\(\int \text {arctanh}(1-i d+d \tan (a+b x)) \, dx\) [323]

Optimal result

Integrand size = 16, antiderivative size = 93 \[ \int \text {arctanh}(1-i d+d \tan (a+b x)) \, dx=\frac {1}{2} i b x^2+x \text {arctanh}(1-i d+d \tan (a+b x))-\frac {1}{2} x \log \left (1+(1-i d) e^{2 i a+2 i b x}\right )+\frac {i \operatorname {PolyLog}\left (2,-\left ((1-i d) e^{2 i a+2 i b x}\right )\right )}{4 b} \] Output:

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

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(766\) vs. \(2(93)=186\).

Time = 7.50 (sec) , antiderivative size = 766, normalized size of antiderivative = 8.24 \[ \int \text {arctanh}(1-i d+d \tan (a+b x)) \, dx=x \text {arctanh}(1-i d+d \tan (a+b x))+\frac {x \left (2 i b x \log (2 \cos (b x) (\cos (b x)-i \sin (b x)))-\log \left (\frac {\sec (b x) (\cos (a)-i \sin (a)) ((2 i+d) \cos (a+b x)+i d \sin (a+b x))}{2 (i+d)}\right ) \log (1-i \tan (b x))+\log \left (\frac {\sec (b x) ((2-i d) \cos (a+b x)+d \sin (a+b x))}{2 \cos (a)-2 i \sin (a)}\right ) \log (1+i \tan (b x))-\operatorname {PolyLog}(2,-\cos (2 b x)+i \sin (2 b x))-\operatorname {PolyLog}\left (2,\frac {\sec (b x) (d \cos (a)+i (2 i+d) \sin (a)) (\cos (a+b x)-i \sin (a+b x))}{2 (i+d)}\right )+\operatorname {PolyLog}\left (2,-\frac {1}{2} (\cos (a)+i \sin (a)) (d \cos (a)+i (2 i+d) \sin (a)) (-i+\tan (b x))\right )\right ) \sec ^2(a+b x) (\cos (b x)+i \sin (b x)) (i \cos (b x)+\sin (b x)) ((2-i d) \cos (a+b x)+d \sin (a+b x))}{((2 i+d) \cos (a+b x)+i d \sin (a+b x)) \left (\frac {i \log (1+i \tan (b x)) \sec (b x) (d \cos (a)+i (2 i+d) \sin (a))}{(2 i+d) \cos (a+b x)+i d \sin (a+b x)}+\frac {\log (1-i \tan (b x)) \sec (b x) (-i d \cos (a)+(2 i+d) \sin (a))}{(2 i+d) \cos (a+b x)+i d \sin (a+b x)}+2 b x (1-i \tan (b x))+\frac {\log \left (\frac {\sec (b x) ((2-i d) \cos (a+b x)+d \sin (a+b x))}{2 \cos (a)-2 i \sin (a)}\right ) \sec ^2(b x)}{-i+\tan (b x)}-\frac {\log \left (1+\frac {1}{2} (\cos (a)+i \sin (a)) (d \cos (a)+i (2 i+d) \sin (a)) (-i+\tan (b x))\right ) \sec ^2(b x)}{-i+\tan (b x)}+\log \left (\frac {\sec (b x) (\cos (a)-i \sin (a)) ((2 i+d) \cos (a+b x)+i d \sin (a+b x))}{2 (i+d)}\right ) (-i+\tan (b x))-\frac {\log \left (\frac {\sec (b x) (\cos (a)-i \sin (a)) ((2 i+d) \cos (a+b x)+i d \sin (a+b x))}{2 (i+d)}\right ) \sec ^2(b x)}{i+\tan (b x)}\right ) (-i+\tan (a+b x)) (2-i d+d \tan (a+b x))} \] Input:

Integrate[ArcTanh[1 - I*d + d*Tan[a + b*x]],x]
 

