3.43 \(\int \frac{\cot ^4(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx\)

Optimal. Leaf size=155 \[ -\frac{(5 A+3 i B) \cot ^3(c+d x)}{6 a d}+\frac{(-B+i A) \cot ^2(c+d x)}{a d}+\frac{(5 A+3 i B) \cot (c+d x)}{2 a d}+\frac{2 (-B+i A) \log (\sin (c+d x))}{a d}+\frac{(A+i B) \cot ^3(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac{x (5 A+3 i B)}{2 a} \]

[Out]

((5*A + (3*I)*B)*x)/(2*a) + ((5*A + (3*I)*B)*Cot[c + d*x])/(2*a*d) + ((I*A - B)*Cot[c + d*x]^2)/(a*d) - ((5*A
+ (3*I)*B)*Cot[c + d*x]^3)/(6*a*d) + (2*(I*A - B)*Log[Sin[c + d*x]])/(a*d) + ((A + I*B)*Cot[c + d*x]^3)/(2*d*(
a + I*a*Tan[c + d*x]))

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Rubi [A]  time = 0.246209, antiderivative size = 155, 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} \[ -\frac{(5 A+3 i B) \cot ^3(c+d x)}{6 a d}+\frac{(-B+i A) \cot ^2(c+d x)}{a d}+\frac{(5 A+3 i B) \cot (c+d x)}{2 a d}+\frac{2 (-B+i A) \log (\sin (c+d x))}{a d}+\frac{(A+i B) \cot ^3(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac{x (5 A+3 i B)}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((5*A + (3*I)*B)*x)/(2*a) + ((5*A + (3*I)*B)*Cot[c + d*x])/(2*a*d) + ((I*A - B)*Cot[c + d*x]^2)/(a*d) - ((5*A
+ (3*I)*B)*Cot[c + d*x]^3)/(6*a*d) + (2*(I*A - B)*Log[Sin[c + d*x]])/(a*d) + ((A + I*B)*Cot[c + d*x]^3)/(2*d*(
a + I*a*Tan[c + d*x]))

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 ^4(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx &=\frac{(A+i B) \cot ^3(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac{\int \cot ^4(c+d x) (a (5 A+3 i B)-4 a (i A-B) \tan (c+d x)) \, dx}{2 a^2}\\ &=-\frac{(5 A+3 i B) \cot ^3(c+d x)}{6 a d}+\frac{(A+i B) \cot ^3(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac{\int \cot ^3(c+d x) (-4 a (i A-B)-a (5 A+3 i B) \tan (c+d x)) \, dx}{2 a^2}\\ &=\frac{(i A-B) \cot ^2(c+d x)}{a d}-\frac{(5 A+3 i B) \cot ^3(c+d x)}{6 a d}+\frac{(A+i B) \cot ^3(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac{\int \cot ^2(c+d x) (-a (5 A+3 i B)+4 a (i A-B) \tan (c+d x)) \, dx}{2 a^2}\\ &=\frac{(5 A+3 i B) \cot (c+d x)}{2 a d}+\frac{(i A-B) \cot ^2(c+d x)}{a d}-\frac{(5 A+3 i B) \cot ^3(c+d x)}{6 a d}+\frac{(A+i B) \cot ^3(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac{\int \cot (c+d x) (4 a (i A-B)+a (5 A+3 i B) \tan (c+d x)) \, dx}{2 a^2}\\ &=\frac{(5 A+3 i B) x}{2 a}+\frac{(5 A+3 i B) \cot (c+d x)}{2 a d}+\frac{(i A-B) \cot ^2(c+d x)}{a d}-\frac{(5 A+3 i B) \cot ^3(c+d x)}{6 a d}+\frac{(A+i B) \cot ^3(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac{(2 (i A-B)) \int \cot (c+d x) \, dx}{a}\\ &=\frac{(5 A+3 i B) x}{2 a}+\frac{(5 A+3 i B) \cot (c+d x)}{2 a d}+\frac{(i A-B) \cot ^2(c+d x)}{a d}-\frac{(5 A+3 i B) \cot ^3(c+d x)}{6 a d}+\frac{2 (i A-B) \log (\sin (c+d x))}{a d}+\frac{(A+i B) \cot ^3(c+d x)}{2 d (a+i a \tan (c+d x))}\\ \end{align*}

