3.16 \(\int \cot ^5(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=139 \[ -\frac{a^2 (4 B+5 i A) \cot ^3(c+d x)}{12 d}+\frac{a^2 (A-i B) \cot ^2(c+d x)}{d}+\frac{2 a^2 (B+i A) \cot (c+d x)}{d}+\frac{2 a^2 (A-i B) \log (\sin (c+d x))}{d}+2 a^2 x (B+i A)-\frac{A \cot ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d} \]

[Out]

2*a^2*(I*A + B)*x + (2*a^2*(I*A + B)*Cot[c + d*x])/d + (a^2*(A - I*B)*Cot[c + d*x]^2)/d - (a^2*((5*I)*A + 4*B)
*Cot[c + d*x]^3)/(12*d) + (2*a^2*(A - I*B)*Log[Sin[c + d*x]])/d - (A*Cot[c + d*x]^4*(a^2 + I*a^2*Tan[c + d*x])
)/(4*d)

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Rubi [A]  time = 0.293271, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.147, Rules used = {3593, 3591, 3529, 3531, 3475} \[ -\frac{a^2 (4 B+5 i A) \cot ^3(c+d x)}{12 d}+\frac{a^2 (A-i B) \cot ^2(c+d x)}{d}+\frac{2 a^2 (B+i A) \cot (c+d x)}{d}+\frac{2 a^2 (A-i B) \log (\sin (c+d x))}{d}+2 a^2 x (B+i A)-\frac{A \cot ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^5*(a + I*a*Tan[c + d*x])^2*(A + B*Tan[c + d*x]),x]

[Out]

2*a^2*(I*A + B)*x + (2*a^2*(I*A + B)*Cot[c + d*x])/d + (a^2*(A - I*B)*Cot[c + d*x]^2)/d - (a^2*((5*I)*A + 4*B)
*Cot[c + d*x]^3)/(12*d) + (2*a^2*(A - I*B)*Log[Sin[c + d*x]])/d - (A*Cot[c + d*x]^4*(a^2 + I*a^2*Tan[c + d*x])
)/(4*d)

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]

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 3529

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

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 3475

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

Rubi steps

\begin{align*} \int \cot ^5(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx &=-\frac{A \cot ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d}+\frac{1}{4} \int \cot ^4(c+d x) (a+i a \tan (c+d x)) (a (5 i A+4 B)-a (3 A-4 i B) \tan (c+d x)) \, dx\\ &=-\frac{a^2 (5 i A+4 B) \cot ^3(c+d x)}{12 d}-\frac{A \cot ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d}+\frac{1}{4} \int \cot ^3(c+d x) \left (-8 a^2 (A-i B)-8 a^2 (i A+B) \tan (c+d x)\right ) \, dx\\ &=\frac{a^2 (A-i B) \cot ^2(c+d x)}{d}-\frac{a^2 (5 i A+4 B) \cot ^3(c+d x)}{12 d}-\frac{A \cot ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d}+\frac{1}{4} \int \cot ^2(c+d x) \left (-8 a^2 (i A+B)+8 a^2 (A-i B) \tan (c+d x)\right ) \, dx\\ &=\frac{2 a^2 (i A+B) \cot (c+d x)}{d}+\frac{a^2 (A-i B) \cot ^2(c+d x)}{d}-\frac{a^2 (5 i A+4 B) \cot ^3(c+d x)}{12 d}-\frac{A \cot ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d}+\frac{1}{4} \int \cot (c+d x) \left (8 a^2 (A-i B)+8 a^2 (i A+B) \tan (c+d x)\right ) \, dx\\ &=2 a^2 (i A+B) x+\frac{2 a^2 (i A+B) \cot (c+d x)}{d}+\frac{a^2 (A-i B) \cot ^2(c+d x)}{d}-\frac{a^2 (5 i A+4 B) \cot ^3(c+d x)}{12 d}-\frac{A \cot ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d}+\left (2 a^2 (A-i B)\right ) \int \cot (c+d x) \, dx\\ &=2 a^2 (i A+B) x+\frac{2 a^2 (i A+B) \cot (c+d x)}{d}+\frac{a^2 (A-i B) \cot ^2(c+d x)}{d}-\frac{a^2 (5 i A+4 B) \cot ^3(c+d x)}{12 d}+\frac{2 a^2 (A-i B) \log (\sin (c+d x))}{d}-\frac{A \cot ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d}\\ \end{align*}

