3.698 \(\int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^2} \, dx\)

Optimal. Leaf size=123 \[ -\frac {4 a^3 (A-2 i B)}{c^2 f (\tan (e+f x)+i)}+\frac {2 a^3 (B+i A)}{c^2 f (\tan (e+f x)+i)^2}-\frac {a^3 (5 B+i A) \log (\cos (e+f x))}{c^2 f}+\frac {a^3 x (A-5 i B)}{c^2}+\frac {i a^3 B \tan (e+f x)}{c^2 f} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 123, 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} \[ -\frac {4 a^3 (A-2 i B)}{c^2 f (\tan (e+f x)+i)}+\frac {2 a^3 (B+i A)}{c^2 f (\tan (e+f x)+i)^2}-\frac {a^3 (5 B+i A) \log (\cos (e+f x))}{c^2 f}+\frac {a^3 x (A-5 i B)}{c^2}+\frac {i a^3 B \tan (e+f x)}{c^2 f} \]

Antiderivative was successfully verified.

[In]

Int[((a + I*a*Tan[e + f*x])^3*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^2,x]

[Out]

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

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

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Mathematica [B]  time = 10.09, size = 1063, normalized size = 8.64 \[ \frac {x \left (\frac {A \cos ^3(e)}{2 c^2}-\frac {5 i B \cos ^3(e)}{2 c^2}-\frac {2 i A \sin (e) \cos ^2(e)}{c^2}-\frac {10 B \sin (e) \cos ^2(e)}{c^2}-\frac {3 A \sin ^2(e) \cos (e)}{c^2}+\frac {15 i B \sin ^2(e) \cos (e)}{c^2}-\frac {A \cos (e)}{2 c^2}+\frac {5 i B \cos (e)}{2 c^2}+\frac {2 i A \sin ^3(e)}{c^2}+\frac {10 B \sin ^3(e)}{c^2}+\frac {i A \sin (e)}{c^2}+\frac {5 B \sin (e)}{c^2}+\frac {A \sin ^3(e) \tan (e)}{2 c^2}-\frac {5 i B \sin ^3(e) \tan (e)}{2 c^2}+\frac {A \sin (e) \tan (e)}{2 c^2}-\frac {5 i B \sin (e) \tan (e)}{2 c^2}+i (A-5 i B) \left (\frac {\cos (3 e)}{c^2}-\frac {i \sin (3 e)}{c^2}\right ) \tan (e)\right ) (i \tan (e+f x) a+a)^3 (A+B \tan (e+f x)) \cos ^4(e+f x)}{(\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))}+\frac {(A-3 i B) \left (\frac {i \sin (e)}{c^2}-\frac {\cos (e)}{c^2}\right ) \sin (2 f x) (i \tan (e+f x) a+a)^3 (A+B \tan (e+f x)) \cos ^4(e+f x)}{f (\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))}+\frac {(A-i B) \left (\frac {\cos (e)}{2 c^2}+\frac {i \sin (e)}{2 c^2}\right ) \sin (4 f x) (i \tan (e+f x) a+a)^3 (A+B \tan (e+f x)) \cos ^4(e+f x)}{f (\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))}+\frac {(i A+3 B) \cos (2 f x) \left (\frac {\cos (e)}{c^2}-\frac {i \sin (e)}{c^2}\right ) (i \tan (e+f x) a+a)^3 (A+B \tan (e+f x)) \cos ^4(e+f x)}{f (\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))}+\frac {(A-i B) \cos (4 f x) \left (\frac {\sin (e)}{2 c^2}-\frac {i \cos (e)}{2 c^2}\right ) (i \tan (e+f x) a+a)^3 (A+B \tan (e+f x)) \cos ^4(e+f x)}{f (\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))}+\frac {(A-5 i B) \left (\frac {f x \cos (3 e)}{c^2}-\frac {i f x \sin (3 e)}{c^2}\right ) (i \tan (e+f x) a+a)^3 (A+B \tan (e+f x)) \cos ^4(e+f x)}{f (\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))}+\frac {(A-5 i B) \left (-\frac {i \cos (3 e) \log \left (\cos ^2(e+f x)\right )}{2 c^2}-\frac {\sin (3 e) \log \left (\cos ^2(e+f x)\right )}{2 c^2}\right ) (i \tan (e+f x) a+a)^3 (A+B \tan (e+f x)) \cos ^4(e+f x)}{f (\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))}+\frac {i B \left (\frac {\cos (3 e)}{c^2}-\frac {i \sin (3 e)}{c^2}\right ) \sin (f x) (i \tan (e+f x) a+a)^3 (A+B \tan (e+f x)) \cos ^3(e+f x)}{f \left (\cos \left (\frac {e}{2}\right )-\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {e}{2}\right )+\sin \left (\frac {e}{2}\right )\right ) (\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + I*a*Tan[e + f*x])^3*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^2,x]

