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

Optimal. Leaf size=157 \[ -\frac{c^4 (-5 B+i A) \tan ^2(e+f x)}{2 a f}+\frac{c^4 (5 A+12 i B) \tan (e+f x)}{a f}-\frac{8 c^4 (A+i B)}{a f (-\tan (e+f x)+i)}-\frac{4 c^4 (-5 B+3 i A) \log (\cos (e+f x))}{a f}-\frac{4 c^4 x (3 A+5 i B)}{a}-\frac{i B c^4 \tan ^3(e+f x)}{3 a f} \]

[Out]

(-4*(3*A + (5*I)*B)*c^4*x)/a - (4*((3*I)*A - 5*B)*c^4*Log[Cos[e + f*x]])/(a*f) - (8*(A + I*B)*c^4)/(a*f*(I - T
an[e + f*x])) + ((5*A + (12*I)*B)*c^4*Tan[e + f*x])/(a*f) - ((I*A - 5*B)*c^4*Tan[e + f*x]^2)/(2*a*f) - ((I/3)*
B*c^4*Tan[e + f*x]^3)/(a*f)

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Rubi [A]  time = 0.215932, antiderivative size = 157, normalized size of antiderivative = 1., 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} \[ -\frac{c^4 (-5 B+i A) \tan ^2(e+f x)}{2 a f}+\frac{c^4 (5 A+12 i B) \tan (e+f x)}{a f}-\frac{8 c^4 (A+i B)}{a f (-\tan (e+f x)+i)}-\frac{4 c^4 (-5 B+3 i A) \log (\cos (e+f x))}{a f}-\frac{4 c^4 x (3 A+5 i B)}{a}-\frac{i B c^4 \tan ^3(e+f x)}{3 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(3*A + (5*I)*B)*c^4*x)/a - (4*((3*I)*A - 5*B)*c^4*Log[Cos[e + f*x]])/(a*f) - (8*(A + I*B)*c^4)/(a*f*(I - T
an[e + f*x])) + ((5*A + (12*I)*B)*c^4*Tan[e + f*x])/(a*f) - ((I*A - 5*B)*c^4*Tan[e + f*x]^2)/(2*a*f) - ((I/3)*
B*c^4*Tan[e + f*x]^3)/(a*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]

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]))))

Rubi steps

\begin{align*} \int \frac{(A+B \tan (e+f x)) (c-i c \tan (e+f x))^4}{a+i a \tan (e+f x)} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(A+B x) (c-i c x)^3}{(a+i a x)^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{(a c) \operatorname{Subst}\left (\int \left (\frac{(5 A+12 i B) c^3}{a^2}+\frac{(-i A+5 B) c^3 x}{a^2}-\frac{i B c^3 x^2}{a^2}-\frac{8 (A+i B) c^3}{a^2 (-i+x)^2}+\frac{4 i (3 A+5 i B) c^3}{a^2 (-i+x)}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{4 (3 A+5 i B) c^4 x}{a}-\frac{4 (3 i A-5 B) c^4 \log (\cos (e+f x))}{a f}-\frac{8 (A+i B) c^4}{a f (i-\tan (e+f x))}+\frac{(5 A+12 i B) c^4 \tan (e+f x)}{a f}-\frac{(i A-5 B) c^4 \tan ^2(e+f x)}{2 a f}-\frac{i B c^4 \tan ^3(e+f x)}{3 a f}\\ \end{align*}

