3.737 \(\int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^4} \, dx\)

Optimal. Leaf size=251 \[ -\frac {5 (3 A+i B)}{128 a^3 c^4 f (-\tan (e+f x)+i)}-\frac {-3 B+5 i A}{128 a^3 c^4 f (-\tan (e+f x)+i)^2}+\frac {B+5 i A}{64 a^3 c^4 f (\tan (e+f x)+i)^2}+\frac {A+i B}{96 a^3 c^4 f (-\tan (e+f x)+i)^3}-\frac {2 A-i B}{48 a^3 c^4 f (\tan (e+f x)+i)^3}-\frac {B+i A}{64 a^3 c^4 f (\tan (e+f x)+i)^4}+\frac {5 x (7 A+i B)}{128 a^3 c^4}+\frac {5 A}{32 a^3 c^4 f (\tan (e+f x)+i)} \]

[Out]

5/128*(7*A+I*B)*x/a^3/c^4+1/96*(A+I*B)/a^3/c^4/f/(-tan(f*x+e)+I)^3+1/128*(-5*I*A+3*B)/a^3/c^4/f/(-tan(f*x+e)+I
)^2-5/128*(3*A+I*B)/a^3/c^4/f/(-tan(f*x+e)+I)+1/64*(-I*A-B)/a^3/c^4/f/(tan(f*x+e)+I)^4+1/48*(-2*A+I*B)/a^3/c^4
/f/(tan(f*x+e)+I)^3+1/64*(5*I*A+B)/a^3/c^4/f/(tan(f*x+e)+I)^2+5/32*A/a^3/c^4/f/(tan(f*x+e)+I)

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Rubi [A]  time = 0.31, antiderivative size = 251, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.073, Rules used = {3588, 77, 203} \[ -\frac {5 (3 A+i B)}{128 a^3 c^4 f (-\tan (e+f x)+i)}-\frac {-3 B+5 i A}{128 a^3 c^4 f (-\tan (e+f x)+i)^2}+\frac {B+5 i A}{64 a^3 c^4 f (\tan (e+f x)+i)^2}+\frac {A+i B}{96 a^3 c^4 f (-\tan (e+f x)+i)^3}-\frac {2 A-i B}{48 a^3 c^4 f (\tan (e+f x)+i)^3}-\frac {B+i A}{64 a^3 c^4 f (\tan (e+f x)+i)^4}+\frac {5 x (7 A+i B)}{128 a^3 c^4}+\frac {5 A}{32 a^3 c^4 f (\tan (e+f x)+i)} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Tan[e + f*x])/((a + I*a*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^4),x]

[Out]

(5*(7*A + I*B)*x)/(128*a^3*c^4) + (A + I*B)/(96*a^3*c^4*f*(I - Tan[e + f*x])^3) - ((5*I)*A - 3*B)/(128*a^3*c^4
*f*(I - Tan[e + f*x])^2) - (5*(3*A + I*B))/(128*a^3*c^4*f*(I - Tan[e + f*x])) - (I*A + B)/(64*a^3*c^4*f*(I + T
an[e + f*x])^4) - (2*A - I*B)/(48*a^3*c^4*f*(I + Tan[e + f*x])^3) + ((5*I)*A + B)/(64*a^3*c^4*f*(I + Tan[e + f
*x])^2) + (5*A)/(32*a^3*c^4*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 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

