3.937 \(\int \frac{1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^2} \, dx\)

Optimal. Leaf size=114 \[ \frac{i \cos ^6(e+f x)}{6 a^3 c^2 f}+\frac{\sin (e+f x) \cos ^5(e+f x)}{6 a^3 c^2 f}+\frac{5 \sin (e+f x) \cos ^3(e+f x)}{24 a^3 c^2 f}+\frac{5 \sin (e+f x) \cos (e+f x)}{16 a^3 c^2 f}+\frac{5 x}{16 a^3 c^2} \]

[Out]

(5*x)/(16*a^3*c^2) + ((I/6)*Cos[e + f*x]^6)/(a^3*c^2*f) + (5*Cos[e + f*x]*Sin[e + f*x])/(16*a^3*c^2*f) + (5*Co
s[e + f*x]^3*Sin[e + f*x])/(24*a^3*c^2*f) + (Cos[e + f*x]^5*Sin[e + f*x])/(6*a^3*c^2*f)

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Rubi [A]  time = 0.125646, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {3522, 3486, 2635, 8} \[ \frac{i \cos ^6(e+f x)}{6 a^3 c^2 f}+\frac{\sin (e+f x) \cos ^5(e+f x)}{6 a^3 c^2 f}+\frac{5 \sin (e+f x) \cos ^3(e+f x)}{24 a^3 c^2 f}+\frac{5 \sin (e+f x) \cos (e+f x)}{16 a^3 c^2 f}+\frac{5 x}{16 a^3 c^2} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + I*a*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^2),x]

[Out]

(5*x)/(16*a^3*c^2) + ((I/6)*Cos[e + f*x]^6)/(a^3*c^2*f) + (5*Cos[e + f*x]*Sin[e + f*x])/(16*a^3*c^2*f) + (5*Co
s[e + f*x]^3*Sin[e + f*x])/(24*a^3*c^2*f) + (Cos[e + f*x]^5*Sin[e + f*x])/(6*a^3*c^2*f)

Rule 3522

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Sec[e + f*x]^(2*m)*(c + d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] &&  !(IGtQ[n, 0] && (LtQ[m, 0] || GtQ[m, n]))

Rule 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^2} \, dx &=\frac{\int \cos ^6(e+f x) (c-i c \tan (e+f x)) \, dx}{a^3 c^3}\\ &=\frac{i \cos ^6(e+f x)}{6 a^3 c^2 f}+\frac{\int \cos ^6(e+f x) \, dx}{a^3 c^2}\\ &=\frac{i \cos ^6(e+f x)}{6 a^3 c^2 f}+\frac{\cos ^5(e+f x) \sin (e+f x)}{6 a^3 c^2 f}+\frac{5 \int \cos ^4(e+f x) \, dx}{6 a^3 c^2}\\ &=\frac{i \cos ^6(e+f x)}{6 a^3 c^2 f}+\frac{5 \cos ^3(e+f x) \sin (e+f x)}{24 a^3 c^2 f}+\frac{\cos ^5(e+f x) \sin (e+f x)}{6 a^3 c^2 f}+\frac{5 \int \cos ^2(e+f x) \, dx}{8 a^3 c^2}\\ &=\frac{i \cos ^6(e+f x)}{6 a^3 c^2 f}+\frac{5 \cos (e+f x) \sin (e+f x)}{16 a^3 c^2 f}+\frac{5 \cos ^3(e+f x) \sin (e+f x)}{24 a^3 c^2 f}+\frac{\cos ^5(e+f x) \sin (e+f x)}{6 a^3 c^2 f}+\frac{5 \int 1 \, dx}{16 a^3 c^2}\\ &=\frac{5 x}{16 a^3 c^2}+\frac{i \cos ^6(e+f x)}{6 a^3 c^2 f}+\frac{5 \cos (e+f x) \sin (e+f x)}{16 a^3 c^2 f}+\frac{5 \cos ^3(e+f x) \sin (e+f x)}{24 a^3 c^2 f}+\frac{\cos ^5(e+f x) \sin (e+f x)}{6 a^3 c^2 f}\\ \end{align*}

Mathematica [A]  time = 1.16503, size = 135, normalized size = 1.18 \[ \frac{\sec ^3(e+f x) (\cos (2 (e+f x))+i \sin (2 (e+f x))) (-120 f x \sin (e+f x)+60 i \sin (e+f x)+45 i \sin (3 (e+f x))+5 i \sin (5 (e+f x))+60 i (2 f x+i) \cos (e+f x)+15 \cos (3 (e+f x))+\cos (5 (e+f x)))}{384 a^3 c^2 f (\tan (e+f x)-i)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + I*a*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^2),x]

[Out]

(Sec[e + f*x]^3*(Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)])*((60*I)*(I + 2*f*x)*Cos[e + f*x] + 15*Cos[3*(e + f*x)]
+ Cos[5*(e + f*x)] + (60*I)*Sin[e + f*x] - 120*f*x*Sin[e + f*x] + (45*I)*Sin[3*(e + f*x)] + (5*I)*Sin[5*(e +
f*x)]))/(384*a^3*c^2*f*(-I + Tan[e + f*x])^3)

