3.920 \(\int \frac {(c-i c \tan (e+f x))^4}{(a+i a \tan (e+f x))^2} \, dx\)

Optimal. Leaf size=101 \[ -\frac {c^4 \tan (e+f x)}{a^2 f}-\frac {12 i c^4}{f \left (a^2+i a^2 \tan (e+f x)\right )}+\frac {6 i c^4 \log (\cos (e+f x))}{a^2 f}+\frac {6 c^4 x}{a^2}+\frac {4 i c^4}{f (a+i a \tan (e+f x))^2} \]

[Out]

6*c^4*x/a^2+6*I*c^4*ln(cos(f*x+e))/a^2/f-c^4*tan(f*x+e)/a^2/f+4*I*c^4/f/(a+I*a*tan(f*x+e))^2-12*I*c^4/f/(a^2+I
*a^2*tan(f*x+e))

________________________________________________________________________________________

Rubi [A]  time = 0.13, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {3522, 3487, 43} \[ -\frac {c^4 \tan (e+f x)}{a^2 f}-\frac {12 i c^4}{f \left (a^2+i a^2 \tan (e+f x)\right )}+\frac {6 i c^4 \log (\cos (e+f x))}{a^2 f}+\frac {6 c^4 x}{a^2}+\frac {4 i c^4}{f (a+i a \tan (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*c^4*x)/a^2 + ((6*I)*c^4*Log[Cos[e + f*x]])/(a^2*f) - (c^4*Tan[e + f*x])/(a^2*f) + ((4*I)*c^4)/(f*(a + I*a*T
an[e + f*x])^2) - ((12*I)*c^4)/(f*(a^2 + I*a^2*Tan[e + f*x]))

Rule 43

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 3487

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 3522

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*} \int \frac {(c-i c \tan (e+f x))^4}{(a+i a \tan (e+f x))^2} \, dx &=\left (a^4 c^4\right ) \int \frac {\sec ^8(e+f x)}{(a+i a \tan (e+f x))^6} \, dx\\ &=-\frac {\left (i c^4\right ) \operatorname {Subst}\left (\int \frac {(a-x)^3}{(a+x)^3} \, dx,x,i a \tan (e+f x)\right )}{a^3 f}\\ &=-\frac {\left (i c^4\right ) \operatorname {Subst}\left (\int \left (-1+\frac {8 a^3}{(a+x)^3}-\frac {12 a^2}{(a+x)^2}+\frac {6 a}{a+x}\right ) \, dx,x,i a \tan (e+f x)\right )}{a^3 f}\\ &=\frac {6 c^4 x}{a^2}+\frac {6 i c^4 \log (\cos (e+f x))}{a^2 f}-\frac {c^4 \tan (e+f x)}{a^2 f}+\frac {4 i c^4}{f (a+i a \tan (e+f x))^2}-\frac {12 i c^4}{f \left (a^2+i a^2 \tan (e+f x)\right )}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 2.63, size = 279, normalized size = 2.76 \[ \frac {c^4 \sec ^2(e+f x) (\cos (f x)+i \sin (f x))^2 \left (-24 f x \sin ^2(e)-12 i f x \sin (2 e)+2 i \sin (2 e) \sin (4 f x)+12 i f x \tan (e)-2 \sin (2 e) \cos (4 f x)+i \sec (e) \cos (2 e-f x) \sec (e+f x)-i \sec (e) \cos (2 e+f x) \sec (e+f x)-\sec (e) \sin (2 e-f x) \sec (e+f x)+\sec (e) \sin (2 e+f x) \sec (e+f x)+6 \sin (2 e) \log \left (\cos ^2(e+f x)\right )-12 (\cos (2 e)+i \sin (2 e)) \tan ^{-1}(\tan (f x))+2 i \cos (2 e) \left (6 f x \tan (e)-3 \log \left (\cos ^2(e+f x)\right )+6 i f x+i \sin (4 f x)-\cos (4 f x)\right )+12 f x+8 \sin (2 f x)+8 i \cos (2 f x)\right )}{2 a^2 f (\tan (e+f x)-i)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^4*Sec[e + f*x]^2*(Cos[f*x] + I*Sin[f*x])^2*(12*f*x + (8*I)*Cos[2*f*x] + I*Cos[2*e - f*x]*Sec[e]*Sec[e + f*x
] - I*Cos[2*e + f*x]*Sec[e]*Sec[e + f*x] - 24*f*x*Sin[e]^2 - 12*ArcTan[Tan[f*x]]*(Cos[2*e] + I*Sin[2*e]) - (12
*I)*f*x*Sin[2*e] - 2*Cos[4*f*x]*Sin[2*e] + 6*Log[Cos[e + f*x]^2]*Sin[2*e] + 8*Sin[2*f*x] + (2*I)*Sin[2*e]*Sin[
4*f*x] - Sec[e]*Sec[e + f*x]*Sin[2*e - f*x] + Sec[e]*Sec[e + f*x]*Sin[2*e + f*x] + (12*I)*f*x*Tan[e] + (2*I)*C
os[2*e]*((6*I)*f*x - Cos[4*f*x] - 3*Log[Cos[e + f*x]^2] + I*Sin[4*f*x] + 6*f*x*Tan[e])))/(2*a^2*f*(-I + Tan[e
+ f*x])^2)

