\(\int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))} \, dx\) [930]

Optimal result

Integrand size = 31, antiderivative size = 124 \[ \int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))} \, dx=\frac {x}{4 a^3 c}-\frac {i}{16 a^3 f (c-i c \tan (e+f x))}+\frac {i c^2}{12 a^3 f (c+i c \tan (e+f x))^3}+\frac {i c}{8 a^3 f (c+i c \tan (e+f x))^2}+\frac {3 i}{16 a^3 f (c+i c \tan (e+f x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {3603, 3568, 46, 212} \[ \int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))} \, dx=\frac {i c^2}{12 a^3 f (c+i c \tan (e+f x))^3}+\frac {i c}{8 a^3 f (c+i c \tan (e+f x))^2}-\frac {i}{16 a^3 f (c-i c \tan (e+f x))}+\frac {3 i}{16 a^3 f (c+i c \tan (e+f x))}+\frac {x}{4 a^3 c} \]

[In]

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

[Out]

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

Rule 46

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] && Lt
Q[m + n + 2, 0])

Rule 212

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

Rule 3568

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 3603

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]))

Rubi steps \begin{align*} \text {integral}& = \frac {\int \cos ^6(e+f x) (c-i c \tan (e+f x))^2 \, dx}{a^3 c^3} \\ & = \frac {\left (i c^4\right ) \text {Subst}\left (\int \frac {1}{(c-x)^4 (c+x)^2} \, dx,x,-i c \tan (e+f x)\right )}{a^3 f} \\ & = \frac {\left (i c^4\right ) \text {Subst}\left (\int \left (\frac {1}{4 c^2 (c-x)^4}+\frac {1}{4 c^3 (c-x)^3}+\frac {3}{16 c^4 (c-x)^2}+\frac {1}{16 c^4 (c+x)^2}+\frac {1}{4 c^4 \left (c^2-x^2\right )}\right ) \, dx,x,-i c \tan (e+f x)\right )}{a^3 f} \\ & = -\frac {i}{16 a^3 f (c-i c \tan (e+f x))}+\frac {i c^2}{12 a^3 f (c+i c \tan (e+f x))^3}+\frac {i c}{8 a^3 f (c+i c \tan (e+f x))^2}+\frac {3 i}{16 a^3 f (c+i c \tan (e+f x))}+\frac {i \text {Subst}\left (\int \frac {1}{c^2-x^2} \, dx,x,-i c \tan (e+f x)\right )}{4 a^3 f} \\ & = \frac {x}{4 a^3 c}-\frac {i}{16 a^3 f (c-i c \tan (e+f x))}+\frac {i c^2}{12 a^3 f (c+i c \tan (e+f x))^3}+\frac {i c}{8 a^3 f (c+i c \tan (e+f x))^2}+\frac {3 i}{16 a^3 f (c+i c \tan (e+f x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.26 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.81 \[ \int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))} \, dx=-\frac {4 i+\tan (e+f x)+6 i \tan ^2(e+f x)-3 \tan ^3(e+f x)-3 \arctan (\tan (e+f x)) (-i+\tan (e+f x))^3 (i+\tan (e+f x))}{12 a^3 c f (-i+\tan (e+f x))^3 (i+\tan (e+f x))} \]

[In]

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

[Out]

-1/12*(4*I + Tan[e + f*x] + (6*I)*Tan[e + f*x]^2 - 3*Tan[e + f*x]^3 - 3*ArcTan[Tan[e + f*x]]*(-I + Tan[e + f*x
])^3*(I + Tan[e + f*x]))/(a^3*c*f*(-I + Tan[e + f*x])^3*(I + Tan[e + f*x]))

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.76

method	result	size
risch	\(\frac {x}{4 a^{3} c}+\frac {i {\mathrm e}^{-4 i \left (f x +e \right )}}{16 a^{3} c f}+\frac {i {\mathrm e}^{-6 i \left (f x +e \right )}}{96 a^{3} c f}+\frac {5 i \cos \left (2 f x +2 e \right )}{32 a^{3} c f}+\frac {7 \sin \left (2 f x +2 e \right )}{32 a^{3} c f}\)	\(94\)
derivativedivides	\(\frac {\arctan \left (\tan \left (f x +e \right )\right )}{4 f \,a^{3} c}+\frac {1}{16 f \,a^{3} c \left (\tan \left (f x +e \right )+i\right )}-\frac {i}{8 f \,a^{3} c \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {1}{12 f \,a^{3} c \left (\tan \left (f x +e \right )-i\right )^{3}}+\frac {3}{16 f \,a^{3} c \left (\tan \left (f x +e \right )-i\right )}\)	\(109\)
default	\(\frac {\arctan \left (\tan \left (f x +e \right )\right )}{4 f \,a^{3} c}+\frac {1}{16 f \,a^{3} c \left (\tan \left (f x +e \right )+i\right )}-\frac {i}{8 f \,a^{3} c \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {1}{12 f \,a^{3} c \left (\tan \left (f x +e \right )-i\right )^{3}}+\frac {3}{16 f \,a^{3} c \left (\tan \left (f x +e \right )-i\right )}\)	\(109\)
norman	\(\frac {\frac {x}{4 a c}+\frac {3 \tan \left (f x +e \right )}{4 a c f}+\frac {2 \left (\tan ^{3}\left (f x +e \right )\right )}{3 a c f}+\frac {\tan ^{5}\left (f x +e \right )}{4 a c f}+\frac {3 x \left (\tan ^{2}\left (f x +e \right )\right )}{4 a c}+\frac {3 x \left (\tan ^{4}\left (f x +e \right )\right )}{4 a c}+\frac {x \left (\tan ^{6}\left (f x +e \right )\right )}{4 a c}+\frac {i}{3 a c f}}{\left (1+\tan ^{2}\left (f x +e \right )\right )^{3} a^{2}}\)	\(145\)

