3.726 \(\int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^5} \, dx\)

Optimal. Leaf size=251 \[ -\frac {3 A+2 i B}{64 a^2 c^5 f (-\tan (e+f x)+i)}+\frac {5 (3 A+i B)}{128 a^2 c^5 f (\tan (e+f x)+i)}-\frac {-B+i A}{128 a^2 c^5 f (-\tan (e+f x)+i)^2}+\frac {-B+5 i A}{64 a^2 c^5 f (\tan (e+f x)+i)^2}-\frac {B+3 i A}{64 a^2 c^5 f (\tan (e+f x)+i)^4}+\frac {A-i B}{40 a^2 c^5 f (\tan (e+f x)+i)^5}+\frac {3 x (7 A+3 i B)}{128 a^2 c^5}-\frac {A}{16 a^2 c^5 f (\tan (e+f x)+i)^3} \]

[Out]

3/128*(7*A+3*I*B)*x/a^2/c^5+1/128*(-I*A+B)/a^2/c^5/f/(-tan(f*x+e)+I)^2+1/64*(-3*A-2*I*B)/a^2/c^5/f/(-tan(f*x+e
)+I)+1/40*(A-I*B)/a^2/c^5/f/(tan(f*x+e)+I)^5+1/64*(-3*I*A-B)/a^2/c^5/f/(tan(f*x+e)+I)^4-1/16*A/a^2/c^5/f/(tan(
f*x+e)+I)^3+1/64*(5*I*A-B)/a^2/c^5/f/(tan(f*x+e)+I)^2+5/128*(3*A+I*B)/a^2/c^5/f/(tan(f*x+e)+I)

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Rubi [A]  time = 0.30, 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} \[ -\frac {3 A+2 i B}{64 a^2 c^5 f (-\tan (e+f x)+i)}+\frac {5 (3 A+i B)}{128 a^2 c^5 f (\tan (e+f x)+i)}-\frac {-B+i A}{128 a^2 c^5 f (-\tan (e+f x)+i)^2}+\frac {-B+5 i A}{64 a^2 c^5 f (\tan (e+f x)+i)^2}-\frac {B+3 i A}{64 a^2 c^5 f (\tan (e+f x)+i)^4}+\frac {A-i B}{40 a^2 c^5 f (\tan (e+f x)+i)^5}+\frac {3 x (7 A+3 i B)}{128 a^2 c^5}-\frac {A}{16 a^2 c^5 f (\tan (e+f x)+i)^3} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Tan[e + f*x])/((a + I*a*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^5),x]

[Out]

