3.22 \(\int \cot ^3(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=123 \[ -\frac {a^3 (4 A-3 i B) \log (\sin (c+d x))}{d}-\frac {(B+2 i A) \cot (c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{d}-4 a^3 x (B+i A)+\frac {i a^3 B \log (\cos (c+d x))}{d}-\frac {a A \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.32, antiderivative size = 123, 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, 3589, 3475, 3531} \[ -\frac {a^3 (4 A-3 i B) \log (\sin (c+d x))}{d}-\frac {(B+2 i A) \cot (c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{d}-4 a^3 x (B+i A)+\frac {i a^3 B \log (\cos (c+d x))}{d}-\frac {a A \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^3*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]

[Out]

-4*a^3*(I*A + B)*x + (I*a^3*B*Log[Cos[c + d*x]])/d - (a^3*(4*A - (3*I)*B)*Log[Sin[c + d*x]])/d - (a*A*Cot[c +
d*x]^2*(a + I*a*Tan[c + d*x])^2)/(2*d) - (((2*I)*A + B)*Cot[c + d*x]*(a^3 + I*a^3*Tan[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 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 ^3(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx &=-\frac {a A \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d}+\frac {1}{2} \int \cot ^2(c+d x) (a+i a \tan (c+d x))^2 (2 a (2 i A+B)+2 i a B \tan (c+d x)) \, dx\\ &=-\frac {a A \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d}-\frac {(2 i A+B) \cot (c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{d}+\frac {1}{2} \int \cot (c+d x) (a+i a \tan (c+d x)) \left (-2 a^2 (4 A-3 i B)-2 a^2 B \tan (c+d x)\right ) \, dx\\ &=-\frac {a A \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d}-\frac {(2 i A+B) \cot (c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{d}+\frac {1}{2} \int \cot (c+d x) \left (-2 a^3 (4 A-3 i B)-8 a^3 (i A+B) \tan (c+d x)\right ) \, dx-\left (i a^3 B\right ) \int \tan (c+d x) \, dx\\ &=-4 a^3 (i A+B) x+\frac {i a^3 B \log (\cos (c+d x))}{d}-\frac {a A \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d}-\frac {(2 i A+B) \cot (c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{d}-\left (a^3 (4 A-3 i B)\right ) \int \cot (c+d x) \, dx\\ &=-4 a^3 (i A+B) x+\frac {i a^3 B \log (\cos (c+d x))}{d}-\frac {a^3 (4 A-3 i B) \log (\sin (c+d x))}{d}-\frac {a A \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d}-\frac {(2 i A+B) \cot (c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{d}\\ \end {align*}