[In]

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

[Out]

1/4*x/a^3/c+1/16*I/a^3/c/f*exp(-4*I*(f*x+e))+1/96*I/a^3/c/f*exp(-6*I*(f*x+e))+5/32*I/a^3/c/f*cos(2*f*x+2*e)+7/
32/a^3/c/f*sin(2*f*x+2*e)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.55 \[ \int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))} \, dx=\frac {{\left (24 \, f x e^{\left (6 i \, f x + 6 i \, e\right )} - 3 i \, e^{\left (8 i \, f x + 8 i \, e\right )} + 18 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 6 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + i\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{96 \, a^{3} c f} \]

[In]

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

[Out]

1/96*(24*f*x*e^(6*I*f*x + 6*I*e) - 3*I*e^(8*I*f*x + 8*I*e) + 18*I*e^(4*I*f*x + 4*I*e) + 6*I*e^(2*I*f*x + 2*I*e
) + I)*e^(-6*I*f*x - 6*I*e)/(a^3*c*f)

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.73 \[ \int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))} \, dx=\begin {cases} \frac {\left (- 24576 i a^{9} c^{3} f^{3} e^{14 i e} e^{2 i f x} + 147456 i a^{9} c^{3} f^{3} e^{10 i e} e^{- 2 i f x} + 49152 i a^{9} c^{3} f^{3} e^{8 i e} e^{- 4 i f x} + 8192 i a^{9} c^{3} f^{3} e^{6 i e} e^{- 6 i f x}\right ) e^{- 12 i e}}{786432 a^{12} c^{4} f^{4}} & \text {for}\: a^{12} c^{4} f^{4} e^{12 i e} \neq 0 \\x \left (\frac {\left (e^{8 i e} + 4 e^{6 i e} + 6 e^{4 i e} + 4 e^{2 i e} + 1\right ) e^{- 6 i e}}{16 a^{3} c} - \frac {1}{4 a^{3} c}\right ) & \text {otherwise} \end {cases} + \frac {x}{4 a^{3} c} \]

[In]

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

[Out]

Piecewise(((-24576*I*a**9*c**3*f**3*exp(14*I*e)*exp(2*I*f*x) + 147456*I*a**9*c**3*f**3*exp(10*I*e)*exp(-2*I*f*
x) + 49152*I*a**9*c**3*f**3*exp(8*I*e)*exp(-4*I*f*x) + 8192*I*a**9*c**3*f**3*exp(6*I*e)*exp(-6*I*f*x))*exp(-12
*I*e)/(786432*a**12*c**4*f**4), Ne(a**12*c**4*f**4*exp(12*I*e), 0)), (x*((exp(8*I*e) + 4*exp(6*I*e) + 6*exp(4*
I*e) + 4*exp(2*I*e) + 1)*exp(-6*I*e)/(16*a**3*c) - 1/(4*a**3*c)), True)) + x/(4*a**3*c)

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))} \, dx=\text {Exception raised: RuntimeError} \]

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negative exponent.

Giac [A] (verification not implemented)

none

Time = 0.49 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.93 \[ \int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))} \, dx=-\frac {-\frac {6 i \, \log \left (\tan \left (f x + e\right ) + i\right )}{a^{3} c} + \frac {6 i \, \log \left (\tan \left (f x + e\right ) - i\right )}{a^{3} c} + \frac {3 \, {\left (2 i \, \tan \left (f x + e\right ) - 3\right )}}{a^{3} c {\left (\tan \left (f x + e\right ) + i\right )}} + \frac {-11 i \, \tan \left (f x + e\right )^{3} - 42 \, \tan \left (f x + e\right )^{2} + 57 i \, \tan \left (f x + e\right ) + 30}{a^{3} c {\left (\tan \left (f x + e\right ) - i\right )}^{3}}}{48 \, f} \]

[In]

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

[Out]

-1/48*(-6*I*log(tan(f*x + e) + I)/(a^3*c) + 6*I*log(tan(f*x + e) - I)/(a^3*c) + 3*(2*I*tan(f*x + e) - 3)/(a^3*
c*(tan(f*x + e) + I)) + (-11*I*tan(f*x + e)^3 - 42*tan(f*x + e)^2 + 57*I*tan(f*x + e) + 30)/(a^3*c*(tan(f*x +
e) - I)^3))/f

Mupad [B] (verification not implemented)

Time = 6.44 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.62 \[ \int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))} \, dx=\frac {x}{4\,a^3\,c}-\frac {\frac {{\mathrm {tan}\left (e+f\,x\right )}^3\,1{}\mathrm {i}}{4}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2}{2}-\frac {\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}{12}+\frac {1}{3}}{a^3\,c\,f\,{\left (1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3\,\left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )} \]

[In]

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

[Out]

x/(4*a^3*c) - (tan(e + f*x)^2/2 - (tan(e + f*x)*1i)/12 + (tan(e + f*x)^3*1i)/4 + 1/3)/(a^3*c*f*(tan(e + f*x)*1
i + 1)^3*(tan(e + f*x) + 1i))