\(\int \arctan (c+(1+i c) \tan (a+b x)) \, dx\) [54]

Optimal result

Integrand size = 17, antiderivative size = 85 \[ \int \arctan (c+(1+i c) \tan (a+b x)) \, dx=-\frac {b x^2}{2}+x \arctan (c+(1+i c) \tan (a+b x))-\frac {1}{2} i x \log \left (1-i c e^{2 i a+2 i b x}\right )-\frac {\operatorname {PolyLog}\left (2,i c e^{2 i a+2 i b x}\right )}{4 b} \] Output:

-1/2*b*x^2+x*arctan(c+(1+I*c)*tan(b*x+a))-1/2*I*x*ln(1-I*c*exp(2*I*a+2*I*b 
*x))-1/4*polylog(2,I*c*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. \(967\) vs. \(2(85)=170\).

Time = 8.55 (sec) , antiderivative size = 967, normalized size of antiderivative = 11.38 \[ \int \arctan (c+(1+i c) \tan (a+b x)) \, dx=x \arctan (c+(1+i c) \tan (a+b x))+\frac {i 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)) ((i+c) \cos (a+b x)+(1+i c) \sin (a+b x))}{2 c}\right ) \log (1-i \tan (b x))+\log \left (\frac {\sec (b x) ((1-i c) \cos (a+b x)+(-i+c) \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) ((-i+c) \cos (a)+i (i+c) \sin (a)) (\cos (a+b x)-i \sin (a+b x))}{2 c}\right )+\operatorname {PolyLog}\left (2,\frac {1}{2} \sec (b x) ((1+i c) \cos (a)-(i+c) \sin (a)) (\cos (a+b x)+i \sin (a+b x))\right )\right ) \sec ^2(a+b x) (\cos (b x)+i \sin (b x)) (i \cos (b x)+\sin (b x)) ((1-i c) \cos (a+b x)+(-i+c) \sin (a+b x))}{((i+c) \cos (a+b x)+(1+i c) \sin (a+b x)) \left (2 b x-i \log \left (1-\frac {\sec (b x) ((-i+c) \cos (a)+i (i+c) \sin (a)) (\cos (a+b x)-i \sin (a+b x))}{2 c}\right )-i \log \left (1+\frac {1}{2} \sec (b x) ((-1-i c) \cos (a)+(i+c) \sin (a)) (\cos (a+b x)+i \sin (a+b x))\right )-\frac {i (-i+c) \cos (a+b x) (\log (1-i \tan (b x))-\log (1+i \tan (b x)))}{(i+c) \cos (a+b x)+(1+i c) \sin (a+b x)}+\frac {(i+c) (\log (1-i \tan (b x))-\log (1+i \tan (b x))) \sin (a+b x)}{(i+c) \cos (a+b x)+(1+i c) \sin (a+b x)}-2 i b x \tan (b x)+\log \left (1-\frac {\sec (b x) ((-i+c) \cos (a)+i (i+c) \sin (a)) (\cos (a+b x)-i \sin (a+b x))}{2 c}\right ) \tan (b x)-\log \left (1+\frac {1}{2} \sec (b x) ((-1-i c) \cos (a)+(i+c) \sin (a)) (\cos (a+b x)+i \sin (a+b x))\right ) \tan (b x)-\log (1-i \tan (b x)) \tan (b x)+\cos ^2(a) \log (1+i \tan (b x)) \tan (b x)+\log (1+i \tan (b x)) \sin ^2(a) \tan (b x)+\frac {\log \left (\frac {\sec (b x) ((1-i c) \cos (a+b x)+(-i+c) \sin (a+b x))}{2 \cos (a)-2 i \sin (a)}\right ) \sec ^2(b x)}{-i+\tan (b x)}-\frac {\log \left (\frac {\sec (b x) (\cos (a)-i \sin (a)) ((i+c) \cos (a+b x)+(1+i c) \sin (a+b x))}{2 c}\right ) \sec ^2(b x)}{i+\tan (b x)}\right ) (-i+\tan (a+b x)) (1-i c+(-i+c) \tan (a+b x))} \] Input:

