\(\int \frac {(a+i a \tan (e+f x))^3}{(c-i c \tan (e+f x))^2} \, dx\) [932]

Optimal result

Integrand size = 31, antiderivative size = 83 \[ \int \frac {(a+i a \tan (e+f x))^3}{(c-i c \tan (e+f x))^2} \, dx=\frac {a^3 x}{c^2}-\frac {i a^3 \log (\cos (e+f x))}{c^2 f}-\frac {2 i a^3}{f (c-i c \tan (e+f x))^2}+\frac {4 i a^3}{f \left (c^2-i c^2 \tan (e+f x)\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 83, 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} \[ \int \frac {(a+i a \tan (e+f x))^3}{(c-i c \tan (e+f x))^2} \, dx=\frac {4 i a^3}{f \left (c^2-i c^2 \tan (e+f x)\right )}-\frac {i a^3 \log (\cos (e+f x))}{c^2 f}+\frac {a^3 x}{c^2}-\frac {2 i a^3}{f (c-i c \tan (e+f x))^2} \]

[In]

Int[(a + I*a*Tan[e + f*x])^3/(c - I*c*Tan[e + f*x])^2,x]

[Out]

(a^3*x)/c^2 - (I*a^3*Log[Cos[e + f*x]])/(c^2*f) - ((2*I)*a^3)/(f*(c - I*c*Tan[e + f*x])^2) + ((4*I)*a^3)/(f*(c
^2 - I*c^2*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))^5} \, dx \\ & = \frac {\left (i a^3\right ) \text {Subst}\left (\int \frac {(c-x)^2}{(c+x)^3} \, dx,x,-i c \tan (e+f x)\right )}{c^2 f} \\ & = \frac {\left (i a^3\right ) \text {Subst}\left (\int \left (\frac {4 c^2}{(c+x)^3}-\frac {4 c}{(c+x)^2}+\frac {1}{c+x}\right ) \, dx,x,-i c \tan (e+f x)\right )}{c^2 f} \\ & = \frac {a^3 x}{c^2}-\frac {i a^3 \log (\cos (e+f x))}{c^2 f}-\frac {2 i a^3}{f (c-i c \tan (e+f x))^2}+\frac {4 i a^3}{f \left (c^2-i c^2 \tan (e+f x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.15 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.60 \[ \int \frac {(a+i a \tan (e+f x))^3}{(c-i c \tan (e+f x))^2} \, dx=\frac {i a^3 \left (\log (i+\tan (e+f x))+\frac {-2+4 i \tan (e+f x)}{(i+\tan (e+f x))^2}\right )}{c^2 f} \]

[In]

Integrate[(a + I*a*Tan[e + f*x])^3/(c - I*c*Tan[e + f*x])^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.96

method	result	size
risch	\(-\frac {i a^{3} {\mathrm e}^{4 i \left (f x +e \right )}}{2 c^{2} f}+\frac {i a^{3} {\mathrm e}^{2 i \left (f x +e \right )}}{c^{2} f}-\frac {2 a^{3} e}{c^{2} f}-\frac {i a^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{c^{2} f}\)	\(80\)
derivativedivides	\(-\frac {4 a^{3}}{f \,c^{2} \left (\tan \left (f x +e \right )+i\right )}+\frac {i a^{3} \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 f \,c^{2}}+\frac {a^{3} \arctan \left (\tan \left (f x +e \right )\right )}{f \,c^{2}}+\frac {2 i a^{3}}{f \,c^{2} \left (\tan \left (f x +e \right )+i\right )^{2}}\)	\(87\)
default	\(-\frac {4 a^{3}}{f \,c^{2} \left (\tan \left (f x +e \right )+i\right )}+\frac {i a^{3} \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 f \,c^{2}}+\frac {a^{3} \arctan \left (\tan \left (f x +e \right )\right )}{f \,c^{2}}+\frac {2 i a^{3}}{f \,c^{2} \left (\tan \left (f x +e \right )+i\right )^{2}}\)	\(87\)
norman	\(\frac {\frac {a^{3} x}{c}+\frac {2 i a^{3}}{c f}+\frac {a^{3} x \left (\tan ^{4}\left (f x +e \right )\right )}{c}+\frac {2 a^{3} x \left (\tan ^{2}\left (f x +e \right )\right )}{c}-\frac {4 a^{3} \left (\tan ^{3}\left (f x +e \right )\right )}{c f}+\frac {6 i a^{3} \left (\tan ^{2}\left (f x +e \right )\right )}{c f}}{c \left (1+\tan ^{2}\left (f x +e \right )\right )^{2}}+\frac {i a^{3} \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 f \,c^{2}}\)	\(134\)

