\(\int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^3} \, dx\) [944]

Optimal result

Integrand size = 31, antiderivative size = 131 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^3} \, dx=\frac {x}{4 a c^3}-\frac {i}{12 a f (c-i c \tan (e+f x))^3}-\frac {i}{8 a c f (c-i c \tan (e+f x))^2}-\frac {3 i}{16 a f \left (c^3-i c^3 \tan (e+f x)\right )}+\frac {i}{16 a f \left (c^3+i c^3 \tan (e+f x)\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 131, 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} \[ \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^3} \, dx=-\frac {3 i}{16 a f \left (c^3-i c^3 \tan (e+f x)\right )}+\frac {i}{16 a f \left (c^3+i c^3 \tan (e+f x)\right )}+\frac {x}{4 a c^3}-\frac {i}{8 a c f (c-i c \tan (e+f x))^2}-\frac {i}{12 a f (c-i c \tan (e+f x))^3} \]

[In]

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

[Out]

x/(4*a*c^3) - (I/12)/(a*f*(c - I*c*Tan[e + f*x])^3) - (I/8)/(a*c*f*(c - I*c*Tan[e + f*x])^2) - ((3*I)/16)/(a*f
*(c^3 - I*c^3*Tan[e + f*x])) + (I/16)/(a*f*(c^3 + I*c^3*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 \frac {\cos ^2(e+f x)}{(c-i c \tan (e+f x))^2} \, dx}{a c} \\ & = \frac {\left (i c^2\right ) \text {Subst}\left (\int \frac {1}{(c-x)^2 (c+x)^4} \, dx,x,-i c \tan (e+f x)\right )}{a f} \\ & = \frac {\left (i c^2\right ) \text {Subst}\left (\int \left (\frac {1}{16 c^4 (c-x)^2}+\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}{4 c^4 \left (c^2-x^2\right )}\right ) \, dx,x,-i c \tan (e+f x)\right )}{a f} \\ & = -\frac {i}{12 a f (c-i c \tan (e+f x))^3}-\frac {i}{8 a c f (c-i c \tan (e+f x))^2}-\frac {3 i}{16 a f \left (c^3-i c^3 \tan (e+f x)\right )}+\frac {i}{16 a f \left (c^3+i c^3 \tan (e+f x)\right )}+\frac {i \text {Subst}\left (\int \frac {1}{c^2-x^2} \, dx,x,-i c \tan (e+f x)\right )}{4 a c^2 f} \\ & = \frac {x}{4 a c^3}-\frac {i}{12 a f (c-i c \tan (e+f x))^3}-\frac {i}{8 a c f (c-i c \tan (e+f x))^2}-\frac {3 i}{16 a f \left (c^3-i c^3 \tan (e+f x)\right )}+\frac {i}{16 a f \left (c^3+i c^3 \tan (e+f x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.96 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.78 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^3} \, dx=\frac {4 i-\tan (e+f x)+6 i \tan ^2(e+f x)+3 \tan ^3(e+f x)+3 \arctan (\tan (e+f x)) (-i+\tan (e+f x)) (i+\tan (e+f x))^3}{12 a c^3 f (-i+\tan (e+f x)) (i+\tan (e+f x))^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.72

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.52 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^3} \, dx=\frac {{\left (24 \, f x e^{\left (2 i \, f x + 2 i \, e\right )} - i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 6 i \, e^{\left (6 i \, f x + 6 i \, e\right )} - 18 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 3 i\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{96 \, a c^{3} f} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.58 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^3} \, dx=\begin {cases} \frac {\left (- 8192 i a^{3} c^{9} f^{3} e^{8 i e} e^{6 i f x} - 49152 i a^{3} c^{9} f^{3} e^{6 i e} e^{4 i f x} - 147456 i a^{3} c^{9} f^{3} e^{4 i e} e^{2 i f x} + 24576 i a^{3} c^{9} f^{3} e^{- 2 i f x}\right ) e^{- 2 i e}}{786432 a^{4} c^{12} f^{4}} & \text {for}\: a^{4} c^{12} f^{4} e^{2 i e} \neq 0 \\x \left (\frac {\left (e^{8 i e} + 4 e^{6 i e} + 6 e^{4 i e} + 4 e^{2 i e} + 1\right ) e^{- 2 i e}}{16 a c^{3}} - \frac {1}{4 a c^{3}}\right ) & \text {otherwise} \end {cases} + \frac {x}{4 a c^{3}} \]

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.46 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.88 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^3} \, dx=-\frac {-\frac {6 i \, \log \left (\tan \left (f x + e\right ) + i\right )}{a c^{3}} + \frac {6 i \, \log \left (\tan \left (f x + e\right ) - i\right )}{a c^{3}} + \frac {3 \, {\left (-2 i \, \tan \left (f x + e\right ) - 3\right )}}{a c^{3} {\left (\tan \left (f x + e\right ) - i\right )}} + \frac {11 i \, \tan \left (f x + e\right )^{3} - 42 \, \tan \left (f x + e\right )^{2} - 57 i \, \tan \left (f x + e\right ) + 30}{a c^{3} {\left (\tan \left (f x + e\right ) + i\right )}^{3}}}{48 \, f} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.69 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.59 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^3} \, dx=\frac {x}{4\,a\,c^3}-\frac {-\frac {{\mathrm {tan}\left (e+f\,x\right )}^3\,1{}\mathrm {i}}{4}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2}{2}+\frac {\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}{12}+\frac {1}{3}}{a\,c^3\,f\,\left (1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,{\left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )}^3} \]

[In]

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

[Out]

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