(3*(7*A + (3*I)*B)*x)/(128*a^2*c^5) - (I*A - B)/(128*a^2*c^5*f*(I - Tan[e + f*x])^2) - (3*A + (2*I)*B)/(64*a^2
*c^5*f*(I - Tan[e + f*x])) + (A - I*B)/(40*a^2*c^5*f*(I + Tan[e + f*x])^5) - ((3*I)*A + B)/(64*a^2*c^5*f*(I +
Tan[e + f*x])^4) - A/(16*a^2*c^5*f*(I + Tan[e + f*x])^3) + ((5*I)*A - B)/(64*a^2*c^5*f*(I + Tan[e + f*x])^2) +
(5*(3*A + I*B))/(128*a^2*c^5*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))^2 (c-i c \tan (e+f x))^5} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {A+B x}{(a+i a x)^3 (c-i c x)^6} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {(a c) \operatorname {Subst}\left (\int \left (\frac {i (A+i B)}{64 a^3 c^6 (-i+x)^3}+\frac {-3 A-2 i B}{64 a^3 c^6 (-i+x)^2}+\frac {-A+i B}{8 a^3 c^6 (i+x)^6}+\frac {3 i A+B}{16 a^3 c^6 (i+x)^5}+\frac {3 A}{16 a^3 c^6 (i+x)^4}+\frac {-5 i A+B}{32 a^3 c^6 (i+x)^3}-\frac {5 (3 A+i B)}{128 a^3 c^6 (i+x)^2}+\frac {3 (7 A+3 i B)}{128 a^3 c^6 \left (1+x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {i A-B}{128 a^2 c^5 f (i-\tan (e+f x))^2}-\frac {3 A+2 i B}{64 a^2 c^5 f (i-\tan (e+f x))}+\frac {A-i B}{40 a^2 c^5 f (i+\tan (e+f x))^5}-\frac {3 i A+B}{64 a^2 c^5 f (i+\tan (e+f x))^4}-\frac {A}{16 a^2 c^5 f (i+\tan (e+f x))^3}+\frac {5 i A-B}{64 a^2 c^5 f (i+\tan (e+f x))^2}+\frac {5 (3 A+i B)}{128 a^2 c^5 f (i+\tan (e+f x))}+\frac {(3 (7 A+3 i B)) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{128 a^2 c^5 f}\\ &=\frac {3 (7 A+3 i B) x}{128 a^2 c^5}-\frac {i A-B}{128 a^2 c^5 f (i-\tan (e+f x))^2}-\frac {3 A+2 i B}{64 a^2 c^5 f (i-\tan (e+f x))}+\frac {A-i B}{40 a^2 c^5 f (i+\tan (e+f x))^5}-\frac {3 i A+B}{64 a^2 c^5 f (i+\tan (e+f x))^4}-\frac {A}{16 a^2 c^5 f (i+\tan (e+f x))^3}+\frac {5 i A-B}{64 a^2 c^5 f (i+\tan (e+f x))^2}+\frac {5 (3 A+i B)}{128 a^2 c^5 f (i+\tan (e+f x))}\\ \end {align*}

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Mathematica [A]  time = 3.54, size = 274, normalized size = 1.09 \[ \frac {\sec ^2(e+f x) (\cos (5 (e+f x))+i \sin (5 (e+f x))) (50 i (21 A+i B) \cos (e+f x)+20 (A (-42 f x+7 i)+3 B (1-6 i f x)) \cos (3 (e+f x))+350 A \sin (e+f x)-140 A \sin (3 (e+f x))+840 i A f x \sin (3 (e+f x))-175 A \sin (5 (e+f x))-14 A \sin (7 (e+f x))-105 i A \cos (5 (e+f x))-6 i A \cos (7 (e+f x))+150 i B \sin (e+f x)+60 i B \sin (3 (e+f x))-360 B f x \sin (3 (e+f x))-75 i B \sin (5 (e+f x))-6 i B \sin (7 (e+f x))+125 B \cos (5 (e+f x))+14 B \cos (7 (e+f x)))}{5120 a^2 c^5 f (\tan (e+f x)-i)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Tan[e + f*x])/((a + I*a*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^5),x]

[Out]

(Sec[e + f*x]^2*(Cos[5*(e + f*x)] + I*Sin[5*(e + f*x)])*((50*I)*(21*A + I*B)*Cos[e + f*x] + 20*(A*(7*I - 42*f*
x) + 3*B*(1 - (6*I)*f*x))*Cos[3*(e + f*x)] - (105*I)*A*Cos[5*(e + f*x)] + 125*B*Cos[5*(e + f*x)] - (6*I)*A*Cos
[7*(e + f*x)] + 14*B*Cos[7*(e + f*x)] + 350*A*Sin[e + f*x] + (150*I)*B*Sin[e + f*x] - 140*A*Sin[3*(e + f*x)] +
(60*I)*B*Sin[3*(e + f*x)] + (840*I)*A*f*x*Sin[3*(e + f*x)] - 360*B*f*x*Sin[3*(e + f*x)] - 175*A*Sin[5*(e + f*
x)] - (75*I)*B*Sin[5*(e + f*x)] - 14*A*Sin[7*(e + f*x)] - (6*I)*B*Sin[7*(e + f*x)]))/(5120*a^2*c^5*f*(-I + Tan
[e + f*x])^2)

