3.71 \(\int \tan ^{-1}(c+(-1-i c) \cot (a+b x)) \, dx\)

Optimal. Leaf size=86 \[ -\frac {\text {Li}_2\left (-i c e^{2 i a+2 i b x}\right )}{4 b}-\frac {1}{2} i x \log \left (1+i c e^{2 i a+2 i b x}\right )+x \tan ^{-1}(c-(1+i c) \cot (a+b x))-\frac {b x^2}{2} \]

[Out]

-1/2*b*x^2-x*arctan(-c+(1+I*c)*cot(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

________________________________________________________________________________________

Rubi [A]  time = 0.13, antiderivative size = 86, normalized size of antiderivative = 1.00, 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 = {5165, 2184, 2190, 2279, 2391} \[ -\frac {\text {PolyLog}\left (2,-i c e^{2 i a+2 i b x}\right )}{4 b}-\frac {1}{2} i x \log \left (1+i c e^{2 i a+2 i b x}\right )+x \tan ^{-1}(c-(1+i c) \cot (a+b x))-\frac {b x^2}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*x^2)/2 + x*ArcTan[c - (1 + I*c)*Cot[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 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]

Rule 5165

Int[ArcTan[(c_.) + Cot[(a_.) + (b_.)*(x_)]*(d_.)], x_Symbol] :> Simp[x*ArcTan[c + d*Cot[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]

Rubi steps

\begin {align*} \int \tan ^{-1}(c+(-1-i c) \cot (a+b x)) \, dx &=x \tan ^{-1}(c-(1+i c) \cot (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) \cot (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) \cot (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) \cot (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) \cot (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 = 15.67, size = 872, normalized size = 10.14 \[ x \tan ^{-1}(c+(-i c-1) \cot (a+b x))+\frac {i x \csc (a+b x) \left (2 b x \log (2 \cos (b x) (\cos (b x)-i \sin (b x)))+i \log \left (\frac {\sec (b x) (\cos (a)-i \sin (a)) ((c-i) \cos (a+b x)+i (c+i) \sin (a+b x))}{2 c}\right ) \log (1-i \tan (b x))-i \log \left (\frac {1}{2} \sec (b x) (\cos (a)+i \sin (a)) ((i c+1) \cos (a+b x)-(c+i) \sin (a+b x))\right ) \log (i \tan (b x)+1)+i \text {Li}_2(i \sin (2 b x)-\cos (2 b x))+i \text {Li}_2\left (\frac {\sec (b x) ((c+i) \cos (a)+(i c+1) \sin (a)) (\cos (a+b x)-i \sin (a+b x))}{2 c}\right )-i \text {Li}_2\left (\frac {1}{2} (\cos (a)+i \sin (a)) ((c+i) \cos (a)+(i c+1) \sin (a)) (\tan (b x)-i)\right )\right ) (\cos (b x)-i \sin (b x)) (\cos (b x)+i \sin (b x))}{(\cot (a+b x)+i) ((c-i) \cos (a+b x)+i (c+i) \sin (a+b x)) \left (\frac {\log \left (\frac {1}{2} \sec (b x) (\cos (a)+i \sin (a)) ((i c+1) \cos (a+b x)-(c+i) \sin (a+b x))\right ) \sec ^2(b x)}{i \tan (b x)+1}+\frac {i \log \left (1-\frac {1}{2} (\cos (a)+i \sin (a)) ((c+i) \cos (a)+(i c+1) \sin (a)) (\tan (b x)-i)\right ) \sec ^2(b x)}{\tan (b x)-i}+\frac {i \log \left (\frac {\sec (b x) (\cos (a)-i \sin (a)) ((c-i) \cos (a+b x)+i (c+i) \sin (a+b x))}{2 c}\right ) \sec ^2(b x)}{\tan (b x)+i}-2 i b x-\log \left (1-\frac {\sec (b x) ((c+i) \cos (a)+(i c+1) \sin (a)) (\cos (a+b x)-i \sin (a+b x))}{2 c}\right )-2 b x \tan (b x)-i \log \left (1-\frac {\sec (b x) ((c+i) \cos (a)+(i c+1) \sin (a)) (\cos (a+b x)-i \sin (a+b x))}{2 c}\right ) \tan (b x)+i \log (1-i \tan (b x)) \tan (b x)-i \log (i \tan (b x)+1) \tan (b x)-\frac {\log (1-i \tan (b x)) ((c+i) \cos (a+b x)+(i c+1) \sin (a+b x))}{(c-i) \cos (a+b x)+i (c+i) \sin (a+b x)}+\frac {\log (i \tan (b x)+1) ((c+i) \cos (a+b x)+(i c+1) \sin (a+b x))}{(c-i) \cos (a+b x)+i (c+i) \sin (a+b x)}\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

