3.33 \(\int \cot ^5(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=177 \[ \frac {a^4 (64 B+67 i A) \cot (c+d x)}{12 d}+\frac {8 a^4 (A-i B) \log (\sin (c+d x))}{d}+\frac {(19 A-16 i B) \cot ^2(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{12 d}+8 a^4 x (B+i A)-\frac {(4 B+7 i A) \cot ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{12 d}-\frac {a A \cot ^4(c+d x) (a+i a \tan (c+d x))^3}{4 d} \]

[Out]

8*a^4*(I*A+B)*x+1/12*a^4*(67*I*A+64*B)*cot(d*x+c)/d+8*a^4*(A-I*B)*ln(sin(d*x+c))/d-1/4*a*A*cot(d*x+c)^4*(a+I*a
*tan(d*x+c))^3/d-1/12*(7*I*A+4*B)*cot(d*x+c)^3*(a^2+I*a^2*tan(d*x+c))^2/d+1/12*(19*A-16*I*B)*cot(d*x+c)^2*(a^4
+I*a^4*tan(d*x+c))/d

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Rubi [A]  time = 0.53, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3593, 3591, 3531, 3475} \[ \frac {a^4 (64 B+67 i A) \cot (c+d x)}{12 d}+\frac {8 a^4 (A-i B) \log (\sin (c+d x))}{d}-\frac {(4 B+7 i A) \cot ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{12 d}+\frac {(19 A-16 i B) \cot ^2(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{12 d}+8 a^4 x (B+i A)-\frac {a A \cot ^4(c+d x) (a+i a \tan (c+d x))^3}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^5*(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*((67*I)*A + 64*B)*Cot[c + d*x])/(12*d) + (8*a^4*(A - I*B)*Log[Sin[c + d*x]])/d - (a*A
*Cot[c + d*x]^4*(a + I*a*Tan[c + d*x])^3)/(4*d) - (((7*I)*A + 4*B)*Cot[c + d*x]^3*(a^2 + I*a^2*Tan[c + d*x])^2
)/(12*d) + ((19*A - (16*I)*B)*Cot[c + d*x]^2*(a^4 + I*a^4*Tan[c + d*x]))/(12*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 3591

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

Rule 3593

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(a^2*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^
(n + 1))/(d*f*(b*c + a*d)*(n + 1)), x] - Dist[a/(d*(b*c + a*d)*(n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c +
d*Tan[e + f*x])^(n + 1)*Simp[A*b*d*(m - n - 2) - B*(b*c*(m - 1) + a*d*(n + 1)) + (a*A*d*(m + n) - B*(a*c*(m -
1) + b*d*(n + 1)))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, 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 ^5(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx &=-\frac {a A \cot ^4(c+d x) (a+i a \tan (c+d x))^3}{4 d}+\frac {1}{4} \int \cot ^4(c+d x) (a+i a \tan (c+d x))^3 (a (7 i A+4 B)-a (A-4 i B) \tan (c+d x)) \, dx\\ &=-\frac {a A \cot ^4(c+d x) (a+i a \tan (c+d x))^3}{4 d}-\frac {(7 i A+4 B) \cot ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{12 d}+\frac {1}{12} \int \cot ^3(c+d x) (a+i a \tan (c+d x))^2 \left (-2 a^2 (19 A-16 i B)-2 a^2 (5 i A+8 B) \tan (c+d x)\right ) \, dx\\ &=-\frac {a A \cot ^4(c+d x) (a+i a \tan (c+d x))^3}{4 d}-\frac {(7 i A+4 B) \cot ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{12 d}+\frac {(19 A-16 i B) \cot ^2(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{12 d}+\frac {1}{24} \int \cot ^2(c+d x) (a+i a \tan (c+d x)) \left (-2 a^3 (67 i A+64 B)+2 a^3 (29 A-32 i B) \tan (c+d x)\right ) \, dx\\ &=\frac {a^4 (67 i A+64 B) \cot (c+d x)}{12 d}-\frac {a A \cot ^4(c+d x) (a+i a \tan (c+d x))^3}{4 d}-\frac {(7 i A+4 B) \cot ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{12 d}+\frac {(19 A-16 i B) \cot ^2(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{12 d}+\frac {1}{24} \int \cot (c+d x) \left (192 a^4 (A-i B)+192 a^4 (i A+B) \tan (c+d x)\right ) \, dx\\ &=8 a^4 (i A+B) x+\frac {a^4 (67 i A+64 B) \cot (c+d x)}{12 d}-\frac {a A \cot ^4(c+d x) (a+i a \tan (c+d x))^3}{4 d}-\frac {(7 i A+4 B) \cot ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{12 d}+\frac {(19 A-16 i B) \cot ^2(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{12 d}+\left (8 a^4 (A-i B)\right ) \int \cot (c+d x) \, dx\\ &=8 a^4 (i A+B) x+\frac {a^4 (67 i A+64 B) \cot (c+d x)}{12 d}+\frac {8 a^4 (A-i B) \log (\sin (c+d x))}{d}-\frac {a A \cot ^4(c+d x) (a+i a \tan (c+d x))^3}{4 d}-\frac {(7 i A+4 B) \cot ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{12 d}+\frac {(19 A-16 i B) \cot ^2(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{12 d}\\ \end {align*}

