3.717 \(\int \frac{(A+B \tan (e+f x)) (c-i c \tan (e+f x))^4}{(a+i a \tan (e+f x))^2} \, dx\)

Optimal. Leaf size=158 \[ -\frac{c^4 (A+6 i B) \tan (e+f x)}{a^2 f}+\frac{4 c^4 (3 A+5 i B)}{a^2 f (-\tan (e+f x)+i)}-\frac{4 c^4 (-B+i A)}{a^2 f (-\tan (e+f x)+i)^2}+\frac{6 c^4 (-3 B+i A) \log (\cos (e+f x))}{a^2 f}+\frac{6 c^4 x (A+3 i B)}{a^2}-\frac{B c^4 \tan ^2(e+f x)}{2 a^2 f} \]

[Out]

(6*(A + (3*I)*B)*c^4*x)/a^2 + (6*(I*A - 3*B)*c^4*Log[Cos[e + f*x]])/(a^2*f) - (4*(I*A - B)*c^4)/(a^2*f*(I - Ta
n[e + f*x])^2) + (4*(3*A + (5*I)*B)*c^4)/(a^2*f*(I - Tan[e + f*x])) - ((A + (6*I)*B)*c^4*Tan[e + f*x])/(a^2*f)
- (B*c^4*Tan[e + f*x]^2)/(2*a^2*f)

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Rubi [A]  time = 0.210053, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.049, Rules used = {3588, 77} \[ -\frac{c^4 (A+6 i B) \tan (e+f x)}{a^2 f}+\frac{4 c^4 (3 A+5 i B)}{a^2 f (-\tan (e+f x)+i)}-\frac{4 c^4 (-B+i A)}{a^2 f (-\tan (e+f x)+i)^2}+\frac{6 c^4 (-3 B+i A) \log (\cos (e+f x))}{a^2 f}+\frac{6 c^4 x (A+3 i B)}{a^2}-\frac{B c^4 \tan ^2(e+f x)}{2 a^2 f} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^4)/(a + I*a*Tan[e + f*x])^2,x]

[Out]

(6*(A + (3*I)*B)*c^4*x)/a^2 + (6*(I*A - 3*B)*c^4*Log[Cos[e + f*x]])/(a^2*f) - (4*(I*A - B)*c^4)/(a^2*f*(I - Ta
n[e + f*x])^2) + (4*(3*A + (5*I)*B)*c^4)/(a^2*f*(I - Tan[e + f*x])) - ((A + (6*I)*B)*c^4*Tan[e + f*x])/(a^2*f)
- (B*c^4*Tan[e + f*x]^2)/(2*a^2*f)

Rule 3588

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(a*c)/f, Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^(n - 1)*(A + B*x), x
], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int \frac{(A+B \tan (e+f x)) (c-i c \tan (e+f x))^4}{(a+i a \tan (e+f x))^2} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(A+B x) (c-i c x)^3}{(a+i a x)^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{(a c) \operatorname{Subst}\left (\int \left (-\frac{(A+6 i B) c^3}{a^3}-\frac{B c^3 x}{a^3}+\frac{8 i (A+i B) c^3}{a^3 (-i+x)^3}+\frac{4 (3 A+5 i B) c^3}{a^3 (-i+x)^2}+\frac{6 (-i A+3 B) c^3}{a^3 (-i+x)}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{6 (A+3 i B) c^4 x}{a^2}+\frac{6 (i A-3 B) c^4 \log (\cos (e+f x))}{a^2 f}-\frac{4 (i A-B) c^4}{a^2 f (i-\tan (e+f x))^2}+\frac{4 (3 A+5 i B) c^4}{a^2 f (i-\tan (e+f x))}-\frac{(A+6 i B) c^4 \tan (e+f x)}{a^2 f}-\frac{B c^4 \tan ^2(e+f x)}{2 a^2 f}\\ \end{align*}

