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*ln(cos(f*x+e))/a^2/f-4*(I*A-B)*c^4/a^2/f/(-tan(f*x+e)+I)^2+4*(3*A+5*I*B)
*c^4/a^2/f/(-tan(f*x+e)+I)-(A+6*I*B)*c^4*tan(f*x+e)/a^2/f-1/2*B*c^4*tan(f*x+e)^2/a^2/f

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Rubi [A]  time = 0.21, antiderivative size = 158, normalized size of antiderivative = 1.00, 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 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]))))

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]

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*}

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Mathematica [B]  time = 9.36, 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*(-1/2*(B*Cos[2*e]) - (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
*Sin[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*((-1/2*I)*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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fricas [A]  time = 0.96, size = 242, normalized size = 1.53 \[ \frac {12 \, {\left (A + 3 i \, B\right )} c^{4} f x e^{\left (8 i \, f x + 8 i \, e\right )} + {\left (-2 i \, A + 6 \, B\right )} c^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (i \, A - B\right )} c^{4} + {\left (24 \, {\left (A + 3 i \, B\right )} c^{4} f x + {\left (-6 i \, A + 18 \, B\right )} c^{4}\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (12 \, {\left (A + 3 i \, B\right )} c^{4} f x + {\left (-9 i \, A + 27 \, B\right )} c^{4}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left ({\left (6 i \, A - 18 \, B\right )} c^{4} e^{\left (8 i \, f x + 8 i \, e\right )} + {\left (12 i \, A - 36 \, B\right )} c^{4} e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (6 i \, A - 18 \, B\right )} c^{4} e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{a^{2} f e^{\left (8 i \, f x + 8 i \, e\right )} + 2 \, a^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} f e^{\left (4 i \, f x + 4 i \, e\right )}} \]

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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giac [B]  time = 2.87, size = 443, normalized size = 2.80 \[ -\frac {\frac {2 \, {\left (-3 i \, A c^{4} + 9 \, B c^{4}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{a^{2}} - \frac {2 \, {\left (-6 i \, A c^{4} + 18 \, B c^{4}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}{a^{2}} - \frac {2 \, {\left (3 i \, A c^{4} - 9 \, B c^{4}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\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} \]

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]

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

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maple [A]  time = 0.23, size = 198, normalized size = 1.25 \[ -\frac {B \,c^{4} \left (\tan ^{2}\left (f x +e \right )\right )}{2 a^{2} f}-\frac {c^{4} A \tan \left (f x +e \right )}{f \,a^{2}}-\frac {6 i c^{4} B \tan \left (f x +e \right )}{f \,a^{2}}-\frac {20 i c^{4} B}{f \,a^{2} \left (\tan \left (f x +e \right )-i\right )}-\frac {12 c^{4} A}{f \,a^{2} \left (\tan \left (f x +e \right )-i\right )}-\frac {4 i c^{4} A}{f \,a^{2} \left (\tan \left (f x +e \right )-i\right )^{2}}+\frac {4 c^{4} B}{f \,a^{2} \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {6 i c^{4} A \ln \left (\tan \left (f x +e \right )-i\right )}{f \,a^{2}}+\frac {18 c^{4} B \ln \left (\tan \left (f x +e \right )-i\right )}{f \,a^{2}} \]

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-1/f*c^4/a^2*A*tan(f*x+e)-6*I/f*c^4/a^2*B*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-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-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)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

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 >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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mupad [B]  time = 8.77, size = 207, normalized size = 1.31 \[ -\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )\,\left (-\frac {18\,B\,c^4}{a^2}+\frac {A\,c^4\,6{}\mathrm {i}}{a^2}\right )}{f}-\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (\frac {c^4\,\left (A+B\,3{}\mathrm {i}\right )}{a^2}+\frac {B\,c^4\,3{}\mathrm {i}}{a^2}\right )}{f}-\frac {\frac {\left (-6\,B\,c^4+A\,c^4\,2{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,a^2}-\frac {\left (-18\,B\,c^4+A\,c^4\,6{}\mathrm {i}\right )\,3{}\mathrm {i}}{2\,a^2}+\mathrm {tan}\left (e+f\,x\right )\,\left (\frac {2\,\left (-18\,B\,c^4+A\,c^4\,6{}\mathrm {i}\right )}{a^2}+\frac {16\,B\,c^4}{a^2}\right )-\frac {B\,c^4\,8{}\mathrm {i}}{a^2}}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )}-\frac {B\,c^4\,{\mathrm {tan}\left (e+f\,x\right )}^2}{2\,a^2\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*tan(e + f*x))*(c - c*tan(e + f*x)*1i)^4)/(a + a*tan(e + f*x)*1i)^2,x)

[Out]

- (log(tan(e + f*x) - 1i)*((A*c^4*6i)/a^2 - (18*B*c^4)/a^2))/f - (tan(e + f*x)*((c^4*(A + B*3i))/a^2 + (B*c^4*
3i)/a^2))/f - (((A*c^4*2i - 6*B*c^4)*1i)/(2*a^2) - ((A*c^4*6i - 18*B*c^4)*3i)/(2*a^2) + tan(e + f*x)*((2*(A*c^
4*6i - 18*B*c^4))/a^2 + (16*B*c^4)/a^2) - (B*c^4*8i)/a^2)/(f*(2*tan(e + f*x) + tan(e + f*x)^2*1i - 1i)) - (B*c
^4*tan(e + f*x)^2)/(2*a^2*f)

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sympy [A]  time = 1.31, size = 382, normalized size = 2.42 \[ \frac {2 i A c^{4} - 12 B c^{4} + \left (2 i A c^{4} e^{2 i e} - 10 B c^{4} e^{2 i e}\right ) e^{2 i f x}}{- a^{2} f e^{4 i e} e^{4 i f x} - 2 a^{2} f e^{2 i e} e^{2 i f x} - a^{2} f} + \begin {cases} \frac {\left (\left (i A a^{2} c^{4} f e^{2 i e} - B a^{2} c^{4} f e^{2 i e}\right ) e^{- 4 i f x} + \left (- 4 i A a^{2} c^{4} f e^{4 i e} + 8 B a^{2} c^{4} f e^{4 i e}\right ) e^{- 2 i f x}\right ) e^{- 6 i e}}{a^{4} f^{2}} & \text {for}\: a^{4} f^{2} e^{6 i e} \neq 0 \\x \left (- \frac {12 A c^{4} + 36 i B c^{4}}{a^{2}} + \frac {i \left (- 12 i A c^{4} e^{4 i e} + 8 i A c^{4} e^{2 i e} - 4 i A c^{4} + 36 B c^{4} e^{4 i e} - 16 B c^{4} e^{2 i e} + 4 B c^{4}\right ) e^{- 4 i e}}{a^{2}}\right ) & \text {otherwise} \end {cases} + \frac {6 i c^{4} \left (A + 3 i B\right ) \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{a^{2} f} - \frac {x \left (- 12 A c^{4} - 36 i B c^{4}\right )}{a^{2}} \]

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

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