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Mathematica [A]  time = 6.36, size = 319, normalized size = 1.80 \[ \frac {a^4 \sin (c+d x) (\cot (c+d x)+i)^4 (A \cot (c+d x)+B) \left (192 d x (A-i B) (\sin (4 c)+i \cos (4 c)) \sin ^4(c+d x)+48 (A-i B) (\cos (4 c)-i \sin (4 c)) \sin ^4(c+d x) \log \left (\sin ^2(c+d x)\right )-96 i (A-i B) (\cos (4 c)-i \sin (4 c)) \sin ^4(c+d x) \tan ^{-1}(\tan (5 c+d x))+4 (\cos (4 c)-i \sin (4 c)) ((-B-4 i A) \cot (c)+12 A-6 i B) \sin ^2(c+d x)-8 i (14 A-11 i B) \csc (c) (\cos (4 c)-i \sin (4 c)) \sin (d x) \sin ^3(c+d x)+4 (B+4 i A) \csc (c) (\cos (4 c)-i \sin (4 c)) \sin (d x) \sin (c+d x)+3 i A \sin (4 c)-3 A \cos (4 c)\right )}{12 d (\cos (d x)+i \sin (d x))^4 (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^5*(a + I*a*Tan[c + d*x])^4*(A + B*Tan[c + d*x]),x]

[Out]

(a^4*(I + Cot[c + d*x])^4*(B + A*Cot[c + d*x])*Sin[c + d*x]*(-3*A*Cos[4*c] + (3*I)*A*Sin[4*c] + 4*((4*I)*A + B
)*Csc[c]*(Cos[4*c] - I*Sin[4*c])*Sin[d*x]*Sin[c + d*x] + 4*(12*A - (6*I)*B + ((-4*I)*A - B)*Cot[c])*(Cos[4*c]
- I*Sin[4*c])*Sin[c + d*x]^2 - (8*I)*(14*A - (11*I)*B)*Csc[c]*(Cos[4*c] - I*Sin[4*c])*Sin[d*x]*Sin[c + d*x]^3
- (96*I)*(A - I*B)*ArcTan[Tan[5*c + d*x]]*(Cos[4*c] - I*Sin[4*c])*Sin[c + d*x]^4 + 48*(A - I*B)*Log[Sin[c + d*
x]^2]*(Cos[4*c] - I*Sin[4*c])*Sin[c + d*x]^4 + 192*(A - I*B)*d*x*(I*Cos[4*c] + Sin[4*c])*Sin[c + d*x]^4))/(12*
d*(Cos[d*x] + I*Sin[d*x])^4*(A*Cos[c + d*x] + B*Sin[c + d*x]))

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fricas [A]  time = 0.57, size = 228, normalized size = 1.29 \[ -\frac {4 \, {\left (6 \, {\left (5 \, A - 3 i \, B\right )} a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} - 9 \, {\left (7 \, A - 5 i \, B\right )} a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, {\left (25 \, A - 19 i \, B\right )} a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} - {\left (14 \, A - 11 i \, B\right )} a^{4} - 6 \, {\left ({\left (A - i \, B\right )} a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, {\left (A - i \, B\right )} a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, {\left (A - i \, B\right )} a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, {\left (A - i \, B\right )} a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (A - i \, B\right )} a^{4}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )\right )}}{3 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