Integrate[ArcTan[c + (1 + I*c)*Tan[a + b*x]],x]
 

Output:

x*ArcTan[c + (1 + I*c)*Tan[a + b*x]] + (I*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])*((I + c)*Cos[a + 
b*x] + (1 + I*c)*Sin[a + b*x]))/(2*c)]*Log[1 - I*Tan[b*x]] + Log[(Sec[b*x] 
*((1 - I*c)*Cos[a + b*x] + (-I + c)*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]*((-I + c)*Cos[a] + I*(I + c)*Sin[a])*(Cos[a + b*x] - I*Sin[a 
 + b*x]))/(2*c)] + PolyLog[2, (Sec[b*x]*((1 + I*c)*Cos[a] - (I + c)*Sin[a] 
)*(Cos[a + b*x] + I*Sin[a + b*x]))/2])*Sec[a + b*x]^2*(Cos[b*x] + I*Sin[b* 
x])*(I*Cos[b*x] + Sin[b*x])*((1 - I*c)*Cos[a + b*x] + (-I + c)*Sin[a + b*x 
]))/(((I + c)*Cos[a + b*x] + (1 + I*c)*Sin[a + b*x])*(2*b*x - I*Log[1 - (S 
ec[b*x]*((-I + c)*Cos[a] + I*(I + c)*Sin[a])*(Cos[a + b*x] - I*Sin[a + b*x 
]))/(2*c)] - I*Log[1 + (Sec[b*x]*((-1 - I*c)*Cos[a] + (I + c)*Sin[a])*(Cos 
[a + b*x] + I*Sin[a + b*x]))/2] - (I*(-I + c)*Cos[a + b*x]*(Log[1 - I*Tan[ 
b*x]] - Log[1 + I*Tan[b*x]]))/((I + c)*Cos[a + b*x] + (1 + I*c)*Sin[a + b* 
x]) + ((I + c)*(Log[1 - I*Tan[b*x]] - Log[1 + I*Tan[b*x]])*Sin[a + b*x])/( 
(I + c)*Cos[a + b*x] + (1 + I*c)*Sin[a + b*x]) - (2*I)*b*x*Tan[b*x] + Log[ 
1 - (Sec[b*x]*((-I + c)*Cos[a] + I*(I + c)*Sin[a])*(Cos[a + b*x] - I*Sin[a 
 + b*x]))/(2*c)]*Tan[b*x] - Log[1 + (Sec[b*x]*((-1 - I*c)*Cos[a] + (I + c) 
*Sin[a])*(Cos[a + b*x] + I*Sin[a + b*x]))/2]*Tan[b*x] - Log[1 - I*Tan[b*x] 
]*Tan[b*x] + Cos[a]^2*Log[1 + I*Tan[b*x]]*Tan[b*x] + Log[1 + I*Tan[b*x]...
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.26, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {5686, 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.

\(\displaystyle \int \arctan (c+(1+i c) \tan (a+b x)) \, dx\)

\(\Big \downarrow \) 5686

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

\(\Big \downarrow \) 2615

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

Input:

Int[ArcTan[c + (1 + I*c)*Tan[a + b*x]],x]
 

Output:

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

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 5686

Int[ArcTan[(c_.) + (d_.)*Tan[(a_.) + (b_.)*(x_)]], x_Symbol] :> Simp[x*ArcT 
an[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. 562 vs. \(2 (69 ) = 138\).