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)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^4} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {A+B x}{(a+i a x)^4 (c-i c x)^5} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {(a c) \operatorname {Subst}\left (\int \left (\frac {A+i B}{32 a^4 c^5 (-i+x)^4}+\frac {i (5 A+3 i B)}{64 a^4 c^5 (-i+x)^3}-\frac {5 (3 A+i B)}{128 a^4 c^5 (-i+x)^2}+\frac {i A+B}{16 a^4 c^5 (i+x)^5}+\frac {2 A-i B}{16 a^4 c^5 (i+x)^4}-\frac {i (5 A-i B)}{32 a^4 c^5 (i+x)^3}-\frac {5 A}{32 a^4 c^5 (i+x)^2}+\frac {5 (7 A+i B)}{128 a^4 c^5 \left (1+x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {A+i B}{96 a^3 c^4 f (i-\tan (e+f x))^3}-\frac {5 i A-3 B}{128 a^3 c^4 f (i-\tan (e+f x))^2}-\frac {5 (3 A+i B)}{128 a^3 c^4 f (i-\tan (e+f x))}-\frac {i A+B}{64 a^3 c^4 f (i+\tan (e+f x))^4}-\frac {2 A-i B}{48 a^3 c^4 f (i+\tan (e+f x))^3}+\frac {5 i A+B}{64 a^3 c^4 f (i+\tan (e+f x))^2}+\frac {5 A}{32 a^3 c^4 f (i+\tan (e+f x))}+\frac {(5 (7 A+i B)) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{128 a^3 c^4 f}\\ &=\frac {5 (7 A+i B) x}{128 a^3 c^4}+\frac {A+i B}{96 a^3 c^4 f (i-\tan (e+f x))^3}-\frac {5 i A-3 B}{128 a^3 c^4 f (i-\tan (e+f x))^2}-\frac {5 (3 A+i B)}{128 a^3 c^4 f (i-\tan (e+f x))}-\frac {i A+B}{64 a^3 c^4 f (i+\tan (e+f x))^4}-\frac {2 A-i B}{48 a^3 c^4 f (i+\tan (e+f x))^3}+\frac {5 i A+B}{64 a^3 c^4 f (i+\tan (e+f x))^2}+\frac {5 A}{32 a^3 c^4 f (i+\tan (e+f x))}\\ \end {align*}

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Mathematica [A]  time = 3.51, size = 267, normalized size = 1.06 \[ \frac {\sec ^3(e+f x) (-\cos (4 (e+f x))-i \sin (4 (e+f x))) (60 (A (-7-14 i f x)+B (2 f x+i)) \cos (e+f x)+18 (7 A+9 i B) \cos (3 (e+f x))-420 i A \sin (e+f x)-840 A f x \sin (e+f x)-378 i A \sin (3 (e+f x))-70 i A \sin (5 (e+f x))-7 i A \sin (7 (e+f x))+14 A \cos (5 (e+f x))+A \cos (7 (e+f x))-60 B \sin (e+f x)-120 i B f x \sin (e+f x)+54 B \sin (3 (e+f x))+10 B \sin (5 (e+f x))+B \sin (7 (e+f x))+50 i B \cos (5 (e+f x))+7 i B \cos (7 (e+f x)))}{3072 a^3 c^4 f (\tan (e+f x)-i)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Tan[e + f*x])/((a + I*a*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^4),x]

[Out]

(Sec[e + f*x]^3*(-Cos[4*(e + f*x)] - I*Sin[4*(e + f*x)])*(60*(A*(-7 - (14*I)*f*x) + B*(I + 2*f*x))*Cos[e + f*x
] + 18*(7*A + (9*I)*B)*Cos[3*(e + f*x)] + 14*A*Cos[5*(e + f*x)] + (50*I)*B*Cos[5*(e + f*x)] + A*Cos[7*(e + f*x
)] + (7*I)*B*Cos[7*(e + f*x)] - (420*I)*A*Sin[e + f*x] - 60*B*Sin[e + f*x] - 840*A*f*x*Sin[e + f*x] - (120*I)*
B*f*x*Sin[e + f*x] - (378*I)*A*Sin[3*(e + f*x)] + 54*B*Sin[3*(e + f*x)] - (70*I)*A*Sin[5*(e + f*x)] + 10*B*Sin
[5*(e + f*x)] - (7*I)*A*Sin[7*(e + f*x)] + B*Sin[7*(e + f*x)]))/(3072*a^3*c^4*f*(-I + Tan[e + f*x])^3)