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fricas [A]  time = 1.18, size = 149, normalized size = 0.59 \[ \frac {{\left (120 \, {\left (7 \, A + 3 i \, B\right )} f x e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (-4 i \, A - 4 \, B\right )} e^{\left (14 i \, f x + 14 i \, e\right )} + {\left (-35 i \, A - 25 \, B\right )} e^{\left (12 i \, f x + 12 i \, e\right )} + {\left (-140 i \, A - 60 \, B\right )} e^{\left (10 i \, f x + 10 i \, e\right )} + {\left (-350 i \, A - 50 \, B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} + {\left (-700 i \, A + 100 \, B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (140 i \, A - 100 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 10 i \, A - 10 \, B\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{5120 \, a^{2} c^{5} f} \]

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))^2/(c-I*c*tan(f*x+e))^5,x, algorithm="fricas")

[Out]

1/5120*(120*(7*A + 3*I*B)*f*x*e^(4*I*f*x + 4*I*e) + (-4*I*A - 4*B)*e^(14*I*f*x + 14*I*e) + (-35*I*A - 25*B)*e^
(12*I*f*x + 12*I*e) + (-140*I*A - 60*B)*e^(10*I*f*x + 10*I*e) + (-350*I*A - 50*B)*e^(8*I*f*x + 8*I*e) + (-700*
I*A + 100*B)*e^(6*I*f*x + 6*I*e) + (140*I*A - 100*B)*e^(2*I*f*x + 2*I*e) + 10*I*A - 10*B)*e^(-4*I*f*x - 4*I*e)
/(a^2*c^5*f)

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giac [A]  time = 4.86, size = 269, normalized size = 1.07 \[ -\frac {\frac {20 \, {\left (-21 i \, A + 9 \, B\right )} \log \left (\tan \left (f x + e\right ) + i\right )}{a^{2} c^{5}} + \frac {20 \, {\left (21 i \, A - 9 \, B\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{a^{2} c^{5}} + \frac {10 \, {\left (63 i \, A \tan \left (f x + e\right )^{2} - 27 \, B \tan \left (f x + e\right )^{2} + 150 \, A \tan \left (f x + e\right ) + 70 i \, B \tan \left (f x + e\right ) - 91 i \, A + 47 \, B\right )}}{a^{2} c^{5} {\left (-i \, \tan \left (f x + e\right ) - 1\right )}^{2}} + \frac {959 i \, A \tan \left (f x + e\right )^{5} - 411 \, B \tan \left (f x + e\right )^{5} - 5395 \, A \tan \left (f x + e\right )^{4} - 2255 i \, B \tan \left (f x + e\right )^{4} - 12390 i \, A \tan \left (f x + e\right )^{3} + 4990 \, B \tan \left (f x + e\right )^{3} + 14710 \, A \tan \left (f x + e\right )^{2} + 5550 i \, B \tan \left (f x + e\right )^{2} + 9275 i \, A \tan \left (f x + e\right ) - 3015 \, B \tan \left (f x + e\right ) - 2647 \, A - 483 i \, B}{a^{2} c^{5} {\left (\tan \left (f x + e\right ) + i\right )}^{5}}}{5120 \, f} \]

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))^2/(c-I*c*tan(f*x+e))^5,x, algorithm="giac")

[Out]

-1/5120*(20*(-21*I*A + 9*B)*log(tan(f*x + e) + I)/(a^2*c^5) + 20*(21*I*A - 9*B)*log(tan(f*x + e) - I)/(a^2*c^5
) + 10*(63*I*A*tan(f*x + e)^2 - 27*B*tan(f*x + e)^2 + 150*A*tan(f*x + e) + 70*I*B*tan(f*x + e) - 91*I*A + 47*B
)/(a^2*c^5*(-I*tan(f*x + e) - 1)^2) + (959*I*A*tan(f*x + e)^5 - 411*B*tan(f*x + e)^5 - 5395*A*tan(f*x + e)^4 -
2255*I*B*tan(f*x + e)^4 - 12390*I*A*tan(f*x + e)^3 + 4990*B*tan(f*x + e)^3 + 14710*A*tan(f*x + e)^2 + 5550*I*
B*tan(f*x + e)^2 + 9275*I*A*tan(f*x + e) - 3015*B*tan(f*x + e) - 2647*A - 483*I*B)/(a^2*c^5*(tan(f*x + e) + I)
^5))/f