-4/3*(6*(5*A - 3*I*B)*a^4*e^(6*I*d*x + 6*I*c) - 9*(7*A - 5*I*B)*a^4*e^(4*I*d*x + 4*I*c) + 2*(25*A - 19*I*B)*a^
4*e^(2*I*d*x + 2*I*c) - (14*A - 11*I*B)*a^4 - 6*((A - I*B)*a^4*e^(8*I*d*x + 8*I*c) - 4*(A - I*B)*a^4*e^(6*I*d*
x + 6*I*c) + 6*(A - I*B)*a^4*e^(4*I*d*x + 4*I*c) - 4*(A - I*B)*a^4*e^(2*I*d*x + 2*I*c) + (A - I*B)*a^4)*log(e^
(2*I*d*x + 2*I*c) - 1))/(d*e^(8*I*d*x + 8*I*c) - 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) - 4*d*e^(2*
I*d*x + 2*I*c) + d)

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giac [B]  time = 7.28, size = 322, normalized size = 1.82 \[ -\frac {3 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 32 i \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 180 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 96 i \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 864 i \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 696 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3072 \, {\left (A a^{4} - i \, B a^{4}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right ) - 1536 \, {\left (A a^{4} - i \, B a^{4}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + \frac {3200 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 3200 i \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 864 i \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 696 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 180 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 96 i \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 32 i \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A a^{4}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{192 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

-1/192*(3*A*a^4*tan(1/2*d*x + 1/2*c)^4 - 32*I*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 8*B*a^4*tan(1/2*d*x + 1/2*c)^3 -
180*A*a^4*tan(1/2*d*x + 1/2*c)^2 + 96*I*B*a^4*tan(1/2*d*x + 1/2*c)^2 + 864*I*A*a^4*tan(1/2*d*x + 1/2*c) + 696*
B*a^4*tan(1/2*d*x + 1/2*c) + 3072*(A*a^4 - I*B*a^4)*log(tan(1/2*d*x + 1/2*c) + I) - 1536*(A*a^4 - I*B*a^4)*log
(tan(1/2*d*x + 1/2*c)) + (3200*A*a^4*tan(1/2*d*x + 1/2*c)^4 - 3200*I*B*a^4*tan(1/2*d*x + 1/2*c)^4 - 864*I*A*a^
4*tan(1/2*d*x + 1/2*c)^3 - 696*B*a^4*tan(1/2*d*x + 1/2*c)^3 - 180*A*a^4*tan(1/2*d*x + 1/2*c)^2 + 96*I*B*a^4*ta
n(1/2*d*x + 1/2*c)^2 + 32*I*A*a^4*tan(1/2*d*x + 1/2*c) + 8*B*a^4*tan(1/2*d*x + 1/2*c) + 3*A*a^4)/tan(1/2*d*x +
1/2*c)^4)/d

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maple [A]  time = 0.45, size = 189, normalized size = 1.07 \[ \frac {7 A \,a^{4} \left (\cot ^{2}\left (d x +c \right )\right )}{2 d}+\frac {8 a^{4} B c}{d}+\frac {7 B \cot \left (d x +c \right ) a^{4}}{d}+8 i A x \,a^{4}+8 a^{4} B x +\frac {8 i A \cot \left (d x +c \right ) a^{4}}{d}+\frac {8 i A \,a^{4} c}{d}-\frac {8 i B \,a^{4} \ln \left (\sin \left (d x +c \right )\right )}{d}-\frac {4 i A \,a^{4} \left (\cot ^{3}\left (d x +c \right )\right )}{3 d}-\frac {2 i B \,a^{4} \left (\cot ^{2}\left (d x +c \right )\right )}{d}-\frac {A \,a^{4} \left (\cot ^{4}\left (d x +c \right )\right )}{4 d}-\frac {a^{4} B \left (\cot ^{3}\left (d x +c \right )\right )}{3 d}+\frac {8 a^{4} A \ln \left (\sin \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x)

[Out]