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fricas [A]  time = 0.74, size = 149, normalized size = 0.59 \[ \frac {{\left (120 \, {\left (7 \, A + i \, B\right )} f x e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (-3 i \, A - 3 \, B\right )} e^{\left (14 i \, f x + 14 i \, e\right )} + {\left (-28 i \, A - 20 \, B\right )} e^{\left (12 i \, f x + 12 i \, e\right )} + {\left (-126 i \, A - 54 \, B\right )} e^{\left (10 i \, f x + 10 i \, e\right )} + {\left (-420 i \, A - 60 \, B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} + {\left (252 i \, A - 108 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (42 i \, A - 30 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 4 i \, A - 4 \, B\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{3072 \, a^{3} c^{4} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^4,x, algorithm="fricas")

[Out]

1/3072*(120*(7*A + I*B)*f*x*e^(6*I*f*x + 6*I*e) + (-3*I*A - 3*B)*e^(14*I*f*x + 14*I*e) + (-28*I*A - 20*B)*e^(1
2*I*f*x + 12*I*e) + (-126*I*A - 54*B)*e^(10*I*f*x + 10*I*e) + (-420*I*A - 60*B)*e^(8*I*f*x + 8*I*e) + (252*I*A
- 108*B)*e^(4*I*f*x + 4*I*e) + (42*I*A - 30*B)*e^(2*I*f*x + 2*I*e) + 4*I*A - 4*B)*e^(-6*I*f*x - 6*I*e)/(a^3*c
^4*f)

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giac [A]  time = 3.85, size = 271, normalized size = 1.08 \[ \frac {\frac {12 \, {\left (35 i \, A - 5 \, B\right )} \log \left (\tan \left (f x + e\right ) + i\right )}{a^{3} c^{4}} - \frac {12 \, {\left (35 i \, A - 5 \, B\right )} \log \left (-i \, \tan \left (f x + e\right ) - 1\right )}{a^{3} c^{4}} + \frac {2 \, {\left (385 \, A \tan \left (f x + e\right )^{3} + 55 i \, B \tan \left (f x + e\right )^{3} - 1335 i \, A \tan \left (f x + e\right )^{2} + 225 \, B \tan \left (f x + e\right )^{2} - 1575 \, A \tan \left (f x + e\right ) - 321 i \, B \tan \left (f x + e\right ) + 641 i \, A - 167 \, B\right )}}{a^{3} c^{4} {\left (i \, \tan \left (f x + e\right ) + 1\right )}^{3}} + \frac {-875 i \, A \tan \left (f x + e\right )^{4} + 125 \, B \tan \left (f x + e\right )^{4} + 3980 \, A \tan \left (f x + e\right )^{3} + 500 i \, B \tan \left (f x + e\right )^{3} + 6930 i \, A \tan \left (f x + e\right )^{2} - 702 \, B \tan \left (f x + e\right )^{2} - 5548 \, A \tan \left (f x + e\right ) - 340 i \, B \tan \left (f x + e\right ) - 1771 i \, A - 35 \, B}{a^{3} c^{4} {\left (\tan \left (f x + e\right ) + i\right )}^{4}}}{3072 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^4,x, algorithm="giac")

[Out]

1/3072*(12*(35*I*A - 5*B)*log(tan(f*x + e) + I)/(a^3*c^4) - 12*(35*I*A - 5*B)*log(-I*tan(f*x + e) - 1)/(a^3*c^
4) + 2*(385*A*tan(f*x + e)^3 + 55*I*B*tan(f*x + e)^3 - 1335*I*A*tan(f*x + e)^2 + 225*B*tan(f*x + e)^2 - 1575*A
*tan(f*x + e) - 321*I*B*tan(f*x + e) + 641*I*A - 167*B)/(a^3*c^4*(I*tan(f*x + e) + 1)^3) + (-875*I*A*tan(f*x +
e)^4 + 125*B*tan(f*x + e)^4 + 3980*A*tan(f*x + e)^3 + 500*I*B*tan(f*x + e)^3 + 6930*I*A*tan(f*x + e)^2 - 702*
B*tan(f*x + e)^2 - 5548*A*tan(f*x + e) - 340*I*B*tan(f*x + e) - 1771*I*A - 35*B)/(a^3*c^4*(tan(f*x + e) + I)^4
))/f

