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

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

[Out]

-1/2*b*x^2+x*arccot(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

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Rubi [A]  time = 0.14, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {5164, 2184, 2190, 2279, 2391} \[ -\frac {\text {PolyLog}\left (2,-i c e^{2 i a+2 i b x}\right )}{4 b}-\frac {1}{2} i x \log \left (1+i c e^{2 i a+2 i b x}\right )+x \cot ^{-1}(c-(1-i c) \tan (a+b x))-\frac {b x^2}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*x^2)/2 + x*ArcCot[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 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 5164

Int[ArcCot[(c_.) + (d_.)*Tan[(a_.) + (b_.)*(x_)]], x_Symbol] :> Simp[x*ArcCot[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]

Rubi steps

\begin {align*} \int \cot ^{-1}(c-(1-i c) \tan (a+b x)) \, dx &=x \cot ^{-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 \cot ^{-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 \cot ^{-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 \cot ^{-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 \cot ^{-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*}

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Mathematica [B]  time = 2.98, size = 847, normalized size = 9.85 \[ x \cot ^{-1}(c+i (c+i) \tan (a+b x))-\frac {i x \left (-2 i b x \log (2 \cos (b x) (\cos (b x)-i \sin (b x)))+\log \left (\frac {\sec (b x) (\cos (a)-i \sin (a)) ((c-i) \cos (a+b x)+i (c+i) \sin (a+b x))}{2 c}\right ) \log (1-i \tan (b x))-\log \left (\frac {1}{2} \sec (b x) (\cos (a)+i \sin (a)) ((i c+1) \cos (a+b x)-(c+i) \sin (a+b x))\right ) \log (i \tan (b x)+1)+\text {Li}_2(i \sin (2 b x)-\cos (2 b x))+\text {Li}_2\left (\frac {\sec (b x) ((c+i) \cos (a)+(i c+1) \sin (a)) (\cos (a+b x)-i \sin (a+b x))}{2 c}\right )-\text {Li}_2\left (\frac {1}{2} (\cos (a)+i \sin (a)) ((c+i) \cos (a)+(i c+1) \sin (a)) (\tan (b x)-i)\right )\right ) \sec (a+b x) (\cos (b x)+i \sin (b x)) (i \cos (b x)+\sin (b x))}{((c-i) \cos (a+b x)+i (c+i) \sin (a+b x)) \left (-\frac {\log \left (\frac {1}{2} \sec (b x) (\cos (a)+i \sin (a)) ((i c+1) \cos (a+b x)-(c+i) \sin (a+b x))\right ) \sec ^2(b x)}{\tan (b x)-i}+\frac {\log \left (1-\frac {1}{2} (\cos (a)+i \sin (a)) ((c+i) \cos (a)+(i c+1) \sin (a)) (\tan (b x)-i)\right ) \sec ^2(b x)}{\tan (b x)-i}+\frac {\log \left (\frac {\sec (b x) (\cos (a)-i \sin (a)) ((c-i) \cos (a+b x)+i (c+i) \sin (a+b x))}{2 c}\right ) \sec ^2(b x)}{\tan (b x)+i}-2 b x+i \log \left (1-\frac {\sec (b x) ((c+i) \cos (a)+(i c+1) \sin (a)) (\cos (a+b x)-i \sin (a+b x))}{2 c}\right )+2 i b x \tan (b x)-\log \left (1-\frac {\sec (b x) ((c+i) \cos (a)+(i c+1) \sin (a)) (\cos (a+b x)-i \sin (a+b x))}{2 c}\right ) \tan (b x)+\log (1-i \tan (b x)) \tan (b x)-\log (i \tan (b x)+1) \tan (b x)+\frac {i (c+i) \cos (a+b x) (\log (1-i \tan (b x))-\log (i \tan (b x)+1))}{(c-i) \cos (a+b x)+i (c+i) \sin (a+b x)}+\frac {(i c+1) (\log (1-i \tan (b x))-\log (i \tan (b x)+1)) \sin (a+b x)}{(-i c-1) \cos (a+b x)+(c+i) \sin (a+b x)}\right ) (\tan (a+b x)-i)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

x*ArcCot[c + I*(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] + I*(I + c)*Sin[a + b*x]))/(2*c)]*Log[1 - I*Tan[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]*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] + (1 + I*c)*Sin[a])*(Cos[a + b*x] - I*Sin[a
+ b*x]))/(2*c)] - PolyLog[2, ((Cos[a] + I*Sin[a])*((I + c)*Cos[a] + (1 + I*c)*Sin[a])*(-I + Tan[b*x]))/2])*Se
c[a + b*x]*(Cos[b*x] + I*Sin[b*x])*(I*Cos[b*x] + Sin[b*x]))/(((-I + c)*Cos[a + b*x] + I*(I + c)*Sin[a + b*x])*
(-2*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)] + (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] + I*(I + c)*Sin[a +
b*x]) + ((1 + I*c)*(Log[1 - I*Tan[b*x]] - Log[1 + I*Tan[b*x]])*Sin[a + b*x])/((-1 - I*c)*Cos[a + b*x] + (I + c
)*Sin[a + b*x]) + (2*I)*b*x*Tan[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)]*Tan[b*x] + Log[1 - I*Tan[b*x]]*Tan[b*x] - Log[1 + I*Tan[b*x]]*Tan[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)/(-I + Tan[b*x]) + (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]) + (
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 + Ta
n[b*x]))*(-I + Tan[a + b*x]))

