3.8.28 \(\int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^5}{(a+i a \tan (e+f x))^3} \, dx\) [728]

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

[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/(I-tan(f*x+e))^3+8*(2*I*A-3
*B)*c^5/a^3/f/(I-tan(f*x+e))^2-8*(3*A+7*I*B)*c^5/a^3/f/(I-tan(f*x+e))+(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 = 0.16, 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 = {3669, 78} \begin {gather*} \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} \end {gather*}

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 78

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 3669

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) \text {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) \text {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 [A]
time = 5.63, size = 347, normalized size = 1.82 \begin {gather*} \frac {c^5 \sec ^2(e+f x) (\cos (f x)+i \sin (f x))^3 \left (12 (-i A+2 B) \cos (4 f x) (\cos (e)-i \sin (e))+36 i (A+3 i B) \cos (2 f x) (\cos (e)+i \sin (e))+48 (-i A+4 B) \log (\cos (e+f x)) \left (\cos \left (\frac {3 e}{2}\right )+i \sin \left (\frac {3 e}{2}\right )\right )^2-48 (A+4 i B) f x (\cos (3 e)+i \sin (3 e))+3 B \sec ^2(e+f x) (\cos (3 e)+i \sin (3 e))+4 (A+i B) \cos (6 f x) (i \cos (3 e)+\sin (3 e))+6 (A+8 i B) \sec (e) \sec (e+f x) (\cos (3 e)+i \sin (3 e)) \sin (f x)+36 (A+3 i B) (\cos (e)+i \sin (e)) \sin (2 f x)-12 (A+2 i B) (\cos (e)-i \sin (e)) \sin (4 f x)+4 (A+i B) (\cos (3 e)-i \sin (3 e)) \sin (6 f x)\right ) (A+B \tan (e+f x))}{6 f (A \cos (e+f x)+B \sin (e+f x)) (a+i a \tan (e+f x))^3} \end {gather*}

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]

(c^5*Sec[e + f*x]^2*(Cos[f*x] + I*Sin[f*x])^3*(12*((-I)*A + 2*B)*Cos[4*f*x]*(Cos[e] - I*Sin[e]) + (36*I)*(A +
(3*I)*B)*Cos[2*f*x]*(Cos[e] + I*Sin[e]) + 48*((-I)*A + 4*B)*Log[Cos[e + f*x]]*(Cos[(3*e)/2] + I*Sin[(3*e)/2])^
2 - 48*(A + (4*I)*B)*f*x*(Cos[3*e] + I*Sin[3*e]) + 3*B*Sec[e + f*x]^2*(Cos[3*e] + I*Sin[3*e]) + 4*(A + I*B)*Co
s[6*f*x]*(I*Cos[3*e] + Sin[3*e]) + 6*(A + (8*I)*B)*Sec[e]*Sec[e + f*x]*(Cos[3*e] + I*Sin[3*e])*Sin[f*x] + 36*(
A + (3*I)*B)*(Cos[e] + I*Sin[e])*Sin[2*f*x] - 12*(A + (2*I)*B)*(Cos[e] - I*Sin[e])*Sin[4*f*x] + 4*(A + I*B)*(C
os[3*e] - I*Sin[3*e])*Sin[6*f*x])*(A + B*Tan[e + f*x]))/(6*f*(A*Cos[e + f*x] + B*Sin[e + f*x])*(a + I*a*Tan[e
+ f*x])^3)

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Maple [A]
time = 0.35, size = 123, normalized size = 0.64

