3.54 \(\int \tan ^{-1}(c+(1+i c) \tan (a+b x)) \, dx\)
Optimal. Leaf size=85 \[ -\frac{\text{PolyLog}\left (2,i c e^{2 i a+2 i b x}\right )}{4 b}-\frac{1}{2} i x \log \left (1-i c e^{2 i a+2 i b x}\right )+x \tan ^{-1}(c+(1+i c) \tan (a+b x))-\frac{b x^2}{2} \]
[Out]
-(b*x^2)/2 + x*ArcTan[c + (1 + I*c)*Tan[a + b*x]] - (I/2)*x*Log[1 - I*c*E^((2*I)*a + (2*I)*b*x)] - PolyLog[2,
I*c*E^((2*I)*a + (2*I)*b*x)]/(4*b)
________________________________________________________________________________________
Rubi [A] time = 0.124773, antiderivative size = 85, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used =
{5163, 2184, 2190, 2279, 2391} \[ -\frac{\text{PolyLog}\left (2,i c e^{2 i a+2 i b x}\right )}{4 b}-\frac{1}{2} i x \log \left (1-i c e^{2 i a+2 i b x}\right )+x \tan ^{-1}(c+(1+i c) \tan (a+b x))-\frac{b x^2}{2} \]
Antiderivative was successfully verified.
[In]
Int[ArcTan[c + (1 + I*c)*Tan[a + b*x]],x]
[Out]
-(b*x^2)/2 + x*ArcTan[c + (1 + I*c)*Tan[a + b*x]] - (I/2)*x*Log[1 - I*c*E^((2*I)*a + (2*I)*b*x)] - PolyLog[2,
I*c*E^((2*I)*a + (2*I)*b*x)]/(4*b)
Rule 5163
Int[ArcTan[(c_.) + (d_.)*Tan[(a_.) + (b_.)*(x_)]], x_Symbol] :> Simp[x*ArcTan[c + d*Tan[a + b*x]], x] - Dist[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]
Rule 2184
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] - Dist[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 2190
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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(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 2279
Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[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 2391
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]
Rubi steps
\begin{align*} \int \tan ^{-1}(c+(1+i c) \tan (a+b x)) \, dx &=x \tan ^{-1}(c+(1+i c) \tan (a+b x))-(i b) \int \frac{x}{i (1+i c)+c+c e^{2 i a+2 i b x}} \, dx\\ &=-\frac{b x^2}{2}+x \tan ^{-1}(c+(1+i c) \tan (a+b x))+(b c) \int \frac{e^{2 i a+2 i b x} x}{i (1+i c)+c+c e^{2 i a+2 i b x}} \, dx\\ &=-\frac{b x^2}{2}+x \tan ^{-1}(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{1}{2} i \int \log \left (1+\frac{c e^{2 i a+2 i b x}}{i (1+i c)+c}\right ) \, dx\\ &=-\frac{b x^2}{2}+x \tan ^{-1}(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{Subst}\left (\int \frac{\log \left (1+\frac{c x}{i (1+i c)+c}\right )}{x} \, dx,x,e^{2 i a+2 i b x}\right )}{4 b}\\ &=-\frac{b x^2}{2}+x \tan ^{-1}(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{\text{Li}_2\left (i c e^{2 i a+2 i b x}\right )}{4 b}\\ \end{align*}
