3.29 \(\int \cot (c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=142 \[ -\frac {(3 A-4 i B) \left (a^4+i a^4 \tan (c+d x)\right )}{d}+\frac {a^4 (7 A-8 i B) \log (\cos (c+d x))}{d}+8 a^4 x (B+i A)+\frac {a^4 A \log (\sin (c+d x))}{d}-\frac {(A-2 i B) \left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}+\frac {i a B (a+i a \tan (c+d x))^3}{3 d} \]

[Out]

8*a^4*(I*A+B)*x+a^4*(7*A-8*I*B)*ln(cos(d*x+c))/d+a^4*A*ln(sin(d*x+c))/d+1/3*I*a*B*(a+I*a*tan(d*x+c))^3/d-1/2*(
A-2*I*B)*(a^2+I*a^2*tan(d*x+c))^2/d-(3*A-4*I*B)*(a^4+I*a^4*tan(d*x+c))/d

________________________________________________________________________________________

Rubi [A]  time = 0.42, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3594, 3589, 3475, 3531} \[ -\frac {(A-2 i B) \left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}-\frac {(3 A-4 i B) \left (a^4+i a^4 \tan (c+d x)\right )}{d}+\frac {a^4 (7 A-8 i B) \log (\cos (c+d x))}{d}+8 a^4 x (B+i A)+\frac {a^4 A \log (\sin (c+d x))}{d}+\frac {i a B (a+i a \tan (c+d x))^3}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]*(a + I*a*Tan[c + d*x])^4*(A + B*Tan[c + d*x]),x]

[Out]

8*a^4*(I*A + B)*x + (a^4*(7*A - (8*I)*B)*Log[Cos[c + d*x]])/d + (a^4*A*Log[Sin[c + d*x]])/d + ((I/3)*a*B*(a +
I*a*Tan[c + d*x])^3)/d - ((A - (2*I)*B)*(a^2 + I*a^2*Tan[c + d*x])^2)/(2*d) - ((3*A - (4*I)*B)*(a^4 + I*a^4*Ta
n[c + d*x]))/d

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3531

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((a*c +
b*d)*x)/(a^2 + b^2), x] + Dist[(b*c - a*d)/(a^2 + b^2), Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[a*c + b*d, 0]

Rule 3589

Int[(((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]))/((a_.) + (b_.)*tan[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Dist[(B*d)/b, Int[Tan[e + f*x], x], x] + Dist[1/b, Int[Simp[A*b*c + (A*b*d + B*(
b*c - a*d))*Tan[e + f*x], x]/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a
*d, 0]

Rule 3594

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*B*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1))/(d*f
*(m + n)), x] + Dist[1/(d*(m + n)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n)
+ B*(a*c*(m - 1) - b*d*(n + 1)) - (B*(b*c - a*d)*(m - 1) - d*(A*b + a*B)*(m + n))*Tan[e + f*x], x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && GtQ[m, 1] &&  !LtQ[n, -1]