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maple [A]  time = 0.48, size = 397, normalized size = 1.58 \[ \frac {5 A}{32 a^{3} c^{4} f \left (\tan \left (f x +e \right )+i\right )}+\frac {5 i A}{64 f \,a^{3} c^{4} \left (\tan \left (f x +e \right )+i\right )^{2}}-\frac {5 \ln \left (\tan \left (f x +e \right )+i\right ) B}{256 f \,a^{3} c^{4}}-\frac {B}{64 f \,a^{3} c^{4} \left (\tan \left (f x +e \right )+i\right )^{4}}+\frac {i B}{48 f \,a^{3} c^{4} \left (\tan \left (f x +e \right )+i\right )^{3}}+\frac {5 i B}{128 f \,a^{3} c^{4} \left (\tan \left (f x +e \right )-i\right )}+\frac {B}{64 f \,a^{3} c^{4} \left (\tan \left (f x +e \right )+i\right )^{2}}-\frac {A}{24 f \,a^{3} c^{4} \left (\tan \left (f x +e \right )+i\right )^{3}}-\frac {i A}{64 f \,a^{3} c^{4} \left (\tan \left (f x +e \right )+i\right )^{4}}-\frac {5 i A}{128 f \,a^{3} c^{4} \left (\tan \left (f x +e \right )-i\right )^{2}}+\frac {3 B}{128 f \,a^{3} c^{4} \left (\tan \left (f x +e \right )-i\right )^{2}}+\frac {15 A}{128 f \,a^{3} c^{4} \left (\tan \left (f x +e \right )-i\right )}+\frac {35 i \ln \left (\tan \left (f x +e \right )+i\right ) A}{256 f \,a^{3} c^{4}}-\frac {A}{96 f \,a^{3} c^{4} \left (\tan \left (f x +e \right )-i\right )^{3}}-\frac {i B}{96 f \,a^{3} c^{4} \left (\tan \left (f x +e \right )-i\right )^{3}}-\frac {35 i \ln \left (\tan \left (f x +e \right )-i\right ) A}{256 f \,a^{3} c^{4}}+\frac {5 \ln \left (\tan \left (f x +e \right )-i\right ) B}{256 f \,a^{3} c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^4,x)

[Out]

