3.728 \(\int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^5}{(a+i a \tan (e+f x))^3} \, dx\)

Optimal. Leaf size=191 \[ \frac {c^5 (A+8 i B) \tan (e+f x)}{a^3 f}-\frac {8 c^5 (3 A+7 i B)}{a^3 f (-\tan (e+f x)+i)}+\frac {8 c^5 (-3 B+2 i A)}{a^3 f (-\tan (e+f x)+i)^2}+\frac {16 c^5 (A+i B)}{3 a^3 f (-\tan (e+f x)+i)^3}-\frac {8 c^5 (-4 B+i A) \log (\cos (e+f x))}{a^3 f}-\frac {8 c^5 x (A+4 i B)}{a^3}+\frac {B c^5 \tan ^2(e+f x)}{2 a^3 f} \]

[Out]

-8*(A+4*I*B)*c^5*x/a^3-8*(I*A-4*B)*c^5*ln(cos(f*x+e))/a^3/f+16/3*(A+I*B)*c^5/a^3/f/(-tan(f*x+e)+I)^3+8*(2*I*A-
3*B)*c^5/a^3/f/(-tan(f*x+e)+I)^2-8*(3*A+7*I*B)*c^5/a^3/f/(-tan(f*x+e)+I)+(A+8*I*B)*c^5*tan(f*x+e)/a^3/f+1/2*B*
c^5*tan(f*x+e)^2/a^3/f

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Rubi [A]  time = 0.24, antiderivative size = 191, 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} \[ \frac {c^5 (A+8 i B) \tan (e+f x)}{a^3 f}-\frac {8 c^5 (3 A+7 i B)}{a^3 f (-\tan (e+f x)+i)}+\frac {8 c^5 (-3 B+2 i A)}{a^3 f (-\tan (e+f x)+i)^2}+\frac {16 c^5 (A+i B)}{3 a^3 f (-\tan (e+f x)+i)^3}-\frac {8 c^5 (-4 B+i A) \log (\cos (e+f x))}{a^3 f}-\frac {8 c^5 x (A+4 i B)}{a^3}+\frac {B c^5 \tan ^2(e+f x)}{2 a^3 f} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^5)/(a + I*a*Tan[e + f*x])^3,x]

[Out]

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

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Mathematica [B]  time = 11.67, size = 1496, normalized size = 7.83 \[ \text {result too large to display} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^5)/(a + I*a*Tan[e + f*x])^3,x]

[Out]