Mathematica [B]  time = 8.43952, size = 902, normalized size = 6.49 \[ a^2 \left (\frac{(\cot (c+d x)+i)^2 (B+A \cot (c+d x)) (A \cos (c)-i B \cos (c)-i A \sin (c)-B \sin (c)) \left (-2 i \tan ^{-1}(\tan (3 c+d x)) \cos (c)-2 \tan ^{-1}(\tan (3 c+d x)) \sin (c)\right ) \sin ^3(c+d x)}{d (\cos (d x)+i \sin (d x))^2 (A \cos (c+d x)+B \sin (c+d x))}+\frac{(\cot (c+d x)+i)^2 (B+A \cot (c+d x)) (A \cos (c)-i B \cos (c)-i A \sin (c)-B \sin (c)) \left (\cos (c) \log \left (\sin ^2(c+d x)\right )-i \log \left (\sin ^2(c+d x)\right ) \sin (c)\right ) \sin ^3(c+d x)}{d (\cos (d x)+i \sin (d x))^2 (A \cos (c+d x)+B \sin (c+d x))}+\frac{x (\cot (c+d x)+i)^2 (B+A \cot (c+d x)) \left (6 i A \cos ^2(c)+6 B \cos ^2(c)-2 A \cot (c) \cos ^2(c)+2 i B \cot (c) \cos ^2(c)+6 A \sin (c) \cos (c)-6 i B \sin (c) \cos (c)-2 i A \sin ^2(c)-2 B \sin ^2(c)+(A-i B) \cot (c) (2 \cos (2 c)-2 i \sin (2 c))\right ) \sin ^3(c+d x)}{(\cos (d x)+i \sin (d x))^2 (A \cos (c+d x)+B \sin (c+d x))}+\frac{(i A+B) (\cot (c+d x)+i)^2 (B+A \cot (c+d x)) (2 d x \cos (2 c)-2 i d x \sin (2 c)) \sin ^3(c+d x)}{d (\cos (d x)+i \sin (d x))^2 (A \cos (c+d x)+B \sin (c+d x))}+\frac{(\cot (c+d x)+i)^2 (B+A \cot (c+d x)) \csc (c) \left (\frac{1}{3} \cos (2 c)-\frac{1}{3} i \sin (2 c)\right ) (-8 i A \sin (d x)-7 B \sin (d x)) \sin ^2(c+d x)}{d (\cos (d x)+i \sin (d x))^2 (A \cos (c+d x)+B \sin (c+d x))}+\frac{(\cot (c+d x)+i)^2 (B+A \cot (c+d x)) \csc (c) (-4 i A \cos (c)-2 B \cos (c)+9 A \sin (c)-6 i B \sin (c)) \left (\frac{1}{6} \cos (2 c)-\frac{1}{6} i \sin (2 c)\right ) \sin (c+d x)}{d (\cos (d x)+i \sin (d x))^2 (A \cos (c+d x)+B \sin (c+d x))}+\frac{(\cot (c+d x)+i)^2 (B+A \cot (c+d x)) \csc (c) \left (\frac{1}{3} \cos (2 c)-\frac{1}{3} i \sin (2 c)\right ) (2 i A \sin (d x)+B \sin (d x))}{d (\cos (d x)+i \sin (d x))^2 (A \cos (c+d x)+B \sin (c+d x))}+\frac{(\cot (c+d x)+i)^2 (B+A \cot (c+d x)) \csc (c+d x) \left (\frac{1}{4} i A \sin (2 c)-\frac{1}{4} A \cos (2 c)\right )}{d (\cos (d x)+i \sin (d x))^2 (A \cos (c+d x)+B \sin (c+d x))}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^5*(a + I*a*Tan[c + d*x])^2*(A + B*Tan[c + d*x]),x]

