 1)) - arctan2((d^2*tan(b*x + a) + (c - 1)*d)/(c^2 + 
d^2 - 2*c + 1), ((c - 1)*d*tan(b*x + a) + c^2 - 2*c + 1)/(c^2 + d^2 - 2*c 
+ 1)))*log(tan(b*x + a)^2 + 1) - (b*x + a)*log((d^2*tan(b*x + a)^2 + 2*(c 
+ 1)*d*tan(b*x + a) + c^2 + 2*c + 1)/(c^2 + d^2 + 2*c + 1)) + (b*x + a)*lo 
g((d^2*tan(b*x + a)^2 + 2*(c - 1)*d*tan(b*x + a) + c^2 - 2*c + 1)/(c^2 + d 
^2 - 2*c + 1)) - I*dilog(-(I*d*tan(b*x + a) - d)/(I*c + d + I)) + I*dilog( 
-(I*d*tan(b*x + a) - d)/(I*c + d - I)) - I*dilog((I*d*tan(b*x + a) + d)/(- 
I*c + d + I)) + I*dilog((I*d*tan(b*x + a) + d)/(-I*c + d - I)))/b
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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