Mathematica [B]  time = 7.36207, size = 1062, normalized size = 6.85 \[ \frac{\csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) (\cos (d x)+i \sin (d x)) \left (\frac{1}{2} i A \cos (c-d x)-\frac{1}{2} i A \cos (c+d x)-\frac{1}{2} A \sin (c-d x)+\frac{1}{2} A \sin (c+d x)\right ) (A+B \tan (c+d x)) \csc ^3(c+d x)}{6 d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)}+\frac{\csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) \left (-\frac{\cos (c)}{12}-\frac{1}{12} i \sin (c)\right ) (2 A \cos (c)-3 i A \sin (c)+3 B \sin (c)) (\cos (d x)+i \sin (d x)) (A+B \tan (c+d x)) \csc ^2(c+d x)}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)}+\frac{\csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) (\cos (d x)+i \sin (d x)) \left (-\frac{7}{2} i A \cos (c-d x)+\frac{3}{2} B \cos (c-d x)+\frac{7}{2} i A \cos (c+d x)-\frac{3}{2} B \cos (c+d x)+\frac{7}{2} A \sin (c-d x)+\frac{3}{2} i B \sin (c-d x)-\frac{7}{2} A \sin (c+d x)-\frac{3}{2} i B \sin (c+d x)\right ) (A+B \tan (c+d x)) \csc (c+d x)}{6 d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)}+\frac{\left (A \cos \left (\frac{c}{2}\right )+i B \cos \left (\frac{c}{2}\right )+i A \sin \left (\frac{c}{2}\right )-B \sin \left (\frac{c}{2}\right )\right ) \left (2 \tan ^{-1}(\tan (d x)) \cos \left (\frac{c}{2}\right )+2 i \tan ^{-1}(\tan (d x)) \sin \left (\frac{c}{2}\right )\right ) (\cos (d x)+i \sin (d x)) (A+B \tan (c+d x))}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)}+\frac{\left (A \cos \left (\frac{c}{2}\right )+i B \cos \left (\frac{c}{2}\right )+i A \sin \left (\frac{c}{2}\right )-B \sin \left (\frac{c}{2}\right )\right ) \left (i \cos \left (\frac{c}{2}\right ) \log \left (\sin ^2(c+d x)\right )-\log \left (\sin ^2(c+d x)\right ) \sin \left (\frac{c}{2}\right )\right ) (\cos (d x)+i \sin (d x)) (A+B \tan (c+d x))}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)}+\frac{x (-2 i A \csc (c)+2 B \csc (c)+i (A+i B) \cot (c) (2 \cos (c)+2 i \sin (c))) (\cos (d x)+i \sin (d x)) (A+B \tan (c+d x))}{(A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)}+\frac{(A+i B) \cos (2 d x) \left (\frac{1}{4} i \cos (c)+\frac{\sin (c)}{4}\right ) (\cos (d x)+i \sin (d x)) (A+B \tan (c+d x))}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)}+\frac{(5 A+3 i B) \left (\frac{1}{2} d x \cos (c)+\frac{1}{2} i d x \sin (c)\right ) (\cos (d x)+i \sin (d x)) (A+B \tan (c+d x))}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)}+\frac{(A+i B) \left (\frac{\cos (c)}{4}-\frac{1}{4} i \sin (c)\right ) (\cos (d x)+i \sin (d x)) \sin (2 d x) (A+B \tan (c+d x))}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A*Cos[c/2] + I*B*Cos[c/2] + I*A*Sin[c/2] - B*Sin[c/2])*(2*ArcTan[Tan[d*x]]*Cos[c/2] + (2*I)*ArcTan[Tan[d*x]]
*Sin[c/2])*(Cos[d*x] + I*Sin[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])) + ((A*Cos[c/2] + I*B*Cos[c/2] + I*A*Sin[c/2] - B*Sin[c/2])*(I*Cos[c/2]*Log[Sin[c + d*x]^2] - Log[Sin[c
+ d*x]^2]*Sin[c/2])*(Cos[d*x] + I*Sin[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])) + (x*((-2*I)*A*Csc[c] + 2*B*Csc[c] + I*(A + I*B)*Cot[c]*(2*Cos[c] + (2*I)*Sin[c]))*(Cos[d*x]
+ I*Sin[d*x])*(A + B*Tan[c + d*x]))/((A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])) + ((A + I*B)*C
os[2*d*x]*((I/4)*Cos[c] + Sin[c]/4)*(Cos[d*x] + I*Sin[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])) + (Csc[c/2]*Csc[c + d*x]^2*Sec[c/2]*(-Cos[c]/12 - (I/12)*Sin[c])*(2*A*Cos[c]
- (3*I)*A*Sin[c] + 3*B*Sin[c])*(Cos[d*x] + I*Sin[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])) + ((5*A + (3*I)*B)*((d*x*Cos[c])/2 + (I/2)*d*x*Sin[c])*(Cos[d*x] + I*Sin[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])) + ((A + I*B)*(Cos[c]/4 - (I/4)
*Sin[c])*(Cos[d*x] + I*Sin[d*x])*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])) + (Csc[c/2]*Csc[c + d*x]^3*Sec[c/2]*(Cos[d*x] + I*Sin[d*x])*((I/2)*A*Cos[c - d*x] - (I/2)*A*C
os[c + d*x] - (A*Sin[c - d*x])/2 + (A*Sin[c + d*x])/2)*(A + B*Tan[c + d*x]))/(6*d*(A*Cos[c + d*x] + B*Sin[c +
d*x])*(a + I*a*Tan[c + d*x])) + (Csc[c/2]*Csc[c + d*x]*Sec[c/2]*(Cos[d*x] + I*Sin[d*x])*(((-7*I)/2)*A*Cos[c -
d*x] + (3*B*Cos[c - d*x])/2 + ((7*I)/2)*A*Cos[c + d*x] - (3*B*Cos[c + d*x])/2 + (7*A*Sin[c - d*x])/2 + ((3*I)/
2)*B*Sin[c - d*x] - (7*A*Sin[c + d*x])/2 - ((3*I)/2)*B*Sin[c + d*x])*(A + B*Tan[c + d*x]))/(6*d*(A*Cos[c + d*x
] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x]))