Rubi steps

\begin {align*} \int \cot (c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx &=\frac {i a B (a+i a \tan (c+d x))^3}{3 d}+\frac {1}{3} \int \cot (c+d x) (a+i a \tan (c+d x))^3 (3 a A+3 a (i A+2 B) \tan (c+d x)) \, dx\\ &=\frac {i a B (a+i a \tan (c+d x))^3}{3 d}-\frac {(A-2 i B) \left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}+\frac {1}{6} \int \cot (c+d x) (a+i a \tan (c+d x))^2 \left (6 a^2 A+6 a^2 (3 i A+4 B) \tan (c+d x)\right ) \, dx\\ &=\frac {i a B (a+i a \tan (c+d x))^3}{3 d}-\frac {(A-2 i B) \left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}-\frac {(3 A-4 i B) \left (a^4+i a^4 \tan (c+d x)\right )}{d}+\frac {1}{6} \int \cot (c+d x) (a+i a \tan (c+d x)) \left (6 a^3 A+6 a^3 (7 i A+8 B) \tan (c+d x)\right ) \, dx\\ &=\frac {i a B (a+i a \tan (c+d x))^3}{3 d}-\frac {(A-2 i B) \left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}-\frac {(3 A-4 i B) \left (a^4+i a^4 \tan (c+d x)\right )}{d}+\frac {1}{6} \int \cot (c+d x) \left (6 a^4 A+48 a^4 (i A+B) \tan (c+d x)\right ) \, dx-\left (a^4 (7 A-8 i B)\right ) \int \tan (c+d x) \, dx\\ &=8 a^4 (i A+B) x+\frac {a^4 (7 A-8 i B) \log (\cos (c+d x))}{d}+\frac {i a B (a+i a \tan (c+d x))^3}{3 d}-\frac {(A-2 i B) \left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}-\frac {(3 A-4 i B) \left (a^4+i a^4 \tan (c+d x)\right )}{d}+\left (a^4 A\right ) \int \cot (c+d x) \, dx\\ &=8 a^4 (i A+B) x+\frac {a^4 (7 A-8 i B) \log (\cos (c+d x))}{d}+\frac {a^4 A \log (\sin (c+d x))}{d}+\frac {i a B (a+i a \tan (c+d x))^3}{3 d}-\frac {(A-2 i B) \left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}-\frac {(3 A-4 i B) \left (a^4+i a^4 \tan (c+d x)\right )}{d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 7.91, size = 429, normalized size = 3.02 \[ \frac {a^4 \sec (c) \sec ^3(c+d x) (\cos (4 d x)+i \sin (4 d x)) \left (3 \cos (d x) \left (3 (7 A-8 i B) \log \left (\cos ^2(c+d x)\right )+3 A \log \left (\sin ^2(c+d x)\right )+48 i A d x+4 A+48 B d x-16 i B\right )+3 \cos (2 c+d x) \left (3 (7 A-8 i B) \log \left (\cos ^2(c+d x)\right )+3 A \log \left (\sin ^2(c+d x)\right )+48 i A d x+4 A+48 B d x-16 i B\right )+48 i A \sin (2 c+d x)-48 i A \sin (2 c+3 d x)+48 i A d x \cos (2 c+3 d x)+48 i A d x \cos (4 c+3 d x)+21 A \cos (2 c+3 d x) \log \left (\cos ^2(c+d x)\right )+21 A \cos (4 c+3 d x) \log \left (\cos ^2(c+d x)\right )+3 A \cos (2 c+3 d x) \log \left (\sin ^2(c+d x)\right )+3 A \cos (4 c+3 d x) \log \left (\sin ^2(c+d x)\right )-96 i A \sin (d x)+96 B \sin (2 c+d x)-88 B \sin (2 c+3 d x)+48 B d x \cos (2 c+3 d x)+48 B d x \cos (4 c+3 d x)-24 i B \cos (2 c+3 d x) \log \left (\cos ^2(c+d x)\right )-24 i B \cos (4 c+3 d x) \log \left (\cos ^2(c+d x)\right )-168 B \sin (d x)\right )}{48 d (\cos (d x)+i \sin (d x))^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]*(a + I*a*Tan[c + d*x])^4*(A + B*Tan[c + d*x]),x]

[Out]

(a^4*Sec[c]*Sec[c + d*x]^3*(Cos[4*d*x] + I*Sin[4*d*x])*((48*I)*A*d*x*Cos[2*c + 3*d*x] + 48*B*d*x*Cos[2*c + 3*d
*x] + (48*I)*A*d*x*Cos[4*c + 3*d*x] + 48*B*d*x*Cos[4*c + 3*d*x] + 21*A*Cos[2*c + 3*d*x]*Log[Cos[c + d*x]^2] -
(24*I)*B*Cos[2*c + 3*d*x]*Log[Cos[c + d*x]^2] + 21*A*Cos[4*c + 3*d*x]*Log[Cos[c + d*x]^2] - (24*I)*B*Cos[4*c +
3*d*x]*Log[Cos[c + d*x]^2] + 3*A*Cos[2*c + 3*d*x]*Log[Sin[c + d*x]^2] + 3*A*Cos[4*c + 3*d*x]*Log[Sin[c + d*x]
^2] + 3*Cos[d*x]*(4*A - (16*I)*B + (48*I)*A*d*x + 48*B*d*x + 3*(7*A - (8*I)*B)*Log[Cos[c + d*x]^2] + 3*A*Log[S
in[c + d*x]^2]) + 3*Cos[2*c + d*x]*(4*A - (16*I)*B + (48*I)*A*d*x + 48*B*d*x + 3*(7*A - (8*I)*B)*Log[Cos[c + d
*x]^2] + 3*A*Log[Sin[c + d*x]^2]) - (96*I)*A*Sin[d*x] - 168*B*Sin[d*x] + (48*I)*A*Sin[2*c + d*x] + 96*B*Sin[2*
c + d*x] - (48*I)*A*Sin[2*c + 3*d*x] - 88*B*Sin[2*c + 3*d*x]))/(48*d*(Cos[d*x] + I*Sin[d*x])^4)

