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

Optimal. Leaf size=194 \[ \frac {c^5 (-7 B+i A) \tan ^2(e+f x)}{2 a^2 f}-\frac {c^5 (7 A+24 i B) \tan (e+f x)}{a^2 f}+\frac {16 c^5 (2 A+3 i B)}{a^2 f (-\tan (e+f x)+i)}-\frac {8 c^5 (-B+i A)}{a^2 f (-\tan (e+f x)+i)^2}+\frac {8 c^5 (-7 B+3 i A) \log (\cos (e+f x))}{a^2 f}+\frac {8 c^5 x (3 A+7 i B)}{a^2}+\frac {i B c^5 \tan ^3(e+f x)}{3 a^2 f} \]

[Out]

8*(3*A+7*I*B)*c^5*x/a^2+8*(3*I*A-7*B)*c^5*ln(cos(f*x+e))/a^2/f-8*(I*A-B)*c^5/a^2/f/(-tan(f*x+e)+I)^2+16*(2*A+3
*I*B)*c^5/a^2/f/(-tan(f*x+e)+I)-(7*A+24*I*B)*c^5*tan(f*x+e)/a^2/f+1/2*(I*A-7*B)*c^5*tan(f*x+e)^2/a^2/f+1/3*I*B
*c^5*tan(f*x+e)^3/a^2/f

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Rubi [A]  time = 0.25, antiderivative size = 194, 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} \[ \frac {c^5 (-7 B+i A) \tan ^2(e+f x)}{2 a^2 f}-\frac {c^5 (7 A+24 i B) \tan (e+f x)}{a^2 f}+\frac {16 c^5 (2 A+3 i B)}{a^2 f (-\tan (e+f x)+i)}-\frac {8 c^5 (-B+i A)}{a^2 f (-\tan (e+f x)+i)^2}+\frac {8 c^5 (-7 B+3 i A) \log (\cos (e+f x))}{a^2 f}+\frac {8 c^5 x (3 A+7 i B)}{a^2}+\frac {i B c^5 \tan ^3(e+f x)}{3 a^2 f} \]

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])^2,x]

[Out]

(8*(3*A + (7*I)*B)*c^5*x)/a^2 + (8*((3*I)*A - 7*B)*c^5*Log[Cos[e + f*x]])/(a^2*f) - (8*(I*A - B)*c^5)/(a^2*f*(
I - Tan[e + f*x])^2) + (16*(2*A + (3*I)*B)*c^5)/(a^2*f*(I - Tan[e + f*x])) - ((7*A + (24*I)*B)*c^5*Tan[e + f*x
])/(a^2*f) + ((I*A - 7*B)*c^5*Tan[e + f*x]^2)/(2*a^2*f) + ((I/3)*B*c^5*Tan[e + f*x]^3)/(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))^5}{(a+i a \tan (e+f x))^2} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {(A+B x) (c-i c x)^4}{(a+i a x)^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {(a c) \operatorname {Subst}\left (\int \left (-\frac {(7 A+24 i B) c^4}{a^3}+\frac {i (A+7 i B) c^4 x}{a^3}+\frac {i B c^4 x^2}{a^3}+\frac {16 i (A+i B) c^4}{a^3 (-i+x)^3}+\frac {16 (2 A+3 i B) c^4}{a^3 (-i+x)^2}+\frac {8 (-3 i A+7 B) c^4}{a^3 (-i+x)}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {8 (3 A+7 i B) c^5 x}{a^2}+\frac {8 (3 i A-7 B) c^5 \log (\cos (e+f x))}{a^2 f}-\frac {8 (i A-B) c^5}{a^2 f (i-\tan (e+f x))^2}+\frac {16 (2 A+3 i B) c^5}{a^2 f (i-\tan (e+f x))}-\frac {(7 A+24 i B) c^5 \tan (e+f x)}{a^2 f}+\frac {(i A-7 B) c^5 \tan ^2(e+f x)}{2 a^2 f}+\frac {i B c^5 \tan ^3(e+f x)}{3 a^2 f}\\ \end {align*}