Output:

x*ArcTanh[1 - I*d + d*Tan[a + b*x]] + (x*((2*I)*b*x*Log[2*Cos[b*x]*(Cos[b* 
x] - I*Sin[b*x])] - Log[(Sec[b*x]*(Cos[a] - I*Sin[a])*((2*I + d)*Cos[a + b 
*x] + I*d*Sin[a + b*x]))/(2*(I + d))]*Log[1 - I*Tan[b*x]] + Log[(Sec[b*x]* 
((2 - I*d)*Cos[a + b*x] + d*Sin[a + b*x]))/(2*Cos[a] - (2*I)*Sin[a])]*Log[ 
1 + I*Tan[b*x]] - PolyLog[2, -Cos[2*b*x] + I*Sin[2*b*x]] - PolyLog[2, (Sec 
[b*x]*(d*Cos[a] + I*(2*I + d)*Sin[a])*(Cos[a + b*x] - I*Sin[a + b*x]))/(2* 
(I + d))] + PolyLog[2, -1/2*((Cos[a] + I*Sin[a])*(d*Cos[a] + I*(2*I + d)*S 
in[a])*(-I + Tan[b*x]))])*Sec[a + b*x]^2*(Cos[b*x] + I*Sin[b*x])*(I*Cos[b* 
x] + Sin[b*x])*((2 - I*d)*Cos[a + b*x] + d*Sin[a + b*x]))/(((2*I + d)*Cos[ 
a + b*x] + I*d*Sin[a + b*x])*((I*Log[1 + I*Tan[b*x]]*Sec[b*x]*(d*Cos[a] + 
I*(2*I + d)*Sin[a]))/((2*I + d)*Cos[a + b*x] + I*d*Sin[a + b*x]) + (Log[1 
- I*Tan[b*x]]*Sec[b*x]*((-I)*d*Cos[a] + (2*I + d)*Sin[a]))/((2*I + d)*Cos[ 
a + b*x] + I*d*Sin[a + b*x]) + 2*b*x*(1 - I*Tan[b*x]) + (Log[(Sec[b*x]*((2 
 - I*d)*Cos[a + b*x] + d*Sin[a + b*x]))/(2*Cos[a] - (2*I)*Sin[a])]*Sec[b*x 
]^2)/(-I + Tan[b*x]) - (Log[1 + ((Cos[a] + I*Sin[a])*(d*Cos[a] + I*(2*I + 
d)*Sin[a])*(-I + Tan[b*x]))/2]*Sec[b*x]^2)/(-I + Tan[b*x]) + Log[(Sec[b*x] 
*(Cos[a] - I*Sin[a])*((2*I + d)*Cos[a + b*x] + I*d*Sin[a + b*x]))/(2*(I + 
d))]*(-I + Tan[b*x]) - (Log[(Sec[b*x]*(Cos[a] - I*Sin[a])*((2*I + d)*Cos[a 
 + b*x] + I*d*Sin[a + b*x]))/(2*(I + d))]*Sec[b*x]^2)/(I + Tan[b*x]))*(-I 
+ Tan[a + b*x])*(2 - I*d + d*Tan[a + b*x]))
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.32, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {6809, 2615, 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.

\(\displaystyle \int \text {arctanh}(d \tan (a+b x)-i d+1) \, dx\)

\(\Big \downarrow \) 6809

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

\(\Big \downarrow \) 2615

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

Input:

Int[ArcTanh[1 - I*d + d*Tan[a + b*x]],x]
 

Output:

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

Defintions of rubi rules used

rule 2615

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

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 6809

Int[ArcTanh[(c_.) + (d_.)*Tan[(a_.) + (b_.)*(x_)]], x_Symbol] :> Simp[x*Arc 
Tanh[c + d*Tan[a + b*x]], x] + Simp[I*b   Int[x/(c + I*d + c*E^(2*I*a + 2*I 
*b*x)), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[(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. 306 vs. \(2 (76 ) = 152\).