Mathematica [B] time = 14.2647, size = 967, normalized size = 11.38 \[ \frac{i x \left (2 i b x \log (2 \cos (b x) (\cos (b x)-i \sin (b x)))-\log \left (\frac{\sec (b x) (\cos (a)-i \sin (a)) ((c+i) \cos (a+b x)+(i c+1) \sin (a+b x))}{2 c}\right ) \log (1-i \tan (b x))+\log \left (\frac{\sec (b x) ((1-i c) \cos (a+b x)+(c-i) \sin (a+b x))}{2 \cos (a)-2 i \sin (a)}\right ) \log (i \tan (b x)+1)-\text{PolyLog}(2,i \sin (2 b x)-\cos (2 b x))-\text{PolyLog}\left (2,\frac{\sec (b x) ((c-i) \cos (a)+i (c+i) \sin (a)) (\cos (a+b x)-i \sin (a+b x))}{2 c}\right )+\text{PolyLog}\left (2,\frac{1}{2} \sec (b x) ((i c+1) \cos (a)-(c+i) \sin (a)) (\cos (a+b x)+i \sin (a+b x))\right )\right ) (\cos (b x)+i \sin (b x)) (i \cos (b x)+\sin (b x)) ((1-i c) \cos (a+b x)+(c-i) \sin (a+b x)) \sec ^2(a+b x)}{((c+i) \cos (a+b x)+(i c+1) \sin (a+b x)) \left (\log (i \tan (b x)+1) \tan (b x) \cos ^2(a)+2 b x-i \log \left (1-\frac{\sec (b x) ((c-i) \cos (a)+i (c+i) \sin (a)) (\cos (a+b x)-i \sin (a+b x))}{2 c}\right )-i \log \left (\frac{1}{2} \sec (b x) ((-i c-1) \cos (a)+(c+i) \sin (a)) (\cos (a+b x)+i \sin (a+b x))+1\right )+\log (i \tan (b x)+1) \sin ^2(a) \tan (b x)-2 i b x \tan (b x)+\log \left (1-\frac{\sec (b x) ((c-i) \cos (a)+i (c+i) \sin (a)) (\cos (a+b x)-i \sin (a+b x))}{2 c}\right ) \tan (b x)-\log \left (\frac{1}{2} \sec (b x) ((-i c-1) \cos (a)+(c+i) \sin (a)) (\cos (a+b x)+i \sin (a+b x))+1\right ) \tan (b x)-\log (1-i \tan (b x)) \tan (b x)-\frac{i (c-i) \cos (a+b x) (\log (1-i \tan (b x))-\log (i \tan (b x)+1))}{(c+i) \cos (a+b x)+(i c+1) \sin (a+b x)}+\frac{(c+i) (\log (1-i \tan (b x))-\log (i \tan (b x)+1)) \sin (a+b x)}{(c+i) \cos (a+b x)+(i c+1) \sin (a+b x)}+\frac{\log \left (\frac{\sec (b x) ((1-i c) \cos (a+b x)+(c-i) \sin (a+b x))}{2 \cos (a)-2 i \sin (a)}\right ) \sec ^2(b x)}{\tan (b x)-i}-\frac{\log \left (\frac{\sec (b x) (\cos (a)-i \sin (a)) ((c+i) \cos (a+b x)+(i c+1) \sin (a+b x))}{2 c}\right ) \sec ^2(b x)}{\tan (b x)+i}\right ) (\tan (a+b x)-i) (-i c+(c-i) \tan (a+b x)+1)}+x \tan ^{-1}(c+(i c+1) \tan (a+b x)) \]
Warning: Unable to verify antiderivative.
