\(\int \frac {(c-i c \tan (e+f x))^3}{a+i a \tan (e+f x)} \, dx\) [909]

Optimal result

Integrand size = 31, antiderivative size = 71 \[ \int \frac {(c-i c \tan (e+f x))^3}{a+i a \tan (e+f x)} \, dx=-\frac {4 c^3 x}{a}-\frac {4 i c^3 \log (\cos (e+f x))}{a f}+\frac {c^3 \tan (e+f x)}{a f}+\frac {4 i c^3}{f (a+i a \tan (e+f x))} \]

[Out]

-4*c^3*x/a-4*I*c^3*ln(cos(f*x+e))/a/f+c^3*tan(f*x+e)/a/f+4*I*c^3/f/(a+I*a*tan(f*x+e))

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {3603, 3568, 45} \[ \int \frac {(c-i c \tan (e+f x))^3}{a+i a \tan (e+f x)} \, dx=\frac {c^3 \tan (e+f x)}{a f}+\frac {4 i c^3}{f (a+i a \tan (e+f x))}-\frac {4 i c^3 \log (\cos (e+f x))}{a f}-\frac {4 c^3 x}{a} \]

[In]

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

[Out]

(-4*c^3*x)/a - ((4*I)*c^3*Log[Cos[e + f*x]])/(a*f) + (c^3*Tan[e + f*x])/(a*f) + ((4*I)*c^3)/(f*(a + I*a*Tan[e
+ f*x]))

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 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}& = \left (a^3 c^3\right ) \int \frac {\sec ^6(e+f x)}{(a+i a \tan (e+f x))^4} \, dx \\ & = -\frac {\left (i c^3\right ) \text {Subst}\left (\int \frac {(a-x)^2}{(a+x)^2} \, dx,x,i a \tan (e+f x)\right )}{a^2 f} \\ & = -\frac {\left (i c^3\right ) \text {Subst}\left (\int \left (1+\frac {4 a^2}{(a+x)^2}-\frac {4 a}{a+x}\right ) \, dx,x,i a \tan (e+f x)\right )}{a^2 f} \\ & = -\frac {4 c^3 x}{a}-\frac {4 i c^3 \log (\cos (e+f x))}{a f}+\frac {c^3 \tan (e+f x)}{a f}+\frac {4 i c^3}{f (a+i a \tan (e+f x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.01 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.77 \[ \int \frac {(c-i c \tan (e+f x))^3}{a+i a \tan (e+f x)} \, dx=-\frac {i c^3 \left (-4 \log (i-\tan (e+f x))+i \tan (e+f x)+\frac {4 i}{-i+\tan (e+f x)}\right )}{a f} \]

[In]

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

[Out]

((-I)*c^3*(-4*Log[I - Tan[e + f*x]] + I*Tan[e + f*x] + (4*I)/(-I + Tan[e + f*x])))/(a*f)

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.14

method	result	size
derivativedivides	\(\frac {c^{3} \tan \left (f x +e \right )}{a f}-\frac {4 c^{3} \arctan \left (\tan \left (f x +e \right )\right )}{f a}+\frac {2 i c^{3} \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{f a}+\frac {4 c^{3}}{f a \left (\tan \left (f x +e \right )-i\right )}\)	\(81\)
default	\(\frac {c^{3} \tan \left (f x +e \right )}{a f}-\frac {4 c^{3} \arctan \left (\tan \left (f x +e \right )\right )}{f a}+\frac {2 i c^{3} \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{f a}+\frac {4 c^{3}}{f a \left (\tan \left (f x +e \right )-i\right )}\)	\(81\)
risch	\(\frac {2 i c^{3} {\mathrm e}^{-2 i \left (f x +e \right )}}{a f}-\frac {8 c^{3} x}{a}-\frac {8 c^{3} e}{a f}+\frac {2 i c^{3}}{f a \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}-\frac {4 i c^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{a f}\)	\(93\)
norman	\(\frac {\frac {4 i c^{3}}{a f}+\frac {c^{3} \left (\tan ^{3}\left (f x +e \right )\right )}{a f}-\frac {4 c^{3} x}{a}-\frac {4 c^{3} x \left (\tan ^{2}\left (f x +e \right )\right )}{a}+\frac {5 c^{3} \tan \left (f x +e \right )}{a f}}{1+\tan ^{2}\left (f x +e \right )}+\frac {2 i c^{3} \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{f a}\)	\(112\)