[Out]

a^2*(((I + Cot[c + d*x])^2*(B + A*Cot[c + d*x])*Csc[c + d*x]*(-(A*Cos[2*c])/4 + (I/4)*A*Sin[2*c]))/(d*(Cos[d*x
] + I*Sin[d*x])^2*(A*Cos[c + d*x] + B*Sin[c + d*x])) + ((I + Cot[c + d*x])^2*(B + A*Cot[c + d*x])*Csc[c]*(Cos[
2*c]/3 - (I/3)*Sin[2*c])*((2*I)*A*Sin[d*x] + B*Sin[d*x]))/(d*(Cos[d*x] + I*Sin[d*x])^2*(A*Cos[c + d*x] + B*Sin
[c + d*x])) + ((I + Cot[c + d*x])^2*(B + A*Cot[c + d*x])*Csc[c]*((-4*I)*A*Cos[c] - 2*B*Cos[c] + 9*A*Sin[c] - (
6*I)*B*Sin[c])*(Cos[2*c]/6 - (I/6)*Sin[2*c])*Sin[c + d*x])/(d*(Cos[d*x] + I*Sin[d*x])^2*(A*Cos[c + d*x] + B*Si
n[c + d*x])) + ((I + Cot[c + d*x])^2*(B + A*Cot[c + d*x])*Csc[c]*(Cos[2*c]/3 - (I/3)*Sin[2*c])*((-8*I)*A*Sin[d
*x] - 7*B*Sin[d*x])*Sin[c + d*x]^2)/(d*(Cos[d*x] + I*Sin[d*x])^2*(A*Cos[c + d*x] + B*Sin[c + d*x])) + ((I + Co
t[c + d*x])^2*(B + A*Cot[c + d*x])*(A*Cos[c] - I*B*Cos[c] - I*A*Sin[c] - B*Sin[c])*((-2*I)*ArcTan[Tan[3*c + d*
x]]*Cos[c] - 2*ArcTan[Tan[3*c + d*x]]*Sin[c])*Sin[c + d*x]^3)/(d*(Cos[d*x] + I*Sin[d*x])^2*(A*Cos[c + d*x] + B
*Sin[c + d*x])) + ((I + Cot[c + d*x])^2*(B + A*Cot[c + d*x])*(A*Cos[c] - I*B*Cos[c] - I*A*Sin[c] - B*Sin[c])*(
Cos[c]*Log[Sin[c + d*x]^2] - I*Log[Sin[c + d*x]^2]*Sin[c])*Sin[c + d*x]^3)/(d*(Cos[d*x] + I*Sin[d*x])^2*(A*Cos
[c + d*x] + B*Sin[c + d*x])) + (x*(I + Cot[c + d*x])^2*(B + A*Cot[c + d*x])*((6*I)*A*Cos[c]^2 + 6*B*Cos[c]^2 -
2*A*Cos[c]^2*Cot[c] + (2*I)*B*Cos[c]^2*Cot[c] + 6*A*Cos[c]*Sin[c] - (6*I)*B*Cos[c]*Sin[c] - (2*I)*A*Sin[c]^2
- 2*B*Sin[c]^2 + (A - I*B)*Cot[c]*(2*Cos[2*c] - (2*I)*Sin[2*c]))*Sin[c + d*x]^3)/((Cos[d*x] + I*Sin[d*x])^2*(A
*Cos[c + d*x] + B*Sin[c + d*x])) + ((I*A + B)*(I + Cot[c + d*x])^2*(B + A*Cot[c + d*x])*(2*d*x*Cos[2*c] - (2*I
)*d*x*Sin[2*c])*Sin[c + d*x]^3)/(d*(Cos[d*x] + I*Sin[d*x])^2*(A*Cos[c + d*x] + B*Sin[c + d*x])))