[In]
Integrate[ArcTan[c + (1 + I*c)*Tan[a + b*x]],x]
[Out]
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*S
in[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 - (Sec[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)*Co
s[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*Lo
g[1 + I*Tan[b*x]]*Tan[b*x] + Log[1 + I*Tan[b*x]]*Sin[a]^2*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])]*Sec[b*x]^2)/(-I + Tan[b*x]) - (Log[(Sec[b*x]*(Cos[a] - I*Si
n[a])*((I + c)*Cos[a + b*x] + (1 + I*c)*Sin[a + b*x]))/(2*c)]*Sec[b*x]^2)/(I + Tan[b*x]))*(-I + Tan[a + b*x])*
(1 - I*c + (-I + c)*Tan[a + b*x]))
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Maple [B] time = 0.132, size = 1489, normalized size = 17.5 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(arctan(c+(1+I*c)*tan(b*x+a)),x)
[Out]
1/2/(1+I*c)/b/(I-c)*ln((c+(1+I*c)*tan(b*x+a)-I)/(-2*I+2*c))*ln(-c+(1+I*c)*tan(b*x+a)+I)*c+1/(1+I*c)/b*arctan(c
+(1+I*c)*tan(b*x+a))/(2*I-2*c)*ln(-c+(1+I*c)*tan(b*x+a)+I)*c^2-1/(1+I*c)/b*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+1/2/(1+I*c)/b/(I-c)*ln(-1/2*I*(c+(1+I*c)*tan(b*x+a)+I))*ln(c+(1+I*c)*tan(b*
x+a)-I)*c-1/4*I/(1+I*c)/b/(I-c)*ln((c+(1+I*c)*tan(b*x+a)-I)/(-2*I+2*c))*ln(-c+(1+I*c)*tan(b*x+a)+I)+1/4*I/(1+I
*c)/b/(I-c)*dilog(-1/2*I*(c+(1+I*c)*tan(b*x+a)+I))*c^2-1/8*I/(1+I*c)/b/(I-c)*ln(c+(1+I*c)*tan(b*x+a)-I)^2*c^2-
1/2/(1+I*c)/b/(I-c)*ln(1/2*(c+(1+I*c)*tan(b*x+a)+I)/c)*ln(-c+(1+I*c)*tan(b*x+a)+I)*c+1/4*I/(1+I*c)/b/(I-c)*dil
og((c+(1+I*c)*tan(b*x+a)-I)/(-2*I+2*c))*c^2-1/4*I/(1+I*c)/b/(I-c)*dilog(1/2*(c+(1+I*c)*tan(b*x+a)+I)/c)*c^2-1/
4*I/(1+I*c)/b/(I-c)*ln(-1/2*I*(c+(1+I*c)*tan(b*x+a)+I))*ln(c+(1+I*c)*tan(b*x+a)-I)+1/4*I/(1+I*c)/b/(I-c)*ln(1/
2*(c+(1+I*c)*tan(b*x+a)+I)/c)*ln(-c+(1+I*c)*tan(b*x+a)+I)-1/4/(1+I*c)/b/(I-c)*ln(c+(1+I*c)*tan(b*x+a)-I)^2*c-1
/2/(1+I*c)/b/(I-c)*dilog(1/2*(c+(1+I*c)*tan(b*x+a)+I)/c)*c+1/2/(1+I*c)/b/(I-c)*dilog((c+(1+I*c)*tan(b*x+a)-I)/
(-2*I+2*c))*c-1/(1+I*c)/b*arctan(c+(1+I*c)*tan(b*x+a))/(2*I-2*c)*ln(-c+(1+I*c)*tan(b*x+a)+I)+1/(1+I*c)/b*arcta
n(c+(1+I*c)*tan(b*x+a))/(2*I-2*c)*ln(c+(1+I*c)*tan(b*x+a)-I)+1/8*I/(1+I*c)/b/(I-c)*ln(c+(1+I*c)*tan(b*x+a)-I)^
2+1/4*I/(1+I*c)/b/(I-c)*dilog(1/2*(c+(1+I*c)*tan(b*x+a)+I)/c)-1/4*I/(1+I*c)/b/(I-c)*dilog((c+(1+I*c)*tan(b*x+a
)-I)/(-2*I+2*c))-1/4*I/(1+I*c)/b/(I-c)*dilog(-1/2*I*(c+(1+I*c)*tan(b*x+a)+I))+1/2/(1+I*c)/b/(I-c)*dilog(-1/2*I
*(c+(1+I*c)*tan(b*x+a)+I))*c+2*I/(1+I*c)/b*arctan(c+(1+I*c)*tan(b*x+a))/(2*I-2*c)*ln(c+(1+I*c)*tan(b*x+a)-I)*c
+1/4*I/(1+I*c)/b/(I-c)*ln(-1/2*I*(c+(1+I*c)*tan(b*x+a)+I))*ln(c+(1+I*c)*tan(b*x+a)-I)*c^2-1/4*I/(1+I*c)/b/(I-c