Time = 3.37 (sec) , antiderivative size = 307, normalized size of antiderivative = 3.30

method	result	size
derivativedivides	\(\frac {-\frac {i \operatorname {arctanh}\left (1-i d +d \tan \left (b x +a \right )\right ) d \ln \left (-i d +d \tan \left (b x +a \right )\right )}{2}+\frac {i \operatorname {arctanh}\left (1-i d +d \tan \left (b x +a \right )\right ) d \ln \left (i d +d \tan \left (b x +a \right )\right )}{2}-\frac {d^{2} \left (-\frac {i \left (\frac {\operatorname {dilog}\left (\frac {i \left (-i d +d \tan \left (b x +a \right )\right )}{2 d}\right )}{2}+\frac {\ln \left (i d +d \tan \left (b x +a \right )\right ) \ln \left (\frac {i \left (-i d +d \tan \left (b x +a \right )\right )}{2 d}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {i \left (i d +d \tan \left (b x +a \right )-i \left (2 i+2 d \right )\right )}{2 i+2 d}\right )}{2}-\frac {\ln \left (i d +d \tan \left (b x +a \right )\right ) \ln \left (\frac {i \left (i d +d \tan \left (b x +a \right )-i \left (2 i+2 d \right )\right )}{2 i+2 d}\right )}{2}\right )}{d}+\frac {i \left (\frac {\ln \left (-i d +d \tan \left (b x +a \right )\right )^{2}}{4}-\frac {\operatorname {dilog}\left (1-\frac {i d}{2}+\frac {d \tan \left (b x +a \right )}{2}\right )}{2}-\frac {\ln \left (-i d +d \tan \left (b x +a \right )\right ) \ln \left (1-\frac {i d}{2}+\frac {d \tan \left (b x +a \right )}{2}\right )}{2}\right )}{d}\right )}{2}}{b d}\)	\(307\)
default	\(\frac {-\frac {i \operatorname {arctanh}\left (1-i d +d \tan \left (b x +a \right )\right ) d \ln \left (-i d +d \tan \left (b x +a \right )\right )}{2}+\frac {i \operatorname {arctanh}\left (1-i d +d \tan \left (b x +a \right )\right ) d \ln \left (i d +d \tan \left (b x +a \right )\right )}{2}-\frac {d^{2} \left (-\frac {i \left (\frac {\operatorname {dilog}\left (\frac {i \left (-i d +d \tan \left (b x +a \right )\right )}{2 d}\right )}{2}+\frac {\ln \left (i d +d \tan \left (b x +a \right )\right ) \ln \left (\frac {i \left (-i d +d \tan \left (b x +a \right )\right )}{2 d}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {i \left (i d +d \tan \left (b x +a \right )-i \left (2 i+2 d \right )\right )}{2 i+2 d}\right )}{2}-\frac {\ln \left (i d +d \tan \left (b x +a \right )\right ) \ln \left (\frac {i \left (i d +d \tan \left (b x +a \right )-i \left (2 i+2 d \right )\right )}{2 i+2 d}\right )}{2}\right )}{d}+\frac {i \left (\frac {\ln \left (-i d +d \tan \left (b x +a \right )\right )^{2}}{4}-\frac {\operatorname {dilog}\left (1-\frac {i d}{2}+\frac {d \tan \left (b x +a \right )}{2}\right )}{2}-\frac {\ln \left (-i d +d \tan \left (b x +a \right )\right ) \ln \left (1-\frac {i d}{2}+\frac {d \tan \left (b x +a \right )}{2}\right )}{2}\right )}{d}\right )}{2}}{b d}\)	\(307\)
risch	\(\text {Expression too large to display}\)	\(1623\)

Input:

int(arctanh(1-I*d+d*tan(b*x+a)),x,method=_RETURNVERBOSE)
 

Output:

1/b/d*(-1/2*I*arctanh(1-I*d+d*tan(b*x+a))*d*ln(-I*d+d*tan(b*x+a))+1/2*I*ar 
ctanh(1-I*d+d*tan(b*x+a))*d*ln(I*d+d*tan(b*x+a))-1/2*d^2*(-I/d*(1/2*dilog( 
1/2*I*(-I*d+d*tan(b*x+a))/d)+1/2*ln(I*d+d*tan(b*x+a))*ln(1/2*I*(-I*d+d*tan 
(b*x+a))/d)-1/2*dilog(I*(I*d+d*tan(b*x+a)-I*(2*I+2*d))/(2*I+2*d))-1/2*ln(I 
*d+d*tan(b*x+a))*ln(I*(I*d+d*tan(b*x+a)-I*(2*I+2*d))/(2*I+2*d)))+I/d*(1/4* 
ln(-I*d+d*tan(b*x+a))^2-1/2*dilog(1-1/2*I*d+1/2*d*tan(b*x+a))-1/2*ln(-I*d+ 
d*tan(b*x+a))*ln(1-1/2*I*d+1/2*d*tan(b*x+a)))))
 

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. 218 vs. \(2 (65) = 130\).

Time = 0.10 (sec) , antiderivative size = 218, normalized size of antiderivative = 2.34 \[ \int \text {arctanh}(1-i d+d \tan (a+b x)) \, dx=\frac {i \, b^{2} x^{2} + b x \log \left (-\frac {{\left ({\left (d + i\right )} e^{\left (2 i \, b x + 2 i \, a\right )} + i\right )} e^{\left (-2 i \, b x - 2 i \, a\right )}}{d}\right ) - i \, a^{2} - {\left (b x + a\right )} \log \left (\frac {1}{2} \, \sqrt {4 i \, d - 4} e^{\left (i \, b x + i \, a\right )} + 1\right ) - {\left (b x + a\right )} \log \left (-\frac {1}{2} \, \sqrt {4 i \, d - 4} e^{\left (i \, b x + i \, a\right )} + 1\right ) + a \log \left (\frac {2 \, {\left (d + i\right )} e^{\left (i \, b x + i \, a\right )} + i \, \sqrt {4 i \, d - 4}}{2 \, {\left (d + i\right )}}\right ) + a \log \left (\frac {2 \, {\left (d + i\right )} e^{\left (i \, b x + i \, a\right )} - i \, \sqrt {4 i \, d - 4}}{2 \, {\left (d + i\right )}}\right ) + i \, {\rm Li}_2\left (\frac {1}{2} \, \sqrt {4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right ) + i \, {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right )}{2 \, b} \] Input:

integrate(arctanh(1-I*d+d*tan(b*x+a)),x, algorithm="fricas")
 

Output:

1/2*(I*b^2*x^2 + b*x*log(-((d + I)*e^(2*I*b*x + 2*I*a) + I)*e^(-2*I*b*x - 
2*I*a)/d) - I*a^2 - (b*x + a)*log(1/2*sqrt(4*I*d - 4)*e^(I*b*x + I*a) + 1) 
 - (b*x + a)*log(-1/2*sqrt(4*I*d - 4)*e^(I*b*x + I*a) + 1) + a*log(1/2*(2* 
(d + I)*e^(I*b*x + I*a) + I*sqrt(4*I*d - 4))/(d + I)) + a*log(1/2*(2*(d + 
I)*e^(I*b*x + I*a) - I*sqrt(4*I*d - 4))/(d + I)) + I*dilog(1/2*sqrt(4*I*d 
- 4)*e^(I*b*x + I*a)) + I*dilog(-1/2*sqrt(4*I*d - 4)*e^(I*b*x + I*a)))/b
 

Sympy [F]

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

integrate(atanh(1-I*d+d*tan(b*x+a)),x)
 

Output:

Integral(atanh(d*tan(a + b*x) - I*d + 1), 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. 260 vs. \(2 (65) = 130\).