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maple [A]  time = 0.45, size = 397, normalized size = 1.58 \[ \frac {15 A}{128 f \,a^{2} c^{5} \left (\tan \left (f x +e \right )+i\right )}+\frac {i B}{32 f \,a^{2} c^{5} \left (\tan \left (f x +e \right )-i\right )}-\frac {21 i \ln \left (\tan \left (f x +e \right )-i\right ) A}{256 f \,a^{2} c^{5}}-\frac {9 \ln \left (\tan \left (f x +e \right )+i\right ) B}{256 f \,a^{2} c^{5}}+\frac {5 i A}{64 f \,a^{2} c^{5} \left (\tan \left (f x +e \right )+i\right )^{2}}-\frac {B}{64 f \,a^{2} c^{5} \left (\tan \left (f x +e \right )+i\right )^{4}}-\frac {i A}{128 f \,a^{2} c^{5} \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {B}{64 f \,a^{2} c^{5} \left (\tan \left (f x +e \right )+i\right )^{2}}+\frac {A}{40 f \,a^{2} c^{5} \left (\tan \left (f x +e \right )+i\right )^{5}}-\frac {3 i A}{64 f \,a^{2} c^{5} \left (\tan \left (f x +e \right )+i\right )^{4}}-\frac {A}{16 a^{2} c^{5} f \left (\tan \left (f x +e \right )+i\right )^{3}}+\frac {B}{128 f \,a^{2} c^{5} \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {i B}{40 f \,a^{2} c^{5} \left (\tan \left (f x +e \right )+i\right )^{5}}+\frac {21 i \ln \left (\tan \left (f x +e \right )+i\right ) A}{256 f \,a^{2} c^{5}}+\frac {3 A}{64 f \,a^{2} c^{5} \left (\tan \left (f x +e \right )-i\right )}+\frac {5 i B}{128 f \,a^{2} c^{5} \left (\tan \left (f x +e \right )+i\right )}+\frac {9 \ln \left (\tan \left (f x +e \right )-i\right ) B}{256 f \,a^{2} c^{5}} \]

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))^2/(c-I*c*tan(f*x+e))^5,x)

[Out]