Mathematica [B]  time = 9.05899, size = 1079, normalized size = 6.83 \[ c^4 \left (\frac{\left (-\frac{1}{2} B \cos (2 e)-\frac{1}{2} i B \sin (2 e)\right ) (\cos (f x)+i \sin (f x))^2 (A+B \tan (e+f x)) \sec ^3(e+f x)}{f (A \cos (e+f x)+B \sin (e+f x)) (i \tan (e+f x) a+a)^2}+\frac{\sec (e) (\cos (f x)+i \sin (f x))^2 \left (-\frac{1}{2} i A \cos (2 e-f x)+3 B \cos (2 e-f x)+\frac{1}{2} i A \cos (2 e+f x)-3 B \cos (2 e+f x)+\frac{1}{2} A \sin (2 e-f x)+3 i B \sin (2 e-f x)-\frac{1}{2} A \sin (2 e+f x)-3 i B \sin (2 e+f x)\right ) (A+B \tan (e+f x)) \sec ^2(e+f x)}{f (A \cos (e+f x)+B \sin (e+f x)) (i \tan (e+f x) a+a)^2}+\frac{x (\cos (f x)+i \sin (f x))^2 (-6 i \tan (e) A-6 A-18 i B+18 B \tan (e)+(3 B-i A) (6 \cos (2 e)+6 i \sin (2 e)) \tan (e)) (A+B \tan (e+f x)) \sec (e+f x)}{(A \cos (e+f x)+B \sin (e+f x)) (i \tan (e+f x) a+a)^2}+\frac{4 (2 B-i A) \cos (2 f x) (\cos (f x)+i \sin (f x))^2 (A+B \tan (e+f x)) \sec (e+f x)}{f (A \cos (e+f x)+B \sin (e+f x)) (i \tan (e+f x) a+a)^2}+\frac{(A \cos (e)+3 i B \cos (e)+i A \sin (e)-3 B \sin (e)) \left (6 \tan ^{-1}(\tan (f x)) \cos (e)+6 i \tan ^{-1}(\tan (f x)) \sin (e)\right ) (\cos (f x)+i \sin (f x))^2 (A+B \tan (e+f x)) \sec (e+f x)}{f (A \cos (e+f x)+B \sin (e+f x)) (i \tan (e+f x) a+a)^2}+\frac{(A \cos (e)+3 i B \cos (e)+i A \sin (e)-3 B \sin (e)) \left (3 i \cos (e) \log \left (\cos ^2(e+f x)\right )-3 \log \left (\cos ^2(e+f x)\right ) \sin (e)\right ) (\cos (f x)+i \sin (f x))^2 (A+B \tan (e+f x)) \sec (e+f x)}{f (A \cos (e+f x)+B \sin (e+f x)) (i \tan (e+f x) a+a)^2}+\frac{(A+i B) \cos (4 f x) (i \cos (2 e)+\sin (2 e)) (\cos (f x)+i \sin (f x))^2 (A+B \tan (e+f x)) \sec (e+f x)}{f (A \cos (e+f x)+B \sin (e+f x)) (i \tan (e+f x) a+a)^2}+\frac{(A+3 i B) (6 f x \cos (2 e)+6 i f x \sin (2 e)) (\cos (f x)+i \sin (f x))^2 (A+B \tan (e+f x)) \sec (e+f x)}{f (A \cos (e+f x)+B \sin (e+f x)) (i \tan (e+f x) a+a)^2}-\frac{4 (A+2 i B) (\cos (f x)+i \sin (f x))^2 \sin (2 f x) (A+B \tan (e+f x)) \sec (e+f x)}{f (A \cos (e+f x)+B \sin (e+f x)) (i \tan (e+f x) a+a)^2}+\frac{(A+i B) (\cos (2 e)-i \sin (2 e)) (\cos (f x)+i \sin (f x))^2 \sin (4 f x) (A+B \tan (e+f x)) \sec (e+f x)}{f (A \cos (e+f x)+B \sin (e+f x)) (i \tan (e+f x) a+a)^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^4)/(a + I*a*Tan[e + f*x])^2,x]

[Out]

