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

Optimal. Leaf size=59 \[ \frac {c (A+i B)}{3 a^3 f (-\tan (e+f x)+i)^3}-\frac {B c}{2 a^3 f (-\tan (e+f x)+i)^2} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {3588, 43} \[ \frac {c (A+i B)}{3 a^3 f (-\tan (e+f x)+i)^3}-\frac {B c}{2 a^3 f (-\tan (e+f x)+i)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A + I*B)*c)/(3*a^3*f*(I - Tan[e + f*x])^3) - (B*c)/(2*a^3*f*(I - Tan[e + f*x])^2)

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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

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Mathematica [A]  time = 1.39, size = 81, normalized size = 1.37 \[ \frac {c (\tan (e+f x)+i) \sec ^2(e+f x) (-2 (A-2 i B) \sin (2 (e+f x))+2 (B+2 i A) \cos (2 (e+f x))+3 i A)}{24 a^3 f (\tan (e+f x)-i)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*Sec[e + f*x]^2*((3*I)*A + 2*((2*I)*A + B)*Cos[2*(e + f*x)] - 2*(A - (2*I)*B)*Sin[2*(e + f*x)])*(I + Tan[e +
f*x]))/(24*a^3*f*(-I + Tan[e + f*x])^3)

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fricas [A]  time = 0.58, size = 58, normalized size = 0.98 \[ \frac {{\left ({\left (3 i \, A + 3 \, B\right )} c e^{\left (4 i \, f x + 4 i \, e\right )} + 3 i \, A c e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (i \, A - B\right )} c\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{24 \, a^{3} f} \]

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

[Out]

1/24*((3*I*A + 3*B)*c*e^(4*I*f*x + 4*I*e) + 3*I*A*c*e^(2*I*f*x + 2*I*e) + (I*A - B)*c)*e^(-6*I*f*x - 6*I*e)/(a
^3*f)

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giac [B]  time = 2.64, size = 149, normalized size = 2.53 \[ -\frac {2 \, {\left (3 \, A c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 6 i \, A c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 3 \, B c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 10 \, A c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 2 i \, B c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 6 i \, A c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, B c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, A c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{3 \, a^{3} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}^{6}} \]

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

[Out]

-2/3*(3*A*c*tan(1/2*f*x + 1/2*e)^5 - 6*I*A*c*tan(1/2*f*x + 1/2*e)^4 - 3*B*c*tan(1/2*f*x + 1/2*e)^4 - 10*A*c*ta
n(1/2*f*x + 1/2*e)^3 + 2*I*B*c*tan(1/2*f*x + 1/2*e)^3 + 6*I*A*c*tan(1/2*f*x + 1/2*e)^2 + 3*B*c*tan(1/2*f*x + 1
/2*e)^2 + 3*A*c*tan(1/2*f*x + 1/2*e))/(a^3*f*(tan(1/2*f*x + 1/2*e) - I)^6)

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maple [A]  time = 0.26, size = 43, normalized size = 0.73 \[ \frac {c \left (-\frac {B}{2 \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {i B +A}{3 \left (\tan \left (f x +e \right )-i\right )^{3}}\right )}{f \,a^{3}} \]

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

[Out]

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

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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))/(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 = 8.84, size = 62, normalized size = 1.05 \[ \frac {\frac {c\,\left (B+A\,2{}\mathrm {i}\right )}{6}+\frac {B\,c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}{2}}{a^3\,f\,\left (-{\mathrm {tan}\left (e+f\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (e+f\,x\right )}^2+\mathrm {tan}\left (e+f\,x\right )\,3{}\mathrm {i}+1\right )} \]

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))/(a + a*tan(e + f*x)*1i)^3,x)

[Out]

((c*(A*2i + B))/6 + (B*c*tan(e + f*x)*1i)/2)/(a^3*f*(tan(e + f*x)*3i - 3*tan(e + f*x)^2 - tan(e + f*x)^3*1i +
1))

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sympy [A]  time = 0.57, size = 211, normalized size = 3.58 \[ \begin {cases} - \frac {\left (- 192 i A a^{6} c f^{2} e^{8 i e} e^{- 4 i f x} + \left (- 64 i A a^{6} c f^{2} e^{6 i e} + 64 B a^{6} c f^{2} e^{6 i e}\right ) e^{- 6 i f x} + \left (- 192 i A a^{6} c f^{2} e^{10 i e} - 192 B a^{6} c f^{2} e^{10 i e}\right ) e^{- 2 i f x}\right ) e^{- 12 i e}}{1536 a^{9} f^{3}} & \text {for}\: 1536 a^{9} f^{3} e^{12 i e} \neq 0 \\\frac {x \left (A c e^{4 i e} + 2 A c e^{2 i e} + A c - i B c e^{4 i e} + i B c\right ) e^{- 6 i e}}{4 a^{3}} & \text {otherwise} \end {cases} \]

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

[Out]

Piecewise((-(-192*I*A*a**6*c*f**2*exp(8*I*e)*exp(-4*I*f*x) + (-64*I*A*a**6*c*f**2*exp(6*I*e) + 64*B*a**6*c*f**
2*exp(6*I*e))*exp(-6*I*f*x) + (-192*I*A*a**6*c*f**2*exp(10*I*e) - 192*B*a**6*c*f**2*exp(10*I*e))*exp(-2*I*f*x)
)*exp(-12*I*e)/(1536*a**9*f**3), Ne(1536*a**9*f**3*exp(12*I*e), 0)), (x*(A*c*exp(4*I*e) + 2*A*c*exp(2*I*e) + A
*c - I*B*c*exp(4*I*e) + I*B*c)*exp(-6*I*e)/(4*a**3), True))

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