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Maple [A]  time = 0.074, size = 188, normalized size = 1.4 \begin{align*}{\frac{{a}^{2}A \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{d}}+2\,{\frac{{a}^{2}A\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}+2\,{a}^{2}Bx+2\,{\frac{\cot \left ( dx+c \right ) B{a}^{2}}{d}}+2\,{\frac{B{a}^{2}c}{d}}+{\frac{2\,iA{a}^{2}\cot \left ( dx+c \right ) }{d}}+2\,iA{a}^{2}x+{\frac{2\,iA{a}^{2}c}{d}}-{\frac{iB{a}^{2} \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{d}}-{\frac{{\frac{2\,i}{3}}A{a}^{2} \left ( \cot \left ( dx+c \right ) \right ) ^{3}}{d}}-{\frac{2\,iB{a}^{2}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{{a}^{2}A \left ( \cot \left ( dx+c \right ) \right ) ^{4}}{4\,d}}-{\frac{{a}^{2}B \left ( \cot \left ( dx+c \right ) \right ) ^{3}}{3\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.71061, size = 182, normalized size = 1.31 \begin{align*} -\frac{24 \,{\left (d x + c\right )}{\left (-i \, A - B\right )} a^{2} + 12 \,{\left (A - i \, B\right )} a^{2} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 12 \,{\left (2 \, A - 2 i \, B\right )} a^{2} \log \left (\tan \left (d x + c\right )\right ) - \frac{24 \,{\left (i \, A + B\right )} a^{2} \tan \left (d x + c\right )^{3} +{\left (12 \, A - 12 i \, B\right )} a^{2} \tan \left (d x + c\right )^{2} + 4 \,{\left (-2 i \, A - B\right )} a^{2} \tan \left (d x + c\right ) - 3 \, A a^{2}}{\tan \left (d x + c\right )^{4}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

-1/12*(24*(d*x + c)*(-I*A - B)*a^2 + 12*(A - I*B)*a^2*log(tan(d*x + c)^2 + 1) - 12*(2*A - 2*I*B)*a^2*log(tan(d
*x + c)) - (24*(I*A + B)*a^2*tan(d*x + c)^3 + (12*A - 12*I*B)*a^2*tan(d*x + c)^2 + 4*(-2*I*A - B)*a^2*tan(d*x
+ c) - 3*A*a^2)/tan(d*x + c)^4)/d

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Fricas [A]  time = 1.43193, size = 620, normalized size = 4.46 \begin{align*} -\frac{2 \,{\left (3 \,{\left (7 \, A - 5 i \, B\right )} a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \,{\left (12 \, A - 11 i \, B\right )} a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (29 \, A - 25 i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} -{\left (8 \, A - 7 i \, B\right )} a^{2} - 3 \,{\left ({\left (A - i \, B\right )} a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \,{\left (A - i \, B\right )} a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \,{\left (A - i \, B\right )} a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \,{\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (A - i \, B\right )} a^{2}\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