[Out]

((I*A + 3*B)*Cos[2*f*x]*Cos[e + f*x]^4*(Cos[e]/c^2 - (I*Sin[e])/c^2)*(a + I*a*Tan[e + f*x])^3*(A + B*Tan[e + f
*x]))/(f*(Cos[f*x] + I*Sin[f*x])^3*(A*Cos[e + f*x] + B*Sin[e + f*x])) + ((A - I*B)*Cos[4*f*x]*Cos[e + f*x]^4*(
((-1/2*I)*Cos[e])/c^2 + Sin[e]/(2*c^2))*(a + I*a*Tan[e + f*x])^3*(A + B*Tan[e + f*x]))/(f*(Cos[f*x] + I*Sin[f*
x])^3*(A*Cos[e + f*x] + B*Sin[e + f*x])) + ((A - (5*I)*B)*Cos[e + f*x]^4*((f*x*Cos[3*e])/c^2 - (I*f*x*Sin[3*e]
)/c^2)*(a + I*a*Tan[e + f*x])^3*(A + B*Tan[e + f*x]))/(f*(Cos[f*x] + I*Sin[f*x])^3*(A*Cos[e + f*x] + B*Sin[e +
f*x])) + ((A - (5*I)*B)*Cos[e + f*x]^4*(((-1/2*I)*Cos[3*e]*Log[Cos[e + f*x]^2])/c^2 - (Log[Cos[e + f*x]^2]*Si
n[3*e])/(2*c^2))*(a + I*a*Tan[e + f*x])^3*(A + B*Tan[e + f*x]))/(f*(Cos[f*x] + I*Sin[f*x])^3*(A*Cos[e + f*x] +
B*Sin[e + f*x])) + (I*B*Cos[e + f*x]^3*(Cos[3*e]/c^2 - (I*Sin[3*e])/c^2)*Sin[f*x]*(a + I*a*Tan[e + f*x])^3*(A
+ B*Tan[e + f*x]))/(f*(Cos[e/2] - Sin[e/2])*(Cos[e/2] + Sin[e/2])*(Cos[f*x] + I*Sin[f*x])^3*(A*Cos[e + f*x] +
B*Sin[e + f*x])) + ((A - (3*I)*B)*Cos[e + f*x]^4*(-(Cos[e]/c^2) + (I*Sin[e])/c^2)*Sin[2*f*x]*(a + I*a*Tan[e +
f*x])^3*(A + B*Tan[e + f*x]))/(f*(Cos[f*x] + I*Sin[f*x])^3*(A*Cos[e + f*x] + B*Sin[e + f*x])) + ((A - I*B)*Co
s[e + f*x]^4*(Cos[e]/(2*c^2) + ((I/2)*Sin[e])/c^2)*Sin[4*f*x]*(a + I*a*Tan[e + f*x])^3*(A + B*Tan[e + f*x]))/(
f*(Cos[f*x] + I*Sin[f*x])^3*(A*Cos[e + f*x] + B*Sin[e + f*x])) + (x*Cos[e + f*x]^4*(-1/2*(A*Cos[e])/c^2 + (((5
*I)/2)*B*Cos[e])/c^2 + (A*Cos[e]^3)/(2*c^2) - (((5*I)/2)*B*Cos[e]^3)/c^2 + (I*A*Sin[e])/c^2 + (5*B*Sin[e])/c^2
- ((2*I)*A*Cos[e]^2*Sin[e])/c^2 - (10*B*Cos[e]^2*Sin[e])/c^2 - (3*A*Cos[e]*Sin[e]^2)/c^2 + ((15*I)*B*Cos[e]*S
in[e]^2)/c^2 + ((2*I)*A*Sin[e]^3)/c^2 + (10*B*Sin[e]^3)/c^2 + (A*Sin[e]*Tan[e])/(2*c^2) - (((5*I)/2)*B*Sin[e]*
Tan[e])/c^2 + (A*Sin[e]^3*Tan[e])/(2*c^2) - (((5*I)/2)*B*Sin[e]^3*Tan[e])/c^2 + I*(A - (5*I)*B)*(Cos[3*e]/c^2
- (I*Sin[3*e])/c^2)*Tan[e])*(a + I*a*Tan[e + f*x])^3*(A + B*Tan[e + f*x]))/((Cos[f*x] + I*Sin[f*x])^3*(A*Cos[e
+ f*x] + B*Sin[e + f*x]))