x*ArcTan[c + (-1 - I*c)*Cot[a + b*x]] + (I*x*Csc[a + b*x]*(2*b*x*Log[2*Cos[b*x]*(Cos[b*x] - I*Sin[b*x])] + I*L
og[(Sec[b*x]*(Cos[a] - I*Sin[a])*((-I + c)*Cos[a + b*x] + I*(I + c)*Sin[a + b*x]))/(2*c)]*Log[1 - I*Tan[b*x]]
- I*Log[(Sec[b*x]*(Cos[a] + I*Sin[a])*((1 + I*c)*Cos[a + b*x] - (I + c)*Sin[a + b*x]))/2]*Log[1 + I*Tan[b*x]]
+ I*PolyLog[2, -Cos[2*b*x] + I*Sin[2*b*x]] + I*PolyLog[2, (Sec[b*x]*((I + c)*Cos[a] + (1 + I*c)*Sin[a])*(Cos[a
+ b*x] - I*Sin[a + b*x]))/(2*c)] - I*PolyLog[2, ((Cos[a] + I*Sin[a])*((I + c)*Cos[a] + (1 + I*c)*Sin[a])*(-I
+ Tan[b*x]))/2])*(Cos[b*x] - I*Sin[b*x])*(Cos[b*x] + I*Sin[b*x]))/((I + Cot[a + b*x])*((-I + c)*Cos[a + b*x] +
I*(I + c)*Sin[a + b*x])*((-2*I)*b*x - Log[1 - (Sec[b*x]*((I + c)*Cos[a] + (1 + I*c)*Sin[a])*(Cos[a + b*x] - I
*Sin[a + b*x]))/(2*c)] - (Log[1 - I*Tan[b*x]]*((I + c)*Cos[a + b*x] + (1 + I*c)*Sin[a + b*x]))/((-I + c)*Cos[a
+ b*x] + I*(I + c)*Sin[a + b*x]) + (Log[1 + I*Tan[b*x]]*((I + c)*Cos[a + b*x] + (1 + I*c)*Sin[a + b*x]))/((-I
+ c)*Cos[a + b*x] + I*(I + c)*Sin[a + b*x]) + (Log[(Sec[b*x]*(Cos[a] + I*Sin[a])*((1 + I*c)*Cos[a + b*x] - (I
+ c)*Sin[a + b*x]))/2]*Sec[b*x]^2)/(1 + I*Tan[b*x]) - 2*b*x*Tan[b*x] - I*Log[1 - (Sec[b*x]*((I + c)*Cos[a] +
(1 + I*c)*Sin[a])*(Cos[a + b*x] - I*Sin[a + b*x]))/(2*c)]*Tan[b*x] + I*Log[1 - I*Tan[b*x]]*Tan[b*x] - I*Log[1
+ I*Tan[b*x]]*Tan[b*x] + (I*Log[1 - ((Cos[a] + I*Sin[a])*((I + c)*Cos[a] + (1 + I*c)*Sin[a])*(-I + Tan[b*x]))/
2]*Sec[b*x]^2)/(-I + Tan[b*x]) + (I*Log[(Sec[b*x]*(Cos[a] - I*Sin[a])*((-I + c)*Cos[a + b*x] + I*(I + c)*Sin[a
+ b*x]))/(2*c)]*Sec[b*x]^2)/(I + Tan[b*x])))

________________________________________________________________________________________

fricas [A]  time = 0.55, size = 112, normalized size = 1.30 \[ -\frac {2 \, b^{2} x^{2} - 2 i \, b x \log \left (-\frac {{\left (c e^{\left (2 i \, b x + 2 i \, a\right )} - i\right )} e^{\left (-2 i \, b x - 2 i \, a\right )}}{c - i}\right ) - 2 \, a^{2} - {\left (-2 i \, b x - 2 i \, a\right )} \log \left (i \, c e^{\left (2 i \, b x + 2 i \, a\right )} + 1\right ) - 2 i \, a \log \left (\frac {c e^{\left (2 i \, b x + 2 i \, a\right )} - i}{c}\right ) + {\rm Li}_2\left (-i \, c e^{\left (2 i \, b x + 2 i \, a\right )}\right )}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\arctan \left (-{\left (-i \, c - 1\right )} \cot \left (b x + a\right ) - c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.70, size = 1753, normalized size = 20.38 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-arctan(-c-(-1-I*c)*cot(b*x+a)),x)