((A + (3*I)*B)*Cos[2*f*x]*Sec[e + f*x]^2*((6*I)*c^5*Cos[e] - 6*c^5*Sin[e])*(Cos[f*x] + I*Sin[f*x])^3*(A + B*Ta
n[e + f*x]))/(f*(A*Cos[e + f*x] + B*Sin[e + f*x])*(a + I*a*Tan[e + f*x])^3) + (((-I)*A + 2*B)*Cos[4*f*x]*Sec[e
+ f*x]^2*(2*c^5*Cos[e] - (2*I)*c^5*Sin[e])*(Cos[f*x] + I*Sin[f*x])^3*(A + B*Tan[e + f*x]))/(f*(A*Cos[e + f*x]
+ B*Sin[e + f*x])*(a + I*a*Tan[e + f*x])^3) + (Sec[e + f*x]^2*((-I)*A*c^5*Cos[(3*e)/2] + 4*B*c^5*Cos[(3*e)/2]
+ A*c^5*Sin[(3*e)/2] + (4*I)*B*c^5*Sin[(3*e)/2])*(8*Cos[(3*e)/2]*Log[Cos[e + f*x]] + (8*I)*Log[Cos[e + f*x]]*
Sin[(3*e)/2])*(Cos[f*x] + I*Sin[f*x])^3*(A + B*Tan[e + f*x]))/(f*(A*Cos[e + f*x] + B*Sin[e + f*x])*(a + I*a*Ta
n[e + f*x])^3) + ((A + I*B)*Cos[6*f*x]*Sec[e + f*x]^2*(((2*I)/3)*c^5*Cos[3*e] + (2*c^5*Sin[3*e])/3)*(Cos[f*x]
+ I*Sin[f*x])^3*(A + B*Tan[e + f*x]))/(f*(A*Cos[e + f*x] + B*Sin[e + f*x])*(a + I*a*Tan[e + f*x])^3) + (Sec[e
+ f*x]^4*((B*c^5*Cos[3*e])/2 + (I/2)*B*c^5*Sin[3*e])*(Cos[f*x] + I*Sin[f*x])^3*(A + B*Tan[e + f*x]))/(f*(A*Cos
[e + f*x] + B*Sin[e + f*x])*(a + I*a*Tan[e + f*x])^3) + ((A + (4*I)*B)*Sec[e + f*x]^2*(-8*c^5*f*x*Cos[3*e] - (
8*I)*c^5*f*x*Sin[3*e])*(Cos[f*x] + I*Sin[f*x])^3*(A + B*Tan[e + f*x]))/(f*(A*Cos[e + f*x] + B*Sin[e + f*x])*(a
+ I*a*Tan[e + f*x])^3) + ((A + (3*I)*B)*Sec[e + f*x]^2*(6*c^5*Cos[e] + (6*I)*c^5*Sin[e])*(Cos[f*x] + I*Sin[f*
x])^3*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])^3) + ((A +
(2*I)*B)*Sec[e + f*x]^2*(-2*c^5*Cos[e] + (2*I)*c^5*Sin[e])*(Cos[f*x] + I*Sin[f*x])^3*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])^3) + ((A + I*B)*Sec[e + f*x]^2*((2*c^5*Cos
[3*e])/3 - ((2*I)/3)*c^5*Sin[3*e])*(Cos[f*x] + I*Sin[f*x])^3*Sin[6*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])^3) + (Sec[e + f*x]^3*(Cos[f*x] + I*Sin[f*x])^3*((I/2)*A*c^5*Cos[3*
e - f*x] - 4*B*c^5*Cos[3*e - f*x] - (I/2)*A*c^5*Cos[3*e + f*x] + 4*B*c^5*Cos[3*e + f*x] - (A*c^5*Sin[3*e - f*x
])/2 - (4*I)*B*c^5*Sin[3*e - f*x] + (A*c^5*Sin[3*e + f*x])/2 + (4*I)*B*c^5*Sin[3*e + f*x])*(A + B*Tan[e + f*x]
))/(f*(Cos[e/2] - Sin[e/2])*(Cos[e/2] + Sin[e/2])*(A*Cos[e + f*x] + B*Sin[e + f*x])*(a + I*a*Tan[e + f*x])^3)
+ (x*Sec[e + f*x]^2*(Cos[f*x] + I*Sin[f*x])^3*(4*A*c^5*Cos[e] + (16*I)*B*c^5*Cos[e] - 4*A*c^5*Cos[e]^3 - (16*I
)*B*c^5*Cos[e]^3 + (8*I)*A*c^5*Sin[e] - 32*B*c^5*Sin[e] - (16*I)*A*c^5*Cos[e]^2*Sin[e] + 64*B*c^5*Cos[e]^2*Sin
[e] + 24*A*c^5*Cos[e]*Sin[e]^2 + (96*I)*B*c^5*Cos[e]*Sin[e]^2 + (16*I)*A*c^5*Sin[e]^3 - 64*B*c^5*Sin[e]^3 - 4*
A*c^5*Sin[e]*Tan[e] - (16*I)*B*c^5*Sin[e]*Tan[e] - 4*A*c^5*Sin[e]^3*Tan[e] - (16*I)*B*c^5*Sin[e]^3*Tan[e] + I*
(A + (4*I)*B)*(8*c^5*Cos[3*e] + (8*I)*c^5*Sin[3*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])^3)

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fricas [A]  time = 0.71, size = 269, normalized size = 1.41 \[ -\frac {48 \, {\left (A + 4 i \, B\right )} c^{5} f x e^{\left (10 i \, f x + 10 i \, e\right )} - {\left (8 i \, A - 32 \, B\right )} c^{5} e^{\left (4 i \, f x + 4 i \, e\right )} - {\left (-2 i \, A + 8 \, B\right )} c^{5} e^{\left (2 i \, f x + 2 i \, e\right )} - {\left (2 i \, A - 2 \, B\right )} c^{5} + {\left (96 \, {\left (A + 4 i \, B\right )} c^{5} f x - {\left (24 i \, A - 96 \, B\right )} c^{5}\right )} e^{\left (8 i \, f x + 8 i \, e\right )} + {\left (48 \, {\left (A + 4 i \, B\right )} c^{5} f x - {\left (36 i \, A - 144 \, B\right )} c^{5}\right )} e^{\left (6 i \, f x + 6 i \, e\right )} - {\left ({\left (-24 i \, A + 96 \, B\right )} c^{5} e^{\left (10 i \, f x + 10 i \, e\right )} + {\left (-48 i \, A + 192 \, B\right )} c^{5} e^{\left (8 i \, f x + 8 i \, e\right )} + {\left (-24 i \, A + 96 \, B\right )} c^{5} e^{\left (6 i \, f x + 6 i \, e\right )}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{3 \, {\left (a^{3} f e^{\left (10 i \, f x + 10 i \, e\right )} + 2 \, a^{3} f e^{\left (8 i \, f x + 8 i \, e\right )} + a^{3} f e^{\left (6 i \, f x + 6 i \, e\right )}\right )}} \]