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Mathematica [B]  time = 11.38, size = 1357, normalized size = 6.99 \[ \frac {4 (5 B-3 i A) \cos (2 f x) \sec (e+f x) (\cos (f x)+i \sin (f x))^2 (A+B \tan (e+f x)) c^5}{f (A \cos (e+f x)+B \sin (e+f x)) (i \tan (e+f x) a+a)^2}-\frac {4 (3 A+5 i B) \sec (e+f x) (\cos (f x)+i \sin (f x))^2 \sin (2 f x) (A+B \tan (e+f x)) c^5}{f (A \cos (e+f x)+B \sin (e+f x)) (i \tan (e+f x) a+a)^2}+\frac {\sec (e) \sec ^4(e+f x) (\cos (f x)+i \sin (f x))^2 \left (-\frac {1}{2} B \cos (2 e-f x) c^5+\frac {1}{2} B \cos (2 e+f x) c^5-\frac {1}{2} i B \sin (2 e-f x) c^5+\frac {1}{2} i B \sin (2 e+f x) c^5\right ) (A+B \tan (e+f x))}{3 f (A \cos (e+f x)+B \sin (e+f x)) (i \tan (e+f x) a+a)^2}+\frac {\sec (e) \sec ^2(e+f x) (\cos (f x)+i \sin (f x))^2 \left (-\frac {21}{2} i A \cos (2 e-f x) c^5+\frac {73}{2} B \cos (2 e-f x) c^5+\frac {21}{2} i A \cos (2 e+f x) c^5-\frac {73}{2} B \cos (2 e+f x) c^5+\frac {21}{2} A \sin (2 e-f x) c^5+\frac {73}{2} i B \sin (2 e-f x) c^5-\frac {21}{2} A \sin (2 e+f x) c^5-\frac {73}{2} i B \sin (2 e+f x) c^5\right ) (A+B \tan (e+f x))}{3 f (A \cos (e+f x)+B \sin (e+f x)) (i \tan (e+f x) a+a)^2}+\frac {x \sec (e+f x) (\cos (f x)+i \sin (f x))^2 \left (-24 A c^5-56 i B c^5-24 i A \tan (e) c^5+56 B \tan (e) c^5+(7 B-3 i A) \left (8 \cos (2 e) c^5+8 i \sin (2 e) c^5\right ) \tan (e)\right ) (A+B \tan (e+f x))}{(A \cos (e+f x)+B \sin (e+f x)) (i \tan (e+f x) a+a)^2}+\frac {\sec (e+f x) \left (3 A \cos (e) c^5+7 i B \cos (e) c^5+3 i A \sin (e) c^5-7 B \sin (e) c^5\right ) \left (8 \tan ^{-1}(\tan (f x)) \cos (e)+8 i \tan ^{-1}(\tan (f x)) \sin (e)\right ) (\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)) (i \tan (e+f x) a+a)^2}+\frac {\sec (e+f x) \left (3 A \cos (e) c^5+7 i B \cos (e) c^5+3 i A \sin (e) c^5-7 B \sin (e) c^5\right ) \left (4 i \cos (e) \log \left (\cos ^2(e+f x)\right )-4 \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))}{f (A \cos (e+f x)+B \sin (e+f x)) (i \tan (e+f x) a+a)^2}+\frac {\sec (e) \sec ^3(e+f x) (3 A \cos (e)+21 i B \cos (e)+2 B \sin (e)) \left (\frac {1}{6} i c^5 \cos (2 e)-\frac {1}{6} c^5 \sin (2 e)\right ) (\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)) (i \tan (e+f x) a+a)^2}+\frac {(A+i B) \cos (4 f x) \sec (e+f x) \left (2 i \cos (2 e) c^5+2 \sin (2 e) c^5\right ) (\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)) (i \tan (e+f x) a+a)^2}+\frac {(3 A+7 i B) \sec (e+f x) \left (8 f x \cos (2 e) c^5+8 i f x \sin (2 e) c^5\right ) (\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)) (i \tan (e+f x) a+a)^2}+\frac {(A+i B) \sec (e+f x) \left (2 c^5 \cos (2 e)-2 i c^5 \sin (2 e)\right ) (\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)) (i \tan (e+f x) a+a)^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(4*((-3*I)*A + 5*B)*c^5*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]*(3*A*c^5*Cos[e] + (7*I)*B*c^5*Cos[e] + (3*I)*
A*c^5*Sin[e] - 7*B*c^5*Sin[e])*(8*ArcTan[Tan[f*x]]*Cos[e] + (8*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]*(3