c^4*((4*((-I)*A + 2*B)*Cos[2*f*x]*Sec[e + f*x]*(Cos[f*x] + I*Sin[f*x])^2*(A + B*Tan[e + f*x]))/(f*(A*Cos[e + f
*x] + B*Sin[e + f*x])*(a + I*a*Tan[e + f*x])^2) + (Sec[e + f*x]*(A*Cos[e] + (3*I)*B*Cos[e] + I*A*Sin[e] - 3*B*
Sin[e])*(6*ArcTan[Tan[f*x]]*Cos[e] + (6*I)*ArcTan[Tan[f*x]]*Sin[e])*(Cos[f*x] + I*Sin[f*x])^2*(A + B*Tan[e + f
*x]))/(f*(A*Cos[e + f*x] + B*Sin[e + f*x])*(a + I*a*Tan[e + f*x])^2) + (Sec[e + f*x]*(A*Cos[e] + (3*I)*B*Cos[e
] + I*A*Sin[e] - 3*B*Sin[e])*((3*I)*Cos[e]*Log[Cos[e + f*x]^2] - 3*Log[Cos[e + f*x]^2]*Sin[e])*(Cos[f*x] + I*S
in[f*x])^2*(A + B*Tan[e + f*x]))/(f*(A*Cos[e + f*x] + B*Sin[e + f*x])*(a + I*a*Tan[e + f*x])^2) + ((A + I*B)*C
os[4*f*x]*Sec[e + f*x]*(I*Cos[2*e] + Sin[2*e])*(Cos[f*x] + I*Sin[f*x])^2*(A + B*Tan[e + f*x]))/(f*(A*Cos[e + f
*x] + B*Sin[e + f*x])*(a + I*a*Tan[e + f*x])^2) + (Sec[e + f*x]^3*(-(B*Cos[2*e])/2 - (I/2)*B*Sin[2*e])*(Cos[f*
x] + I*Sin[f*x])^2*(A + B*Tan[e + f*x]))/(f*(A*Cos[e + f*x] + B*Sin[e + f*x])*(a + I*a*Tan[e + f*x])^2) + ((A
+ (3*I)*B)*Sec[e + f*x]*(6*f*x*Cos[2*e] + (6*I)*f*x*Sin[2*e])*(Cos[f*x] + I*Sin[f*x])^2*(A + B*Tan[e + f*x]))/
(f*(A*Cos[e + f*x] + B*Sin[e + f*x])*(a + I*a*Tan[e + f*x])^2) - (4*(A + (2*I)*B)*Sec[e + f*x]*(Cos[f*x] + I*S
in[f*x])^2*Sin[2*f*x]*(A + B*Tan[e + f*x]))/(f*(A*Cos[e + f*x] + B*Sin[e + f*x])*(a + I*a*Tan[e + f*x])^2) + (
(A + I*B)*Sec[e + f*x]*(Cos[2*e] - I*Sin[2*e])*(Cos[f*x] + I*Sin[f*x])^2*Sin[4*f*x]*(A + B*Tan[e + f*x]))/(f*(
A*Cos[e + f*x] + B*Sin[e + f*x])*(a + I*a*Tan[e + f*x])^2) + (Sec[e]*Sec[e + f*x]^2*(Cos[f*x] + I*Sin[f*x])^2*
((-I/2)*A*Cos[2*e - f*x] + 3*B*Cos[2*e - f*x] + (I/2)*A*Cos[2*e + f*x] - 3*B*Cos[2*e + f*x] + (A*Sin[2*e - f*x
])/2 + (3*I)*B*Sin[2*e - f*x] - (A*Sin[2*e + f*x])/2 - (3*I)*B*Sin[2*e + f*x])*(A + B*Tan[e + f*x]))/(f*(A*Cos
[e + f*x] + B*Sin[e + f*x])*(a + I*a*Tan[e + f*x])^2) + (x*Sec[e + f*x]*(Cos[f*x] + I*Sin[f*x])^2*(-6*A - (18*
I)*B - (6*I)*A*Tan[e] + 18*B*Tan[e] + ((-I)*A + 3*B)*(6*Cos[2*e] + (6*I)*Sin[2*e])*Tan[e])*(A + B*Tan[e + f*x]
))/((A*Cos[e + f*x] + B*Sin[e + f*x])*(a + I*a*Tan[e + f*x])^2))

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Maple [A]  time = 0.043, size = 198, normalized size = 1.3 \begin{align*} -{\frac{B{c}^{4} \left ( \tan \left ( fx+e \right ) \right ) ^{2}}{2\,{a}^{2}f}}-{\frac{6\,i{c}^{4}B\tan \left ( fx+e \right ) }{{a}^{2}f}}-{\frac{A{c}^{4}\tan \left ( fx+e \right ) }{{a}^{2}f}}-{\frac{20\,i{c}^{4}B}{{a}^{2}f \left ( \tan \left ( fx+e \right ) -i \right ) }}-12\,{\frac{A{c}^{4}}{{a}^{2}f \left ( \tan \left ( fx+e \right ) -i \right ) }}-{\frac{6\,i{c}^{4}A\ln \left ( \tan \left ( fx+e \right ) -i \right ) }{{a}^{2}f}}+18\,{\frac{B{c}^{4}\ln \left ( \tan \left ( fx+e \right ) -i \right ) }{{a}^{2}f}}-{\frac{4\,iA{c}^{4}}{{a}^{2}f \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}+4\,{\frac{B{c}^{4}}{{a}^{2}f \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^4/(a+I*a*tan(f*x+e))^2,x)