________________________________________________________________________________________

fricas [B]  time = 1.04, size = 246, normalized size = 1.73 \[ \frac {6 \, {\left (5 \, A - 12 i \, B\right )} a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 54 \, {\left (A - 2 i \, B\right )} a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + 4 \, {\left (6 \, A - 11 i \, B\right )} a^{4} + 3 \, {\left ({\left (7 \, A - 8 i \, B\right )} a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, {\left (7 \, A - 8 i \, B\right )} a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, {\left (7 \, A - 8 i \, B\right )} a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (7 \, A - 8 i \, B\right )} a^{4}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 3 \, {\left (A a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, A a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, A a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + A a^{4}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}{3 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/3*(6*(5*A - 12*I*B)*a^4*e^(4*I*d*x + 4*I*c) + 54*(A - 2*I*B)*a^4*e^(2*I*d*x + 2*I*c) + 4*(6*A - 11*I*B)*a^4
+ 3*((7*A - 8*I*B)*a^4*e^(6*I*d*x + 6*I*c) + 3*(7*A - 8*I*B)*a^4*e^(4*I*d*x + 4*I*c) + 3*(7*A - 8*I*B)*a^4*e^(
2*I*d*x + 2*I*c) + (7*A - 8*I*B)*a^4)*log(e^(2*I*d*x + 2*I*c) + 1) + 3*(A*a^4*e^(6*I*d*x + 6*I*c) + 3*A*a^4*e^
(4*I*d*x + 4*I*c) + 3*A*a^4*e^(2*I*d*x + 2*I*c) + A*a^4)*log(e^(2*I*d*x + 2*I*c) - 1))/(d*e^(6*I*d*x + 6*I*c)
+ 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)

________________________________________________________________________________________

giac [B]  time = 2.14, size = 333, normalized size = 2.35 \[ \frac {6 \, A a^{4} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 6 \, {\left (7 \, A a^{4} - 8 i \, B a^{4}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - 12 \, {\left (8 \, A a^{4} - 8 i \, B a^{4}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right ) + 6 \, {\left (7 \, A a^{4} - 8 i \, B a^{4}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right ) - \frac {77 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 88 i \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 48 i \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 84 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 243 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 312 i \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 96 i \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 184 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 243 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 312 i \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 48 i \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 84 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 77 \, A a^{4} + 88 i \, B a^{4}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

1/6*(6*A*a^4*log(tan(1/2*d*x + 1/2*c)) + 6*(7*A*a^4 - 8*I*B*a^4)*log(tan(1/2*d*x + 1/2*c) + 1) - 12*(8*A*a^4 -
8*I*B*a^4)*log(tan(1/2*d*x + 1/2*c) + I) + 6*(7*A*a^4 - 8*I*B*a^4)*log(tan(1/2*d*x + 1/2*c) - 1) - (77*A*a^4*
tan(1/2*d*x + 1/2*c)^6 - 88*I*B*a^4*tan(1/2*d*x + 1/2*c)^6 - 48*I*A*a^4*tan(1/2*d*x + 1/2*c)^5 - 84*B*a^4*tan(
1/2*d*x + 1/2*c)^5 - 243*A*a^4*tan(1/2*d*x + 1/2*c)^4 + 312*I*B*a^4*tan(1/2*d*x + 1/2*c)^4 + 96*I*A*a^4*tan(1/
2*d*x + 1/2*c)^3 + 184*B*a^4*tan(1/2*d*x + 1/2*c)^3 + 243*A*a^4*tan(1/2*d*x + 1/2*c)^2 - 312*I*B*a^4*tan(1/2*d
*x + 1/2*c)^2 - 48*I*A*a^4*tan(1/2*d*x + 1/2*c) - 84*B*a^4*tan(1/2*d*x + 1/2*c) - 77*A*a^4 + 88*I*B*a^4)/(tan(
1/2*d*x + 1/2*c)^2 - 1)^3)/d

________________________________________________________________________________________

maple [A]  time = 0.47, size = 169, normalized size = 1.19 \[ \frac {a^{4} A \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {7 A \,a^{4} \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {a^{4} B \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}-\frac {7 a^{4} B \tan \left (d x +c \right )}{d}+8 a^{4} B x +\frac {8 a^{4} B c}{d}-\frac {8 i B \,a^{4} \ln \left (\cos \left (d x +c \right )\right )}{d}-\frac {2 i a^{4} B \left (\tan ^{2}\left (d x +c \right )\right )}{d}+8 i A x \,a^{4}+\frac {8 i A \,a^{4} c}{d}-\frac {4 i a^{4} A \tan \left (d x +c \right )}{d}+\frac {a^{4} A \ln \left (\sin \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x)

[Out]

1/2/d*a^4*A*tan(d*x+c)^2+7/d*A*a^4*ln(cos(d*x+c))+1/3/d*a^4*B*tan(d*x+c)^3-7/d*a^4*B*tan(d*x+c)+8*a^4*B*x+8/d*
a^4*B*c-8*I/d*B*a^4*ln(cos(d*x+c))-2*I/d*B*a^4*tan(d*x+c)^2+8*I*A*x*a^4+8*I/d*A*a^4*c-4*I/d*A*tan(d*x+c)*a^4+a
^4*A*ln(sin(d*x+c))/d