*A*c^5*Cos[e] + (7*I)*B*c^5*Cos[e] + (3*I)*A*c^5*Sin[e] - 7*B*c^5*Sin[e])*((4*I)*Cos[e]*Log[Cos[e + f*x]^2] -
4*Log[Cos[e + f*x]^2]*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]*Sec[e + f*x]^3*(3*A*Cos[e] + (21*I)*B*Cos[e] + 2*B*Sin[e])*((I/6)*c^5
*Cos[2*e] - (c^5*Sin[2*e])/6)*(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 + I*B)*Cos[4*f*x]*Sec[e + f*x]*((2*I)*c^5*Cos[2*e] + 2*c^5*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) + (
(3*A + (7*I)*B)*Sec[e + f*x]*(8*c^5*f*x*Cos[2*e] + (8*I)*c^5*f*x*Sin[2*e])*(Cos[f*x] + I*Sin[f*x])^2*(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])^2) - (4*(3*A + (5*I)*B)*c^5*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*T
an[e + f*x])^2) + ((A + I*B)*Sec[e + f*x]*(2*c^5*Cos[2*e] - (2*I)*c^5*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]^4*(Cos[f*x] + I*Sin[f*x])^2*(-1/2*(B*c^5*Cos[2*e - f*x]) + (B*c^5*Cos[2*e + f*x])/2 - (I/2)*B*c^5*Sin[2*e
- f*x] + (I/2)*B*c^5*Sin[2*e + f*x])*(A + B*Tan[e + f*x]))/(3*f*(A*Cos[e + f*x] + B*Sin[e + f*x])*(a + I*a*Ta
n[e + f*x])^2) + (Sec[e]*Sec[e + f*x]^2*(Cos[f*x] + I*Sin[f*x])^2*(((-21*I)/2)*A*c^5*Cos[2*e - f*x] + (73*B*c^
5*Cos[2*e - f*x])/2 + ((21*I)/2)*A*c^5*Cos[2*e + f*x] - (73*B*c^5*Cos[2*e + f*x])/2 + (21*A*c^5*Sin[2*e - f*x]
)/2 + ((73*I)/2)*B*c^5*Sin[2*e - f*x] - (21*A*c^5*Sin[2*e + f*x])/2 - ((73*I)/2)*B*c^5*Sin[2*e + f*x])*(A + B*
Tan[e + f*x]))/(3*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*(-24*A*c^5 - (56*I)*B*c^5 - (24*I)*A*c^5*Tan[e] + 56*B*c^5*Tan[e] + ((-3*I)*A + 7*B)*(8*c^5*Cos[
2*e] + (8*I)*c^5*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 = 0.59, size = 320, normalized size = 1.65 \[ \frac {48 \, {\left (3 \, A + 7 i \, B\right )} c^{5} f x e^{\left (10 i \, f x + 10 i \, e\right )} + {\left (-18 i \, A + 42 \, B\right )} c^{5} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (6 i \, A - 6 \, B\right )} c^{5} + {\left (144 \, {\left (3 \, A + 7 i \, B\right )} c^{5} f x + {\left (-72 i \, A + 168 \, B\right )} c^{5}\right )} e^{\left (8 i \, f x + 8 i \, e\right )} + {\left (144 \, {\left (3 \, A + 7 i \, B\right )} c^{5} f x + {\left (-180 i \, A + 420 \, B\right )} c^{5}\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (48 \, {\left (3 \, A + 7 i \, B\right )} c^{5} f x + {\left (-132 i \, A + 308 \, B\right )} c^{5}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left ({\left (72 i \, A - 168 \, B\right )} c^{5} e^{\left (10 i \, f x + 10 i \, e\right )} + {\left (216 i \, A - 504 \, B\right )} c^{5} e^{\left (8 i \, f x + 8 i \, e\right )} + {\left (216 i \, A - 504 \, B\right )} c^{5} e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (72 i \, A - 168 \, B\right )} c^{5} e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{3 \, {\left (a^{2} f e^{\left (10 i \, f x + 10 i \, e\right )} + 3 \, a^{2} f e^{\left (8 i \, f x + 8 i \, e\right )} + 3 \, a^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )}} \]