[Out]

-1/2*B*c^4*tan(f*x+e)^2/a^2/f-6*I/f*c^4/a^2*B*tan(f*x+e)-1/f*c^4/a^2*A*tan(f*x+e)-20*I/f*c^4/a^2/(tan(f*x+e)-I
)*B-12/f*c^4/a^2/(tan(f*x+e)-I)*A-6*I/f*c^4/a^2*A*ln(tan(f*x+e)-I)+18/f*c^4/a^2*B*ln(tan(f*x+e)-I)-4*I/f*c^4/a
^2/(tan(f*x+e)-I)^2*A+4/f*c^4/a^2/(tan(f*x+e)-I)^2*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((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^4/(a+I*a*tan(f*x+e))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.06549, size = 651, normalized size = 4.12 \begin{align*} \frac{12 \,{\left (A + 3 i \, B\right )} c^{4} f x e^{\left (8 i \, f x + 8 i \, e\right )} +{\left (-2 i \, A + 6 \, B\right )} c^{4} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (i \, A - B\right )} c^{4} +{\left (24 \,{\left (A + 3 i \, B\right )} c^{4} f x +{\left (-6 i \, A + 18 \, B\right )} c^{4}\right )} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (12 \,{\left (A + 3 i \, B\right )} c^{4} f x +{\left (-9 i \, A + 27 \, B\right )} c^{4}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left ({\left (6 i \, A - 18 \, B\right )} c^{4} e^{\left (8 i \, f x + 8 i \, e\right )} +{\left (12 i \, A - 36 \, B\right )} c^{4} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (6 i \, A - 18 \, B\right )} c^{4} e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{a^{2} f e^{\left (8 i \, f x + 8 i \, e\right )} + 2 \, a^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} f e^{\left (4 i \, f x + 4 i \, e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^4/(a+I*a*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

(12*(A + 3*I*B)*c^4*f*x*e^(8*I*f*x + 8*I*e) + (-2*I*A + 6*B)*c^4*e^(2*I*f*x + 2*I*e) + (I*A - B)*c^4 + (24*(A
+ 3*I*B)*c^4*f*x + (-6*I*A + 18*B)*c^4)*e^(6*I*f*x + 6*I*e) + (12*(A + 3*I*B)*c^4*f*x + (-9*I*A + 27*B)*c^4)*e
^(4*I*f*x + 4*I*e) + ((6*I*A - 18*B)*c^4*e^(8*I*f*x + 8*I*e) + (12*I*A - 36*B)*c^4*e^(6*I*f*x + 6*I*e) + (6*I*
A - 18*B)*c^4*e^(4*I*f*x + 4*I*e))*log(e^(2*I*f*x + 2*I*e) + 1))/(a^2*f*e^(8*I*f*x + 8*I*e) + 2*a^2*f*e^(6*I*f
*x + 6*I*e) + a^2*f*e^(4*I*f*x + 4*I*e))

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Sympy [A]  time = 8.28398, size = 332, normalized size = 2.1 \begin{align*} \frac{- \frac{\left (2 i A c^{4} - 12 B c^{4}\right ) e^{- 4 i e}}{a^{2} f} - \frac{\left (2 i A c^{4} - 10 B c^{4}\right ) e^{- 2 i e} e^{2 i f x}}{a^{2} f}}{e^{4 i f x} + 2 e^{- 2 i e} e^{2 i f x} + e^{- 4 i e}} + \frac{6 c^{4} \left (i A - 3 B\right ) \log{\left (e^{2 i f x} + e^{- 2 i e} \right )}}{a^{2} f} + \frac{\left (\begin{cases} 12 A c^{4} x e^{4 i e} - \frac{4 i A c^{4} e^{2 i e} e^{- 2 i f x}}{f} + \frac{i A c^{4} e^{- 4 i f x}}{f} + 36 i B c^{4} x e^{4 i e} + \frac{8 B c^{4} e^{2 i e} e^{- 2 i f x}}{f} - \frac{B c^{4} e^{- 4 i f x}}{f} & \text{for}\: f \neq 0 \\x \left (12 A c^{4} e^{4 i e} - 8 A c^{4} e^{2 i e} + 4 A c^{4} + 36 i B c^{4} e^{4 i e} - 16 i B c^{4} e^{2 i e} + 4 i B c^{4}\right ) & \text{otherwise} \end{cases}\right ) e^{- 4 i e}}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))**4/(a+I*a*tan(f*x+e))**2,x)