-2/3*(3*(7*A - 5*I*B)*a^2*e^(6*I*d*x + 6*I*c) - 3*(12*A - 11*I*B)*a^2*e^(4*I*d*x + 4*I*c) + (29*A - 25*I*B)*a^
2*e^(2*I*d*x + 2*I*c) - (8*A - 7*I*B)*a^2 - 3*((A - I*B)*a^2*e^(8*I*d*x + 8*I*c) - 4*(A - I*B)*a^2*e^(6*I*d*x
+ 6*I*c) + 6*(A - I*B)*a^2*e^(4*I*d*x + 4*I*c) - 4*(A - I*B)*a^2*e^(2*I*d*x + 2*I*c) + (A - I*B)*a^2)*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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Sympy [A]  time = 22.4911, size = 221, normalized size = 1.59 \begin{align*} \frac{2 a^{2} \left (A - i B\right ) \log{\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac{- \frac{\left (14 A a^{2} - 10 i B a^{2}\right ) e^{- 2 i c} e^{6 i d x}}{d} + \frac{\left (16 A a^{2} - 14 i B a^{2}\right ) e^{- 8 i c}}{3 d} + \frac{\left (24 A a^{2} - 22 i B a^{2}\right ) e^{- 4 i c} e^{4 i d x}}{d} - \frac{\left (58 A a^{2} - 50 i B a^{2}\right ) e^{- 6 i c} e^{2 i d x}}{3 d}}{e^{8 i d x} - 4 e^{- 2 i c} e^{6 i d x} + 6 e^{- 4 i c} e^{4 i d x} - 4 e^{- 6 i c} e^{2 i d x} + e^{- 8 i c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**5*(a+I*a*tan(d*x+c))**2*(A+B*tan(d*x+c)),x)

[Out]

2*a**2*(A - I*B)*log(exp(2*I*d*x) - exp(-2*I*c))/d + (-(14*A*a**2 - 10*I*B*a**2)*exp(-2*I*c)*exp(6*I*d*x)/d +
(16*A*a**2 - 14*I*B*a**2)*exp(-8*I*c)/(3*d) + (24*A*a**2 - 22*I*B*a**2)*exp(-4*I*c)*exp(4*I*d*x)/d - (58*A*a**
2 - 50*I*B*a**2)*exp(-6*I*c)*exp(2*I*d*x)/(3*d))/(exp(8*I*d*x) - 4*exp(-2*I*c)*exp(6*I*d*x) + 6*exp(-4*I*c)*ex
p(4*I*d*x) - 4*exp(-6*I*c)*exp(2*I*d*x) + exp(-8*I*c))

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Giac [B]  time = 1.6494, size = 437, normalized size = 3.14 \begin{align*} -\frac{3 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 16 i \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 8 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 60 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 48 i \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 240 i \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 216 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 384 \,{\left (2 \, A a^{2} - 2 i \, B a^{2}\right )} \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + i\right ) - 384 \,{\left (A a^{2} - i \, B a^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + \frac{800 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 800 i \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 240 i \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 216 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 60 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 48 i \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 16 i \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 8 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, A a^{2}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4}}}{192 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

-1/192*(3*A*a^2*tan(1/2*d*x + 1/2*c)^4 - 16*I*A*a^2*tan(1/2*d*x + 1/2*c)^3 - 8*B*a^2*tan(1/2*d*x + 1/2*c)^3 -
60*A*a^2*tan(1/2*d*x + 1/2*c)^2 + 48*I*B*a^2*tan(1/2*d*x + 1/2*c)^2 + 240*I*A*a^2*tan(1/2*d*x + 1/2*c) + 216*B
*a^2*tan(1/2*d*x + 1/2*c) + 384*(2*A*a^2 - 2*I*B*a^2)*log(tan(1/2*d*x + 1/2*c) + I) - 384*(A*a^2 - I*B*a^2)*lo
g(abs(tan(1/2*d*x + 1/2*c))) + (800*A*a^2*tan(1/2*d*x + 1/2*c)^4 - 800*I*B*a^2*tan(1/2*d*x + 1/2*c)^4 - 240*I*
A*a^2*tan(1/2*d*x + 1/2*c)^3 - 216*B*a^2*tan(1/2*d*x + 1/2*c)^3 - 60*A*a^2*tan(1/2*d*x + 1/2*c)^2 + 48*I*B*a^2
*tan(1/2*d*x + 1/2*c)^2 + 16*I*A*a^2*tan(1/2*d*x + 1/2*c) + 8*B*a^2*tan(1/2*d*x + 1/2*c) + 3*A*a^2)/tan(1/2*d*
x + 1/2*c)^4)/d