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))^2,x, algorithm="fricas")

[Out]

1/3*(48*(3*A + 7*I*B)*c^5*f*x*e^(10*I*f*x + 10*I*e) + (-18*I*A + 42*B)*c^5*e^(2*I*f*x + 2*I*e) + (6*I*A - 6*B)
*c^5 + (144*(3*A + 7*I*B)*c^5*f*x + (-72*I*A + 168*B)*c^5)*e^(8*I*f*x + 8*I*e) + (144*(3*A + 7*I*B)*c^5*f*x +
(-180*I*A + 420*B)*c^5)*e^(6*I*f*x + 6*I*e) + (48*(3*A + 7*I*B)*c^5*f*x + (-132*I*A + 308*B)*c^5)*e^(4*I*f*x +
4*I*e) + ((72*I*A - 168*B)*c^5*e^(10*I*f*x + 10*I*e) + (216*I*A - 504*B)*c^5*e^(8*I*f*x + 8*I*e) + (216*I*A -
504*B)*c^5*e^(6*I*f*x + 6*I*e) + (72*I*A - 168*B)*c^5*e^(4*I*f*x + 4*I*e))*log(e^(2*I*f*x + 2*I*e) + 1))/(a^2
*f*e^(10*I*f*x + 10*I*e) + 3*a^2*f*e^(8*I*f*x + 8*I*e) + 3*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 = 3.66, size = 515, normalized size = 2.65 \[ -\frac {2 \, {\left (\frac {3 \, {\left (-12 i \, A c^{5} + 28 \, B c^{5}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{a^{2}} - \frac {3 \, {\left (-24 i \, A c^{5} + 56 \, B c^{5}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}{a^{2}} - \frac {3 \, {\left (12 i \, A c^{5} - 28 \, B c^{5}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}{a^{2}} + \frac {66 i \, A c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 154 \, B c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 21 \, A c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 72 i \, B c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 201 i \, A c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 483 \, B c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 42 \, A c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 148 i \, B c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 201 i \, A c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 483 \, B c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 21 \, A c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 72 i \, B c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 66 i \, A c^{5} + 154 \, B c^{5}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{3} a^{2}} + \frac {-150 i \, A c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 350 \, B c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 648 \, A c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 1496 i \, B c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 1044 i \, A c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2340 \, B c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 648 \, A c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1496 i \, B c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 150 i \, A c^{5} + 350 \, B c^{5}}{a^{2} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}^{4}}\right )}}{3 \, 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))^2,x, algorithm="giac")

[Out]

-2/3*(3*(-12*I*A*c^5 + 28*B*c^5)*log(tan(1/2*f*x + 1/2*e) + 1)/a^2 - 3*(-24*I*A*c^5 + 56*B*c^5)*log(tan(1/2*f*
x + 1/2*e) - I)/a^2 - 3*(12*I*A*c^5 - 28*B*c^5)*log(tan(1/2*f*x + 1/2*e) - 1)/a^2 + (66*I*A*c^5*tan(1/2*f*x +
1/2*e)^6 - 154*B*c^5*tan(1/2*f*x + 1/2*e)^6 - 21*A*c^5*tan(1/2*f*x + 1/2*e)^5 - 72*I*B*c^5*tan(1/2*f*x + 1/2*e
)^5 - 201*I*A*c^5*tan(1/2*f*x + 1/2*e)^4 + 483*B*c^5*tan(1/2*f*x + 1/2*e)^4 + 42*A*c^5*tan(1/2*f*x + 1/2*e)^3
+ 148*I*B*c^5*tan(1/2*f*x + 1/2*e)^3 + 201*I*A*c^5*tan(1/2*f*x + 1/2*e)^2 - 483*B*c^5*tan(1/2*f*x + 1/2*e)^2 -
21*A*c^5*tan(1/2*f*x + 1/2*e) - 72*I*B*c^5*tan(1/2*f*x + 1/2*e) - 66*I*A*c^5 + 154*B*c^5)/((tan(1/2*f*x + 1/2
*e)^2 - 1)^3*a^2) + (-150*I*A*c^5*tan(1/2*f*x + 1/2*e)^4 + 350*B*c^5*tan(1/2*f*x + 1/2*e)^4 - 648*A*c^5*tan(1/
2*f*x + 1/2*e)^3 - 1496*I*B*c^5*tan(1/2*f*x + 1/2*e)^3 + 1044*I*A*c^5*tan(1/2*f*x + 1/2*e)^2 - 2340*B*c^5*tan(
1/2*f*x + 1/2*e)^2 + 648*A*c^5*tan(1/2*f*x + 1/2*e) + 1496*I*B*c^5*tan(1/2*f*x + 1/2*e) - 150*I*A*c^5 + 350*B*
c^5)/(a^2*(tan(1/2*f*x + 1/2*e) - I)^4))/f

