3.953 \(\int \frac{1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^4} \, dx\)

Optimal. Leaf size=162 \[ -\frac{i}{8 a f \left (c^4-i c^4 \tan (e+f x)\right )}+\frac{i}{32 a f \left (c^4+i c^4 \tan (e+f x)\right )}-\frac{3 i}{32 a f \left (c^2-i c^2 \tan (e+f x)\right )^2}+\frac{5 x}{32 a c^4}-\frac{i}{12 a c f (c-i c \tan (e+f x))^3}-\frac{i}{16 a f (c-i c \tan (e+f x))^4} \]

[Out]

(5*x)/(32*a*c^4) - (I/16)/(a*f*(c - I*c*Tan[e + f*x])^4) - (I/12)/(a*c*f*(c - I*c*Tan[e + f*x])^3) - ((3*I)/32
)/(a*f*(c^2 - I*c^2*Tan[e + f*x])^2) - (I/8)/(a*f*(c^4 - I*c^4*Tan[e + f*x])) + (I/32)/(a*f*(c^4 + I*c^4*Tan[e
+ f*x]))

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Rubi [A]  time = 0.16858, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {3522, 3487, 44, 206} \[ -\frac{i}{8 a f \left (c^4-i c^4 \tan (e+f x)\right )}+\frac{i}{32 a f \left (c^4+i c^4 \tan (e+f x)\right )}-\frac{3 i}{32 a f \left (c^2-i c^2 \tan (e+f x)\right )^2}+\frac{5 x}{32 a c^4}-\frac{i}{12 a c f (c-i c \tan (e+f x))^3}-\frac{i}{16 a f (c-i c \tan (e+f x))^4} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + I*a*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^4),x]

[Out]

(5*x)/(32*a*c^4) - (I/16)/(a*f*(c - I*c*Tan[e + f*x])^4) - (I/12)/(a*c*f*(c - I*c*Tan[e + f*x])^3) - ((3*I)/32
)/(a*f*(c^2 - I*c^2*Tan[e + f*x])^2) - (I/8)/(a*f*(c^4 - I*c^4*Tan[e + f*x])) + (I/32)/(a*f*(c^4 + I*c^4*Tan[e
+ f*x]))

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 3487

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^4} \, dx &=\frac{\int \frac{\cos ^2(e+f x)}{(c-i c \tan (e+f x))^3} \, dx}{a c}\\ &=\frac{\left (i c^2\right ) \operatorname{Subst}\left (\int \frac{1}{(c-x)^2 (c+x)^5} \, dx,x,-i c \tan (e+f x)\right )}{a f}\\ &=\frac{\left (i c^2\right ) \operatorname{Subst}\left (\int \left (\frac{1}{32 c^5 (c-x)^2}+\frac{1}{4 c^2 (c+x)^5}+\frac{1}{4 c^3 (c+x)^4}+\frac{3}{16 c^4 (c+x)^3}+\frac{1}{8 c^5 (c+x)^2}+\frac{5}{32 c^5 \left (c^2-x^2\right )}\right ) \, dx,x,-i c \tan (e+f x)\right )}{a f}\\ &=-\frac{i}{16 a f (c-i c \tan (e+f x))^4}-\frac{i}{12 a c f (c-i c \tan (e+f x))^3}-\frac{3 i}{32 a f \left (c^2-i c^2 \tan (e+f x)\right )^2}-\frac{i}{8 a f \left (c^4-i c^4 \tan (e+f x)\right )}+\frac{i}{32 a f \left (c^4+i c^4 \tan (e+f x)\right )}+\frac{(5 i) \operatorname{Subst}\left (\int \frac{1}{c^2-x^2} \, dx,x,-i c \tan (e+f x)\right )}{32 a c^3 f}\\ &=\frac{5 x}{32 a c^4}-\frac{i}{16 a f (c-i c \tan (e+f x))^4}-\frac{i}{12 a c f (c-i c \tan (e+f x))^3}-\frac{3 i}{32 a f \left (c^2-i c^2 \tan (e+f x)\right )^2}-\frac{i}{8 a f \left (c^4-i c^4 \tan (e+f x)\right )}+\frac{i}{32 a f \left (c^4+i c^4 \tan (e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.965107, size = 134, normalized size = 0.83 \[ \frac{\sec (e+f x) (\cos (4 (e+f x))+i \sin (4 (e+f x))) (60 i \sin (e+f x)-120 f x \sin (3 (e+f x))-20 i \sin (3 (e+f x))-15 i \sin (5 (e+f x))-180 \cos (e+f x)+(-20-120 i f x) \cos (3 (e+f x))+9 \cos (5 (e+f x)))}{768 a c^4 f (\tan (e+f x)-i)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + I*a*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^4),x]

[Out]

(Sec[e + f*x]*(Cos[4*(e + f*x)] + I*Sin[4*(e + f*x)])*(-180*Cos[e + f*x] + (-20 - (120*I)*f*x)*Cos[3*(e + f*x)
] + 9*Cos[5*(e + f*x)] + (60*I)*Sin[e + f*x] - (20*I)*Sin[3*(e + f*x)] - 120*f*x*Sin[3*(e + f*x)] - (15*I)*Sin
[5*(e + f*x)]))/(768*a*c^4*f*(-I + Tan[e + f*x]))

