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

Optimal. Leaf size=163 \[ -\frac{a^4 (7 B+8 i A) \log (\sin (c+d x))}{d}-\frac{(B+2 i A) \cot ^2(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}+\frac{(4 A-3 i B) \cot (c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{d}+8 a^4 x (A-i B)-\frac{a^4 B \log (\cos (c+d x))}{d}-\frac{a A \cot ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.452709, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {3593, 3589, 3475, 3531} \[ -\frac{a^4 (7 B+8 i A) \log (\sin (c+d x))}{d}-\frac{(B+2 i A) \cot ^2(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}+\frac{(4 A-3 i B) \cot (c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{d}+8 a^4 x (A-i B)-\frac{a^4 B \log (\cos (c+d x))}{d}-\frac{a A \cot ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

Mathematica [B]  time = 9.59529, size = 1138, normalized size = 6.98 \[ a^4 \left (\frac{x (\cot (c+d x)+i)^4 (B+A \cot (c+d x)) \left (40 A \cos ^4(c)-\frac{71}{2} i B \cos ^4(c)+8 i A \cot (c) \cos ^4(c)+7 B \cot (c) \cos ^4(c)-80 i A \sin (c) \cos ^3(c)-\frac{145}{2} B \sin (c) \cos ^3(c)-80 A \sin ^2(c) \cos ^2(c)+75 i B \sin ^2(c) \cos ^2(c)+\frac{1}{2} i B \cos ^2(c)+40 i A \sin ^3(c) \cos (c)+40 B \sin ^3(c) \cos (c)+\frac{3}{2} B \sin (c) \cos (c)+8 A \sin ^4(c)-\frac{19}{2} i B \sin ^4(c)-\frac{3}{2} i B \sin ^2(c)-i (4 \cos (2 c) A+4 A-3 i B-4 i B \cos (2 c)) \csc (c) \sec (c) (\cos (4 c)-i \sin (4 c))-\frac{1}{2} B \sin ^4(c) \tan (c)-\frac{1}{2} B \sin ^2(c) \tan (c)\right ) \sin ^5(c+d x)}{(\cos (d x)+i \sin (d x))^4 (A \cos (c+d x)+B \sin (c+d x))}-\frac{B \cos (4 c) (\cot (c+d x)+i)^4 (B+A \cot (c+d x)) \log \left (\cos ^2(c+d x)\right ) \sin ^5(c+d x)}{2 d (\cos (d x)+i \sin (d x))^4 (A \cos (c+d x)+B \sin (c+d x))}+\frac{(\cot (c+d x)+i)^4 (B+A \cot (c+d x)) (8 A \cos (2 c)-7 i B \cos (2 c)-8 i A \sin (2 c)-7 B \sin (2 c)) \left (i \tan ^{-1}(\tan (5 c+d x)) \sin (2 c)-\tan ^{-1}(\tan (5 c+d x)) \cos (2 c)\right ) \sin ^5(c+d x)}{d (\cos (d x)+i \sin (d x))^4 (A \cos (c+d x)+B \sin (c+d x))}+\frac{(\cot (c+d x)+i)^4 (B+A \cot (c+d x)) (8 A \cos (2 c)-7 i B \cos (2 c)-8 i A \sin (2 c)-7 B \sin (2 c)) \left (-\frac{1}{2} i \cos (2 c) \log \left (\sin ^2(c+d x)\right )-\frac{1}{2} \sin (2 c) \log \left (\sin ^2(c+d x)\right )\right ) \sin ^5(c+d x)}{d (\cos (d x)+i \sin (d x))^4 (A \cos (c+d x)+B \sin (c+d x))}+\frac{i B (\cot (c+d x)+i)^4 (B+A \cot (c+d x)) \log \left (\cos ^2(c+d x)\right ) \sin (4 c) \sin ^5(c+d x)}{2 d (\cos (d x)+i \sin (d x))^4 (A \cos (c+d x)+B \sin (c+d x))}+\frac{(A-i B) (\cot (c+d x)+i)^4 (B+A \cot (c+d x)) (8 d x \cos (4 c)-8 i d x \sin (4 c)) \sin ^5(c+d x)}{d (\cos (d x)+i \sin (d x))^4 (A \cos (c+d x)+B \sin (c+d x))}+\frac{(\cot (c+d x)+i)^4 (B+A \cot (c+d x)) \csc (c) \left (\frac{2}{3} i \sin (4 c)-\frac{2}{3} \cos (4 c)\right ) (11 A \sin (d x)-6 i B \sin (d x)) \sin ^4(c+d x)}{d (\cos (d x)+i \sin (d x))^4 (A \cos (c+d x)+B \sin (c+d x))}+\frac{(\cot (c+d x)+i)^4 (B+A \cot (c+d x)) \csc (c) (-2 A \cos (c)-12 i A \sin (c)-3 B \sin (c)) \left (\frac{1}{6} \cos (4 c)-\frac{1}{6} i \sin (4 c)\right ) \sin ^3(c+d x)}{d (\cos (d x)+i \sin (d x))^4 (A \cos (c+d x)+B \sin (c+d x))}+\frac{A (\cot (c+d x)+i)^4 (B+A \cot (c+d x)) \csc (c) \left (\frac{1}{3} \cos (4 c)-\frac{1}{3} i \sin (4 c)\right ) \sin (d x) \sin ^2(c+d x)}{d (\cos (d x)+i \sin (d x))^4 (A \cos (c+d x)+B \sin (c+d x))}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^4*((A*(I + Cot[c + d*x])^4*(B + A*Cot[c + d*x])*Csc[c]*(Cos[4*c]/3 - (I/3)*Sin[4*c])*Sin[d*x]*Sin[c + d*x]^2
)/(d*(Cos[d*x] + I*Sin[d*x])^4*(A*Cos[c + d*x] + B*Sin[c + d*x])) + ((I + Cot[c + d*x])^4*(B + A*Cot[c + d*x])
*Csc[c]*(-2*A*Cos[c] - (12*I)*A*Sin[c] - 3*B*Sin[c])*(Cos[4*c]/6 - (I/6)*Sin[4*c])*Sin[c + d*x]^3)/(d*(Cos[d*x
] + I*Sin[d*x])^4*(A*Cos[c + d*x] + B*Sin[c + d*x])) + ((I + Cot[c + d*x])^4*(B + A*Cot[c + d*x])*Csc[c]*((-2*
Cos[4*c])/3 + ((2*I)/3)*Sin[4*c])*(11*A*Sin[d*x] - (6*I)*B*Sin[d*x])*Sin[c + d*x]^4)/(d*(Cos[d*x] + I*Sin[d*x]
)^4*(A*Cos[c + d*x] + B*Sin[c + d*x])) - (B*Cos[4*c]*(I + Cot[c + d*x])^4*(B + A*Cot[c + d*x])*Log[Cos[c + d*x
]^2]*Sin[c + d*x]^5)/(2*d*(Cos[d*x] + I*Sin[d*x])^4*(A*Cos[c + d*x] + B*Sin[c + d*x])) + ((I + Cot[c + d*x])^4
*(B + A*Cot[c + d*x])*(8*A*Cos[2*c] - (7*I)*B*Cos[2*c] - (8*I)*A*Sin[2*c] - 7*B*Sin[2*c])*(-(ArcTan[Tan[5*c +
d*x]]*Cos[2*c]) + I*ArcTan[Tan[5*c + d*x]]*Sin[2*c])*Sin[c + d*x]^5)/(d*(Cos[d*x] + I*Sin[d*x])^4*(A*Cos[c + d
*x] + B*Sin[c + d*x])) + ((I + Cot[c + d*x])^4*(B + A*Cot[c + d*x])*(8*A*Cos[2*c] - (7*I)*B*Cos[2*c] - (8*I)*A
*Sin[2*c] - 7*B*Sin[2*c])*((-I/2)*Cos[2*c]*Log[Sin[c + d*x]^2] - (Log[Sin[c + d*x]^2]*Sin[2*c])/2)*Sin[c + d*x
]^5)/(d*(Cos[d*x] + I*Sin[d*x])^4*(A*Cos[c + d*x] + B*Sin[c + d*x])) + ((I/2)*B*(I + Cot[c + d*x])^4*(B + A*Co
t[c + d*x])*Log[Cos[c + d*x]^2]*Sin[4*c]*Sin[c + d*x]^5)/(d*(Cos[d*x] + I*Sin[d*x])^4*(A*Cos[c + d*x] + B*Sin[
c + d*x])) + ((A - I*B)*(I + Cot[c + d*x])^4*(B + A*Cot[c + d*x])*(8*d*x*Cos[4*c] - (8*I)*d*x*Sin[4*c])*Sin[c
+ d*x]^5)/(d*(Cos[d*x] + I*Sin[d*x])^4*(A*Cos[c + d*x] + B*Sin[c + d*x])) + (x*(I + Cot[c + d*x])^4*(B + A*Cot
[c + d*x])*Sin[c + d*x]^5*((I/2)*B*Cos[c]^2 + 40*A*Cos[c]^4 - ((71*I)/2)*B*Cos[c]^4 + (8*I)*A*Cos[c]^4*Cot[c]
+ 7*B*Cos[c]^4*Cot[c] + (3*B*Cos[c]*Sin[c])/2 - (80*I)*A*Cos[c]^3*Sin[c] - (145*B*Cos[c]^3*Sin[c])/2 - ((3*I)/
2)*B*Sin[c]^2 - 80*A*Cos[c]^2*Sin[c]^2 + (75*I)*B*Cos[c]^2*Sin[c]^2 + (40*I)*A*Cos[c]*Sin[c]^3 + 40*B*Cos[c]*S
in[c]^3 + 8*A*Sin[c]^4 - ((19*I)/2)*B*Sin[c]^4 - I*(4*A - (3*I)*B + 4*A*Cos[2*c] - (4*I)*B*Cos[2*c])*Csc[c]*Se
c[c]*(Cos[4*c] - I*Sin[4*c]) - (B*Sin[c]^2*Tan[c])/2 - (B*Sin[c]^4*Tan[c])/2))/((Cos[d*x] + I*Sin[d*x])^4*(A*C
os[c + d*x] + B*Sin[c + d*x])))

