3.50 \(\int \frac{\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=170 \[ -\frac{(2 A+i B) \cot ^2(c+d x)}{a^2 d}+\frac{3 (-3 B+5 i A) \cot (c+d x)}{4 a^2 d}-\frac{2 (2 A+i B) \log (\sin (c+d x))}{a^2 d}+\frac{(5 A+3 i B) \cot ^2(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac{3 x (-3 B+5 i A)}{4 a^2}+\frac{(A+i B) \cot ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

[Out]

(3*((5*I)*A - 3*B)*x)/(4*a^2) + (3*((5*I)*A - 3*B)*Cot[c + d*x])/(4*a^2*d) - ((2*A + I*B)*Cot[c + d*x]^2)/(a^2
*d) - (2*(2*A + I*B)*Log[Sin[c + d*x]])/(a^2*d) + ((5*A + (3*I)*B)*Cot[c + d*x]^2)/(4*a^2*d*(1 + I*Tan[c + d*x
])) + ((A + I*B)*Cot[c + d*x]^2)/(4*d*(a + I*a*Tan[c + d*x])^2)

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Rubi [A]  time = 0.404879, antiderivative size = 170, normalized size of antiderivative = 1., 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 = {3596, 3529, 3531, 3475} \[ -\frac{(2 A+i B) \cot ^2(c+d x)}{a^2 d}+\frac{3 (-3 B+5 i A) \cot (c+d x)}{4 a^2 d}-\frac{2 (2 A+i B) \log (\sin (c+d x))}{a^2 d}+\frac{(5 A+3 i B) \cot ^2(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac{3 x (-3 B+5 i A)}{4 a^2}+\frac{(A+i B) \cot ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[(Cot[c + d*x]^3*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^2,x]

[Out]

(3*((5*I)*A - 3*B)*x)/(4*a^2) + (3*((5*I)*A - 3*B)*Cot[c + d*x])/(4*a^2*d) - ((2*A + I*B)*Cot[c + d*x]^2)/(a^2
*d) - (2*(2*A + I*B)*Log[Sin[c + d*x]])/(a^2*d) + ((5*A + (3*I)*B)*Cot[c + d*x]^2)/(4*a^2*d*(1 + I*Tan[c + d*x
])) + ((A + I*B)*Cot[c + d*x]^2)/(4*d*(a + I*a*Tan[c + d*x])^2)

Rule 3596

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*A + b*B)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(2
*f*m*(b*c - a*d)), x] + Dist[1/(2*a*m*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Si
mp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*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] && LtQ[m, 0] &&  !GtQ[n,
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 \frac{\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx &=\frac{(A+i B) \cot ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{\int \frac{\cot ^3(c+d x) (2 a (3 A+i B)-4 a (i A-B) \tan (c+d x))}{a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=\frac{(5 A+3 i B) \cot ^2(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac{(A+i B) \cot ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{\int \cot ^3(c+d x) \left (16 a^2 (2 A+i B)-6 a^2 (5 i A-3 B) \tan (c+d x)\right ) \, dx}{8 a^4}\\ &=-\frac{(2 A+i B) \cot ^2(c+d x)}{a^2 d}+\frac{(5 A+3 i B) \cot ^2(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac{(A+i B) \cot ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{\int \cot ^2(c+d x) \left (-6 a^2 (5 i A-3 B)-16 a^2 (2 A+i B) \tan (c+d x)\right ) \, dx}{8 a^4}\\ &=\frac{3 (5 i A-3 B) \cot (c+d x)}{4 a^2 d}-\frac{(2 A+i B) \cot ^2(c+d x)}{a^2 d}+\frac{(5 A+3 i B) \cot ^2(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac{(A+i B) \cot ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{\int \cot (c+d x) \left (-16 a^2 (2 A+i B)+6 a^2 (5 i A-3 B) \tan (c+d x)\right ) \, dx}{8 a^4}\\ &=\frac{3 (5 i A-3 B) x}{4 a^2}+\frac{3 (5 i A-3 B) \cot (c+d x)}{4 a^2 d}-\frac{(2 A+i B) \cot ^2(c+d x)}{a^2 d}+\frac{(5 A+3 i B) \cot ^2(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac{(A+i B) \cot ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac{(2 (2 A+i B)) \int \cot (c+d x) \, dx}{a^2}\\ &=\frac{3 (5 i A-3 B) x}{4 a^2}+\frac{3 (5 i A-3 B) \cot (c+d x)}{4 a^2 d}-\frac{(2 A+i B) \cot ^2(c+d x)}{a^2 d}-\frac{2 (2 A+i B) \log (\sin (c+d x))}{a^2 d}+\frac{(5 A+3 i B) \cot ^2(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac{(A+i B) \cot ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}\\ \end{align*}