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Maple [A]  time = 0.045, size = 158, normalized size = 1. \begin{align*}{\frac{-{\frac{5\,i}{64}}\ln \left ( \tan \left ( fx+e \right ) -i \right ) }{fa{c}^{4}}}+{\frac{1}{32\,fa{c}^{4} \left ( \tan \left ( fx+e \right ) -i \right ) }}+{\frac{{\frac{3\,i}{32}}}{fa{c}^{4} \left ( \tan \left ( fx+e \right ) +i \right ) ^{2}}}-{\frac{{\frac{i}{16}}}{fa{c}^{4} \left ( \tan \left ( fx+e \right ) +i \right ) ^{4}}}+{\frac{{\frac{5\,i}{64}}\ln \left ( \tan \left ( fx+e \right ) +i \right ) }{fa{c}^{4}}}-{\frac{1}{12\,fa{c}^{4} \left ( \tan \left ( fx+e \right ) +i \right ) ^{3}}}+{\frac{1}{8\,fa{c}^{4} \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))/(c-I*c*tan(f*x+e))^4,x)

[Out]

-5/64*I/f/a/c^4*ln(tan(f*x+e)-I)+1/32/f/a/c^4/(tan(f*x+e)-I)+3/32*I/f/a/c^4/(tan(f*x+e)+I)^2-1/16*I/f/a/c^4/(t
an(f*x+e)+I)^4+5/64*I/f/a/c^4*ln(tan(f*x+e)+I)-1/12/f/a/c^4/(tan(f*x+e)+I)^3+1/8/f/a/c^4/(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))/(c-I*c*tan(f*x+e))^4,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.49811, size = 250, normalized size = 1.54 \begin{align*} \frac{{\left (120 \, f x e^{\left (2 i \, f x + 2 i \, e\right )} - 3 i \, e^{\left (10 i \, f x + 10 i \, e\right )} - 20 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 60 i \, e^{\left (6 i \, f x + 6 i \, e\right )} - 120 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 12 i\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{768 \, a c^{4} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))^4,x, algorithm="fricas")

[Out]

1/768*(120*f*x*e^(2*I*f*x + 2*I*e) - 3*I*e^(10*I*f*x + 10*I*e) - 20*I*e^(8*I*f*x + 8*I*e) - 60*I*e^(6*I*f*x +
6*I*e) - 120*I*e^(4*I*f*x + 4*I*e) + 12*I)*e^(-2*I*f*x - 2*I*e)/(a*c^4*f)

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Sympy [A]  time = 1.41021, size = 248, normalized size = 1.53 \begin{align*} \begin{cases} \frac{\left (- 25165824 i a^{4} c^{16} f^{4} e^{10 i e} e^{8 i f x} - 167772160 i a^{4} c^{16} f^{4} e^{8 i e} e^{6 i f x} - 503316480 i a^{4} c^{16} f^{4} e^{6 i e} e^{4 i f x} - 1006632960 i a^{4} c^{16} f^{4} e^{4 i e} e^{2 i f x} + 100663296 i a^{4} c^{16} f^{4} e^{- 2 i f x}\right ) e^{- 2 i e}}{6442450944 a^{5} c^{20} f^{5}} & \text{for}\: 6442450944 a^{5} c^{20} f^{5} e^{2 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^{- 2 i e}}{32 a c^{4}} - \frac{5}{32 a c^{4}}\right ) & \text{otherwise} \end{cases} + \frac{5 x}{32 a c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))**4,x)

[Out]

Piecewise(((-25165824*I*a**4*c**16*f**4*exp(10*I*e)*exp(8*I*f*x) - 167772160*I*a**4*c**16*f**4*exp(8*I*e)*exp(
6*I*f*x) - 503316480*I*a**4*c**16*f**4*exp(6*I*e)*exp(4*I*f*x) - 1006632960*I*a**4*c**16*f**4*exp(4*I*e)*exp(2
*I*f*x) + 100663296*I*a**4*c**16*f**4*exp(-2*I*f*x))*exp(-2*I*e)/(6442450944*a**5*c**20*f**5), Ne(6442450944*a
**5*c**20*f**5*exp(2*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(-2*I*e)/(32*a*c**4) - 5/(32*a*c**4)), True)) + 5*x/(32*a*c**4)

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Giac [A]  time = 1.32037, size = 189, normalized size = 1.17 \begin{align*} -\frac{\frac{60 i \, \log \left (i \, \tan \left (f x + e\right ) + 1\right )}{a c^{4}} - \frac{60 i \, \log \left (i \, \tan \left (f x + e\right ) - 1\right )}{a c^{4}} - \frac{12 \,{\left (5 \, \tan \left (f x + e\right ) - 7 i\right )}}{a c^{4}{\left (-i \, \tan \left (f x + e\right ) - 1\right )}} + \frac{125 i \, \tan \left (f x + e\right )^{4} - 596 \, \tan \left (f x + e\right )^{3} - 1110 i \, \tan \left (f x + e\right )^{2} + 996 \, \tan \left (f x + e\right ) + 405 i}{a c^{4}{\left (\tan \left (f x + e\right ) + i\right )}^{4}}}{768 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))^4,x, algorithm="giac")

[Out]

-1/768*(60*I*log(I*tan(f*x + e) + 1)/(a*c^4) - 60*I*log(I*tan(f*x + e) - 1)/(a*c^4) - 12*(5*tan(f*x + e) - 7*I
)/(a*c^4*(-I*tan(f*x + e) - 1)) + (125*I*tan(f*x + e)^4 - 596*tan(f*x + e)^3 - 1110*I*tan(f*x + e)^2 + 996*tan
(f*x + e) + 405*I)/(a*c^4*(tan(f*x + e) + I)^4))/f