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))^5/(a+I*a*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

-1/3*(48*(A + 4*I*B)*c^5*f*x*e^(10*I*f*x + 10*I*e) - (8*I*A - 32*B)*c^5*e^(4*I*f*x + 4*I*e) - (-2*I*A + 8*B)*c
^5*e^(2*I*f*x + 2*I*e) - (2*I*A - 2*B)*c^5 + (96*(A + 4*I*B)*c^5*f*x - (24*I*A - 96*B)*c^5)*e^(8*I*f*x + 8*I*e
) + (48*(A + 4*I*B)*c^5*f*x - (36*I*A - 144*B)*c^5)*e^(6*I*f*x + 6*I*e) - ((-24*I*A + 96*B)*c^5*e^(10*I*f*x +
10*I*e) + (-48*I*A + 192*B)*c^5*e^(8*I*f*x + 8*I*e) + (-24*I*A + 96*B)*c^5*e^(6*I*f*x + 6*I*e))*log(e^(2*I*f*x
+ 2*I*e) + 1))/(a^3*f*e^(10*I*f*x + 10*I*e) + 2*a^3*f*e^(8*I*f*x + 8*I*e) + a^3*f*e^(6*I*f*x + 6*I*e))

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giac [B]  time = 5.71, size = 515, normalized size = 2.70 \[ -\frac {2 \, {\left (\frac {15 \, {\left (4 i \, A c^{5} - 16 \, B c^{5}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{a^{3}} - \frac {15 \, {\left (8 i \, A c^{5} - 32 \, B c^{5}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}{a^{3}} - \frac {15 \, {\left (-4 i \, A c^{5} + 16 \, B c^{5}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}{a^{3}} - \frac {15 \, {\left (6 i \, A c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 24 \, B c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - A c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 8 i \, B c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 12 i \, A c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 49 \, B c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + A c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 8 i \, B c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 6 i \, A c^{5} - 24 \, B c^{5}\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{2} a^{3}} + \frac {294 i \, A c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 1176 \, B c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 1884 \, A c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 7416 i \, B c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 4890 i \, A c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 19320 \, B c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 6920 \, A c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 26480 i \, B c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 4890 i \, A c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 19320 \, B c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1884 \, A c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 7416 i \, B c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 294 i \, A c^{5} + 1176 \, B c^{5}}{a^{3} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}^{6}}\right )}}{15 \, 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))^5/(a+I*a*tan(f*x+e))^3,x, algorithm="giac")

[Out]

-2/15*(15*(4*I*A*c^5 - 16*B*c^5)*log(tan(1/2*f*x + 1/2*e) + 1)/a^3 - 15*(8*I*A*c^5 - 32*B*c^5)*log(tan(1/2*f*x
+ 1/2*e) - I)/a^3 - 15*(-4*I*A*c^5 + 16*B*c^5)*log(tan(1/2*f*x + 1/2*e) - 1)/a^3 - 15*(6*I*A*c^5*tan(1/2*f*x
+ 1/2*e)^4 - 24*B*c^5*tan(1/2*f*x + 1/2*e)^4 - A*c^5*tan(1/2*f*x + 1/2*e)^3 - 8*I*B*c^5*tan(1/2*f*x + 1/2*e)^3
- 12*I*A*c^5*tan(1/2*f*x + 1/2*e)^2 + 49*B*c^5*tan(1/2*f*x + 1/2*e)^2 + A*c^5*tan(1/2*f*x + 1/2*e) + 8*I*B*c^
5*tan(1/2*f*x + 1/2*e) + 6*I*A*c^5 - 24*B*c^5)/((tan(1/2*f*x + 1/2*e)^2 - 1)^2*a^3) + (294*I*A*c^5*tan(1/2*f*x
+ 1/2*e)^6 - 1176*B*c^5*tan(1/2*f*x + 1/2*e)^6 + 1884*A*c^5*tan(1/2*f*x + 1/2*e)^5 + 7416*I*B*c^5*tan(1/2*f*x
+ 1/2*e)^5 - 4890*I*A*c^5*tan(1/2*f*x + 1/2*e)^4 + 19320*B*c^5*tan(1/2*f*x + 1/2*e)^4 - 6920*A*c^5*tan(1/2*f*
x + 1/2*e)^3 - 26480*I*B*c^5*tan(1/2*f*x + 1/2*e)^3 + 4890*I*A*c^5*tan(1/2*f*x + 1/2*e)^2 - 19320*B*c^5*tan(1/
2*f*x + 1/2*e)^2 + 1884*A*c^5*tan(1/2*f*x + 1/2*e) + 7416*I*B*c^5*tan(1/2*f*x + 1/2*e) - 294*I*A*c^5 + 1176*B*
c^5)/(a^3*(tan(1/2*f*x + 1/2*e) - I)^6))/f