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fricas [B]  time = 0.58, size = 201, normalized size = 2.34 \[ -\frac {b^{2} x^{2} + i \, b x \log \left (\frac {{\left (c + i\right )} e^{\left (2 i \, b x + 2 i \, a\right )}}{c e^{\left (2 i \, b x + 2 i \, a\right )} - i}\right ) - a^{2} - {\left (-i \, b x - i \, a\right )} \log \left (\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (i \, b x + i \, a\right )} + 1\right ) - {\left (-i \, b x - i \, a\right )} \log \left (-\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (i \, b x + i \, a\right )} + 1\right ) - i \, a \log \left (\frac {2 \, c e^{\left (i \, b x + i \, a\right )} + i \, \sqrt {-4 i \, c}}{2 \, c}\right ) - i \, a \log \left (\frac {2 \, c e^{\left (i \, b x + i \, a\right )} - i \, \sqrt {-4 i \, c}}{2 \, c}\right ) + {\rm Li}_2\left (\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) + {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (i \, b x + i \, a\right )}\right )}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.61, size = 1681, normalized size = 19.55 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*I/b/(-1+I*c)/(I+c)*ln(-(-1+I*c)*tan(b*x+a)+c+I)*ln((-(-1+I*c)*tan(b*x+a)-c-I)/(-2*I-2*c))-1/b/(-1+I*c)*arc
cot(c+(-1+I*c)*tan(b*x+a))/(2*I+2*c)*ln(I+(-1+I*c)*tan(b*x+a)+c)+1/b/(-1+I*c)*arccot(c+(-1+I*c)*tan(b*x+a))/(2
*I+2*c)*ln(-(-1+I*c)*tan(b*x+a)+c+I)-1/2/b/(-1+I*c)/(I+c)*dilog(-1/2*I*(I+(-1+I*c)*tan(b*x+a)+c))*c-1/4/b/(-1+
I*c)/(I+c)*ln(I+(-1+I*c)*tan(b*x+a)+c)^2*c+1/2/b/(-1+I*c)/(I+c)*dilog((-(-1+I*c)*tan(b*x+a)-c-I)/(-2*I-2*c))*c
-1/4*I/b/(-1+I*c)/(I+c)*dilog(-1/2*(-(-1+I*c)*tan(b*x+a)-c+I)/c)-1/8*I/b/(-1+I*c)/(I+c)*ln(I+(-1+I*c)*tan(b*x+
a)+c)^2+1/4*I/b/(-1+I*c)/(I+c)*dilog((-(-1+I*c)*tan(b*x+a)-c-I)/(-2*I-2*c))-1/4*I/b/(-1+I*c)/(I+c)*dilog(-1/2*
I*(I+(-1+I*c)*tan(b*x+a)+c))+1/8*I/b/(-1+I*c)/(I+c)*ln(I+(-1+I*c)*tan(b*x+a)+c)^2*c^2+1/4*I/b/(-1+I*c)/(I+c)*d
ilog(-1/2*(-(-1+I*c)*tan(b*x+a)-c+I)/c)*c^2-1/2/b/(-1+I*c)/(I+c)*ln(-(-1+I*c)*tan(b*x+a)+c+I)*ln(-1/2*(-(-1+I*
c)*tan(b*x+a)-c+I)/c)*c+1/2/b/(-1+I*c)/(I+c)*ln(I+(-1+I*c)*tan(b*x+a)+c)*ln(-1/2*I*(-(-1+I*c)*tan(b*x+a)-c+I))
*c-1/4*I/b/(-1+I*c)/(I+c)*ln(-(-1+I*c)*tan(b*x+a)+c+I)*ln(-1/2*(-(-1+I*c)*tan(b*x+a)-c+I)/c)+1/4*I/b/(-1+I*c)/
(I+c)*ln(I+(-1+I*c)*tan(b*x+a)+c)*ln(-1/2*I*(-(-1+I*c)*tan(b*x+a)-c+I))-1/4*I/b/(-1+I*c)/(I+c)*ln(-1/2*I*(I+(-
1+I*c)*tan(b*x+a)+c))*ln(-1/2*I*(-(-1+I*c)*tan(b*x+a)-c+I))-1/4*I/b/(-1+I*c)/(I+c)*dilog((-(-1+I*c)*tan(b*x+a)
-c-I)/(-2*I-2*c))*c^2+1/2/b/(-1+I*c)/(I+c)*ln(-(-1+I*c)*tan(b*x+a)+c+I)*ln((-(-1+I*c)*tan(b*x+a)-c-I)/(-2*I-2*
c))*c+1/4*I/b/(-1+I*c)/(I+c)*dilog(-1/2*I*(I+(-1+I*c)*tan(b*x+a)+c))*c^2+1/4*I/b/(-1+I*c)/(I+c)*ln(-1/2*I*(I+(
-1+I*c)*tan(b*x+a)+c))*ln(-1/2*I*(-(-1+I*c)*tan(b*x+a)-c+I))*c^2-1/4*I/b/(-1+I*c)/(I+c)*ln(-(-1+I*c)*tan(b*x+a
)+c+I)*ln((-(-1+I*c)*tan(b*x+a)-c-I)/(-2*I-2*c))*c^2+1/4*I/b/(-1+I*c)/(I+c)*ln(-(-1+I*c)*tan(b*x+a)+c+I)*ln(-1
/2*(-(-1+I*c)*tan(b*x+a)-c+I)/c)*c^2+2*I/b/(-1+I*c)*arccot(c+(-1+I*c)*tan(b*x+a))/(2*I+2*c)*ln(I+(-1+I*c)*tan(
b*x+a)+c)*c-2*I/b/(-1+I*c)*arccot(c+(-1+I*c)*tan(b*x+a))/(2*I+2*c)*ln(-(-1+I*c)*tan(b*x+a)+c+I)*c-1/4*I/b/(-1+
I*c)/(I+c)*ln(I+(-1+I*c)*tan(b*x+a)+c)*ln(-1/2*I*(-(-1+I*c)*tan(b*x+a)-c+I))*c^2-1/2/b/(-1+I*c)/(I+c)*dilog(-1
/2*(-(-1+I*c)*tan(b*x+a)-c+I)/c)*c-1/2/b/(-1+I*c)/(I+c)*ln(-1/2*I*(I+(-1+I*c)*tan(b*x+a)+c))*ln(-1/2*I*(-(-1+I
*c)*tan(b*x+a)-c+I))*c+1/b/(-1+I*c)*arccot(c+(-1+I*c)*tan(b*x+a))/(2*I+2*c)*ln(I+(-1+I*c)*tan(b*x+a)+c)*c^2-1/
b/(-1+I*c)*arccot(c+(-1+I*c)*tan(b*x+a))/(2*I+2*c)*ln(-(-1+I*c)*tan(b*x+a)+c+I)*c^2

