\(\int \frac {(c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^2} \, dx\) [910]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.14 (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} \[ \int \frac {(c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^2} \, dx=-\frac {4 i c^3}{f \left (a^2+i a^2 \tan (e+f x)\right )}+\frac {i c^3 \log (\cos (e+f x))}{a^2 f}+\frac {c^3 x}{a^2}+\frac {2 i c^3}{f (a+i a \tan (e+f x))^2} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 5.39 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.63 \[ \int \frac {(c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^2} \, dx=-\frac {i c^3 \left (\log (i-\tan (e+f x))+\frac {-2-4 i \tan (e+f x)}{(-i+\tan (e+f x))^2}\right )}{a^2 f} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.05

method	result	size
derivativedivides	\(-\frac {2 i c^{3}}{f \,a^{2} \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {4 c^{3}}{f \,a^{2} \left (\tan \left (f x +e \right )-i\right )}-\frac {i c^{3} \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 f \,a^{2}}+\frac {c^{3} \arctan \left (\tan \left (f x +e \right )\right )}{f \,a^{2}}\)	\(87\)
default	\(-\frac {2 i c^{3}}{f \,a^{2} \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {4 c^{3}}{f \,a^{2} \left (\tan \left (f x +e \right )-i\right )}-\frac {i c^{3} \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 f \,a^{2}}+\frac {c^{3} \arctan \left (\tan \left (f x +e \right )\right )}{f \,a^{2}}\)	\(87\)
risch	\(-\frac {i c^{3} {\mathrm e}^{-2 i \left (f x +e \right )}}{a^{2} f}+\frac {i c^{3} {\mathrm e}^{-4 i \left (f x +e \right )}}{2 a^{2} f}+\frac {2 c^{3} x}{a^{2}}+\frac {2 c^{3} e}{a^{2} f}+\frac {i c^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{a^{2} f}\)	\(89\)
norman	\(\frac {\frac {c^{3} x}{a}-\frac {2 i c^{3}}{a f}+\frac {c^{3} x \left (\tan ^{4}\left (f x +e \right )\right )}{a}-\frac {4 c^{3} \left (\tan ^{3}\left (f x +e \right )\right )}{a f}+\frac {2 c^{3} x \left (\tan ^{2}\left (f x +e \right )\right )}{a}-\frac {6 i c^{3} \left (\tan ^{2}\left (f x +e \right )\right )}{a f}}{a \left (1+\tan ^{2}\left (f x +e \right )\right )^{2}}-\frac {i c^{3} \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 f \,a^{2}}\)	\(134\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.95 \[ \int \frac {(c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^2} \, dx=\frac {{\left (4 \, c^{3} f x e^{\left (4 i \, f x + 4 i \, e\right )} + 2 i \, c^{3} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - 2 i \, c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c^{3}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{2 \, a^{2} f} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 167, normalized size of antiderivative = 2.01 \[ \int \frac {(c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^2} \, dx=\begin {cases} \frac {\left (- 2 i a^{2} c^{3} f e^{4 i e} e^{- 2 i f x} + i a^{2} c^{3} f e^{2 i e} e^{- 4 i f x}\right ) e^{- 6 i e}}{2 a^{4} f^{2}} & \text {for}\: a^{4} f^{2} e^{6 i e} \neq 0 \\x \left (- \frac {2 c^{3}}{a^{2}} + \frac {\left (2 c^{3} e^{4 i e} - 2 c^{3} e^{2 i e} + 2 c^{3}\right ) e^{- 4 i e}}{a^{2}}\right ) & \text {otherwise} \end {cases} + \frac {2 c^{3} x}{a^{2}} + \frac {i c^{3} \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{a^{2} f} \]

[In]

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

[Out]

Piecewise(((-2*I*a**2*c**3*f*exp(4*I*e)*exp(-2*I*f*x) + I*a**2*c**3*f*exp(2*I*e)*exp(-4*I*f*x))*exp(-6*I*e)/(2
*a**4*f**2), Ne(a**4*f**2*exp(6*I*e), 0)), (x*(-2*c**3/a**2 + (2*c**3*exp(4*I*e) - 2*c**3*exp(2*I*e) + 2*c**3)
*exp(-4*I*e)/a**2), True)) + 2*c**3*x/a**2 + I*c**3*log(exp(2*I*f*x) + exp(-2*I*e))/(a**2*f)

Maxima [F(-2)]

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

[In]

integrate((c-I*c*tan(f*x+e))^3/(a+I*a*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 = 0.53 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.82 \[ \int \frac {(c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^2} \, dx=-\frac {-\frac {6 i \, c^{3} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{a^{2}} + \frac {12 i \, c^{3} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}{a^{2}} - \frac {6 i \, c^{3} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}{a^{2}} + \frac {-25 i \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 100 \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 198 i \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 100 \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 25 i \, c^{3}}{a^{2} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}^{4}}}{6 \, f} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.61 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.92 \[ \int \frac {(c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^2} \, dx=-\frac {\frac {2\,c^3}{a^2}+\frac {c^3\,\mathrm {tan}\left (e+f\,x\right )\,4{}\mathrm {i}}{a^2}}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )}-\frac {c^3\,\ln \left (\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )\,1{}\mathrm {i}}{a^2\,f} \]

[In]

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

[Out]

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