15/128/f/a^2/c^5/(tan(f*x+e)+I)*A+1/32*I/f/a^2/c^5/(tan(f*x+e)-I)*B-21/256*I/f/a^2/c^5*ln(tan(f*x+e)-I)*A-9/25
6/f/a^2/c^5*ln(tan(f*x+e)+I)*B+5/64*I/f/a^2/c^5/(tan(f*x+e)+I)^2*A-1/64/f/a^2/c^5/(tan(f*x+e)+I)^4*B-1/128*I/f
/a^2/c^5/(tan(f*x+e)-I)^2*A-1/64/f/a^2/c^5/(tan(f*x+e)+I)^2*B+1/40/f/a^2/c^5/(tan(f*x+e)+I)^5*A-3/64*I/f/a^2/c
^5/(tan(f*x+e)+I)^4*A-1/16*A/a^2/c^5/f/(tan(f*x+e)+I)^3+1/128/f/a^2/c^5/(tan(f*x+e)-I)^2*B-1/40*I/f/a^2/c^5/(t
an(f*x+e)+I)^5*B+21/256*I/f/a^2/c^5*ln(tan(f*x+e)+I)*A+3/64/f/a^2/c^5/(tan(f*x+e)-I)*A+5/128*I/f/a^2/c^5/(tan(
f*x+e)+I)*B+9/256/f/a^2/c^5*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))^2/(c-I*c*tan(f*x+e))^5,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 = 10.24, size = 291, normalized size = 1.16 \[ \frac {\mathrm {tan}\left (e+f\,x\right )\,\left (-\frac {3\,B}{640\,a^2\,c^5}+\frac {A\,7{}\mathrm {i}}{640\,a^2\,c^5}\right )+{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (\frac {7\,A}{32\,a^2\,c^5}+\frac {B\,3{}\mathrm {i}}{32\,a^2\,c^5}\right )-{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (-\frac {9\,B}{32\,a^2\,c^5}+\frac {A\,21{}\mathrm {i}}{32\,a^2\,c^5}\right )-{\mathrm {tan}\left (e+f\,x\right )}^6\,\left (\frac {21\,A}{128\,a^2\,c^5}+\frac {B\,9{}\mathrm {i}}{128\,a^2\,c^5}\right )-{\mathrm {tan}\left (e+f\,x\right )}^5\,\left (-\frac {27\,B}{128\,a^2\,c^5}+\frac {A\,63{}\mathrm {i}}{128\,a^2\,c^5}\right )+{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (\frac {469\,A}{640\,a^2\,c^5}+\frac {B\,201{}\mathrm {i}}{640\,a^2\,c^5}\right )+\frac {11\,A}{40\,a^2\,c^5}-\frac {B\,1{}\mathrm {i}}{40\,a^2\,c^5}}{f\,\left (-{\mathrm {tan}\left (e+f\,x\right )}^7-{\mathrm {tan}\left (e+f\,x\right )}^6\,3{}\mathrm {i}+{\mathrm {tan}\left (e+f\,x\right )}^5-{\mathrm {tan}\left (e+f\,x\right )}^4\,5{}\mathrm {i}+5\,{\mathrm {tan}\left (e+f\,x\right )}^3-{\mathrm {tan}\left (e+f\,x\right )}^2\,1{}\mathrm {i}+3\,\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )}+\frac {3\,x\,\left (7\,A+B\,3{}\mathrm {i}\right )}{128\,a^2\,c^5} \]

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)^2*(c - c*tan(e + f*x)*1i)^5),x)

[Out]

(tan(e + f*x)*((A*7i)/(640*a^2*c^5) - (3*B)/(640*a^2*c^5)) + tan(e + f*x)^4*((7*A)/(32*a^2*c^5) + (B*3i)/(32*a
^2*c^5)) - tan(e + f*x)^3*((A*21i)/(32*a^2*c^5) - (9*B)/(32*a^2*c^5)) - tan(e + f*x)^6*((21*A)/(128*a^2*c^5) +
(B*9i)/(128*a^2*c^5)) - tan(e + f*x)^5*((A*63i)/(128*a^2*c^5) - (27*B)/(128*a^2*c^5)) + tan(e + f*x)^2*((469*
A)/(640*a^2*c^5) + (B*201i)/(640*a^2*c^5)) + (11*A)/(40*a^2*c^5) - (B*1i)/(40*a^2*c^5))/(f*(3*tan(e + f*x) - t
an(e + f*x)^2*1i + 5*tan(e + f*x)^3 - tan(e + f*x)^4*5i + tan(e + f*x)^5 - tan(e + f*x)^6*3i - tan(e + f*x)^7
+ 1i)) + (3*x*(7*A + B*3i))/(128*a^2*c^5)