Mathematica [B]  time = 7.08443, size = 1112, normalized size = 6.54 \[ \frac{\csc (c) \csc (c+d x) \sec (c+d x) \left (A \cos (2 c-d x)+\frac{1}{2} i B \cos (2 c-d x)-A \cos (2 c+d x)-\frac{1}{2} i B \cos (2 c+d x)+i A \sin (2 c-d x)-\frac{1}{2} B \sin (2 c-d x)-i A \sin (2 c+d x)+\frac{1}{2} B \sin (2 c+d x)\right ) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^2}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^2}-\frac{(4 A+3 i B) \cos (2 d x) \sec (c+d x) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^2}{4 d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^2}+\frac{\sec (c+d x) (2 A \cos (c)+i B \cos (c)+2 i A \sin (c)-B \sin (c)) \left (2 i \tan ^{-1}(\tan (d x)) \cos (c)-2 \tan ^{-1}(\tan (d x)) \sin (c)\right ) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^2}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^2}+\frac{\sec (c+d x) (2 A \cos (c)+i B \cos (c)+2 i A \sin (c)-B \sin (c)) \left (-\cos (c) \log \left (\sin ^2(c+d x)\right )-i \sin (c) \log \left (\sin ^2(c+d x)\right )\right ) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^2}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^2}+\frac{x \sec (c+d x) (4 \cot (c) A+4 i A-2 B+2 i B \cot (c)+(2 A+i B) \cot (c) (-2 \cos (2 c)-2 i \sin (2 c))) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^2}{(A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^2}+\frac{(A+i B) \cos (4 d x) \sec (c+d x) \left (\frac{1}{16} i \sin (2 c)-\frac{1}{16} \cos (2 c)\right ) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^2}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^2}+\frac{\csc ^2(c+d x) \sec (c+d x) \left (-\frac{1}{2} A \cos (2 c)-\frac{1}{2} i A \sin (2 c)\right ) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^2}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^2}+\frac{(5 A+3 i B) \sec (c+d x) \left (\frac{3}{4} i d x \cos (2 c)-\frac{3}{4} d x \sin (2 c)\right ) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^2}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^2}+\frac{i (4 A+3 i B) \sec (c+d x) \sin (2 d x) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^2}{4 d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^2}+\frac{(A+i B) \sec (c+d x) \left (\frac{1}{16} i \cos (2 c)+\frac{1}{16} \sin (2 c)\right ) \sin (4 d x) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^2}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cot[c + d*x]^3*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^2,x]

[Out]

-((4*A + (3*I)*B)*Cos[2*d*x]*Sec[c + d*x]*(Cos[d*x] + I*Sin[d*x])^2*(A + B*Tan[c + d*x]))/(4*d*(A*Cos[c + d*x]
+ B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])^2) + (Sec[c + d*x]*(2*A*Cos[c] + I*B*Cos[c] + (2*I)*A*Sin[c] - B*Sin
[c])*((2*I)*ArcTan[Tan[d*x]]*Cos[c] - 2*ArcTan[Tan[d*x]]*Sin[c])*(Cos[d*x] + I*Sin[d*x])^2*(A + B*Tan[c + d*x]
))/(d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])^2) + (Sec[c + d*x]*(2*A*Cos[c] + I*B*Cos[c] + (
2*I)*A*Sin[c] - B*Sin[c])*(-(Cos[c]*Log[Sin[c + d*x]^2]) - I*Log[Sin[c + d*x]^2]*Sin[c])*(Cos[d*x] + I*Sin[d*x
])^2*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])^2) + (x*Sec[c + d*x]*((
4*I)*A - 2*B + 4*A*Cot[c] + (2*I)*B*Cot[c] + (2*A + I*B)*Cot[c]*(-2*Cos[2*c] - (2*I)*Sin[2*c]))*(Cos[d*x] + I*
Sin[d*x])^2*(A + B*Tan[c + d*x]))/((A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])^2) + ((A + I*B)*Co
s[4*d*x]*Sec[c + d*x]*(-Cos[2*c]/16 + (I/16)*Sin[2*c])*(Cos[d*x] + I*Sin[d*x])^2*(A + B*Tan[c + d*x]))/(d*(A*C
os[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])^2) + (Csc[c + d*x]^2*Sec[c + d*x]*(-(A*Cos[2*c])/2 - (I/2
)*A*Sin[2*c])*(Cos[d*x] + I*Sin[d*x])^2*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Ta
n[c + d*x])^2) + ((5*A + (3*I)*B)*Sec[c + d*x]*(((3*I)/4)*d*x*Cos[2*c] - (3*d*x*Sin[2*c])/4)*(Cos[d*x] + I*Sin
[d*x])^2*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])^2) + ((I/4)*(4*A +
(3*I)*B)*Sec[c + d*x]*(Cos[d*x] + I*Sin[d*x])^2*Sin[2*d*x]*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c
+ d*x])*(a + I*a*Tan[c + d*x])^2) + ((A + I*B)*Sec[c + d*x]*((I/16)*Cos[2*c] + Sin[2*c]/16)*(Cos[d*x] + I*Sin[
d*x])^2*Sin[4*d*x]*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])^2) + (Csc
[c]*Csc[c + d*x]*Sec[c + d*x]*(Cos[d*x] + I*Sin[d*x])^2*(A*Cos[2*c - d*x] + (I/2)*B*Cos[2*c - d*x] - A*Cos[2*c
+ d*x] - (I/2)*B*Cos[2*c + d*x] + I*A*Sin[2*c - d*x] - (B*Sin[2*c - d*x])/2 - I*A*Sin[2*c + d*x] + (B*Sin[2*c
+ d*x])/2)*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])^2)