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maple [A]  time = 0.25, size = 244, normalized size = 1.28 \[ \frac {c^{5} A \tan \left (f x +e \right )}{f \,a^{3}}+\frac {8 i c^{5} B \tan \left (f x +e \right )}{f \,a^{3}}+\frac {B \,c^{5} \left (\tan ^{2}\left (f x +e \right )\right )}{2 a^{3} f}+\frac {16 i c^{5} A}{f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {24 c^{5} B}{f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{2}}+\frac {56 i c^{5} B}{f \,a^{3} \left (\tan \left (f x +e \right )-i\right )}+\frac {24 c^{5} A}{f \,a^{3} \left (\tan \left (f x +e \right )-i\right )}-\frac {16 c^{5} A}{3 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{3}}-\frac {16 i c^{5} B}{3 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{3}}+\frac {8 i c^{5} A \ln \left (\tan \left (f x +e \right )-i\right )}{f \,a^{3}}-\frac {32 c^{5} B \ln \left (\tan \left (f x +e \right )-i\right )}{f \,a^{3}} \]

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))^5/(a+I*a*tan(f*x+e))^3,x)

[Out]

1/f*c^5/a^3*A*tan(f*x+e)+8*I/f*c^5/a^3*B*tan(f*x+e)+1/2*B*c^5*tan(f*x+e)^2/a^3/f+16*I/f*c^5/a^3/(tan(f*x+e)-I)
^2*A-24/f*c^5/a^3/(tan(f*x+e)-I)^2*B+56*I/f*c^5/a^3/(tan(f*x+e)-I)*B+24/f*c^5/a^3/(tan(f*x+e)-I)*A-16/3/f*c^5/
a^3/(tan(f*x+e)-I)^3*A-16/3*I/f*c^5/a^3/(tan(f*x+e)-I)^3*B+8*I/f*c^5/a^3*A*ln(tan(f*x+e)-I)-32/f*c^5/a^3*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))^5/(a+I*a*tan(f*x+e))^3,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 = 8.98, size = 233, normalized size = 1.22 \[ \frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )\,\left (-\frac {32\,B\,c^5}{a^3}+\frac {A\,c^5\,8{}\mathrm {i}}{a^3}\right )}{f}+\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (\frac {c^5\,\left (A+B\,4{}\mathrm {i}\right )}{a^3}+\frac {B\,c^5\,4{}\mathrm {i}}{a^3}\right )}{f}+\frac {\frac {5\,\left (-32\,B\,c^5+A\,c^5\,8{}\mathrm {i}\right )}{3\,a^3}+\mathrm {tan}\left (e+f\,x\right )\,\left (\frac {\left (-32\,B\,c^5+A\,c^5\,8{}\mathrm {i}\right )\,4{}\mathrm {i}}{a^3}+\frac {B\,c^5\,40{}\mathrm {i}}{a^3}\right )-{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (\frac {3\,\left (-32\,B\,c^5+A\,c^5\,8{}\mathrm {i}\right )}{a^3}+\frac {40\,B\,c^5}{a^3}\right )+\frac {16\,B\,c^5}{a^3}}{f\,\left (-{\mathrm {tan}\left (e+f\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (e+f\,x\right )}^2+\mathrm {tan}\left (e+f\,x\right )\,3{}\mathrm {i}+1\right )}+\frac {B\,c^5\,{\mathrm {tan}\left (e+f\,x\right )}^2}{2\,a^3\,f} \]

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)^5)/(a + a*tan(e + f*x)*1i)^3,x)

[Out]