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maple [A]  time = 0.24, size = 240, normalized size = 1.24 \[ \frac {i B \,c^{5} \left (\tan ^{3}\left (f x +e \right )\right )}{3 a^{2} f}+\frac {i c^{5} A \left (\tan ^{2}\left (f x +e \right )\right )}{2 f \,a^{2}}-\frac {24 i c^{5} B \tan \left (f x +e \right )}{f \,a^{2}}-\frac {7 c^{5} B \left (\tan ^{2}\left (f x +e \right )\right )}{2 f \,a^{2}}-\frac {7 c^{5} A \tan \left (f x +e \right )}{f \,a^{2}}-\frac {8 i c^{5} A}{f \,a^{2} \left (\tan \left (f x +e \right )-i\right )^{2}}+\frac {8 c^{5} B}{f \,a^{2} \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {48 i c^{5} B}{f \,a^{2} \left (\tan \left (f x +e \right )-i\right )}-\frac {32 c^{5} A}{f \,a^{2} \left (\tan \left (f x +e \right )-i\right )}-\frac {24 i c^{5} A \ln \left (\tan \left (f x +e \right )-i\right )}{f \,a^{2}}+\frac {56 c^{5} B \ln \left (\tan \left (f x +e \right )-i\right )}{f \,a^{2}} \]

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

[Out]

1/3*I*B*c^5*tan(f*x+e)^3/a^2/f+1/2*I/f*c^5/a^2*A*tan(f*x+e)^2-24*I/f*c^5/a^2*B*tan(f*x+e)-7/2/f*c^5/a^2*B*tan(
f*x+e)^2-7/f*c^5/a^2*A*tan(f*x+e)-8*I/f*c^5/a^2/(tan(f*x+e)-I)^2*A+8/f*c^5/a^2/(tan(f*x+e)-I)^2*B-48*I/f*c^5/a
^2/(tan(f*x+e)-I)*B-32/f*c^5/a^2/(tan(f*x+e)-I)*A-24*I/f*c^5/a^2*A*ln(tan(f*x+e)-I)+56/f*c^5/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))^5/(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 = 8.81, size = 282, normalized size = 1.45 \[ -\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )\,\left (-\frac {56\,B\,c^5}{a^2}+\frac {A\,c^5\,24{}\mathrm {i}}{a^2}\right )}{f}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (-\frac {3\,B\,c^5}{2\,a^2}+\frac {c^5\,\left (A+B\,4{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,a^2}\right )}{f}-\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (\frac {3\,c^5\,\left (A+B\,4{}\mathrm {i}\right )}{a^2}+\frac {B\,c^5\,6{}\mathrm {i}}{a^2}-\frac {c^5\,\left (-3\,B+A\,2{}\mathrm {i}\right )\,2{}\mathrm {i}}{a^2}\right )}{f}+\frac {-\frac {\left (-24\,B\,c^5+A\,c^5\,8{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,a^2}+\frac {16\,A\,c^5+B\,c^5\,64{}\mathrm {i}}{2\,a^2}+\frac {\left (-56\,B\,c^5+A\,c^5\,24{}\mathrm {i}\right )\,3{}\mathrm {i}}{2\,a^2}+\mathrm {tan}\left (e+f\,x\right )\,\left (\frac {\left (16\,A\,c^5+B\,c^5\,64{}\mathrm {i}\right )\,1{}\mathrm {i}}{a^2}-\frac {2\,\left (-56\,B\,c^5+A\,c^5\,24{}\mathrm {i}\right )}{a^2}\right )}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )}+\frac {B\,c^5\,{\mathrm {tan}\left (e+f\,x\right )}^3\,1{}\mathrm {i}}{3\,a^2\,f} \]

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)^5)/(a + a*tan(e + f*x)*1i)^2,x)

[Out]