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Mathematica [B]  time = 9.45, size = 1010, normalized size = 8.21 \[ a^3 \left (\frac {x (\cot (c+d x)+i)^3 (B+A \cot (c+d x)) \left (-16 i A \cos ^3(c)-\frac {25}{2} B \cos ^3(c)+4 A \cot (c) \cos ^3(c)-3 i B \cot (c) \cos ^3(c)-24 A \sin (c) \cos ^2(c)+20 i B \sin (c) \cos ^2(c)+16 i A \sin ^2(c) \cos (c)+15 B \sin ^2(c) \cos (c)+\frac {1}{2} B \cos (c)+4 A \sin ^3(c)-5 i B \sin ^3(c)-i B \sin (c)+(2 \cos (2 c) A+2 A-i B-2 i B \cos (2 c)) \csc (c) \sec (c) (i \sin (3 c)-\cos (3 c))-\frac {1}{2} B \sin ^3(c) \tan (c)-\frac {1}{2} B \sin (c) \tan (c)\right ) \sin ^4(c+d x)}{(\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}+\frac {i B \cos (3 c) (\cot (c+d x)+i)^3 (B+A \cot (c+d x)) \log \left (\cos ^2(c+d x)\right ) \sin ^4(c+d x)}{2 d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}+\frac {(\cot (c+d x)+i)^3 (B+A \cot (c+d x)) \left (4 A \cos \left (\frac {3 c}{2}\right )-3 i B \cos \left (\frac {3 c}{2}\right )-4 i A \sin \left (\frac {3 c}{2}\right )-3 B \sin \left (\frac {3 c}{2}\right )\right ) \left (i \tan ^{-1}(\tan (4 c+d x)) \cos \left (\frac {3 c}{2}\right )+\tan ^{-1}(\tan (4 c+d x)) \sin \left (\frac {3 c}{2}\right )\right ) \sin ^4(c+d x)}{d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}+\frac {(\cot (c+d x)+i)^3 (B+A \cot (c+d x)) \left (4 A \cos \left (\frac {3 c}{2}\right )-3 i B \cos \left (\frac {3 c}{2}\right )-4 i A \sin \left (\frac {3 c}{2}\right )-3 B \sin \left (\frac {3 c}{2}\right )\right ) \left (\frac {1}{2} i \log \left (\sin ^2(c+d x)\right ) \sin \left (\frac {3 c}{2}\right )-\frac {1}{2} \cos \left (\frac {3 c}{2}\right ) \log \left (\sin ^2(c+d x)\right )\right ) \sin ^4(c+d x)}{d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}+\frac {B (\cot (c+d x)+i)^3 (B+A \cot (c+d x)) \log \left (\cos ^2(c+d x)\right ) \sin (3 c) \sin ^4(c+d x)}{2 d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}+\frac {(A-i B) (\cot (c+d x)+i)^3 (B+A \cot (c+d x)) (-4 i d x \cos (3 c)-4 d x \sin (3 c)) \sin ^4(c+d x)}{d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}+\frac {(\cot (c+d x)+i)^3 (B+A \cot (c+d x)) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \left (\frac {1}{2} \cos (3 c)-\frac {1}{2} i \sin (3 c)\right ) (3 i A \sin (d x)+B \sin (d x)) \sin ^3(c+d x)}{d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}+\frac {(\cot (c+d x)+i)^3 (B+A \cot (c+d x)) \left (\frac {1}{2} i A \sin (3 c)-\frac {1}{2} A \cos (3 c)\right ) \sin ^2(c+d x)}{d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^3*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]

[Out]