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Maple [A]  time = 0.106, size = 241, normalized size = 1.6 \begin{align*}{\frac{A}{2\,ad \left ( \tan \left ( dx+c \right ) -i \right ) }}+{\frac{{\frac{i}{2}}B}{ad \left ( \tan \left ( dx+c \right ) -i \right ) }}-{\frac{{\frac{9\,i}{4}}\ln \left ( \tan \left ( dx+c \right ) -i \right ) A}{ad}}+{\frac{7\,\ln \left ( \tan \left ( dx+c \right ) -i \right ) B}{4\,ad}}+{\frac{B\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{4\,ad}}+{\frac{{\frac{i}{4}}A\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{ad}}+{\frac{{\frac{i}{2}}A}{ad \left ( \tan \left ( dx+c \right ) \right ) ^{2}}}-{\frac{B}{2\,ad \left ( \tan \left ( dx+c \right ) \right ) ^{2}}}+{\frac{iB}{ad\tan \left ( dx+c \right ) }}+2\,{\frac{A}{ad\tan \left ( dx+c \right ) }}-{\frac{A}{3\,ad \left ( \tan \left ( dx+c \right ) \right ) ^{3}}}+{\frac{2\,iA\ln \left ( \tan \left ( dx+c \right ) \right ) }{ad}}-2\,{\frac{B\ln \left ( \tan \left ( dx+c \right ) \right ) }{ad}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d/a/(tan(d*x+c)-I)*A+1/2*I/d/a/(tan(d*x+c)-I)*B-9/4*I/d/a*ln(tan(d*x+c)-I)*A+7/4/d/a*ln(tan(d*x+c)-I)*B+1/
4/d/a*B*ln(tan(d*x+c)+I)+1/4*I/d/a*A*ln(tan(d*x+c)+I)+1/2*I/d/a/tan(d*x+c)^2*A-1/2/d/a/tan(d*x+c)^2*B+I/d/a/ta
n(d*x+c)*B+2/d/a*A/tan(d*x+c)-1/3/d/a*A/tan(d*x+c)^3+2*I/d/a*A*ln(tan(d*x+c))-2/d/a*B*ln(tan(d*x+c))

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.53407, size = 724, normalized size = 4.67 \begin{align*} \frac{6 \,{\left (9 \, A + 7 i \, B\right )} d x e^{\left (8 i \, d x + 8 i \, c\right )} -{\left (18 \,{\left (9 \, A + 7 i \, B\right )} d x - 51 i \, A + 3 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (18 \,{\left (9 \, A + 7 i \, B\right )} d x - 81 i \, A + 33 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} -{\left (6 \,{\left (9 \, A + 7 i \, B\right )} d x - 65 i \, A + 33 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left ({\left (24 i \, A - 24 \, B\right )} e^{\left (8 i \, d x + 8 i \, c\right )} +{\left (-72 i \, A + 72 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (72 i \, A - 72 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (-24 i \, A + 24 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) - 3 i \, A + 3 \, B}{12 \,{\left (a d e^{\left (8 i \, d x + 8 i \, c\right )} - 3 \, a d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} - a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(6*(9*A + 7*I*B)*d*x*e^(8*I*d*x + 8*I*c) - (18*(9*A + 7*I*B)*d*x - 51*I*A + 3*B)*e^(6*I*d*x + 6*I*c) + (1
8*(9*A + 7*I*B)*d*x - 81*I*A + 33*B)*e^(4*I*d*x + 4*I*c) - (6*(9*A + 7*I*B)*d*x - 65*I*A + 33*B)*e^(2*I*d*x +
2*I*c) + ((24*I*A - 24*B)*e^(8*I*d*x + 8*I*c) + (-72*I*A + 72*B)*e^(6*I*d*x + 6*I*c) + (72*I*A - 72*B)*e^(4*I*
d*x + 4*I*c) + (-24*I*A + 24*B)*e^(2*I*d*x + 2*I*c))*log(e^(2*I*d*x + 2*I*c) - 1) - 3*I*A + 3*B)/(a*d*e^(8*I*d
*x + 8*I*c) - 3*a*d*e^(6*I*d*x + 6*I*c) + 3*a*d*e^(4*I*d*x + 4*I*c) - a*d*e^(2*I*d*x + 2*I*c))