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Maple [A]  time = 0.074, size = 170, normalized size = 1. \begin{align*} 8\,A{a}^{4}x+8\,{\frac{A{a}^{4}c}{d}}-{\frac{B{a}^{4}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}-8\,iBx{a}^{4}-{\frac{8\,iB{a}^{4}c}{d}}-{\frac{2\,iA{a}^{4} \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{d}}+7\,{\frac{A\cot \left ( dx+c \right ){a}^{4}}{d}}-7\,{\frac{B{a}^{4}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{8\,iA{a}^{4}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{4\,iB\cot \left ( dx+c \right ){a}^{4}}{d}}-{\frac{A{a}^{4} \left ( \cot \left ( dx+c \right ) \right ) ^{3}}{3\,d}}-{\frac{B{a}^{4} \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 2.37657, size = 159, normalized size = 0.98 \begin{align*} \frac{6 \,{\left (d x + c\right )}{\left (8 \, A - 8 i \, B\right )} a^{4} - 24 \,{\left (-i \, A - B\right )} a^{4} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 6 \,{\left (-8 i \, A - 7 \, B\right )} a^{4} \log \left (\tan \left (d x + c\right )\right ) + \frac{{\left (42 \, A - 24 i \, B\right )} a^{4} \tan \left (d x + c\right )^{2} + 3 \,{\left (-4 i \, A - B\right )} a^{4} \tan \left (d x + c\right ) - 2 \, A a^{4}}{\tan \left (d x + c\right )^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*(d*x + c)*(8*A - 8*I*B)*a^4 - 24*(-I*A - B)*a^4*log(tan(d*x + c)^2 + 1) + 6*(-8*I*A - 7*B)*a^4*log(tan(
d*x + c)) + ((42*A - 24*I*B)*a^4*tan(d*x + c)^2 + 3*(-4*I*A - B)*a^4*tan(d*x + c) - 2*A*a^4)/tan(d*x + c)^3)/d

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Fricas [A]  time = 1.53453, size = 679, normalized size = 4.17 \begin{align*} \frac{{\left (72 i \, A + 30 \, B\right )} a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (-108 i \, A - 54 \, B\right )} a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (44 i \, A + 24 \, B\right )} a^{4} - 3 \,{\left (B a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, B a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, B a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} - B a^{4}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) +{\left ({\left (-24 i \, A - 21 \, B\right )} a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (72 i \, A + 63 \, B\right )} a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (-72 i \, A - 63 \, B\right )} a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (24 i \, A + 21 \, B\right )} 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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*((72*I*A + 30*B)*a^4*e^(4*I*d*x + 4*I*c) + (-108*I*A - 54*B)*a^4*e^(2*I*d*x + 2*I*c) + (44*I*A + 24*B)*a^4
- 3*(B*a^4*e^(6*I*d*x + 6*I*c) - 3*B*a^4*e^(4*I*d*x + 4*I*c) + 3*B*a^4*e^(2*I*d*x + 2*I*c) - B*a^4)*log(e^(2*
I*d*x + 2*I*c) + 1) + ((-24*I*A - 21*B)*a^4*e^(6*I*d*x + 6*I*c) + (72*I*A + 63*B)*a^4*e^(4*I*d*x + 4*I*c) + (-
72*I*A - 63*B)*a^4*e^(2*I*d*x + 2*I*c) + (24*I*A + 21*B)*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)