a^3*(((I + Cot[c + d*x])^3*(B + A*Cot[c + d*x])*(-1/2*(A*Cos[3*c]) + (I/2)*A*Sin[3*c])*Sin[c + d*x]^2)/(d*(Cos
[d*x] + I*Sin[d*x])^3*(A*Cos[c + d*x] + B*Sin[c + d*x])) + ((I + Cot[c + d*x])^3*(B + A*Cot[c + d*x])*Csc[c/2]
*Sec[c/2]*(Cos[3*c]/2 - (I/2)*Sin[3*c])*((3*I)*A*Sin[d*x] + B*Sin[d*x])*Sin[c + d*x]^3)/(d*(Cos[d*x] + I*Sin[d
*x])^3*(A*Cos[c + d*x] + B*Sin[c + d*x])) + ((I/2)*B*Cos[3*c]*(I + Cot[c + d*x])^3*(B + A*Cot[c + d*x])*Log[Co
s[c + d*x]^2]*Sin[c + d*x]^4)/(d*(Cos[d*x] + I*Sin[d*x])^3*(A*Cos[c + d*x] + B*Sin[c + d*x])) + ((I + Cot[c +
d*x])^3*(B + A*Cot[c + d*x])*(4*A*Cos[(3*c)/2] - (3*I)*B*Cos[(3*c)/2] - (4*I)*A*Sin[(3*c)/2] - 3*B*Sin[(3*c)/2
])*(I*ArcTan[Tan[4*c + d*x]]*Cos[(3*c)/2] + ArcTan[Tan[4*c + d*x]]*Sin[(3*c)/2])*Sin[c + d*x]^4)/(d*(Cos[d*x]
+ I*Sin[d*x])^3*(A*Cos[c + d*x] + B*Sin[c + d*x])) + ((I + Cot[c + d*x])^3*(B + A*Cot[c + d*x])*(4*A*Cos[(3*c)
/2] - (3*I)*B*Cos[(3*c)/2] - (4*I)*A*Sin[(3*c)/2] - 3*B*Sin[(3*c)/2])*(-1/2*(Cos[(3*c)/2]*Log[Sin[c + d*x]^2])
+ (I/2)*Log[Sin[c + d*x]^2]*Sin[(3*c)/2])*Sin[c + d*x]^4)/(d*(Cos[d*x] + I*Sin[d*x])^3*(A*Cos[c + d*x] + B*Si
n[c + d*x])) + (B*(I + Cot[c + d*x])^3*(B + A*Cot[c + d*x])*Log[Cos[c + d*x]^2]*Sin[3*c]*Sin[c + d*x]^4)/(2*d*
(Cos[d*x] + I*Sin[d*x])^3*(A*Cos[c + d*x] + B*Sin[c + d*x])) + ((A - I*B)*(I + Cot[c + d*x])^3*(B + A*Cot[c +
d*x])*((-4*I)*d*x*Cos[3*c] - 4*d*x*Sin[3*c])*Sin[c + d*x]^4)/(d*(Cos[d*x] + I*Sin[d*x])^3*(A*Cos[c + d*x] + B*
Sin[c + d*x])) + (x*(I + Cot[c + d*x])^3*(B + A*Cot[c + d*x])*Sin[c + d*x]^4*((B*Cos[c])/2 - (16*I)*A*Cos[c]^3
- (25*B*Cos[c]^3)/2 + 4*A*Cos[c]^3*Cot[c] - (3*I)*B*Cos[c]^3*Cot[c] - I*B*Sin[c] - 24*A*Cos[c]^2*Sin[c] + (20
*I)*B*Cos[c]^2*Sin[c] + (16*I)*A*Cos[c]*Sin[c]^2 + 15*B*Cos[c]*Sin[c]^2 + 4*A*Sin[c]^3 - (5*I)*B*Sin[c]^3 + (2
*A - I*B + 2*A*Cos[2*c] - (2*I)*B*Cos[2*c])*Csc[c]*Sec[c]*(-Cos[3*c] + I*Sin[3*c]) - (B*Sin[c]*Tan[c])/2 - (B*
Sin[c]^3*Tan[c])/2))/((Cos[d*x] + I*Sin[d*x])^3*(A*Cos[c + d*x] + B*Sin[c + d*x])))

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fricas [A]  time = 0.46, size = 179, normalized size = 1.46 \[ \frac {2 \, {\left (4 \, A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 2 \, {\left (3 \, A - i \, B\right )} a^{3} + {\left (i \, B a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 2 i \, B a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, B a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - {\left ({\left (4 \, A - 3 i \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, {\left (4 \, A - 3 i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (4 \, A - 3 i \, B\right )} a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}{d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 2.71, size = 223, normalized size = 1.81 \[ -\frac {A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 i \, B a^{3} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - 8 i \, B a^{3} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right ) - 12 i \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 64 \, {\left (A a^{3} - i \, B a^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right ) + 8 \, {\left (4 \, A a^{3} - 3 i \, B a^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - \frac {48 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 36 i \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 i \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - A a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

-1/8*(A*a^3*tan(1/2*d*x + 1/2*c)^2 - 8*I*B*a^3*log(tan(1/2*d*x + 1/2*c) + 1) - 8*I*B*a^3*log(tan(1/2*d*x + 1/2
*c) - 1) - 12*I*A*a^3*tan(1/2*d*x + 1/2*c) - 4*B*a^3*tan(1/2*d*x + 1/2*c) - 64*(A*a^3 - I*B*a^3)*log(tan(1/2*d
*x + 1/2*c) + I) + 8*(4*A*a^3 - 3*I*B*a^3)*log(tan(1/2*d*x + 1/2*c)) - (48*A*a^3*tan(1/2*d*x + 1/2*c)^2 - 36*I
*B*a^3*tan(1/2*d*x + 1/2*c)^2 - 12*I*A*a^3*tan(1/2*d*x + 1/2*c) - 4*B*a^3*tan(1/2*d*x + 1/2*c) - A*a^3)/tan(1/
2*d*x + 1/2*c)^2)/d