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maxima [B]  time = 0.47, size = 450, normalized size = 5.23 \[ \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} + 2 i \, c - 1\right )} \tan \left (b x + a\right ) + 2 i}{2 i \, c^{2} - 2 \, {\left (c^{2} + 2 i \, c - 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} + 2 i \, c - 1\right )} \tan \left (b x + a\right ) + 2 \, c + i\right ) - \log \left (-i \, c^{2} + {\left (c^{2} + 2 i \, c - 1\right )} \tan \left (b x + a\right ) - i\right )\right )} + i \, \log \left (-i \, c^{2} + {\left (c^{2} + 2 i \, c - 1\right )} \tan \left (b x + a\right ) + 2 \, c + i\right )^{2} - 2 i \, \log \left (-i \, c^{2} + {\left (c^{2} + 2 i \, c - 1\right )} \tan \left (b x + a\right ) - i\right ) \log \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 \, \log \left (-i \, c^{2} + {\left (c^{2} + 2 i \, c - 1\right )} \tan \left (b x + a\right ) - 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} + 2 i \, c - 1\right )} \tan \left (b x + a\right ) + 2 \, c + i\right ) \log \left (-\frac {1}{2} i \, \tan \left (b x + a\right ) + \frac {1}{2}\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 )} \operatorname {arccot}\left ({\left (-i \, c + 1\right )} \tan \left (b x + a\right ) - c\right ) + 4 \, {\left (-i \, b x - i \, a\right )} \log \left (\frac {2 i \, c^{2} - 2 \, {\left (c^{2} + 2 i \, c - 1\right )} \tan \left (b x + a\right ) + 2 i}{2 i \, c^{2} - 2 \, {\left (c^{2} + 2 i \, c - 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(arccot(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) + 2*I)/(2*I*c^2 - 2*(c^2 + 2*I*c
- 1)*tan(b*x + a) - 4*c - 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)*arccot((-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) + 2*I)/(2*I*c^2 - 2*(c^2 + 2*I*c - 1)*tan(b*x + a) - 4*c - 2*I)))/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(acot(c-(1-I*c)*tan(b*x+a)),x)

[Out]

Exception raised: CoercionFailed

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