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

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maxima [A]  time = 0.75, size = 134, normalized size = 0.76 \[ \frac {12 \, {\left (d x + c\right )} {\left (8 i \, A + 8 \, B\right )} a^{4} - 48 \, {\left (A - i \, B\right )} a^{4} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 96 \, {\left (A - i \, B\right )} a^{4} \log \left (\tan \left (d x + c\right )\right ) - \frac {{\left (-96 i \, A - 84 \, B\right )} a^{4} \tan \left (d x + c\right )^{3} - 6 \, {\left (7 \, A - 4 i \, B\right )} a^{4} \tan \left (d x + c\right )^{2} + {\left (16 i \, A + 4 \, B\right )} a^{4} \tan \left (d x + c\right ) + 3 \, A a^{4}}{\tan \left (d x + c\right )^{4}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

1/12*(12*(d*x + c)*(8*I*A + 8*B)*a^4 - 48*(A - I*B)*a^4*log(tan(d*x + c)^2 + 1) + 96*(A - I*B)*a^4*log(tan(d*x
+ c)) - ((-96*I*A - 84*B)*a^4*tan(d*x + c)^3 - 6*(7*A - 4*I*B)*a^4*tan(d*x + c)^2 + (16*I*A + 4*B)*a^4*tan(d*
x + c) + 3*A*a^4)/tan(d*x + c)^4)/d

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mupad [B]  time = 6.57, size = 114, normalized size = 0.64 \[ \frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {7\,A\,a^4}{2}-B\,a^4\,2{}\mathrm {i}\right )+{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (7\,B\,a^4+A\,a^4\,8{}\mathrm {i}\right )-\frac {A\,a^4}{4}-\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {B\,a^4}{3}+\frac {A\,a^4\,4{}\mathrm {i}}{3}\right )}{d\,{\mathrm {tan}\left (c+d\,x\right )}^4}+\frac {16\,a^4\,\mathrm {atan}\left (2\,\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^5*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^4,x)

[Out]

(tan(c + d*x)^2*((7*A*a^4)/2 - B*a^4*2i) + tan(c + d*x)^3*(A*a^4*8i + 7*B*a^4) - (A*a^4)/4 - tan(c + d*x)*((A*
a^4*4i)/3 + (B*a^4)/3))/(d*tan(c + d*x)^4) + (16*a^4*atan(2*tan(c + d*x) + 1i)*(A*1i + B))/d

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sympy [A]  time = 1.96, size = 235, normalized size = 1.33 \[ \frac {8 a^{4} \left (A - i B\right ) \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac {- 56 A a^{4} + 44 i B a^{4} + \left (200 A a^{4} e^{2 i c} - 152 i B a^{4} e^{2 i c}\right ) e^{2 i d x} + \left (- 252 A a^{4} e^{4 i c} + 180 i B a^{4} e^{4 i c}\right ) e^{4 i d x} + \left (120 A a^{4} e^{6 i c} - 72 i B a^{4} e^{6 i c}\right ) e^{6 i d x}}{- 3 d e^{8 i c} e^{8 i d x} + 12 d e^{6 i c} e^{6 i d x} - 18 d e^{4 i c} e^{4 i d x} + 12 d e^{2 i c} e^{2 i d x} - 3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**5*(a+I*a*tan(d*x+c))**4*(A+B*tan(d*x+c)),x)

[Out]

8*a**4*(A - I*B)*log(exp(2*I*d*x) - exp(-2*I*c))/d + (-56*A*a**4 + 44*I*B*a**4 + (200*A*a**4*exp(2*I*c) - 152*
I*B*a**4*exp(2*I*c))*exp(2*I*d*x) + (-252*A*a**4*exp(4*I*c) + 180*I*B*a**4*exp(4*I*c))*exp(4*I*d*x) + (120*A*a
**4*exp(6*I*c) - 72*I*B*a**4*exp(6*I*c))*exp(6*I*d*x))/(-3*d*exp(8*I*c)*exp(8*I*d*x) + 12*d*exp(6*I*c)*exp(6*I
*d*x) - 18*d*exp(4*I*c)*exp(4*I*d*x) + 12*d*exp(2*I*c)*exp(2*I*d*x) - 3*d)

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