5/32*A/a^3/c^4/f/(tan(f*x+e)+I)+5/64*I/f/a^3/c^4/(tan(f*x+e)+I)^2*A-5/256/f/a^3/c^4*ln(tan(f*x+e)+I)*B-1/64/f/
a^3/c^4/(tan(f*x+e)+I)^4*B+1/48*I/f/a^3/c^4/(tan(f*x+e)+I)^3*B+5/128*I/f/a^3/c^4/(tan(f*x+e)-I)*B+1/64/f/a^3/c
^4/(tan(f*x+e)+I)^2*B-1/24/f/a^3/c^4/(tan(f*x+e)+I)^3*A-1/64*I/f/a^3/c^4/(tan(f*x+e)+I)^4*A-5/128*I/f/a^3/c^4/
(tan(f*x+e)-I)^2*A+3/128/f/a^3/c^4/(tan(f*x+e)-I)^2*B+15/128/f/a^3/c^4/(tan(f*x+e)-I)*A+35/256*I/f/a^3/c^4*ln(
tan(f*x+e)+I)*A-1/96/f/a^3/c^4/(tan(f*x+e)-I)^3*A-1/96*I/f/a^3/c^4/(tan(f*x+e)-I)^3*B-35/256*I/f/a^3/c^4*ln(ta
n(f*x+e)-I)*A+5/256/f/a^3/c^4*ln(tan(f*x+e)-I)*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+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^4,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 = 10.40, size = 286, normalized size = 1.14 \[ \frac {\mathrm {tan}\left (e+f\,x\right )\,\left (-\frac {11\,B}{128\,a^3\,c^4}+\frac {A\,77{}\mathrm {i}}{128\,a^3\,c^4}\right )+{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (-\frac {5\,B}{48\,a^3\,c^4}+\frac {A\,35{}\mathrm {i}}{48\,a^3\,c^4}\right )+{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (\frac {35\,A}{48\,a^3\,c^4}+\frac {B\,5{}\mathrm {i}}{48\,a^3\,c^4}\right )+{\mathrm {tan}\left (e+f\,x\right )}^5\,\left (-\frac {5\,B}{128\,a^3\,c^4}+\frac {A\,35{}\mathrm {i}}{128\,a^3\,c^4}\right )+{\mathrm {tan}\left (e+f\,x\right )}^6\,\left (\frac {35\,A}{128\,a^3\,c^4}+\frac {B\,5{}\mathrm {i}}{128\,a^3\,c^4}\right )+{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (\frac {77\,A}{128\,a^3\,c^4}+\frac {B\,11{}\mathrm {i}}{128\,a^3\,c^4}\right )+\frac {A}{8\,a^3\,c^4}-\frac {B\,1{}\mathrm {i}}{8\,a^3\,c^4}}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^7+{\mathrm {tan}\left (e+f\,x\right )}^6\,1{}\mathrm {i}+3\,{\mathrm {tan}\left (e+f\,x\right )}^5+{\mathrm {tan}\left (e+f\,x\right )}^4\,3{}\mathrm {i}+3\,{\mathrm {tan}\left (e+f\,x\right )}^3+{\mathrm {tan}\left (e+f\,x\right )}^2\,3{}\mathrm {i}+\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )}+\frac {5\,x\,\left (7\,A+B\,1{}\mathrm {i}\right )}{128\,a^3\,c^4} \]

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)^4),x)

[Out]

(tan(e + f*x)*((A*77i)/(128*a^3*c^4) - (11*B)/(128*a^3*c^4)) + tan(e + f*x)^3*((A*35i)/(48*a^3*c^4) - (5*B)/(4
8*a^3*c^4)) + tan(e + f*x)^4*((35*A)/(48*a^3*c^4) + (B*5i)/(48*a^3*c^4)) + tan(e + f*x)^5*((A*35i)/(128*a^3*c^
4) - (5*B)/(128*a^3*c^4)) + tan(e + f*x)^6*((35*A)/(128*a^3*c^4) + (B*5i)/(128*a^3*c^4)) + tan(e + f*x)^2*((77
*A)/(128*a^3*c^4) + (B*11i)/(128*a^3*c^4)) + A/(8*a^3*c^4) - (B*1i)/(8*a^3*c^4))/(f*(tan(e + f*x) + tan(e + f*
x)^2*3i + 3*tan(e + f*x)^3 + tan(e + f*x)^4*3i + 3*tan(e + f*x)^5 + tan(e + f*x)^6*1i + tan(e + f*x)^7 + 1i))
+ (5*x*(7*A + B*1i))/(128*a^3*c^4)