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Maple [A]  time = 0.118, size = 247, normalized size = 1.5 \begin{align*}{\frac{A}{4\,{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}+{\frac{{\frac{i}{4}}B}{{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}-{\frac{5\,B}{4\,{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) }}+{\frac{{\frac{7\,i}{4}}A}{{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) }}+{\frac{31\,\ln \left ( \tan \left ( dx+c \right ) -i \right ) A}{8\,{a}^{2}d}}+{\frac{{\frac{17\,i}{8}}\ln \left ( \tan \left ( dx+c \right ) -i \right ) B}{{a}^{2}d}}+{\frac{A\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{8\,{a}^{2}d}}-{\frac{{\frac{i}{8}}B\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{{a}^{2}d}}-{\frac{A}{2\,{a}^{2}d \left ( \tan \left ( dx+c \right ) \right ) ^{2}}}-{\frac{2\,iB\ln \left ( \tan \left ( dx+c \right ) \right ) }{{a}^{2}d}}-4\,{\frac{A\ln \left ( \tan \left ( dx+c \right ) \right ) }{{a}^{2}d}}+{\frac{2\,iA}{{a}^{2}d\tan \left ( dx+c \right ) }}-{\frac{B}{{a}^{2}d\tan \left ( dx+c \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^2,x)

[Out]

1/4/d/a^2/(tan(d*x+c)-I)^2*A+1/4*I/d/a^2/(tan(d*x+c)-I)^2*B-5/4/d/a^2/(tan(d*x+c)-I)*B+7/4*I/d/a^2/(tan(d*x+c)
-I)*A+31/8/d/a^2*ln(tan(d*x+c)-I)*A+17/8*I/d/a^2*ln(tan(d*x+c)-I)*B+1/8/d/a^2*A*ln(tan(d*x+c)+I)-1/8*I/d/a^2*B
*ln(tan(d*x+c)+I)-1/2/d/a^2*A/tan(d*x+c)^2-2*I/d/a^2*B*ln(tan(d*x+c))-4/d/a^2*A*ln(tan(d*x+c))+2*I/d/a^2/tan(d
*x+c)*A-1/d/a^2/tan(d*x+c)*B

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.56098, size = 601, normalized size = 3.54 \begin{align*} \frac{{\left (124 i \, A - 68 \, B\right )} d x e^{\left (8 i \, d x + 8 i \, c\right )} +{\left ({\left (-248 i \, A + 136 \, B\right )} d x - 48 \, A - 44 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left ({\left (124 i \, A - 68 \, B\right )} d x + 95 \, A + 55 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \,{\left (7 \, A + 5 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 32 \,{\left ({\left (2 \, A + i \, B\right )} e^{\left (8 i \, d x + 8 i \, c\right )} - 2 \,{\left (2 \, A + i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (2 \, A + i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) - A - i \, B}{16 \,{\left (a^{2} d e^{\left (8 i \, d x + 8 i \, c\right )} - 2 \, a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/16*((124*I*A - 68*B)*d*x*e^(8*I*d*x + 8*I*c) + ((-248*I*A + 136*B)*d*x - 48*A - 44*I*B)*e^(6*I*d*x + 6*I*c)
+ ((124*I*A - 68*B)*d*x + 95*A + 55*I*B)*e^(4*I*d*x + 4*I*c) - 2*(7*A + 5*I*B)*e^(2*I*d*x + 2*I*c) - 32*((2*A
+ I*B)*e^(8*I*d*x + 8*I*c) - 2*(2*A + I*B)*e^(6*I*d*x + 6*I*c) + (2*A + I*B)*e^(4*I*d*x + 4*I*c))*log(e^(2*I*d
*x + 2*I*c) - 1) - A - I*B)/(a^2*d*e^(8*I*d*x + 8*I*c) - 2*a^2*d*e^(6*I*d*x + 6*I*c) + a^2*d*e^(4*I*d*x + 4*I*
c))