Time = 0.12 (sec) , antiderivative size = 260, normalized size of antiderivative = 2.80 \[ \int \text {arctanh}(1-i d+d \tan (a+b x)) \, dx=-\frac {4 \, {\left (b x + a\right )} d {\left (\frac {\log \left (d \tan \left (b x + a\right ) - i \, d + 2\right )}{d} - \frac {\log \left (\tan \left (b x + a\right ) - i\right )}{d}\right )} + d {\left (-\frac {2 i \, {\left (\log \left (d \tan \left (b x + a\right ) - i \, d + 2\right ) \log \left (-\frac {i \, d \tan \left (b x + a\right ) + d + 2 i}{2 \, {\left (d + i\right )}} + 1\right ) + {\rm Li}_2\left (\frac {i \, d \tan \left (b x + a\right ) + d + 2 i}{2 \, {\left (d + i\right )}}\right )\right )}}{d} + \frac {2 i \, \log \left (d \tan \left (b x + a\right ) - i \, d + 2\right ) \log \left (\tan \left (b x + a\right ) - i\right ) - i \, \log \left (\tan \left (b x + a\right ) - i\right )^{2}}{d} - \frac {2 i \, {\left (\log \left (\frac {1}{2} \, d \tan \left (b x + a\right ) - \frac {1}{2} i \, d + 1\right ) \log \left (\tan \left (b x + a\right ) - i\right ) + {\rm Li}_2\left (-\frac {1}{2} \, d \tan \left (b x + a\right ) + \frac {1}{2} i \, d\right )\right )}}{d} + \frac {2 i \, {\left (\log \left (\tan \left (b x + a\right ) - i\right ) \log \left (-\frac {1}{2} i \, \tan \left (b x + a\right ) + \frac {1}{2}\right ) + {\rm Li}_2\left (\frac {1}{2} i \, \tan \left (b x + a\right ) + \frac {1}{2}\right )\right )}}{d}\right )} - 8 \, {\left (b x + a\right )} \operatorname {artanh}\left (d \tan \left (b x + a\right ) - i \, d + 1\right )}{8 \, b} \] Input:

integrate(arctanh(1-I*d+d*tan(b*x+a)),x, algorithm="maxima")
 

Output:

-1/8*(4*(b*x + a)*d*(log(d*tan(b*x + a) - I*d + 2)/d - log(tan(b*x + a) - 
I)/d) + d*(-2*I*(log(d*tan(b*x + a) - I*d + 2)*log(-1/2*(I*d*tan(b*x + a) 
+ d + 2*I)/(d + I) + 1) + dilog(1/2*(I*d*tan(b*x + a) + d + 2*I)/(d + I))) 
/d + (2*I*log(d*tan(b*x + a) - I*d + 2)*log(tan(b*x + a) - I) - I*log(tan( 
b*x + a) - I)^2)/d - 2*I*(log(1/2*d*tan(b*x + a) - 1/2*I*d + 1)*log(tan(b* 
x + a) - I) + dilog(-1/2*d*tan(b*x + a) + 1/2*I*d))/d + 2*I*(log(tan(b*x + 
 a) - I)*log(-1/2*I*tan(b*x + a) + 1/2) + dilog(1/2*I*tan(b*x + a) + 1/2)) 
/d) - 8*(b*x + a)*arctanh(d*tan(b*x + a) - I*d + 1))/b
 

Giac [F]

\[ \int \text {arctanh}(1-i d+d \tan (a+b x)) \, dx=\int { \operatorname {artanh}\left (d \tan \left (b x + a\right ) - i \, d + 1\right ) \,d x } \] Input:

integrate(arctanh(1-I*d+d*tan(b*x+a)),x, algorithm="giac")
 

Output:

integrate(arctanh(d*tan(b*x + a) - I*d + 1), x)
 

Mupad [F(-1)]

Timed out. \[ \int \text {arctanh}(1-i d+d \tan (a+b x)) \, dx=\int \mathrm {atanh}\left (d\,\mathrm {tan}\left (a+b\,x\right )+1-d\,1{}\mathrm {i}\right ) \,d x \] Input:

int(atanh(d*tan(a + b*x) - d*1i + 1),x)
 

Output:

int(atanh(d*tan(a + b*x) - d*1i + 1), x)
 

Reduce [F]

\[ \int \text {arctanh}(1-i d+d \tan (a+b x)) \, dx=\int \mathit {atanh} \left (\tan \left (b x +a \right ) d -d i +1\right )d x \] Input:

int(atanh(1-I*d+d*tan(b*x+a)),x)
 

Output:

int(atanh(tan(a + b*x)*d - d*i + 1),x)