(log(tan(e + f*x) - 1i)*((A*c^5*8i)/a^3 - (32*B*c^5)/a^3))/f + (tan(e + f*x)*((c^5*(A + B*4i))/a^3 + (B*c^5*4i
)/a^3))/f + ((5*(A*c^5*8i - 32*B*c^5))/(3*a^3) + tan(e + f*x)*(((A*c^5*8i - 32*B*c^5)*4i)/a^3 + (B*c^5*40i)/a^
3) - tan(e + f*x)^2*((3*(A*c^5*8i - 32*B*c^5))/a^3 + (40*B*c^5)/a^3) + (16*B*c^5)/a^3)/(f*(tan(e + f*x)*3i - 3
*tan(e + f*x)^2 - tan(e + f*x)^3*1i + 1)) + (B*c^5*tan(e + f*x)^2)/(2*a^3*f)

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sympy [A]  time = 1.82, size = 476, normalized size = 2.49 \[ \frac {2 i A c^{5} - 16 B c^{5} + \left (2 i A c^{5} e^{2 i e} - 14 B c^{5} e^{2 i e}\right ) e^{2 i f x}}{a^{3} f e^{4 i e} e^{4 i f x} + 2 a^{3} f e^{2 i e} e^{2 i f x} + a^{3} f} + \begin {cases} - \frac {\left (\left (- 2 i A a^{6} c^{5} f^{2} e^{6 i e} + 2 B a^{6} c^{5} f^{2} e^{6 i e}\right ) e^{- 6 i f x} + \left (6 i A a^{6} c^{5} f^{2} e^{8 i e} - 12 B a^{6} c^{5} f^{2} e^{8 i e}\right ) e^{- 4 i f x} + \left (- 18 i A a^{6} c^{5} f^{2} e^{10 i e} + 54 B a^{6} c^{5} f^{2} e^{10 i e}\right ) e^{- 2 i f x}\right ) e^{- 12 i e}}{3 a^{9} f^{3}} & \text {for}\: 3 a^{9} f^{3} e^{12 i e} \neq 0 \\x \left (- \frac {- 16 A c^{5} - 64 i B c^{5}}{a^{3}} + \frac {i \left (16 i A c^{5} e^{6 i e} - 12 i A c^{5} e^{4 i e} + 8 i A c^{5} e^{2 i e} - 4 i A c^{5} - 64 B c^{5} e^{6 i e} + 36 B c^{5} e^{4 i e} - 16 B c^{5} e^{2 i e} + 4 B c^{5}\right ) e^{- 6 i e}}{a^{3}}\right ) & \text {otherwise} \end {cases} - \frac {8 i c^{5} \left (A + 4 i B\right ) \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{a^{3} f} - \frac {x \left (16 A c^{5} + 64 i B c^{5}\right )}{a^{3}} \]

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))**5/(a+I*a*tan(f*x+e))**3,x)

[Out]

(2*I*A*c**5 - 16*B*c**5 + (2*I*A*c**5*exp(2*I*e) - 14*B*c**5*exp(2*I*e))*exp(2*I*f*x))/(a**3*f*exp(4*I*e)*exp(
4*I*f*x) + 2*a**3*f*exp(2*I*e)*exp(2*I*f*x) + a**3*f) + Piecewise((-((-2*I*A*a**6*c**5*f**2*exp(6*I*e) + 2*B*a
**6*c**5*f**2*exp(6*I*e))*exp(-6*I*f*x) + (6*I*A*a**6*c**5*f**2*exp(8*I*e) - 12*B*a**6*c**5*f**2*exp(8*I*e))*e
xp(-4*I*f*x) + (-18*I*A*a**6*c**5*f**2*exp(10*I*e) + 54*B*a**6*c**5*f**2*exp(10*I*e))*exp(-2*I*f*x))*exp(-12*I
*e)/(3*a**9*f**3), Ne(3*a**9*f**3*exp(12*I*e), 0)), (x*(-(-16*A*c**5 - 64*I*B*c**5)/a**3 + I*(16*I*A*c**5*exp(
6*I*e) - 12*I*A*c**5*exp(4*I*e) + 8*I*A*c**5*exp(2*I*e) - 4*I*A*c**5 - 64*B*c**5*exp(6*I*e) + 36*B*c**5*exp(4*
I*e) - 16*B*c**5*exp(2*I*e) + 4*B*c**5)*exp(-6*I*e)/a**3), True)) - 8*I*c**5*(A + 4*I*B)*log(exp(2*I*f*x) + ex
p(-2*I*e))/(a**3*f) - x*(16*A*c**5 + 64*I*B*c**5)/a**3

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