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fricas [A]  time = 1.20, size = 136, normalized size = 1.11 \[ \frac {{\left (-i \, A - B\right )} a^{3} e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (i \, A + 5 \, B\right )} a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (2 i \, A + 6 \, B\right )} a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} - 4 \, B a^{3} + {\left ({\left (-2 i \, A - 10 \, B\right )} a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-2 i \, A - 10 \, B\right )} a^{3}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{2 \, {\left (c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2} f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((-I*A - B)*a^3*e^(6*I*f*x + 6*I*e) + (I*A + 5*B)*a^3*e^(4*I*f*x + 4*I*e) + (2*I*A + 6*B)*a^3*e^(2*I*f*x +
2*I*e) - 4*B*a^3 + ((-2*I*A - 10*B)*a^3*e^(2*I*f*x + 2*I*e) + (-2*I*A - 10*B)*a^3)*log(e^(2*I*f*x + 2*I*e) +
1))/(c^2*f*e^(2*I*f*x + 2*I*e) + c^2*f)

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giac [B]  time = 2.31, size = 357, normalized size = 2.90 \[ \frac {\frac {6 \, {\left (-i \, A a^{3} - 5 \, B a^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{c^{2}} + \frac {12 \, {\left (i \, A a^{3} + 5 \, B a^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}{c^{2}} - \frac {6 \, {\left (i \, A a^{3} + 5 \, B a^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}{c^{2}} - \frac {6 \, {\left (-i \, A a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 5 \, B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 i \, B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i \, A a^{3} + 5 \, B a^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} c^{2}} - \frac {25 i \, A a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 125 \, B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 100 \, A a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 548 i \, B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 198 i \, A a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 894 \, B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 100 \, A a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 548 i \, B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 25 i \, A a^{3} + 125 \, B a^{3}}{c^{2} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}^{4}}}{6 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*(-I*A*a^3 - 5*B*a^3)*log(tan(1/2*f*x + 1/2*e) + 1)/c^2 + 12*(I*A*a^3 + 5*B*a^3)*log(tan(1/2*f*x + 1/2*e
) + I)/c^2 - 6*(I*A*a^3 + 5*B*a^3)*log(tan(1/2*f*x + 1/2*e) - 1)/c^2 - 6*(-I*A*a^3*tan(1/2*f*x + 1/2*e)^2 - 5*
B*a^3*tan(1/2*f*x + 1/2*e)^2 + 2*I*B*a^3*tan(1/2*f*x + 1/2*e) + I*A*a^3 + 5*B*a^3)/((tan(1/2*f*x + 1/2*e)^2 -
1)*c^2) - (25*I*A*a^3*tan(1/2*f*x + 1/2*e)^4 + 125*B*a^3*tan(1/2*f*x + 1/2*e)^4 - 100*A*a^3*tan(1/2*f*x + 1/2*
e)^3 + 548*I*B*a^3*tan(1/2*f*x + 1/2*e)^3 - 198*I*A*a^3*tan(1/2*f*x + 1/2*e)^2 - 894*B*a^3*tan(1/2*f*x + 1/2*e
)^2 + 100*A*a^3*tan(1/2*f*x + 1/2*e) - 548*I*B*a^3*tan(1/2*f*x + 1/2*e) + 25*I*A*a^3 + 125*B*a^3)/(c^2*(tan(1/
2*f*x + 1/2*e) + I)^4))/f

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maple [A]  time = 0.19, size = 160, normalized size = 1.30 \[ \frac {i a^{3} B \tan \left (f x +e \right )}{c^{2} f}+\frac {8 i a^{3} B}{f \,c^{2} \left (\tan \left (f x +e \right )+i\right )}-\frac {4 a^{3} A}{f \,c^{2} \left (\tan \left (f x +e \right )+i\right )}+\frac {i a^{3} A \ln \left (\tan \left (f x +e \right )+i\right )}{f \,c^{2}}+\frac {5 a^{3} B \ln \left (\tan \left (f x +e \right )+i\right )}{f \,c^{2}}+\frac {2 i a^{3} A}{f \,c^{2} \left (\tan \left (f x +e \right )+i\right )^{2}}+\frac {2 a^{3} B}{f \,c^{2} \left (\tan \left (f x +e \right )+i\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^2,x)