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Maple [A]  time = 0.046, size = 158, normalized size = 1.4 \begin{align*}{\frac{-{\frac{5\,i}{32}}\ln \left ( \tan \left ( fx+e \right ) -i \right ) }{f{a}^{3}{c}^{2}}}-{\frac{{\frac{3\,i}{32}}}{f{a}^{3}{c}^{2} \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}-{\frac{1}{24\,f{a}^{3}{c}^{2} \left ( \tan \left ( fx+e \right ) -i \right ) ^{3}}}+{\frac{3}{16\,f{a}^{3}{c}^{2} \left ( \tan \left ( fx+e \right ) -i \right ) }}+{\frac{{\frac{i}{32}}}{f{a}^{3}{c}^{2} \left ( \tan \left ( fx+e \right ) +i \right ) ^{2}}}+{\frac{{\frac{5\,i}{32}}\ln \left ( \tan \left ( fx+e \right ) +i \right ) }{f{a}^{3}{c}^{2}}}+{\frac{1}{8\,f{a}^{3}{c}^{2} \left ( \tan \left ( fx+e \right ) +i \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/32*I/f/a^3/c^2*ln(tan(f*x+e)-I)-3/32*I/f/a^3/c^2/(tan(f*x+e)-I)^2-1/24/f/a^3/c^2/(tan(f*x+e)-I)^3+3/16/f/a^
3/c^2/(tan(f*x+e)-I)+1/32*I/f/a^3/c^2/(tan(f*x+e)+I)^2+5/32*I/f/a^3/c^2*ln(tan(f*x+e)+I)+1/8/f/a^3/c^2/(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(1/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.41098, size = 250, normalized size = 2.19 \begin{align*} \frac{{\left (120 \, f x e^{\left (6 i \, f x + 6 i \, e\right )} - 3 i \, e^{\left (10 i \, f x + 10 i \, e\right )} - 30 i \, e^{\left (8 i \, f x + 8 i \, e\right )} + 60 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 15 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{384 \, a^{3} c^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

1/384*(120*f*x*e^(6*I*f*x + 6*I*e) - 3*I*e^(10*I*f*x + 10*I*e) - 30*I*e^(8*I*f*x + 8*I*e) + 60*I*e^(4*I*f*x +
4*I*e) + 15*I*e^(2*I*f*x + 2*I*e) + 2*I)*e^(-6*I*f*x - 6*I*e)/(a^3*c^2*f)

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Sympy [A]  time = 1.5718, size = 260, normalized size = 2.28 \begin{align*} \begin{cases} \frac{\left (- 50331648 i a^{12} c^{8} f^{4} e^{16 i e} e^{4 i f x} - 503316480 i a^{12} c^{8} f^{4} e^{14 i e} e^{2 i f x} + 1006632960 i a^{12} c^{8} f^{4} e^{10 i e} e^{- 2 i f x} + 251658240 i a^{12} c^{8} f^{4} e^{8 i e} e^{- 4 i f x} + 33554432 i a^{12} c^{8} f^{4} e^{6 i e} e^{- 6 i f x}\right ) e^{- 12 i e}}{6442450944 a^{15} c^{10} f^{5}} & \text{for}\: 6442450944 a^{15} c^{10} f^{5} e^{12 i e} \neq 0 \\x \left (\frac{\left (e^{10 i e} + 5 e^{8 i e} + 10 e^{6 i e} + 10 e^{4 i e} + 5 e^{2 i e} + 1\right ) e^{- 6 i e}}{32 a^{3} c^{2}} - \frac{5}{16 a^{3} c^{2}}\right ) & \text{otherwise} \end{cases} + \frac{5 x}{16 a^{3} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))**3/(c-I*c*tan(f*x+e))**2,x)

[Out]

Piecewise(((-50331648*I*a**12*c**8*f**4*exp(16*I*e)*exp(4*I*f*x) - 503316480*I*a**12*c**8*f**4*exp(14*I*e)*exp
(2*I*f*x) + 1006632960*I*a**12*c**8*f**4*exp(10*I*e)*exp(-2*I*f*x) + 251658240*I*a**12*c**8*f**4*exp(8*I*e)*ex
p(-4*I*f*x) + 33554432*I*a**12*c**8*f**4*exp(6*I*e)*exp(-6*I*f*x))*exp(-12*I*e)/(6442450944*a**15*c**10*f**5),
Ne(6442450944*a**15*c**10*f**5*exp(12*I*e), 0)), (x*((exp(10*I*e) + 5*exp(8*I*e) + 10*exp(6*I*e) + 10*exp(4*I
*e) + 5*exp(2*I*e) + 1)*exp(-6*I*e)/(32*a**3*c**2) - 5/(16*a**3*c**2)), True)) + 5*x/(16*a**3*c**2)

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Giac [A]  time = 1.39622, size = 185, normalized size = 1.62 \begin{align*} -\frac{-\frac{30 i \, \log \left (\tan \left (f x + e\right ) + i\right )}{a^{3} c^{2}} + \frac{30 i \, \log \left (\tan \left (f x + e\right ) - i\right )}{a^{3} c^{2}} + \frac{3 \,{\left (-15 i \, \tan \left (f x + e\right )^{2} + 38 \, \tan \left (f x + e\right ) + 25 i\right )}}{a^{3} c^{2}{\left (-i \, \tan \left (f x + e\right ) + 1\right )}^{2}} - \frac{55 i \, \tan \left (f x + e\right )^{3} + 201 \, \tan \left (f x + e\right )^{2} - 255 i \, \tan \left (f x + e\right ) - 117}{a^{3} c^{2}{\left (\tan \left (f x + e\right ) - i\right )}^{3}}}{192 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^2,x, algorithm="giac")

[Out]

-1/192*(-30*I*log(tan(f*x + e) + I)/(a^3*c^2) + 30*I*log(tan(f*x + e) - I)/(a^3*c^2) + 3*(-15*I*tan(f*x + e)^2
+ 38*tan(f*x + e) + 25*I)/(a^3*c^2*(-I*tan(f*x + e) + 1)^2) - (55*I*tan(f*x + e)^3 + 201*tan(f*x + e)^2 - 255
*I*tan(f*x + e) - 117)/(a^3*c^2*(tan(f*x + e) - I)^3))/f