[In]

int((a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.65 \[ \int \frac {(a+i a \tan (e+f x))^3}{(c-i c \tan (e+f x))^2} \, dx=\frac {-i \, a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 i \, a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} - 2 i \, a^{3} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{2 \, c^{2} f} \]

[In]

integrate((a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.47 \[ \int \frac {(a+i a \tan (e+f x))^3}{(c-i c \tan (e+f x))^2} \, dx=- \frac {i a^{3} \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{c^{2} f} + \begin {cases} \frac {- i a^{3} c^{2} f e^{4 i e} e^{4 i f x} + 2 i a^{3} c^{2} f e^{2 i e} e^{2 i f x}}{2 c^{4} f^{2}} & \text {for}\: c^{4} f^{2} \neq 0 \\\frac {x \left (2 a^{3} e^{4 i e} - 2 a^{3} e^{2 i e}\right )}{c^{2}} & \text {otherwise} \end {cases} \]

[In]

integrate((a+I*a*tan(f*x+e))**3/(c-I*c*tan(f*x+e))**2,x)

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {(a+i a \tan (e+f x))^3}{(c-i c \tan (e+f x))^2} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate((a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^2,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. 151 vs. \(2 (73) = 146\).

Time = 0.51 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.82 \[ \int \frac {(a+i a \tan (e+f x))^3}{(c-i c \tan (e+f x))^2} \, dx=-\frac {\frac {6 i \, a^{3} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{c^{2}} - \frac {12 i \, a^{3} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}{c^{2}} + \frac {6 i \, a^{3} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}{c^{2}} + \frac {25 i \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 100 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 198 i \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 100 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 25 i \, a^{3}}{c^{2} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}^{4}}}{6 \, f} \]

[In]

integrate((a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^2,x, algorithm="giac")

[Out]

-1/6*(6*I*a^3*log(tan(1/2*f*x + 1/2*e) + 1)/c^2 - 12*I*a^3*log(tan(1/2*f*x + 1/2*e) + I)/c^2 + 6*I*a^3*log(tan
(1/2*f*x + 1/2*e) - 1)/c^2 + (25*I*a^3*tan(1/2*f*x + 1/2*e)^4 - 100*a^3*tan(1/2*f*x + 1/2*e)^3 - 198*I*a^3*tan
(1/2*f*x + 1/2*e)^2 + 100*a^3*tan(1/2*f*x + 1/2*e) + 25*I*a^3)/(c^2*(tan(1/2*f*x + 1/2*e) + I)^4))/f

Mupad [B] (verification not implemented)

Time = 6.24 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.88 \[ \int \frac {(a+i a \tan (e+f x))^3}{(c-i c \tan (e+f x))^2} \, dx=-\frac {\frac {4\,a^3\,\mathrm {tan}\left (e+f\,x\right )}{c^2}+\frac {a^3\,2{}\mathrm {i}}{c^2}}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^2+\mathrm {tan}\left (e+f\,x\right )\,2{}\mathrm {i}-1\right )}+\frac {a^3\,\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{c^2\,f} \]

[In]

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

[Out]

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