[In]

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

[Out]

c^3*tan(f*x+e)/a/f-4/f*c^3/a*arctan(tan(f*x+e))+2*I/f*c^3/a*ln(1+tan(f*x+e)^2)+4/f*c^3/a/(tan(f*x+e)-I)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.66 \[ \int \frac {(c-i c \tan (e+f x))^3}{a+i a \tan (e+f x)} \, dx=-\frac {2 \, {\left (4 \, c^{3} f x e^{\left (4 i \, f x + 4 i \, e\right )} - i \, c^{3} + 2 \, {\left (2 \, c^{3} f x - i \, c^{3}\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 2 \, {\left (i \, c^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + i \, c^{3} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )\right )}}{a f e^{\left (4 i \, f x + 4 i \, e\right )} + a f e^{\left (2 i \, f x + 2 i \, e\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.87 \[ \int \frac {(c-i c \tan (e+f x))^3}{a+i a \tan (e+f x)} \, dx=\frac {2 i c^{3}}{a f e^{2 i e} e^{2 i f x} + a f} + \begin {cases} \frac {2 i c^{3} e^{- 2 i e} e^{- 2 i f x}}{a f} & \text {for}\: a f e^{2 i e} \neq 0 \\x \left (\frac {8 c^{3}}{a} + \frac {\left (- 8 c^{3} e^{2 i e} + 4 c^{3}\right ) e^{- 2 i e}}{a}\right ) & \text {otherwise} \end {cases} - \frac {8 c^{3} x}{a} - \frac {4 i c^{3} \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{a f} \]

[In]

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

[Out]

2*I*c**3/(a*f*exp(2*I*e)*exp(2*I*f*x) + a*f) + Piecewise((2*I*c**3*exp(-2*I*e)*exp(-2*I*f*x)/(a*f), Ne(a*f*exp
(2*I*e), 0)), (x*(8*c**3/a + (-8*c**3*exp(2*I*e) + 4*c**3)*exp(-2*I*e)/a), True)) - 8*c**3*x/a - 4*I*c**3*log(
exp(2*I*f*x) + exp(-2*I*e))/(a*f)

Maxima [F(-2)]

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

[In]

integrate((c-I*c*tan(f*x+e))^3/(a+I*a*tan(f*x+e)),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. 175 vs. \(2 (65) = 130\).

Time = 0.47 (sec) , antiderivative size = 175, normalized size of antiderivative = 2.46 \[ \int \frac {(c-i c \tan (e+f x))^3}{a+i a \tan (e+f x)} \, dx=\frac {2 \, {\left (-\frac {2 i \, c^{3} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{a} + \frac {4 i \, c^{3} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}{a} - \frac {2 i \, c^{3} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}{a} + \frac {2 i \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 i \, c^{3}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} a} - \frac {2 \, {\left (3 i \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 8 \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 3 i \, c^{3}\right )}}{a {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}^{2}}\right )}}{f} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 6.03 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.90 \[ \int \frac {(c-i c \tan (e+f x))^3}{a+i a \tan (e+f x)} \, dx=\frac {c^3\,\mathrm {tan}\left (e+f\,x\right )}{a\,f}+\frac {c^3\,4{}\mathrm {i}}{a\,f\,\left (1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}+\frac {c^3\,\ln \left (\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )\,4{}\mathrm {i}}{a\,f} \]

[In]

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

[Out]

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