Mathematica [A]  time = 3.76031, size = 260, normalized size = 1.66 \[ \frac{c^4 (\cos (f x)+i \sin (f x)) (A+B \tan (e+f x)) \left (24 (A+i B) (\sin (e)+i \cos (e)) \cos (2 f x)+24 (A+i B) (\cos (e)-i \sin (e)) \sin (2 f x)+12 (5 B-3 i A) \left (\cos \left (\frac{e}{2}\right )+i \sin \left (\frac{e}{2}\right )\right )^2 \log \left (\cos ^2(e+f x)\right )-24 (3 A+5 i B) (\cos (e)+i \sin (e)) \tan ^{-1}(\tan (f x))+\cos (e) (\tan (e)-i) (2 B \tan (e)+3 (A+5 i B)) \sec ^2(e+f x)+2 (15 A+37 i B) (1+i \tan (e)) \sin (f x) \sec (e+f x)+2 B (\tan (e)-i) \sin (f x) \sec ^3(e+f x)\right )}{6 f (a+i a \tan (e+f x)) (A \cos (e+f x)+B \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^4*(Cos[f*x] + I*Sin[f*x])*(12*((-3*I)*A + 5*B)*Log[Cos[e + f*x]^2]*(Cos[e/2] + I*Sin[e/2])^2 - 24*(3*A + (5
*I)*B)*ArcTan[Tan[f*x]]*(Cos[e] + I*Sin[e]) + 24*(A + I*B)*Cos[2*f*x]*(I*Cos[e] + Sin[e]) + 24*(A + I*B)*(Cos[
e] - I*Sin[e])*Sin[2*f*x] + 2*(15*A + (37*I)*B)*Sec[e + f*x]*Sin[f*x]*(1 + I*Tan[e]) + 2*B*Sec[e + f*x]^3*Sin[
f*x]*(-I + Tan[e]) + Cos[e]*Sec[e + f*x]^2*(-I + Tan[e])*(3*(A + (5*I)*B) + 2*B*Tan[e]))*(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]))

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Maple [A]  time = 0.046, size = 193, normalized size = 1.2 \begin{align*}{\frac{5\,B{c}^{4} \left ( \tan \left ( fx+e \right ) \right ) ^{2}}{2\,af}}-{\frac{{\frac{i}{3}}B{c}^{4} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{af}}+5\,{\frac{A{c}^{4}\tan \left ( fx+e \right ) }{af}}-{\frac{{\frac{i}{2}}{c}^{4}A \left ( \tan \left ( fx+e \right ) \right ) ^{2}}{af}}+{\frac{12\,i{c}^{4}B\tan \left ( fx+e \right ) }{af}}+{\frac{8\,iB{c}^{4}}{af \left ( \tan \left ( fx+e \right ) -i \right ) }}+8\,{\frac{A{c}^{4}}{af \left ( \tan \left ( fx+e \right ) -i \right ) }}+{\frac{12\,i{c}^{4}A\ln \left ( \tan \left ( fx+e \right ) -i \right ) }{af}}-20\,{\frac{B{c}^{4}\ln \left ( \tan \left ( fx+e \right ) -i \right ) }{af}} \end{align*}

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))^4/(a+I*a*tan(f*x+e)),x)

[Out]

5/2/f*c^4/a*B*tan(f*x+e)^2-1/3*I*B*c^4*tan(f*x+e)^3/a/f+5/f*c^4/a*A*tan(f*x+e)-1/2*I/f*c^4/a*A*tan(f*x+e)^2+12
*I/f*c^4/a*B*tan(f*x+e)+8*I/f*c^4/a/(tan(f*x+e)-I)*B+8/f*c^4/a/(tan(f*x+e)-I)*A+12*I/f*c^4/a*A*ln(tan(f*x+e)-I
)-20/f*c^4/a*B*ln(tan(f*x+e)-I)