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sympy [A]  time = 1.26, size = 604, normalized size = 2.41 \[ \begin {cases} - \frac {\left (\left (- 11258999068426240 i A a^{12} c^{30} f^{6} e^{2 i e} + 11258999068426240 B a^{12} c^{30} f^{6} e^{2 i e}\right ) e^{- 4 i f x} + \left (- 157625986957967360 i A a^{12} c^{30} f^{6} e^{4 i e} + 112589990684262400 B a^{12} c^{30} f^{6} e^{4 i e}\right ) e^{- 2 i f x} + \left (788129934789836800 i A a^{12} c^{30} f^{6} e^{8 i e} - 112589990684262400 B a^{12} c^{30} f^{6} e^{8 i e}\right ) e^{2 i f x} + \left (394064967394918400 i A a^{12} c^{30} f^{6} e^{10 i e} + 56294995342131200 B a^{12} c^{30} f^{6} e^{10 i e}\right ) e^{4 i f x} + \left (157625986957967360 i A a^{12} c^{30} f^{6} e^{12 i e} + 67553994410557440 B a^{12} c^{30} f^{6} e^{12 i e}\right ) e^{6 i f x} + \left (39406496739491840 i A a^{12} c^{30} f^{6} e^{14 i e} + 28147497671065600 B a^{12} c^{30} f^{6} e^{14 i e}\right ) e^{8 i f x} + \left (4503599627370496 i A a^{12} c^{30} f^{6} e^{16 i e} + 4503599627370496 B a^{12} c^{30} f^{6} e^{16 i e}\right ) e^{10 i f x}\right ) e^{- 6 i e}}{5764607523034234880 a^{14} c^{35} f^{7}} & \text {for}\: 5764607523034234880 a^{14} c^{35} f^{7} e^{6 i e} \neq 0 \\x \left (- \frac {21 A + 9 i B}{128 a^{2} c^{5}} + \frac {\left (A e^{14 i e} + 7 A e^{12 i e} + 21 A e^{10 i e} + 35 A e^{8 i e} + 35 A e^{6 i e} + 21 A e^{4 i e} + 7 A e^{2 i e} + A - i B e^{14 i e} - 5 i B e^{12 i e} - 9 i B e^{10 i e} - 5 i B e^{8 i e} + 5 i B e^{6 i e} + 9 i B e^{4 i e} + 5 i B e^{2 i e} + i B\right ) e^{- 4 i e}}{128 a^{2} c^{5}}\right ) & \text {otherwise} \end {cases} - \frac {x \left (- 21 A - 9 i B\right )}{128 a^{2} c^{5}} \]

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))**2/(c-I*c*tan(f*x+e))**5,x)

[Out]

Piecewise((-((-11258999068426240*I*A*a**12*c**30*f**6*exp(2*I*e) + 11258999068426240*B*a**12*c**30*f**6*exp(2*
I*e))*exp(-4*I*f*x) + (-157625986957967360*I*A*a**12*c**30*f**6*exp(4*I*e) + 112589990684262400*B*a**12*c**30*
f**6*exp(4*I*e))*exp(-2*I*f*x) + (788129934789836800*I*A*a**12*c**30*f**6*exp(8*I*e) - 112589990684262400*B*a*
*12*c**30*f**6*exp(8*I*e))*exp(2*I*f*x) + (394064967394918400*I*A*a**12*c**30*f**6*exp(10*I*e) + 5629499534213
1200*B*a**12*c**30*f**6*exp(10*I*e))*exp(4*I*f*x) + (157625986957967360*I*A*a**12*c**30*f**6*exp(12*I*e) + 675
53994410557440*B*a**12*c**30*f**6*exp(12*I*e))*exp(6*I*f*x) + (39406496739491840*I*A*a**12*c**30*f**6*exp(14*I
*e) + 28147497671065600*B*a**12*c**30*f**6*exp(14*I*e))*exp(8*I*f*x) + (4503599627370496*I*A*a**12*c**30*f**6*
exp(16*I*e) + 4503599627370496*B*a**12*c**30*f**6*exp(16*I*e))*exp(10*I*f*x))*exp(-6*I*e)/(5764607523034234880
*a**14*c**35*f**7), Ne(5764607523034234880*a**14*c**35*f**7*exp(6*I*e), 0)), (x*(-(21*A + 9*I*B)/(128*a**2*c**
5) + (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(-4*I*e)/(128*a**2*c**5)), True)) - x*(-21*A - 9*I*B)/
(128*a**2*c**5)

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