(tan(e + f*x)^2*((c^5*(A + B*4i)*1i)/(2*a^2) - (3*B*c^5)/(2*a^2)))/f - (log(tan(e + f*x) - 1i)*((A*c^5*24i)/a^
2 - (56*B*c^5)/a^2))/f - (tan(e + f*x)*((3*c^5*(A + B*4i))/a^2 + (B*c^5*6i)/a^2 - (c^5*(A*2i - 3*B)*2i)/a^2))/
f + ((16*A*c^5 + B*c^5*64i)/(2*a^2) - ((A*c^5*8i - 24*B*c^5)*1i)/(2*a^2) + ((A*c^5*24i - 56*B*c^5)*3i)/(2*a^2)
+ tan(e + f*x)*(((16*A*c^5 + B*c^5*64i)*1i)/a^2 - (2*(A*c^5*24i - 56*B*c^5))/a^2))/(f*(2*tan(e + f*x) + tan(e
+ f*x)^2*1i - 1i)) + (B*c^5*tan(e + f*x)^3*1i)/(3*a^2*f)

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sympy [A]  time = 1.49, size = 461, normalized size = 2.38 \[ \frac {- 42 A c^{5} - 146 i B c^{5} + \left (- 78 A c^{5} e^{2 i e} - 246 i B c^{5} e^{2 i e}\right ) e^{2 i f x} + \left (- 36 A c^{5} e^{4 i e} - 108 i B c^{5} e^{4 i e}\right ) e^{4 i f x}}{- 3 i a^{2} f e^{6 i e} e^{6 i f x} - 9 i a^{2} f e^{4 i e} e^{4 i f x} - 9 i a^{2} f e^{2 i e} e^{2 i f x} - 3 i a^{2} f} + \begin {cases} \frac {\left (\left (2 i A a^{2} c^{5} f e^{2 i e} - 2 B a^{2} c^{5} f e^{2 i e}\right ) e^{- 4 i f x} + \left (- 12 i A a^{2} c^{5} f e^{4 i e} + 20 B a^{2} c^{5} f e^{4 i e}\right ) e^{- 2 i f x}\right ) e^{- 6 i e}}{a^{4} f^{2}} & \text {for}\: a^{4} f^{2} e^{6 i e} \neq 0 \\x \left (- \frac {48 A c^{5} + 112 i B c^{5}}{a^{2}} + \frac {i \left (- 48 i A c^{5} e^{4 i e} + 24 i A c^{5} e^{2 i e} - 8 i A c^{5} + 112 B c^{5} e^{4 i e} - 40 B c^{5} e^{2 i e} + 8 B c^{5}\right ) e^{- 4 i e}}{a^{2}}\right ) & \text {otherwise} \end {cases} + \frac {8 i c^{5} \left (3 A + 7 i B\right ) \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{a^{2} f} - \frac {x \left (- 48 A c^{5} - 112 i B c^{5}\right )}{a^{2}} \]

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

[Out]

(-42*A*c**5 - 146*I*B*c**5 + (-78*A*c**5*exp(2*I*e) - 246*I*B*c**5*exp(2*I*e))*exp(2*I*f*x) + (-36*A*c**5*exp(
4*I*e) - 108*I*B*c**5*exp(4*I*e))*exp(4*I*f*x))/(-3*I*a**2*f*exp(6*I*e)*exp(6*I*f*x) - 9*I*a**2*f*exp(4*I*e)*e
xp(4*I*f*x) - 9*I*a**2*f*exp(2*I*e)*exp(2*I*f*x) - 3*I*a**2*f) + Piecewise((((2*I*A*a**2*c**5*f*exp(2*I*e) - 2
*B*a**2*c**5*f*exp(2*I*e))*exp(-4*I*f*x) + (-12*I*A*a**2*c**5*f*exp(4*I*e) + 20*B*a**2*c**5*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*(-(48*A*c**5 + 112*I*B*c**5)/a**2 + I*(-4
8*I*A*c**5*exp(4*I*e) + 24*I*A*c**5*exp(2*I*e) - 8*I*A*c**5 + 112*B*c**5*exp(4*I*e) - 40*B*c**5*exp(2*I*e) + 8
*B*c**5)*exp(-4*I*e)/a**2), True)) + 8*I*c**5*(3*A + 7*I*B)*log(exp(2*I*f*x) + exp(-2*I*e))/(a**2*f) - x*(-48*
A*c**5 - 112*I*B*c**5)/a**2

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