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

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))^4/(a+I*a*tan(f*x+e)),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 1.44351, size = 817, normalized size = 5.2 \begin{align*} -\frac{24 \,{\left (3 \, A + 5 i \, B\right )} c^{4} f x e^{\left (8 i \, f x + 8 i \, e\right )} -{\left (12 i \, A - 12 \, B\right )} c^{4} +{\left (72 \,{\left (3 \, A + 5 i \, B\right )} c^{4} f x -{\left (36 i \, A - 60 \, B\right )} c^{4}\right )} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (72 \,{\left (3 \, A + 5 i \, B\right )} c^{4} f x -{\left (90 i \, A - 150 \, B\right )} c^{4}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (24 \,{\left (3 \, A + 5 i \, B\right )} c^{4} f x -{\left (66 i \, A - 110 \, B\right )} c^{4}\right )} e^{\left (2 i \, f x + 2 i \, e\right )} -{\left ({\left (-36 i \, A + 60 \, B\right )} c^{4} e^{\left (8 i \, f x + 8 i \, e\right )} +{\left (-108 i \, A + 180 \, B\right )} c^{4} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (-108 i \, A + 180 \, B\right )} c^{4} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (-36 i \, A + 60 \, B\right )} c^{4} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{3 \,{\left (a f e^{\left (8 i \, f x + 8 i \, e\right )} + 3 \, a f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, a f e^{\left (4 i \, f x + 4 i \, e\right )} + a f e^{\left (2 i \, f x + 2 i \, e\right )}\right )}} \end{align*}

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))^4/(a+I*a*tan(f*x+e)),x, algorithm="fricas")

[Out]

-1/3*(24*(3*A + 5*I*B)*c^4*f*x*e^(8*I*f*x + 8*I*e) - (12*I*A - 12*B)*c^4 + (72*(3*A + 5*I*B)*c^4*f*x - (36*I*A
- 60*B)*c^4)*e^(6*I*f*x + 6*I*e) + (72*(3*A + 5*I*B)*c^4*f*x - (90*I*A - 150*B)*c^4)*e^(4*I*f*x + 4*I*e) + (2
4*(3*A + 5*I*B)*c^4*f*x - (66*I*A - 110*B)*c^4)*e^(2*I*f*x + 2*I*e) - ((-36*I*A + 60*B)*c^4*e^(8*I*f*x + 8*I*e
) + (-108*I*A + 180*B)*c^4*e^(6*I*f*x + 6*I*e) + (-108*I*A + 180*B)*c^4*e^(4*I*f*x + 4*I*e) + (-36*I*A + 60*B)
*c^4*e^(2*I*f*x + 2*I*e))*log(e^(2*I*f*x + 2*I*e) + 1))/(a*f*e^(8*I*f*x + 8*I*e) + 3*a*f*e^(6*I*f*x + 6*I*e) +
3*a*f*e^(4*I*f*x + 4*I*e) + a*f*e^(2*I*f*x + 2*I*e))

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Sympy [A]  time = 11.4177, size = 301, normalized size = 1.92 \begin{align*} \frac{\frac{\left (8 i A c^{4} - 16 B c^{4}\right ) e^{- 2 i e} e^{4 i f x}}{a f} + \frac{\left (18 i A c^{4} - 38 B c^{4}\right ) e^{- 4 i e} e^{2 i f x}}{a f} + \frac{\left (30 i A c^{4} - 74 B c^{4}\right ) e^{- 6 i e}}{3 a f}}{e^{6 i f x} + 3 e^{- 2 i e} e^{4 i f x} + 3 e^{- 4 i e} e^{2 i f x} + e^{- 6 i e}} + \frac{c^{4} \left (- 12 i A + 20 B\right ) \log{\left (e^{2 i f x} + e^{- 2 i e} \right )}}{a f} - \frac{\left (\begin{cases} 24 A c^{4} x e^{2 i e} - \frac{4 i A c^{4} e^{- 2 i f x}}{f} + 40 i B c^{4} x e^{2 i e} + \frac{4 B c^{4} e^{- 2 i f x}}{f} & \text{for}\: f \neq 0 \\x \left (24 A c^{4} e^{2 i e} - 8 A c^{4} + 40 i B c^{4} e^{2 i e} - 8 i B c^{4}\right ) & \text{otherwise} \end{cases}\right ) e^{- 2 i e}}{a} \end{align*}

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

[Out]

