\(\int \frac {(a+i a \tan (e+f x))^3}{c-i c \tan (e+f x)} \, dx\) [925]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.16 (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} \[ \int \frac {(a+i a \tan (e+f x))^3}{c-i c \tan (e+f x)} \, dx=\frac {a^3 \tan (e+f x)}{c f}-\frac {4 i a^3}{f (c-i c \tan (e+f x))}+\frac {4 i a^3 \log (\cos (e+f x))}{c f}-\frac {4 a^3 x}{c} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 3.40 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.75 \[ \int \frac {(a+i a \tan (e+f x))^3}{c-i c \tan (e+f x)} \, dx=\frac {i a^3 \left (-4 \log (i+\tan (e+f x))-i \tan (e+f x)-\frac {4 i}{i+\tan (e+f x)}\right )}{c f} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.14

method	result	size
derivativedivides	\(\frac {a^{3} \tan \left (f x +e \right )}{c f}-\frac {4 a^{3} \arctan \left (\tan \left (f x +e \right )\right )}{f c}-\frac {2 i a^{3} \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{f c}+\frac {4 a^{3}}{f c \left (\tan \left (f x +e \right )+i\right )}\)	\(81\)
default	\(\frac {a^{3} \tan \left (f x +e \right )}{c f}-\frac {4 a^{3} \arctan \left (\tan \left (f x +e \right )\right )}{f c}-\frac {2 i a^{3} \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{f c}+\frac {4 a^{3}}{f c \left (\tan \left (f x +e \right )+i\right )}\)	\(81\)
risch	\(-\frac {2 i a^{3} {\mathrm e}^{2 i \left (f x +e \right )}}{c f}+\frac {8 a^{3} e}{c f}+\frac {2 i a^{3}}{f c \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}+\frac {4 i a^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{f c}\)	\(84\)
norman	\(\frac {-\frac {4 i a^{3}}{c f}+\frac {a^{3} \left (\tan ^{3}\left (f x +e \right )\right )}{c f}-\frac {4 a^{3} x}{c}-\frac {4 a^{3} x \left (\tan ^{2}\left (f x +e \right )\right )}{c}+\frac {5 a^{3} \tan \left (f x +e \right )}{c f}}{1+\tan ^{2}\left (f x +e \right )}-\frac {2 i a^{3} \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{f c}\)	\(112\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.24 \[ \int \frac {(a+i a \tan (e+f x))^3}{c-i c \tan (e+f x)} \, dx=-\frac {2 \, {\left (i \, a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + i \, a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a^{3} + 2 \, {\left (-i \, a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a^{3}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )\right )}}{c f e^{\left (2 i \, f x + 2 i \, e\right )} + c f} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.41 \[ \int \frac {(a+i a \tan (e+f x))^3}{c-i c \tan (e+f x)} \, dx=\frac {2 i a^{3}}{c f e^{2 i e} e^{2 i f x} + c f} + \frac {4 i a^{3} \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{c f} + \begin {cases} - \frac {2 i a^{3} e^{2 i e} e^{2 i f x}}{c f} & \text {for}\: c f \neq 0 \\\frac {4 a^{3} x e^{2 i e}}{c} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

integrate((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 [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 \[ \int \frac {(a+i a \tan (e+f x))^3}{c-i c \tan (e+f x)} \, dx=\frac {2 \, {\left (\frac {2 i \, a^{3} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{c} - \frac {4 i \, a^{3} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}{c} + \frac {2 i \, a^{3} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}{c} + \frac {-2 i \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 i \, a^{3}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} c} - \frac {2 \, {\left (-3 i \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 8 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 3 i \, a^{3}\right )}}{c {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}^{2}}\right )}}{f} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.35 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.86 \[ \int \frac {(a+i a \tan (e+f x))^3}{c-i c \tan (e+f x)} \, dx=\frac {a^3\,\mathrm {tan}\left (e+f\,x\right )}{c\,f}+\frac {4\,a^3}{c\,f\,\left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )}-\frac {a^3\,\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,4{}\mathrm {i}}{c\,f} \]

[In]

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

[Out]

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