)*ln(1/2*(c+(1+I*c)*tan(b*x+a)+I)/c)*ln(-c+(1+I*c)*tan(b*x+a)+I)*c^2+1/4*I/(1+I*c)/b/(I-c)*ln((c+(1+I*c)*tan(b
*x+a)-I)/(-2*I+2*c))*ln(-c+(1+I*c)*tan(b*x+a)+I)*c^2-2*I/(1+I*c)/b*arctan(c+(1+I*c)*tan(b*x+a))/(2*I-2*c)*ln(-
c+(1+I*c)*tan(b*x+a)+I)*c
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Maxima [B] time = 1.59028, size = 605, normalized size = 7.12 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(arctan(c+(1+I*c)*tan(b*x+a)),x, algorithm="maxima")
[Out]
-1/8*((I*c + 1)*(4*I*(b*x + a)*log((2*I*c^2 - 2*(c^2 - 2*I*c - 1)*tan(b*x + a) + 4*c - 2*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 - 2*(
c^2 - 2*I*c - 1)*tan(b*x + a) + 4*c - 2*I)/(2*I*c^2 - 2*(c^2 - 2*I*c - 1)*tan(b*x + a) + 2*I)))/b
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Fricas [B] time = 2.30371, size = 547, normalized size = 6.44 \begin{align*} -\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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(arctan(c+(1+I*c)*tan(b*x+a)),x, algorithm="fricas")
[Out]
-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)*lo
g(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
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} b \int \frac{x}{i c e^{2 i a} e^{2 i b x} - 1}\, dx + \frac{i x \log{\left (- i c + \frac{i c}{e^{2 i a} e^{2 i b x} + 1} - \frac{i c e^{i a} e^{i b x}}{e^{i a} e^{i b x} + e^{- i a} e^{- i b x}} + 1 + \frac{1}{e^{2 i a} e^{2 i b x} + 1} - \frac{e^{i a} e^{i b x}}{e^{i a} e^{i b x} + e^{- i a} e^{- i b x}} \right )}}{2} - \frac{i x \log{\left (i c - \frac{i c}{e^{2 i a} e^{2 i b x} + 1} + \frac{i c e^{i a} e^{i b x}}{e^{i a} e^{i b x} + e^{- i a} e^{- i b x}} + 1 - \frac{1}{e^{2 i a} e^{2 i b x} + 1} + \frac{e^{i a} e^{i b x}}{e^{i a} e^{i b x} + e^{- i a} e^{- i b x}} \right )}}{2} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(atan(c+(1+I*c)*tan(b*x+a)),x)
[Out]
b*Integral(x/(I*c*exp(2*I*a)*exp(2*I*b*x) - 1), x) + I*x*log(-I*c + I*c/(exp(2*I*a)*exp(2*I*b*x) + 1) - I*c*ex
p(I*a)*exp(I*b*x)/(exp(I*a)*exp(I*b*x) + exp(-I*a)*exp(-I*b*x)) + 1 + 1/(exp(2*I*a)*exp(2*I*b*x) + 1) - exp(I*
a)*exp(I*b*x)/(exp(I*a)*exp(I*b*x) + exp(-I*a)*exp(-I*b*x)))/2 - I*x*log(I*c - I*c/(exp(2*I*a)*exp(2*I*b*x) +
1) + I*c*exp(I*a)*exp(I*b*x)/(exp(I*a)*exp(I*b*x) + exp(-I*a)*exp(-I*b*x)) + 1 - 1/(exp(2*I*a)*exp(2*I*b*x) +
1) + exp(I*a)*exp(I*b*x)/(exp(I*a)*exp(I*b*x) + exp(-I*a)*exp(-I*b*x)))/2
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \arctan \left ({\left (i \, c + 1\right )} \tan \left (b x + a\right ) + c\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(arctan(c+(1+I*c)*tan(b*x+a)),x, algorithm="giac")
[Out]
integrate(arctan((I*c + 1)*tan(b*x + a) + c), x)