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maple [A]  time = 0.50, size = 136, normalized size = 1.11 \[ -4 i A x \,a^{3}-\frac {4 i A \,a^{3} c}{d}+\frac {i a^{3} B \ln \left (\cos \left (d x +c \right )\right )}{d}-\frac {4 a^{3} A \ln \left (\sin \left (d x +c \right )\right )}{d}-4 a^{3} B x -\frac {4 a^{3} B c}{d}-\frac {3 i A \cot \left (d x +c \right ) a^{3}}{d}+\frac {3 i a^{3} B \ln \left (\sin \left (d x +c \right )\right )}{d}-\frac {A \,a^{3} \left (\cot ^{2}\left (d x +c \right )\right )}{2 d}-\frac {B \cot \left (d x +c \right ) a^{3}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x)

[Out]

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

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maxima [A]  time = 0.63, size = 96, normalized size = 0.78 \[ \frac {2 \, {\left (d x + c\right )} {\left (-4 i \, A - 4 \, B\right )} a^{3} + 4 \, {\left (A - i \, B\right )} a^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 2 \, {\left (4 \, A - 3 i \, B\right )} a^{3} \log \left (\tan \left (d x + c\right )\right ) - \frac {{\left (6 i \, A + 2 \, B\right )} a^{3} \tan \left (d x + c\right ) + A a^{3}}{\tan \left (d x + c\right )^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 6.44, size = 88, normalized size = 0.72 \[ -\frac {\frac {A\,a^3}{2}+\mathrm {tan}\left (c+d\,x\right )\,\left (B\,a^3+A\,a^3\,3{}\mathrm {i}\right )}{d\,{\mathrm {tan}\left (c+d\,x\right )}^2}-\frac {a^3\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (4\,A-B\,3{}\mathrm {i}\right )}{d}+\frac {4\,a^3\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^3*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^3,x)

[Out]

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

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sympy [B]  time = 2.55, size = 233, normalized size = 1.89 \[ \frac {i B a^{3} \log {\left (\frac {2 A a^{3} - i B a^{3}}{2 A a^{3} e^{2 i c} - i B a^{3} e^{2 i c}} + e^{2 i d x} \right )}}{d} - \frac {a^{3} \left (4 A - 3 i B\right ) \log {\left (e^{2 i d x} + \frac {2 A a^{3} - 2 i B a^{3} - a^{3} \left (4 A - 3 i B\right )}{2 A a^{3} e^{2 i c} - i B a^{3} e^{2 i c}} \right )}}{d} + \frac {6 i A a^{3} + 2 B a^{3} + \left (- 8 i A a^{3} e^{2 i c} - 2 B a^{3} e^{2 i c}\right ) e^{2 i d x}}{- i d e^{4 i c} e^{4 i d x} + 2 i d e^{2 i c} e^{2 i d x} - i d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**3*(a+I*a*tan(d*x+c))**3*(A+B*tan(d*x+c)),x)

[Out]

I*B*a**3*log((2*A*a**3 - I*B*a**3)/(2*A*a**3*exp(2*I*c) - I*B*a**3*exp(2*I*c)) + exp(2*I*d*x))/d - a**3*(4*A -
3*I*B)*log(exp(2*I*d*x) + (2*A*a**3 - 2*I*B*a**3 - a**3*(4*A - 3*I*B))/(2*A*a**3*exp(2*I*c) - I*B*a**3*exp(2*
I*c)))/d + (6*I*A*a**3 + 2*B*a**3 + (-8*I*A*a**3*exp(2*I*c) - 2*B*a**3*exp(2*I*c))*exp(2*I*d*x))/(-I*d*exp(4*I
*c)*exp(4*I*d*x) + 2*I*d*exp(2*I*c)*exp(2*I*d*x) - I*d)

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