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Sympy [A]  time = 25.1523, size = 243, normalized size = 1.57 \begin{align*} \frac{\frac{4 i A e^{- 2 i c} e^{4 i d x}}{a d} - \frac{\left (6 i A - 2 B\right ) e^{- 4 i c} e^{2 i d x}}{a d} + \frac{\left (14 i A - 6 B\right ) e^{- 6 i c}}{3 a 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}} + \frac{\left (\begin{cases} 9 A x e^{2 i c} + \frac{i A e^{- 2 i d x}}{2 d} + 7 i B x e^{2 i c} - \frac{B e^{- 2 i d x}}{2 d} & \text{for}\: d \neq 0 \\x \left (9 A e^{2 i c} + A + 7 i B e^{2 i c} + i B\right ) & \text{otherwise} \end{cases}\right ) e^{- 2 i c}}{2 a} + \frac{2 \left (i A - B\right ) \log{\left (e^{2 i d x} - e^{- 2 i c} \right )}}{a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*I*A*exp(-2*I*c)*exp(4*I*d*x)/(a*d) - (6*I*A - 2*B)*exp(-4*I*c)*exp(2*I*d*x)/(a*d) + (14*I*A - 6*B)*exp(-6*I
*c)/(3*a*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)) + Piecewis
e((9*A*x*exp(2*I*c) + I*A*exp(-2*I*d*x)/(2*d) + 7*I*B*x*exp(2*I*c) - B*exp(-2*I*d*x)/(2*d), Ne(d, 0)), (x*(9*A
*exp(2*I*c) + A + 7*I*B*exp(2*I*c) + I*B), True))*exp(-2*I*c)/(2*a) + 2*(I*A - B)*log(exp(2*I*d*x) - exp(-2*I*
c))/(a*d)

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Giac [A]  time = 1.44846, size = 252, normalized size = 1.63 \begin{align*} -\frac{\frac{3 \,{\left (9 i \, A - 7 \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a} + \frac{3 \,{\left (-i \, A - B\right )} \log \left (-i \, \tan \left (d x + c\right ) + 1\right )}{a} + \frac{24 \,{\left (-i \, A + B\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a} + \frac{3 \,{\left (-9 i \, A \tan \left (d x + c\right ) + 7 \, B \tan \left (d x + c\right ) - 11 \, A - 9 i \, B\right )}}{a{\left (\tan \left (d x + c\right ) - i\right )}} + \frac{2 i \,{\left (22 \, A \tan \left (d x + c\right )^{3} + 22 i \, B \tan \left (d x + c\right )^{3} + 12 i \, A \tan \left (d x + c\right )^{2} - 6 \, B \tan \left (d x + c\right )^{2} - 3 \, A \tan \left (d x + c\right ) - 3 i \, B \tan \left (d x + c\right ) - 2 i \, A\right )}}{a \tan \left (d x + c\right )^{3}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*(9*I*A - 7*B)*log(tan(d*x + c) - I)/a + 3*(-I*A - B)*log(-I*tan(d*x + c) + 1)/a + 24*(-I*A + B)*log(a
bs(tan(d*x + c)))/a + 3*(-9*I*A*tan(d*x + c) + 7*B*tan(d*x + c) - 11*A - 9*I*B)/(a*(tan(d*x + c) - I)) + 2*I*(
22*A*tan(d*x + c)^3 + 22*I*B*tan(d*x + c)^3 + 12*I*A*tan(d*x + c)^2 - 6*B*tan(d*x + c)^2 - 3*A*tan(d*x + c) -
3*I*B*tan(d*x + c) - 2*I*A)/(a*tan(d*x + c)^3))/d