Time = 1.55 (sec) , antiderivative size = 563, normalized size of antiderivative = 6.62

method	result	size
derivativedivides	\(\frac {\frac {\arctan \left (c +\left (i c +1\right ) \tan \left (b x +a \right )\right ) \ln \left (-c +\left (i c +1\right ) \tan \left (b x +a \right )+i\right ) c^{2}}{2 i-2 c}-\frac {2 i \arctan \left (c +\left (i c +1\right ) \tan \left (b x +a \right )\right ) \ln \left (-c +\left (i c +1\right ) \tan \left (b x +a \right )+i\right ) c}{2 i-2 c}-\frac {\arctan \left (c +\left (i c +1\right ) \tan \left (b x +a \right )\right ) \ln \left (-c +\left (i c +1\right ) \tan \left (b x +a \right )+i\right )}{2 i-2 c}-\frac {\arctan \left (c +\left (i c +1\right ) \tan \left (b x +a \right )\right ) \ln \left (-i+c +\left (i c +1\right ) \tan \left (b x +a \right )\right ) c^{2}}{2 i-2 c}+\frac {2 i \arctan \left (c +\left (i c +1\right ) \tan \left (b x +a \right )\right ) \ln \left (-i+c +\left (i c +1\right ) \tan \left (b x +a \right )\right ) c}{2 i-2 c}+\frac {\arctan \left (c +\left (i c +1\right ) \tan \left (b x +a \right )\right ) \ln \left (-i+c +\left (i c +1\right ) \tan \left (b x +a \right )\right )}{2 i-2 c}-\left (i c +1\right )^{2} \left (\frac {-\frac {i \ln \left (-i+c +\left (i c +1\right ) \tan \left (b x +a \right )\right )^{2}}{4}+\frac {i \left (\operatorname {dilog}\left (-\frac {i \left (c +\left (i c +1\right ) \tan \left (b x +a \right )+i\right )}{2}\right )+\ln \left (-i+c +\left (i c +1\right ) \tan \left (b x +a \right )\right ) \ln \left (-\frac {i \left (c +\left (i c +1\right ) \tan \left (b x +a \right )+i\right )}{2}\right )\right )}{2}}{2 i-2 c}-\frac {\frac {i \left (\operatorname {dilog}\left (\frac {c +\left (i c +1\right ) \tan \left (b x +a \right )+i}{2 c}\right )+\ln \left (-c +\left (i c +1\right ) \tan \left (b x +a \right )+i\right ) \ln \left (\frac {c +\left (i c +1\right ) \tan \left (b x +a \right )+i}{2 c}\right )\right )}{2}-\frac {i \left (\operatorname {dilog}\left (\frac {-i+c +\left (i c +1\right ) \tan \left (b x +a \right )}{-2 i+2 c}\right )+\ln \left (-c +\left (i c +1\right ) \tan \left (b x +a \right )+i\right ) \ln \left (\frac {-i+c +\left (i c +1\right ) \tan \left (b x +a \right )}{-2 i+2 c}\right )\right )}{2}}{2 \left (i-c \right )}\right )}{b \left (i c +1\right )}\)	\(563\)
default	\(\frac {\frac {\arctan \left (c +\left (i c +1\right ) \tan \left (b x +a \right )\right ) \ln \left (-c +\left (i c +1\right ) \tan \left (b x +a \right )+i\right ) c^{2}}{2 i-2 c}-\frac {2 i \arctan \left (c +\left (i c +1\right ) \tan \left (b x +a \right )\right ) \ln \left (-c +\left (i c +1\right ) \tan \left (b x +a \right )+i\right ) c}{2 i-2 c}-\frac {\arctan \left (c +\left (i c +1\right ) \tan \left (b x +a \right )\right ) \ln \left (-c +\left (i c +1\right ) \tan \left (b x +a \right )+i\right )}{2 i-2 c}-\frac {\arctan \left (c +\left (i c +1\right ) \tan \left (b x +a \right )\right ) \ln \left (-i+c +\left (i c +1\right ) \tan \left (b x +a \right )\right ) c^{2}}{2 i-2 c}+\frac {2 i \arctan \left (c +\left (i c +1\right ) \tan \left (b x +a \right )\right ) \ln \left (-i+c +\left (i c +1\right ) \tan \left (b x +a \right )\right ) c}{2 i-2 c}+\frac {\arctan \left (c +\left (i c +1\right ) \tan \left (b x +a \right )\right ) \ln \left (-i+c +\left (i c +1\right ) \tan \left (b x +a \right )\right )}{2 i-2 c}-\left (i c +1\right )^{2} \left (\frac {-\frac {i \ln \left (-i+c +\left (i c +1\right ) \tan \left (b x +a \right )\right )^{2}}{4}+\frac {i \left (\operatorname {dilog}\left (-\frac {i \left (c +\left (i c +1\right ) \tan \left (b x +a \right )+i\right )}{2}\right )+\ln \left (-i+c +\left (i c +1\right ) \tan \left (b x +a \right )\right ) \ln \left (-\frac {i \left (c +\left (i c +1\right ) \tan \left (b x +a \right )+i\right )}{2}\right )\right )}{2}}{2 i-2 c}-\frac {\frac {i \left (\operatorname {dilog}\left (\frac {c +\left (i c +1\right ) \tan \left (b x +a \right )+i}{2 c}\right )+\ln \left (-c +\left (i c +1\right ) \tan \left (b x +a \right )+i\right ) \ln \left (\frac {c +\left (i c +1\right ) \tan \left (b x +a \right )+i}{2 c}\right )\right )}{2}-\frac {i \left (\operatorname {dilog}\left (\frac {-i+c +\left (i c +1\right ) \tan \left (b x +a \right )}{-2 i+2 c}\right )+\ln \left (-c +\left (i c +1\right ) \tan \left (b x +a \right )+i\right ) \ln \left (\frac {-i+c +\left (i c +1\right ) \tan \left (b x +a \right )}{-2 i+2 c}\right )\right )}{2}}{2 \left (i-c \right )}\right )}{b \left (i c +1\right )}\)	\(563\)
risch	\(\text {Expression too large to display}\)	\(1248\)

