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

Optimal. Leaf size=121 \[ \frac {c^3 (A+4 i B) \tan (e+f x)}{a f}-\frac {4 c^3 (A+i B)}{a f (-\tan (e+f x)+i)}-\frac {4 c^3 (-2 B+i A) \log (\cos (e+f x))}{a f}-\frac {4 c^3 x (A+2 i B)}{a}+\frac {B c^3 \tan ^2(e+f x)}{2 a f} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 121, 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} \[ \frac {c^3 (A+4 i B) \tan (e+f x)}{a f}-\frac {4 c^3 (A+i B)}{a f (-\tan (e+f x)+i)}-\frac {4 c^3 (-2 B+i A) \log (\cos (e+f x))}{a f}-\frac {4 c^3 x (A+2 i B)}{a}+\frac {B c^3 \tan ^2(e+f x)}{2 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 6.20, size = 212, normalized size = 1.75 \[ \frac {c^3 (\cos (f x)+i \sin (f x)) (A+B \tan (e+f x)) \left (4 (A+i B) (\sin (e)+i \cos (e)) \cos (2 f x)+4 (A+i B) (\cos (e)-i \sin (e)) \sin (2 f x)+4 (2 B-i A) (\cos (e)+i \sin (e)) \log \left (\cos ^2(e+f x)\right )-8 (A+2 i B) (\cos (e)+i \sin (e)) \tan ^{-1}(\tan (f x))+2 (A+4 i B) (1+i \tan (e)) \sin (f x) \sec (e+f x)+B (\cos (e)+i \sin (e)) \sec ^2(e+f x)\right )}{2 f (a+i a \tan (e+f x)) (A \cos (e+f x)+B \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^3*(Cos[f*x] + I*Sin[f*x])*(-8*(A + (2*I)*B)*ArcTan[Tan[f*x]]*(Cos[e] + I*Sin[e]) + 4*((-I)*A + 2*B)*Log[Cos
[e + f*x]^2]*(Cos[e] + I*Sin[e]) + B*Sec[e + f*x]^2*(Cos[e] + I*Sin[e]) + 4*(A + I*B)*Cos[2*f*x]*(I*Cos[e] + S
in[e]) + 4*(A + I*B)*(Cos[e] - I*Sin[e])*Sin[2*f*x] + 2*(A + (4*I)*B)*Sec[e + f*x]*Sin[f*x]*(1 + I*Tan[e]))*(A
+ B*Tan[e + f*x]))/(2*f*(A*Cos[e + f*x] + B*Sin[e + f*x])*(a + I*a*Tan[e + f*x]))

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fricas [B]  time = 1.53, size = 221, normalized size = 1.83 \[ -\frac {8 \, {\left (A + 2 i \, B\right )} c^{3} f x e^{\left (6 i \, f x + 6 i \, e\right )} - {\left (2 i \, A - 2 \, B\right )} c^{3} + {\left (16 \, {\left (A + 2 i \, B\right )} c^{3} f x - {\left (4 i \, A - 8 \, B\right )} c^{3}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (8 \, {\left (A + 2 i \, B\right )} c^{3} f x - {\left (6 i \, A - 12 \, B\right )} c^{3}\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - {\left ({\left (-4 i \, A + 8 \, B\right )} c^{3} e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (-8 i \, A + 16 \, B\right )} c^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (-4 i \, A + 8 \, B\right )} c^{3} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{a f e^{\left (6 i \, f x + 6 i \, e\right )} + 2 \, a f e^{\left (4 i \, f x + 4 i \, e\right )} + a f e^{\left (2 i \, f x + 2 i \, e\right )}} \]

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

[Out]

-(8*(A + 2*I*B)*c^3*f*x*e^(6*I*f*x + 6*I*e) - (2*I*A - 2*B)*c^3 + (16*(A + 2*I*B)*c^3*f*x - (4*I*A - 8*B)*c^3)
*e^(4*I*f*x + 4*I*e) + (8*(A + 2*I*B)*c^3*f*x - (6*I*A - 12*B)*c^3)*e^(2*I*f*x + 2*I*e) - ((-4*I*A + 8*B)*c^3*
e^(6*I*f*x + 6*I*e) + (-8*I*A + 16*B)*c^3*e^(4*I*f*x + 4*I*e) + (-4*I*A + 8*B)*c^3*e^(2*I*f*x + 2*I*e))*log(e^
(2*I*f*x + 2*I*e) + 1))/(a*f*e^(6*I*f*x + 6*I*e) + 2*a*f*e^(4*I*f*x + 4*I*e) + a*f*e^(2*I*f*x + 2*I*e))

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giac [B]  time = 1.79, size = 321, normalized size = 2.65 \[ -\frac {2 \, {\left (\frac {{\left (2 i \, A c^{3} - 4 \, B c^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{a} - \frac {{\left (4 i \, A c^{3} - 8 \, B c^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}{a} - \frac {{\left (-2 i \, A c^{3} + 4 \, B c^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}{a} + \frac {5 \, A c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 8 i \, B c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 2 i \, A c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 7 \, B c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 10 \, A c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 14 i \, B c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 2 i \, A c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 7 \, B c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 5 \, A c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 8 i \, B c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - i \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}^{2} a}\right )}}{f} \]

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

[Out]