method	result	size
derivativedivides	\(\frac {c^{5} \left (A \tan \left (f x +e \right )+8 i B \tan \left (f x +e \right )+\frac {B \left (\tan ^{2}\left (f x +e \right )\right )}{2}-\frac {16 i B +16 A}{3 \left (-i+\tan \left (f x +e \right )\right )^{3}}+\left (8 i A -32 B \right ) \ln \left (-i+\tan \left (f x +e \right )\right )-\frac {-56 i B -24 A}{-i+\tan \left (f x +e \right )}-\frac {-32 i A +48 B}{2 \left (-i+\tan \left (f x +e \right )\right )^{2}}\right )}{f \,a^{3}}\)	\(123\)
default	\(\frac {c^{5} \left (A \tan \left (f x +e \right )+8 i B \tan \left (f x +e \right )+\frac {B \left (\tan ^{2}\left (f x +e \right )\right )}{2}-\frac {16 i B +16 A}{3 \left (-i+\tan \left (f x +e \right )\right )^{3}}+\left (8 i A -32 B \right ) \ln \left (-i+\tan \left (f x +e \right )\right )-\frac {-56 i B -24 A}{-i+\tan \left (f x +e \right )}-\frac {-32 i A +48 B}{2 \left (-i+\tan \left (f x +e \right )\right )^{2}}\right )}{f \,a^{3}}\)	\(123\)
risch	\(-\frac {18 c^{5} {\mathrm e}^{-2 i \left (f x +e \right )} B}{a^{3} f}+\frac {6 i c^{5} {\mathrm e}^{-2 i \left (f x +e \right )} A}{a^{3} f}+\frac {4 c^{5} {\mathrm e}^{-4 i \left (f x +e \right )} B}{a^{3} f}-\frac {2 i c^{5} {\mathrm e}^{-4 i \left (f x +e \right )} A}{a^{3} f}-\frac {2 c^{5} {\mathrm e}^{-6 i \left (f x +e \right )} B}{3 a^{3} f}+\frac {2 i c^{5} {\mathrm e}^{-6 i \left (f x +e \right )} A}{3 a^{3} f}-\frac {64 i c^{5} B x}{a^{3}}-\frac {16 c^{5} A x}{a^{3}}-\frac {64 i c^{5} B e}{f \,a^{3}}-\frac {16 c^{5} A e}{f \,a^{3}}-\frac {2 c^{5} \left (-i A \,{\mathrm e}^{2 i \left (f x +e \right )}+7 B \,{\mathrm e}^{2 i \left (f x +e \right )}-i A +8 B \right )}{f \,a^{3} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2}}+\frac {32 c^{5} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) B}{f \,a^{3}}-\frac {8 i c^{5} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) A}{f \,a^{3}}\)	\(285\)
norman	\(\frac {\frac {\left (8 i c^{5} B +A \,c^{5}\right ) \left (\tan ^{7}\left (f x +e \right )\right )}{a f}+\frac {\left (32 i c^{5} B +9 A \,c^{5}\right ) \tan \left (f x +e \right )}{a f}+\frac {\left (80 i c^{5} B +27 A \,c^{5}\right ) \left (\tan ^{5}\left (f x +e \right )\right )}{a f}-\frac {8 \left (4 i c^{5} B +A \,c^{5}\right ) x}{a}-\frac {-80 i c^{5} A +233 B \,c^{5}}{6 a f}-\frac {24 \left (4 i c^{5} B +A \,c^{5}\right ) x \left (\tan ^{2}\left (f x +e \right )\right )}{a}-\frac {24 \left (4 i c^{5} B +A \,c^{5}\right ) x \left (\tan ^{4}\left (f x +e \right )\right )}{a}-\frac {8 \left (4 i c^{5} B +A \,c^{5}\right ) x \left (\tan ^{6}\left (f x +e \right )\right )}{a}+\frac {\left (248 i c^{5} B +41 A \,c^{5}\right ) \left (\tan ^{3}\left (f x +e \right )\right )}{3 a f}-\frac {\left (-32 i c^{5} A +100 B \,c^{5}\right ) \left (\tan ^{2}\left (f x +e \right )\right )}{a f}-\frac {\left (-40 i c^{5} A +83 B \,c^{5}\right ) \left (\tan ^{4}\left (f x +e \right )\right )}{a f}+\frac {B \,c^{5} \left (\tan ^{8}\left (f x +e \right )\right )}{2 a f}}{a^{2} \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}-\frac {4 \left (-i c^{5} A +4 B \,c^{5}\right ) \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{a^{3} f}\)	\(368\)

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,method=_RETURNVERBOSE)

[Out]

1/f*c^5/a^3*(A*tan(f*x+e)+8*I*B*tan(f*x+e)+1/2*B*tan(f*x+e)^2-1/3*(16*A+16*I*B)/(-I+tan(f*x+e))^3+(8*I*A-32*B)
*ln(-I+tan(f*x+e))-(-24*A-56*I*B)/(-I+tan(f*x+e))-1/2*(-32*I*A+48*B)/(-I+tan(f*x+e))^2)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