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sympy [A]  time = 1.26, size = 604, normalized size = 2.41 \[ \begin {cases} - \frac {\left (\left (- 13510798882111488 i A a^{18} c^{24} f^{6} e^{6 i e} + 13510798882111488 B a^{18} c^{24} f^{6} e^{6 i e}\right ) e^{- 6 i f x} + \left (- 141863388262170624 i A a^{18} c^{24} f^{6} e^{8 i e} + 101330991615836160 B a^{18} c^{24} f^{6} e^{8 i e}\right ) e^{- 4 i f x} + \left (- 851180329573023744 i A a^{18} c^{24} f^{6} e^{10 i e} + 364791569817010176 B a^{18} c^{24} f^{6} e^{10 i e}\right ) e^{- 2 i f x} + \left (1418633882621706240 i A a^{18} c^{24} f^{6} e^{14 i e} + 202661983231672320 B a^{18} c^{24} f^{6} e^{14 i e}\right ) e^{2 i f x} + \left (425590164786511872 i A a^{18} c^{24} f^{6} e^{16 i e} + 182395784908505088 B a^{18} c^{24} f^{6} e^{16 i e}\right ) e^{4 i f x} + \left (94575592174780416 i A a^{18} c^{24} f^{6} e^{18 i e} + 67553994410557440 B a^{18} c^{24} f^{6} e^{18 i e}\right ) e^{6 i f x} + \left (10133099161583616 i A a^{18} c^{24} f^{6} e^{20 i e} + 10133099161583616 B a^{18} c^{24} f^{6} e^{20 i e}\right ) e^{8 i f x}\right ) e^{- 12 i e}}{10376293541461622784 a^{21} c^{28} f^{7}} & \text {for}\: 10376293541461622784 a^{21} c^{28} f^{7} e^{12 i e} \neq 0 \\x \left (- \frac {35 A + 5 i B}{128 a^{3} c^{4}} + \frac {\left (A e^{14 i e} + 7 A e^{12 i e} + 21 A e^{10 i e} + 35 A e^{8 i e} + 35 A e^{6 i e} + 21 A e^{4 i e} + 7 A e^{2 i e} + A - i B e^{14 i e} - 5 i B e^{12 i e} - 9 i B e^{10 i e} - 5 i B e^{8 i e} + 5 i B e^{6 i e} + 9 i B e^{4 i e} + 5 i B e^{2 i e} + i B\right ) e^{- 6 i e}}{128 a^{3} c^{4}}\right ) & \text {otherwise} \end {cases} - \frac {x \left (- 35 A - 5 i B\right )}{128 a^{3} c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))**3/(c-I*c*tan(f*x+e))**4,x)

[Out]

Piecewise((-((-13510798882111488*I*A*a**18*c**24*f**6*exp(6*I*e) + 13510798882111488*B*a**18*c**24*f**6*exp(6*
I*e))*exp(-6*I*f*x) + (-141863388262170624*I*A*a**18*c**24*f**6*exp(8*I*e) + 101330991615836160*B*a**18*c**24*
f**6*exp(8*I*e))*exp(-4*I*f*x) + (-851180329573023744*I*A*a**18*c**24*f**6*exp(10*I*e) + 364791569817010176*B*
a**18*c**24*f**6*exp(10*I*e))*exp(-2*I*f*x) + (1418633882621706240*I*A*a**18*c**24*f**6*exp(14*I*e) + 20266198
3231672320*B*a**18*c**24*f**6*exp(14*I*e))*exp(2*I*f*x) + (425590164786511872*I*A*a**18*c**24*f**6*exp(16*I*e)
+ 182395784908505088*B*a**18*c**24*f**6*exp(16*I*e))*exp(4*I*f*x) + (94575592174780416*I*A*a**18*c**24*f**6*e
xp(18*I*e) + 67553994410557440*B*a**18*c**24*f**6*exp(18*I*e))*exp(6*I*f*x) + (10133099161583616*I*A*a**18*c**
24*f**6*exp(20*I*e) + 10133099161583616*B*a**18*c**24*f**6*exp(20*I*e))*exp(8*I*f*x))*exp(-12*I*e)/(1037629354
1461622784*a**21*c**28*f**7), Ne(10376293541461622784*a**21*c**28*f**7*exp(12*I*e), 0)), (x*(-(35*A + 5*I*B)/(
128*a**3*c**4) + (A*exp(14*I*e) + 7*A*exp(12*I*e) + 21*A*exp(10*I*e) + 35*A*exp(8*I*e) + 35*A*exp(6*I*e) + 21*
A*exp(4*I*e) + 7*A*exp(2*I*e) + A - I*B*exp(14*I*e) - 5*I*B*exp(12*I*e) - 9*I*B*exp(10*I*e) - 5*I*B*exp(8*I*e)
+ 5*I*B*exp(6*I*e) + 9*I*B*exp(4*I*e) + 5*I*B*exp(2*I*e) + I*B)*exp(-6*I*e)/(128*a**3*c**4)), True)) - x*(-35
*A - 5*I*B)/(128*a**3*c**4)

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