________________________________________________________________________________________

maxima [A]  time = 0.79, size = 106, normalized size = 0.75 \[ \frac {2 \, B a^{4} \tan \left (d x + c\right )^{3} + 3 \, {\left (A - 4 i \, B\right )} a^{4} \tan \left (d x + c\right )^{2} + 6 \, {\left (d x + c\right )} {\left (8 i \, A + 8 \, B\right )} a^{4} - 24 \, {\left (A - i \, B\right )} a^{4} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 6 \, A a^{4} \log \left (\tan \left (d x + c\right )\right ) + {\left (-24 i \, A - 42 \, B\right )} a^{4} \tan \left (d x + c\right )}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

1/6*(2*B*a^4*tan(d*x + c)^3 + 3*(A - 4*I*B)*a^4*tan(d*x + c)^2 + 6*(d*x + c)*(8*I*A + 8*B)*a^4 - 24*(A - I*B)*
a^4*log(tan(d*x + c)^2 + 1) + 6*A*a^4*log(tan(d*x + c)) + (-24*I*A - 42*B)*a^4*tan(d*x + c))/d

________________________________________________________________________________________

mupad [B]  time = 6.22, size = 133, normalized size = 0.94 \[ \frac {A\,a^4\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )}{d}-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (a^4\,\left (A-B\,1{}\mathrm {i}\right )\,3{}\mathrm {i}+B\,a^4+a^4\,\left (3\,B+A\,1{}\mathrm {i}\right )\right )}{d}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {B\,a^4\,1{}\mathrm {i}}{2}+\frac {a^4\,\left (3\,B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}\right )}{d}-\frac {8\,a^4\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )}{d}+\frac {B\,a^4\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^4,x)

[Out]

(A*a^4*log(tan(c + d*x)))/d - (tan(c + d*x)*(a^4*(A - B*1i)*3i + B*a^4 + a^4*(A*1i + 3*B)))/d - (tan(c + d*x)^
2*((B*a^4*1i)/2 + (a^4*(A*1i + 3*B)*1i)/2))/d - (8*a^4*log(tan(c + d*x) + 1i)*(A - B*1i))/d + (B*a^4*tan(c + d
*x)^3)/(3*d)

________________________________________________________________________________________

sympy [B]  time = 3.76, size = 289, normalized size = 2.04 \[ \frac {A a^{4} \log {\left (\frac {- 3 A a^{4} + 4 i B a^{4}}{3 A a^{4} e^{2 i c} - 4 i B a^{4} e^{2 i c}} + e^{2 i d x} \right )}}{d} + \frac {a^{4} \left (7 A - 8 i B\right ) \log {\left (e^{2 i d x} + \frac {- 4 A a^{4} + 4 i B a^{4} + a^{4} \left (7 A - 8 i B\right )}{3 A a^{4} e^{2 i c} - 4 i B a^{4} e^{2 i c}} \right )}}{d} + \frac {24 A a^{4} - 44 i B a^{4} + \left (54 A a^{4} e^{2 i c} - 108 i B a^{4} e^{2 i c}\right ) e^{2 i d x} + \left (30 A a^{4} e^{4 i c} - 72 i B a^{4} e^{4 i c}\right ) e^{4 i d x}}{3 d e^{6 i c} e^{6 i d x} + 9 d e^{4 i c} e^{4 i d x} + 9 d e^{2 i c} e^{2 i d x} + 3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)*(a+I*a*tan(d*x+c))**4*(A+B*tan(d*x+c)),x)

[Out]

A*a**4*log((-3*A*a**4 + 4*I*B*a**4)/(3*A*a**4*exp(2*I*c) - 4*I*B*a**4*exp(2*I*c)) + exp(2*I*d*x))/d + a**4*(7*
A - 8*I*B)*log(exp(2*I*d*x) + (-4*A*a**4 + 4*I*B*a**4 + a**4*(7*A - 8*I*B))/(3*A*a**4*exp(2*I*c) - 4*I*B*a**4*
exp(2*I*c)))/d + (24*A*a**4 - 44*I*B*a**4 + (54*A*a**4*exp(2*I*c) - 108*I*B*a**4*exp(2*I*c))*exp(2*I*d*x) + (3
0*A*a**4*exp(4*I*c) - 72*I*B*a**4*exp(4*I*c))*exp(4*I*d*x))/(3*d*exp(6*I*c)*exp(6*I*d*x) + 9*d*exp(4*I*c)*exp(
4*I*d*x) + 9*d*exp(2*I*c)*exp(2*I*d*x) + 3*d)

________________________________________________________________________________________