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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Fricas [A]
time = 3.30, size = 279, normalized size = 1.46 \begin {gather*} -\frac {2 \, {\left (24 \, {\left (A + 4 i \, B\right )} c^{5} f x e^{\left (10 i \, f x + 10 i \, e\right )} + 4 \, {\left (-i \, A + 4 \, B\right )} c^{5} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (i \, A - 4 \, B\right )} c^{5} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-i \, A + B\right )} c^{5} + 12 \, {\left (4 \, {\left (A + 4 i \, B\right )} c^{5} f x + {\left (-i \, A + 4 \, B\right )} c^{5}\right )} e^{\left (8 i \, f x + 8 i \, e\right )} + 6 \, {\left (4 \, {\left (A + 4 i \, B\right )} c^{5} f x + 3 \, {\left (-i \, A + 4 \, B\right )} c^{5}\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + 12 \, {\left ({\left (i \, A - 4 \, B\right )} c^{5} e^{\left (10 i \, f x + 10 i \, e\right )} + 2 \, {\left (i \, A - 4 \, B\right )} c^{5} e^{\left (8 i \, f x + 8 i \, e\right )} + {\left (i \, A - 4 \, 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 )\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 )}} \end {gather*}

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]

-2/3*(24*(A + 4*I*B)*c^5*f*x*e^(10*I*f*x + 10*I*e) + 4*(-I*A + 4*B)*c^5*e^(4*I*f*x + 4*I*e) + (I*A - 4*B)*c^5*
e^(2*I*f*x + 2*I*e) + (-I*A + B)*c^5 + 12*(4*(A + 4*I*B)*c^5*f*x + (-I*A + 4*B)*c^5)*e^(8*I*f*x + 8*I*e) + 6*(
4*(A + 4*I*B)*c^5*f*x + 3*(-I*A + 4*B)*c^5)*e^(6*I*f*x + 6*I*e) + 12*((I*A - 4*B)*c^5*e^(10*I*f*x + 10*I*e) +
2*(I*A - 4*B)*c^5*e^(8*I*f*x + 8*I*e) + (I*A - 4*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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Sympy [A]
time = 0.94, size = 473, normalized size = 2.48 \begin {gather*} \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}\: 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 {\left (- 16 A c^{5} e^{6 i e} + 12 A c^{5} e^{4 i e} - 8 A c^{5} e^{2 i e} + 4 A c^{5} - 64 i B c^{5} e^{6 i e} + 36 i B c^{5} e^{4 i e} - 16 i B c^{5} e^{2 i e} + 4 i 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}} \end {gather*}

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

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Giac [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 516 vs. \(2 (169) = 338\).
time = 1.30, size = 516, normalized size = 2.70 \begin {gather*} \frac {2 \, {\left (\frac {60 \, {\left (-i \, A c^{5} + 4 \, B c^{5}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{a^{3}} - \frac {120 \, {\left (-i \, A c^{5} + 4 \, B c^{5}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}{a^{3}} - \frac {60 \, {\left (i \, A c^{5} - 4 \, 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 {2 \, {\left (147 i \, A c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 588 \, B c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 942 \, A c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 3708 i \, B c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 2445 i \, A c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 9660 \, B c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 3460 \, A c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 13240 i \, B c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 2445 i \, A c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 9660 \, B c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 942 \, A c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 3708 i \, B c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 147 i \, A c^{5} + 588 \, B c^{5}\right )}}{a^{3} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}^{6}}\right )}}{15 \, f} \end {gather*}

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*(60*(-I*A*c^5 + 4*B*c^5)*log(tan(1/2*f*x + 1/2*e) + 1)/a^3 - 120*(-I*A*c^5 + 4*B*c^5)*log(tan(1/2*f*x + 1
/2*e) - I)/a^3 - 60*(I*A*c^5 - 4*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) - 2*(147*I*A*c^5*tan(1/2*f*x + 1/2
*e)^6 - 588*B*c^5*tan(1/2*f*x + 1/2*e)^6 + 942*A*c^5*tan(1/2*f*x + 1/2*e)^5 + 3708*I*B*c^5*tan(1/2*f*x + 1/2*e
)^5 - 2445*I*A*c^5*tan(1/2*f*x + 1/2*e)^4 + 9660*B*c^5*tan(1/2*f*x + 1/2*e)^4 - 3460*A*c^5*tan(1/2*f*x + 1/2*e
)^3 - 13240*I*B*c^5*tan(1/2*f*x + 1/2*e)^3 + 2445*I*A*c^5*tan(1/2*f*x + 1/2*e)^2 - 9660*B*c^5*tan(1/2*f*x + 1/
2*e)^2 + 942*A*c^5*tan(1/2*f*x + 1/2*e) + 3708*I*B*c^5*tan(1/2*f*x + 1/2*e) - 147*I*A*c^5 + 588*B*c^5)/(a^3*(t
an(1/2*f*x + 1/2*e) - I)^6))/f

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Mupad [B]
time = 8.98, size = 233, normalized size = 1.22 \begin {gather*} \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} \end {gather*}

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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