[Out]

I*a^3*B*tan(f*x+e)/c^2/f+8*I/f*a^3/c^2/(tan(f*x+e)+I)*B-4/f*a^3/c^2/(tan(f*x+e)+I)*A+I/f*a^3/c^2*A*ln(tan(f*x+
e)+I)+5/f*a^3/c^2*B*ln(tan(f*x+e)+I)+2*I/f*a^3/c^2/(tan(f*x+e)+I)^2*A+2/f*a^3/c^2/(tan(f*x+e)+I)^2*B

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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+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(c-I*c*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 = 9.16, size = 182, normalized size = 1.48 \[ -\frac {a^3\,\left (7\,B\,\mathrm {tan}\left (e+f\,x\right )+B\,6{}\mathrm {i}+A\,\mathrm {tan}\left (e+f\,x\right )\,4{}\mathrm {i}-2\,A+B\,{\mathrm {tan}\left (e+f\,x\right )}^2\,2{}\mathrm {i}+B\,{\mathrm {tan}\left (e+f\,x\right )}^3-A\,\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )+B\,\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,5{}\mathrm {i}+A\,\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,\mathrm {tan}\left (e+f\,x\right )\,2{}\mathrm {i}+10\,B\,\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,\mathrm {tan}\left (e+f\,x\right )+A\,\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^2-B\,\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^2\,5{}\mathrm {i}\right )\,1{}\mathrm {i}}{c^2\,f\,{\left (-1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*tan(e + f*x))*(a + a*tan(e + f*x)*1i)^3)/(c - c*tan(e + f*x)*1i)^2,x)

[Out]

-(a^3*(B*6i - 2*A + A*tan(e + f*x)*4i + 7*B*tan(e + f*x) + B*tan(e + f*x)^2*2i + B*tan(e + f*x)^3 - A*log(tan(
e + f*x) + 1i) + B*log(tan(e + f*x) + 1i)*5i + A*log(tan(e + f*x) + 1i)*tan(e + f*x)*2i + 10*B*log(tan(e + f*x
) + 1i)*tan(e + f*x) + A*log(tan(e + f*x) + 1i)*tan(e + f*x)^2 - B*log(tan(e + f*x) + 1i)*tan(e + f*x)^2*5i)*1
i)/(c^2*f*(tan(e + f*x)*1i - 1)^2)

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sympy [A]  time = 1.00, size = 241, normalized size = 1.96 \[ \frac {2 B a^{3}}{- c^{2} f e^{2 i e} e^{2 i f x} - c^{2} f} - \frac {i a^{3} \left (A - 5 i B\right ) \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{c^{2} f} + \begin {cases} \frac {\left (2 i A a^{3} c^{2} f e^{2 i e} + 6 B a^{3} c^{2} f e^{2 i e}\right ) e^{2 i f x} + \left (- i A a^{3} c^{2} f e^{4 i e} - B a^{3} c^{2} f e^{4 i e}\right ) e^{4 i f x}}{2 c^{4} f^{2}} & \text {for}\: 2 c^{4} f^{2} \neq 0 \\- \frac {x \left (- 2 A a^{3} e^{4 i e} + 2 A a^{3} e^{2 i e} + 2 i B a^{3} e^{4 i e} - 6 i B a^{3} e^{2 i e}\right )}{c^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**2,x)

[Out]

2*B*a**3/(-c**2*f*exp(2*I*e)*exp(2*I*f*x) - c**2*f) - I*a**3*(A - 5*I*B)*log(exp(2*I*f*x) + exp(-2*I*e))/(c**2
*f) + Piecewise((((2*I*A*a**3*c**2*f*exp(2*I*e) + 6*B*a**3*c**2*f*exp(2*I*e))*exp(2*I*f*x) + (-I*A*a**3*c**2*f
*exp(4*I*e) - B*a**3*c**2*f*exp(4*I*e))*exp(4*I*f*x))/(2*c**4*f**2), Ne(2*c**4*f**2, 0)), (-x*(-2*A*a**3*exp(4
*I*e) + 2*A*a**3*exp(2*I*e) + 2*I*B*a**3*exp(4*I*e) - 6*I*B*a**3*exp(2*I*e))/c**2, True))

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