-2*((2*I*A*c^3 - 4*B*c^3)*log(tan(1/2*f*x + 1/2*e) + 1)/a - (4*I*A*c^3 - 8*B*c^3)*log(tan(1/2*f*x + 1/2*e) - I
)/a - (-2*I*A*c^3 + 4*B*c^3)*log(tan(1/2*f*x + 1/2*e) - 1)/a + (5*A*c^3*tan(1/2*f*x + 1/2*e)^5 + 8*I*B*c^3*tan
(1/2*f*x + 1/2*e)^5 - 2*I*A*c^3*tan(1/2*f*x + 1/2*e)^4 + 7*B*c^3*tan(1/2*f*x + 1/2*e)^4 - 10*A*c^3*tan(1/2*f*x
+ 1/2*e)^3 - 14*I*B*c^3*tan(1/2*f*x + 1/2*e)^3 + 2*I*A*c^3*tan(1/2*f*x + 1/2*e)^2 - 7*B*c^3*tan(1/2*f*x + 1/2
*e)^2 + 5*A*c^3*tan(1/2*f*x + 1/2*e) + 8*I*B*c^3*tan(1/2*f*x + 1/2*e))/((tan(1/2*f*x + 1/2*e)^3 - I*tan(1/2*f*
x + 1/2*e)^2 - tan(1/2*f*x + 1/2*e) + I)^2*a))/f

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maple [A]  time = 0.19, size = 150, normalized size = 1.24 \[ \frac {c^{3} A \tan \left (f x +e \right )}{f a}+\frac {4 i c^{3} B \tan \left (f x +e \right )}{f a}+\frac {B \,c^{3} \left (\tan ^{2}\left (f x +e \right )\right )}{2 a f}+\frac {4 i c^{3} B}{f a \left (\tan \left (f x +e \right )-i\right )}+\frac {4 c^{3} A}{f a \left (\tan \left (f x +e \right )-i\right )}+\frac {4 i c^{3} A \ln \left (\tan \left (f x +e \right )-i\right )}{f a}-\frac {8 c^{3} B \ln \left (\tan \left (f x +e \right )-i\right )}{f a} \]

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

[Out]

1/f*c^3/a*A*tan(f*x+e)+4*I/f*c^3/a*B*tan(f*x+e)+1/2*B*c^3*tan(f*x+e)^2/a/f+4*I/f*c^3/a/(tan(f*x+e)-I)*B+4/f*c^
3/a/(tan(f*x+e)-I)*A+4*I/f*c^3/a*A*ln(tan(f*x+e)-I)-8/f*c^3/a*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))^3/(a+I*a*tan(f*x+e)),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 = 8.70, size = 136, normalized size = 1.12 \[ \frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )\,\left (-\frac {8\,B\,c^3}{a}+\frac {A\,c^3\,4{}\mathrm {i}}{a}\right )}{f}+\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (\frac {c^3\,\left (A+B\,2{}\mathrm {i}\right )}{a}+\frac {B\,c^3\,2{}\mathrm {i}}{a}\right )}{f}+\frac {\frac {4\,B\,c^3}{a}+\frac {\left (4\,A\,c^3+B\,c^3\,8{}\mathrm {i}\right )\,1{}\mathrm {i}}{a}}{f\,\left (1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}+\frac {B\,c^3\,{\mathrm {tan}\left (e+f\,x\right )}^2}{2\,a\,f} \]

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)^3)/(a + a*tan(e + f*x)*1i),x)

[Out]

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

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sympy [A]  time = 0.95, size = 269, normalized size = 2.22 \[ \frac {- 2 i A c^{3} + 8 B c^{3} + \left (- 2 i A c^{3} e^{2 i e} + 6 B c^{3} e^{2 i e}\right ) e^{2 i f x}}{- a f e^{4 i e} e^{4 i f x} - 2 a f e^{2 i e} e^{2 i f x} - a f} + \begin {cases} - \frac {\left (- 2 i A c^{3} + 2 B c^{3}\right ) e^{- 2 i e} e^{- 2 i f x}}{a f} & \text {for}\: a f e^{2 i e} \neq 0 \\x \left (- \frac {- 8 A c^{3} - 16 i B c^{3}}{a} + \frac {i \left (8 i A c^{3} e^{2 i e} - 4 i A c^{3} - 16 B c^{3} e^{2 i e} + 4 B c^{3}\right ) e^{- 2 i e}}{a}\right ) & \text {otherwise} \end {cases} - \frac {4 i c^{3} \left (A + 2 i B\right ) \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{a f} - \frac {x \left (8 A c^{3} + 16 i B c^{3}\right )}{a} \]

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

[Out]

(-2*I*A*c**3 + 8*B*c**3 + (-2*I*A*c**3*exp(2*I*e) + 6*B*c**3*exp(2*I*e))*exp(2*I*f*x))/(-a*f*exp(4*I*e)*exp(4*
I*f*x) - 2*a*f*exp(2*I*e)*exp(2*I*f*x) - a*f) + Piecewise((-(-2*I*A*c**3 + 2*B*c**3)*exp(-2*I*e)*exp(-2*I*f*x)
/(a*f), Ne(a*f*exp(2*I*e), 0)), (x*(-(-8*A*c**3 - 16*I*B*c**3)/a + I*(8*I*A*c**3*exp(2*I*e) - 4*I*A*c**3 - 16*
B*c**3*exp(2*I*e) + 4*B*c**3)*exp(-2*I*e)/a), True)) - 4*I*c**3*(A + 2*I*B)*log(exp(2*I*f*x) + exp(-2*I*e))/(a
*f) - x*(8*A*c**3 + 16*I*B*c**3)/a

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