Input:

int(arctan(c+(1+I*c)*tan(b*x+a)),x,method=_RETURNVERBOSE)
 

Output:

1/b/(1+I*c)*(arctan(c+(1+I*c)*tan(b*x+a))/(2*I-2*c)*ln(-c+(1+I*c)*tan(b*x+ 
a)+I)*c^2-2*I*arctan(c+(1+I*c)*tan(b*x+a))/(2*I-2*c)*ln(-c+(1+I*c)*tan(b*x 
+a)+I)*c-arctan(c+(1+I*c)*tan(b*x+a))/(2*I-2*c)*ln(-c+(1+I*c)*tan(b*x+a)+I 
)-arctan(c+(1+I*c)*tan(b*x+a))/(2*I-2*c)*ln(-I+c+(1+I*c)*tan(b*x+a))*c^2+2 
*I*arctan(c+(1+I*c)*tan(b*x+a))/(2*I-2*c)*ln(-I+c+(1+I*c)*tan(b*x+a))*c+ar 
ctan(c+(1+I*c)*tan(b*x+a))/(2*I-2*c)*ln(-I+c+(1+I*c)*tan(b*x+a))-(1+I*c)^2 
*(1/2/(I-c)*(-1/4*I*ln(-I+c+(1+I*c)*tan(b*x+a))^2+1/2*I*(dilog(-1/2*I*(c+( 
1+I*c)*tan(b*x+a)+I))+ln(-I+c+(1+I*c)*tan(b*x+a))*ln(-1/2*I*(c+(1+I*c)*tan 
(b*x+a)+I))))-1/2/(I-c)*(1/2*I*(dilog(1/2*(c+(1+I*c)*tan(b*x+a)+I)/c)+ln(- 
c+(1+I*c)*tan(b*x+a)+I)*ln(1/2*(c+(1+I*c)*tan(b*x+a)+I)/c))-1/2*I*(dilog(( 
-I+c+(1+I*c)*tan(b*x+a))/(-2*I+2*c))+ln(-c+(1+I*c)*tan(b*x+a)+I)*ln((-I+c+ 
(1+I*c)*tan(b*x+a))/(-2*I+2*c))))))
 

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. 202 vs. \(2 (60) = 120\).