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Sympy [A]  time = 28.4718, size = 262, normalized size = 1.61 \begin{align*} \frac{\frac{\left (24 i A a^{4} + 10 B a^{4}\right ) e^{- 2 i c} e^{4 i d x}}{d} - \frac{\left (36 i A a^{4} + 18 B a^{4}\right ) e^{- 4 i c} e^{2 i d x}}{d} + \frac{\left (44 i A a^{4} + 24 B a^{4}\right ) e^{- 6 i c}}{3 d}}{e^{6 i d x} - 3 e^{- 2 i c} e^{4 i d x} + 3 e^{- 4 i c} e^{2 i d x} - e^{- 6 i c}} + \operatorname{RootSum}{\left (z^{2} d^{2} + z \left (8 i A a^{4} d + 8 B a^{4} d\right ) + 8 i A B a^{8} + 7 B^{2} a^{8}, \left ( i \mapsto i \log{\left (- \frac{i i d}{4 A a^{4} e^{2 i c} - 3 i B a^{4} e^{2 i c}} + \frac{4 A - 4 i B}{4 A e^{2 i c} - 3 i B e^{2 i c}} + e^{2 i d x} \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((24*I*A*a**4 + 10*B*a**4)*exp(-2*I*c)*exp(4*I*d*x)/d - (36*I*A*a**4 + 18*B*a**4)*exp(-4*I*c)*exp(2*I*d*x)/d +
(44*I*A*a**4 + 24*B*a**4)*exp(-6*I*c)/(3*d))/(exp(6*I*d*x) - 3*exp(-2*I*c)*exp(4*I*d*x) + 3*exp(-4*I*c)*exp(2
*I*d*x) - exp(-6*I*c)) + RootSum(_z**2*d**2 + _z*(8*I*A*a**4*d + 8*B*a**4*d) + 8*I*A*B*a**8 + 7*B**2*a**8, Lam
bda(_i, _i*log(-_i*I*d/(4*A*a**4*exp(2*I*c) - 3*I*B*a**4*exp(2*I*c)) + (4*A - 4*I*B)/(4*A*exp(2*I*c) - 3*I*B*e
xp(2*I*c)) + exp(2*I*d*x))))

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Giac [B]  time = 1.90507, size = 397, normalized size = 2.44 \begin{align*} \frac{A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 i \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 3 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 24 \, B a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 24 \, B a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - 87 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 48 i \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 384 \,{\left (-i \, A a^{4} - B a^{4}\right )} \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + i\right ) - 24 \,{\left (8 i \, A a^{4} + 7 \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - \frac{-352 i \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 308 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 87 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 48 i \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 12 i \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + A a^{4}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(A*a^4*tan(1/2*d*x + 1/2*c)^3 - 12*I*A*a^4*tan(1/2*d*x + 1/2*c)^2 - 3*B*a^4*tan(1/2*d*x + 1/2*c)^2 - 24*B
*a^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 24*B*a^4*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 87*A*a^4*tan(1/2*d*x +
1/2*c) + 48*I*B*a^4*tan(1/2*d*x + 1/2*c) - 384*(-I*A*a^4 - B*a^4)*log(tan(1/2*d*x + 1/2*c) + I) - 24*(8*I*A*a
^4 + 7*B*a^4)*log(abs(tan(1/2*d*x + 1/2*c))) - (-352*I*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 308*B*a^4*tan(1/2*d*x +
1/2*c)^3 - 87*A*a^4*tan(1/2*d*x + 1/2*c)^2 + 48*I*B*a^4*tan(1/2*d*x + 1/2*c)^2 + 12*I*A*a^4*tan(1/2*d*x + 1/2*
c) + 3*B*a^4*tan(1/2*d*x + 1/2*c) + A*a^4)/tan(1/2*d*x + 1/2*c)^3)/d