[Out]

(-(2*I*A*c**4 - 12*B*c**4)*exp(-4*I*e)/(a**2*f) - (2*I*A*c**4 - 10*B*c**4)*exp(-2*I*e)*exp(2*I*f*x)/(a**2*f))/
(exp(4*I*f*x) + 2*exp(-2*I*e)*exp(2*I*f*x) + exp(-4*I*e)) + 6*c**4*(I*A - 3*B)*log(exp(2*I*f*x) + exp(-2*I*e))
/(a**2*f) + Piecewise((12*A*c**4*x*exp(4*I*e) - 4*I*A*c**4*exp(2*I*e)*exp(-2*I*f*x)/f + I*A*c**4*exp(-4*I*f*x)
/f + 36*I*B*c**4*x*exp(4*I*e) + 8*B*c**4*exp(2*I*e)*exp(-2*I*f*x)/f - B*c**4*exp(-4*I*f*x)/f, Ne(f, 0)), (x*(1
2*A*c**4*exp(4*I*e) - 8*A*c**4*exp(2*I*e) + 4*A*c**4 + 36*I*B*c**4*exp(4*I*e) - 16*I*B*c**4*exp(2*I*e) + 4*I*B
*c**4), True))*exp(-4*I*e)/a**2

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Giac [B]  time = 2.09514, size = 601, normalized size = 3.8 \begin{align*} -\frac{\frac{12 \,{\left (i \, A c^{4} - 3 \, B c^{4}\right )} \log \left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - i\right )}{a^{2}} - \frac{6 \,{\left (i \, A c^{4} - 3 \, B c^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{a^{2}} + \frac{6 \,{\left (-i \, A c^{4} + 3 \, B c^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{a^{2}} + \frac{9 i \, A c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 27 \, B c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 2 \, A c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 12 i \, B c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 18 i \, A c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 56 \, B c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 2 \, A c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 12 i \, B c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 9 i \, A c^{4} - 27 \, B c^{4}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right )}^{2} a^{2}} + \frac{-25 i \, A c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 75 \, B c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 108 \, A c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 324 i \, B c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 182 i \, A c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 514 \, B c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 108 \, A c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 324 i \, B c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 25 i \, A c^{4} + 75 \, B c^{4}}{a^{2}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - i\right )}^{4}}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^4/(a+I*a*tan(f*x+e))^2,x, algorithm="giac")

[Out]

-(12*(I*A*c^4 - 3*B*c^4)*log(tan(1/2*f*x + 1/2*e) - I)/a^2 - 6*(I*A*c^4 - 3*B*c^4)*log(abs(tan(1/2*f*x + 1/2*e
) + 1))/a^2 + 6*(-I*A*c^4 + 3*B*c^4)*log(abs(tan(1/2*f*x + 1/2*e) - 1))/a^2 + (9*I*A*c^4*tan(1/2*f*x + 1/2*e)^
4 - 27*B*c^4*tan(1/2*f*x + 1/2*e)^4 - 2*A*c^4*tan(1/2*f*x + 1/2*e)^3 - 12*I*B*c^4*tan(1/2*f*x + 1/2*e)^3 - 18*
I*A*c^4*tan(1/2*f*x + 1/2*e)^2 + 56*B*c^4*tan(1/2*f*x + 1/2*e)^2 + 2*A*c^4*tan(1/2*f*x + 1/2*e) + 12*I*B*c^4*t
an(1/2*f*x + 1/2*e) + 9*I*A*c^4 - 27*B*c^4)/((tan(1/2*f*x + 1/2*e)^2 - 1)^2*a^2) + (-25*I*A*c^4*tan(1/2*f*x +
1/2*e)^4 + 75*B*c^4*tan(1/2*f*x + 1/2*e)^4 - 108*A*c^4*tan(1/2*f*x + 1/2*e)^3 - 324*I*B*c^4*tan(1/2*f*x + 1/2*
e)^3 + 182*I*A*c^4*tan(1/2*f*x + 1/2*e)^2 - 514*B*c^4*tan(1/2*f*x + 1/2*e)^2 + 108*A*c^4*tan(1/2*f*x + 1/2*e)
+ 324*I*B*c^4*tan(1/2*f*x + 1/2*e) - 25*I*A*c^4 + 75*B*c^4)/(a^2*(tan(1/2*f*x + 1/2*e) - I)^4))/f