Time = 0.10 (sec) , antiderivative size = 202, normalized size of antiderivative = 2.38 \[ \int \arctan (c+(1+i c) \tan (a+b x)) \, dx=-\frac {b^{2} x^{2} - i \, b x \log \left (-\frac {{\left (c e^{\left (2 i \, b x + 2 i \, a\right )} + i\right )} e^{\left (-2 i \, b x - 2 i \, a\right )}}{c - i}\right ) - a^{2} - {\left (-i \, b x - i \, a\right )} \log \left (\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (i \, b x + i \, a\right )} + 1\right ) - {\left (-i \, b x - i \, a\right )} \log \left (-\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (i \, b x + i \, a\right )} + 1\right ) - i \, a \log \left (\frac {2 \, c e^{\left (i \, b x + i \, a\right )} + i \, \sqrt {4 i \, c}}{2 \, c}\right ) - i \, a \log \left (\frac {2 \, c e^{\left (i \, b x + i \, a\right )} - i \, \sqrt {4 i \, c}}{2 \, c}\right ) + {\rm Li}_2\left (\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) + {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (i \, b x + i \, a\right )}\right )}{2 \, b} \] Input:

integrate(arctan(c+(1+I*c)*tan(b*x+a)),x, algorithm="fricas")
 

Output:

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

Sympy [F(-2)]

Exception generated. \[ \int \arctan (c+(1+i c) \tan (a+b x)) \, dx=\text {Exception raised: CoercionFailed} \] Input:

integrate(atan(c+(1+I*c)*tan(b*x+a)),x)
 

Output:

Exception raised: CoercionFailed >> Cannot convert _t0**2*exp(2*I*a) + 1 o 
f type <class 'sympy.core.add.Add'> to QQ_I[b,_t0,exp(I*a)]
 

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. 448 vs. \(2 (60) = 120\).

Time = 0.17 (sec) , antiderivative size = 448, normalized size of antiderivative = 5.27 \[ \int \arctan (c+(1+i c) \tan (a+b x)) \, dx =\text {Too large to display} \] Input:

integrate(arctan(c+(1+I*c)*tan(b*x+a)),x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int \arctan (c+(1+i c) \tan (a+b x)) \, dx=\int { \arctan \left ({\left (i \, c + 1\right )} \tan \left (b x + a\right ) + c\right ) \,d x } \] Input:

integrate(arctan(c+(1+I*c)*tan(b*x+a)),x, algorithm="giac")
 

Output:

integrate(arctan((I*c + 1)*tan(b*x + a) + c), x)
 

Mupad [F(-1)]

Timed out. \[ \int \arctan (c+(1+i c) \tan (a+b x)) \, dx=\int \mathrm {atan}\left (c+\mathrm {tan}\left (a+b\,x\right )\,\left (1+c\,1{}\mathrm {i}\right )\right ) \,d x \] Input:

int(atan(c + tan(a + b*x)*(c*1i + 1)),x)
 

Output:

int(atan(c + tan(a + b*x)*(c*1i + 1)), x)
 

Reduce [F]

\[ \int \arctan (c+(1+i c) \tan (a+b x)) \, dx=\int \mathit {atan} \left (\tan \left (b x +a \right ) c i +\tan \left (b x +a \right )+c \right )d x \] Input:

int(atan(c+(1+I*c)*tan(b*x+a)),x)
 

Output:

int(atan(tan(a + b*x)*c*i + tan(a + b*x) + c),x)