\(\int \frac {c-i c \tan (e+f x)}{(a+i a \tan (e+f x))^3} \, dx\) [895]

Optimal result

Integrand size = 29, antiderivative size = 25 \[ \int \frac {c-i c \tan (e+f x)}{(a+i a \tan (e+f x))^3} \, dx=\frac {i c}{3 f (a+i a \tan (e+f x))^3} \]

[Out]

1/3*I*c/f/(a+I*a*tan(f*x+e))^3

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {3603, 3568, 32} \[ \int \frac {c-i c \tan (e+f x)}{(a+i a \tan (e+f x))^3} \, dx=\frac {i c}{3 f (a+i a \tan (e+f x))^3} \]

[In]

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

[Out]

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

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

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}& = (a c) \int \frac {\sec ^2(e+f x)}{(a+i a \tan (e+f x))^4} \, dx \\ & = -\frac {(i c) \text {Subst}\left (\int \frac {1}{(a+x)^4} \, dx,x,i a \tan (e+f x)\right )}{f} \\ & = \frac {i c}{3 f (a+i a \tan (e+f x))^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.51 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {c-i c \tan (e+f x)}{(a+i a \tan (e+f x))^3} \, dx=-\frac {c}{3 a^3 f (-i+\tan (e+f x))^3} \]

[In]

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

[Out]

-1/3*c/(a^3*f*(-I + Tan[e + f*x])^3)

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84

method	result	size
derivativedivides	\(-\frac {c}{3 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{3}}\)	\(21\)
default	\(-\frac {c}{3 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{3}}\)	\(21\)
risch	\(\frac {i c \,{\mathrm e}^{-2 i \left (f x +e \right )}}{8 a^{3} f}+\frac {i c \,{\mathrm e}^{-4 i \left (f x +e \right )}}{8 a^{3} f}+\frac {i c \,{\mathrm e}^{-6 i \left (f x +e \right )}}{24 a^{3} f}\)	\(59\)
norman	\(\frac {\frac {c \tan \left (f x +e \right )}{a f}-\frac {i c \left (\tan ^{2}\left (f x +e \right )\right )}{a f}+\frac {i c}{3 a f}-\frac {c \left (\tan ^{3}\left (f x +e \right )\right )}{3 a f}}{a^{2} \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}\)	\(77\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (19) = 38\).

Time = 0.24 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.80 \[ \int \frac {c-i c \tan (e+f x)}{(a+i a \tan (e+f x))^3} \, dx=\frac {{\left (3 i \, c e^{\left (4 i \, f x + 4 i \, e\right )} + 3 i \, c e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{24 \, a^{3} f} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 141 vs. \(2 (19) = 38\).

Time = 0.20 (sec) , antiderivative size = 141, normalized size of antiderivative = 5.64 \[ \int \frac {c-i c \tan (e+f x)}{(a+i a \tan (e+f x))^3} \, dx=\begin {cases} \frac {\left (192 i a^{6} c f^{2} e^{10 i e} e^{- 2 i f x} + 192 i a^{6} c f^{2} e^{8 i e} e^{- 4 i f x} + 64 i a^{6} c f^{2} e^{6 i e} e^{- 6 i f x}\right ) e^{- 12 i e}}{1536 a^{9} f^{3}} & \text {for}\: a^{9} f^{3} e^{12 i e} \neq 0 \\\frac {x \left (c e^{4 i e} + 2 c e^{2 i e} + c\right ) e^{- 6 i e}}{4 a^{3}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 90 vs. \(2 (19) = 38\).

Time = 0.54 (sec) , antiderivative size = 90, normalized size of antiderivative = 3.60 \[ \int \frac {c-i c \tan (e+f x)}{(a+i a \tan (e+f x))^3} \, dx=-\frac {2 \, {\left (3 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 6 i \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 10 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 6 i \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{3 \, a^{3} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}^{6}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 6.04 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {c-i c \tan (e+f x)}{(a+i a \tan (e+f x))^3} \, dx=-\frac {c}{3\,a^3\,f\,{\left (\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )}^3} \]

[In]

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

[Out]

-c/(3*a^3*f*(tan(e + f*x) - 1i)^3)