((8*I*A*c**4 - 16*B*c**4)*exp(-2*I*e)*exp(4*I*f*x)/(a*f) + (18*I*A*c**4 - 38*B*c**4)*exp(-4*I*e)*exp(2*I*f*x)/
(a*f) + (30*I*A*c**4 - 74*B*c**4)*exp(-6*I*e)/(3*a*f))/(exp(6*I*f*x) + 3*exp(-2*I*e)*exp(4*I*f*x) + 3*exp(-4*I
*e)*exp(2*I*f*x) + exp(-6*I*e)) + c**4*(-12*I*A + 20*B)*log(exp(2*I*f*x) + exp(-2*I*e))/(a*f) - Piecewise((24*
A*c**4*x*exp(2*I*e) - 4*I*A*c**4*exp(-2*I*f*x)/f + 40*I*B*c**4*x*exp(2*I*e) + 4*B*c**4*exp(-2*I*f*x)/f, Ne(f,
0)), (x*(24*A*c**4*exp(2*I*e) - 8*A*c**4 + 40*I*B*c**4*exp(2*I*e) - 8*I*B*c**4), True))*exp(-2*I*e)/a

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Giac [B]  time = 1.53566, size = 602, normalized size = 3.83 \begin{align*} -\frac{2 \,{\left (\frac{12 \,{\left (-3 i \, A c^{4} + 5 \, B c^{4}\right )} \log \left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - i\right )}{a} - \frac{6 \,{\left (-3 i \, A c^{4} + 5 \, B c^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{a} + \frac{6 \,{\left (3 i \, A c^{4} - 5 \, B c^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{a} - \frac{3 \,{\left (-18 i \, A c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 30 \, B c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 44 \, A c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 68 i \, B c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 18 i \, A c^{4} - 30 \, B c^{4}\right )}}{a{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - i\right )}^{2}} + \frac{-33 i \, A c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} + 55 \, B c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} + 15 \, A c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 36 i \, B c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 102 i \, A c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 180 \, B c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 30 \, A c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 76 i \, B c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 102 i \, A c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 180 \, B c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 15 \, A c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 36 i \, B c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 33 i \, A c^{4} - 55 \, B c^{4}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right )}^{3} a}\right )}}{3 \, f} \end{align*}

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))^4/(a+I*a*tan(f*x+e)),x, algorithm="giac")

[Out]

-2/3*(12*(-3*I*A*c^4 + 5*B*c^4)*log(tan(1/2*f*x + 1/2*e) - I)/a - 6*(-3*I*A*c^4 + 5*B*c^4)*log(abs(tan(1/2*f*x
+ 1/2*e) + 1))/a + 6*(3*I*A*c^4 - 5*B*c^4)*log(abs(tan(1/2*f*x + 1/2*e) - 1))/a - 3*(-18*I*A*c^4*tan(1/2*f*x
+ 1/2*e)^2 + 30*B*c^4*tan(1/2*f*x + 1/2*e)^2 - 44*A*c^4*tan(1/2*f*x + 1/2*e) - 68*I*B*c^4*tan(1/2*f*x + 1/2*e)
+ 18*I*A*c^4 - 30*B*c^4)/(a*(tan(1/2*f*x + 1/2*e) - I)^2) + (-33*I*A*c^4*tan(1/2*f*x + 1/2*e)^6 + 55*B*c^4*ta
n(1/2*f*x + 1/2*e)^6 + 15*A*c^4*tan(1/2*f*x + 1/2*e)^5 + 36*I*B*c^4*tan(1/2*f*x + 1/2*e)^5 + 102*I*A*c^4*tan(1
/2*f*x + 1/2*e)^4 - 180*B*c^4*tan(1/2*f*x + 1/2*e)^4 - 30*A*c^4*tan(1/2*f*x + 1/2*e)^3 - 76*I*B*c^4*tan(1/2*f*
x + 1/2*e)^3 - 102*I*A*c^4*tan(1/2*f*x + 1/2*e)^2 + 180*B*c^4*tan(1/2*f*x + 1/2*e)^2 + 15*A*c^4*tan(1/2*f*x +
1/2*e) + 36*I*B*c^4*tan(1/2*f*x + 1/2*e) + 33*I*A*c^4 - 55*B*c^4)/((tan(1/2*f*x + 1/2*e)^2 - 1)^3*a))/f