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Sympy [A]  time = 35.1293, size = 274, normalized size = 1.61 \begin{align*} \frac{- \frac{\left (2 A + 2 i B\right ) e^{- 2 i c} e^{2 i d x}}{a^{2} d} + \frac{\left (4 A + 2 i B\right ) e^{- 4 i c}}{a^{2} d}}{e^{4 i d x} - 2 e^{- 2 i c} e^{2 i d x} + e^{- 4 i c}} + \frac{\left (\begin{cases} 31 i A x e^{4 i c} - \frac{4 A e^{2 i c} e^{- 2 i d x}}{d} - \frac{A e^{- 4 i d x}}{4 d} - 17 B x e^{4 i c} - \frac{3 i B e^{2 i c} e^{- 2 i d x}}{d} - \frac{i B e^{- 4 i d x}}{4 d} & \text{for}\: d \neq 0 \\x \left (31 i A e^{4 i c} + 8 i A e^{2 i c} + i A - 17 B e^{4 i c} - 6 B e^{2 i c} - B\right ) & \text{otherwise} \end{cases}\right ) e^{- 4 i c}}{4 a^{2}} - \frac{\left (4 A + 2 i B\right ) \log{\left (e^{2 i d x} - e^{- 2 i c} \right )}}{a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**2,x)

[Out]

(-(2*A + 2*I*B)*exp(-2*I*c)*exp(2*I*d*x)/(a**2*d) + (4*A + 2*I*B)*exp(-4*I*c)/(a**2*d))/(exp(4*I*d*x) - 2*exp(
-2*I*c)*exp(2*I*d*x) + exp(-4*I*c)) + Piecewise((31*I*A*x*exp(4*I*c) - 4*A*exp(2*I*c)*exp(-2*I*d*x)/d - A*exp(
-4*I*d*x)/(4*d) - 17*B*x*exp(4*I*c) - 3*I*B*exp(2*I*c)*exp(-2*I*d*x)/d - I*B*exp(-4*I*d*x)/(4*d), Ne(d, 0)), (
x*(31*I*A*exp(4*I*c) + 8*I*A*exp(2*I*c) + I*A - 17*B*exp(4*I*c) - 6*B*exp(2*I*c) - B), True))*exp(-4*I*c)/(4*a
**2) - (4*A + 2*I*B)*log(exp(2*I*d*x) - exp(-2*I*c))/(a**2*d)

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Giac [A]  time = 1.42164, size = 239, normalized size = 1.41 \begin{align*} \frac{\frac{4 \,{\left (A - i \, B\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{a^{2}} + \frac{4 \,{\left (31 \, A + 17 i \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a^{2}} - \frac{64 \,{\left (2 \, A + i \, B\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{2}} + \frac{3 \, A \tan \left (d x + c\right )^{4} - 3 i \, B \tan \left (d x + c\right )^{4} + 114 i \, A \tan \left (d x + c\right )^{3} - 78 \, B \tan \left (d x + c\right )^{3} + 173 \, A \tan \left (d x + c\right )^{2} + 115 i \, B \tan \left (d x + c\right )^{2} - 32 i \, A \tan \left (d x + c\right ) + 32 \, B \tan \left (d x + c\right ) + 16 \, A}{{\left (\tan \left (d x + c\right )^{2} - i \, \tan \left (d x + c\right )\right )}^{2} a^{2}}}{32 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

1/32*(4*(A - I*B)*log(tan(d*x + c) + I)/a^2 + 4*(31*A + 17*I*B)*log(tan(d*x + c) - I)/a^2 - 64*(2*A + I*B)*log
(abs(tan(d*x + c)))/a^2 + (3*A*tan(d*x + c)^4 - 3*I*B*tan(d*x + c)^4 + 114*I*A*tan(d*x + c)^3 - 78*B*tan(d*x +
c)^3 + 173*A*tan(d*x + c)^2 + 115*I*B*tan(d*x + c)^2 - 32*I*A*tan(d*x + c) + 32*B*tan(d*x + c) + 16*A)/((tan(
d*x + c)^2 - I*tan(d*x + c))^2*a^2))/d