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 0.47, size = 456, normalized size = 5.30 \[ -\frac {{\left (i \, c + 1\right )} {\left (\frac {4 i \, {\left (b x + a\right )} \log \left (\frac {-2 i \, c^{2} + 2 \, {\left (c^{2} + 1\right )} \tan \left (b x + a\right ) - 2 i}{-2 i \, c^{2} + 2 \, {\left (c^{2} + 1\right )} \tan \left (b x + a\right ) - 4 \, c + 2 i}\right )}{i \, c + 1} + \frac {i \, {\left (4 \, {\left (b x + a\right )} {\left (\log \left (-i \, c^{2} + {\left (c^{2} + 1\right )} \tan \left (b x + a\right ) - 2 \, c + i\right ) - \log \left (-i \, c^{2} + {\left (c^{2} + 1\right )} \tan \left (b x + a\right ) - i\right )\right )} - 2 i \, \log \left (-i \, c^{2} + {\left (c^{2} + 1\right )} \tan \left (b x + a\right ) - 2 \, c + i\right ) \log \left (-\frac {{\left (i \, c - 1\right )} \tan \left (b x + a\right ) + c - i}{2 \, c} + 1\right ) + 2 i \, \log \left (-i \, c^{2} + {\left (c^{2} + 1\right )} \tan \left (b x + a\right ) - 2 \, c + i\right ) \log \left (\tan \left (b x + a\right ) - i\right ) - 2 i \, \log \left (-\frac {1}{2} \, {\left (c + i\right )} \tan \left (b x + a\right ) + \frac {1}{2} i \, c + \frac {1}{2}\right ) \log \left (\tan \left (b x + a\right ) - i\right ) - i \, \log \left (\tan \left (b x + a\right ) - i\right )^{2} - 2 i \, \log \left (c^{2} + 1\right ) \log \left (i \, \tan \left (b x + a\right ) + 1\right ) + 2 i \, \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 ) + 2 i \, \log \left (c^{2} + 1\right ) \log \left (-i \, \tan \left (b x + a\right ) + 1\right ) - 2 i \, {\rm Li}_2\left (\frac {1}{2} \, {\left (c + i\right )} \tan \left (b x + a\right ) - \frac {1}{2} i \, c + \frac {1}{2}\right ) - 2 i \, {\rm Li}_2\left (\frac {{\left (i \, c - 1\right )} \tan \left (b x + a\right ) + c - i}{2 \, c}\right ) + 2 i \, {\rm Li}_2\left (\frac {1}{2} i \, \tan \left (b x + a\right ) + \frac {1}{2}\right )\right )}}{i \, c + 1}\right )} - 8 \, {\left (b x + a\right )} \arctan \left (c + \frac {-i \, c - 1}{\tan \left (b x + a\right )}\right ) + 4 \, {\left (-i \, b x - i \, a\right )} \log \left (\frac {-2 i \, c^{2} + 2 \, {\left (c^{2} + 1\right )} \tan \left (b x + a\right ) - 2 i}{-2 i \, c^{2} + 2 \, {\left (c^{2} + 1\right )} \tan \left (b x + a\right ) - 4 \, c + 2 i}\right )}{8 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-arctan(-c-(-1-I*c)*cot(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 + 1)*tan(b*x + a) - 2*I)/(-2*I*c^2 + 2*(c^2 + 1)*tan(b*x
+ a) - 4*c + 2*I))/(I*c + 1) + I*(4*(b*x + a)*(log(-I*c^2 + (c^2 + 1)*tan(b*x + a) - 2*c + I) - log(-I*c^2 +
(c^2 + 1)*tan(b*x + a) - I)) - 2*I*log(-I*c^2 + (c^2 + 1)*tan(b*x + a) - 2*c + I)*log(-1/2*((I*c - 1)*tan(b*x
+ a) + c - I)/c + 1) + 2*I*log(-I*c^2 + (c^2 + 1)*tan(b*x + a) - 2*c + I)*log(tan(b*x + a) - I) - 2*I*log(-1/2
*(c + I)*tan(b*x + a) + 1/2*I*c + 1/2)*log(tan(b*x + a) - I) - I*log(tan(b*x + a) - I)^2 - 2*I*log(c^2 + 1)*lo
g(I*tan(b*x + a) + 1) + 2*I*log(tan(b*x + a) - I)*log(-1/2*I*tan(b*x + a) + 1/2) + 2*I*log(c^2 + 1)*log(-I*tan
(b*x + a) + 1) - 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(c + (-I*c - 1)/tan(b*x + a))
+ 4*(-I*b*x - I*a)*log((-2*I*c^2 + 2*(c^2 + 1)*tan(b*x + a) - 2*I)/(-2*I*c^2 + 2*(c^2 + 1)*tan(b*x + a) - 4*c
+ 2*I)))/b

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \mathrm {atan}\left (c-\mathrm {cot}\left (a+b\,x\right )\,\left (1+c\,1{}\mathrm {i}\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: CoercionFailed} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: CoercionFailed

________________________________________________________________________________________