________________________________________________________________________________________

fricas [A]  time = 0.44, size = 133, normalized size = 1.32 \[ \frac {12 \, c^{4} f x e^{\left (6 i \, f x + 6 i \, e\right )} - 3 i \, c^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c^{4} + {\left (12 \, c^{4} f x - 6 i \, c^{4}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (6 i \, c^{4} e^{\left (6 i \, f x + 6 i \, e\right )} + 6 i \, c^{4} e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{a^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} f e^{\left (4 i \, f x + 4 i \, e\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))^4/(a+I*a*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [B]  time = 1.51, size = 217, normalized size = 2.15 \[ -\frac {-\frac {6 i \, c^{4} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{a^{2}} + \frac {12 i \, c^{4} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}{a^{2}} - \frac {6 i \, c^{4} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}{a^{2}} - \frac {2 \, {\left (-3 i \, c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 3 i \, c^{4}\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} a^{2}} + \frac {-25 i \, c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 108 \, c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 182 i \, c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 108 \, c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 25 i \, c^{4}}{a^{2} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}^{4}}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))^4/(a+I*a*tan(f*x+e))^2,x, algorithm="giac")

[Out]

-(-6*I*c^4*log(tan(1/2*f*x + 1/2*e) + 1)/a^2 + 12*I*c^4*log(tan(1/2*f*x + 1/2*e) - I)/a^2 - 6*I*c^4*log(tan(1/
2*f*x + 1/2*e) - 1)/a^2 - 2*(-3*I*c^4*tan(1/2*f*x + 1/2*e)^2 + c^4*tan(1/2*f*x + 1/2*e) + 3*I*c^4)/((tan(1/2*f
*x + 1/2*e)^2 - 1)*a^2) + (-25*I*c^4*tan(1/2*f*x + 1/2*e)^4 - 108*c^4*tan(1/2*f*x + 1/2*e)^3 + 182*I*c^4*tan(1
/2*f*x + 1/2*e)^2 + 108*c^4*tan(1/2*f*x + 1/2*e) - 25*I*c^4)/(a^2*(tan(1/2*f*x + 1/2*e) - I)^4))/f

________________________________________________________________________________________

maple [A]  time = 0.18, size = 86, normalized size = 0.85 \[ -\frac {c^{4} \tan \left (f x +e \right )}{a^{2} f}-\frac {12 c^{4}}{f \,a^{2} \left (\tan \left (f x +e \right )-i\right )}-\frac {4 i c^{4}}{f \,a^{2} \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {6 i c^{4} \ln \left (\tan \left (f x +e \right )-i\right )}{f \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-I*c*tan(f*x+e))^4/(a+I*a*tan(f*x+e))^2,x)

[Out]

-c^4*tan(f*x+e)/a^2/f-12/f*c^4/a^2/(tan(f*x+e)-I)-4*I/f*c^4/a^2/(tan(f*x+e)-I)^2-6*I/f*c^4/a^2*ln(tan(f*x+e)-I
)

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))^4/(a+I*a*tan(f*x+e))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

________________________________________________________________________________________

mupad [B]  time = 4.84, size = 93, normalized size = 0.92 \[ -\frac {\frac {8\,c^4}{a^2}+\frac {c^4\,\mathrm {tan}\left (e+f\,x\right )\,12{}\mathrm {i}}{a^2}}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )}-\frac {c^4\,\mathrm {tan}\left (e+f\,x\right )}{a^2\,f}-\frac {c^4\,\ln \left (\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )\,6{}\mathrm {i}}{a^2\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- ((8*c^4)/a^2 + (c^4*tan(e + f*x)*12i)/a^2)/(f*(2*tan(e + f*x) + tan(e + f*x)^2*1i - 1i)) - (c^4*tan(e + f*x)
)/(a^2*f) - (c^4*log(tan(e + f*x) - 1i)*6i)/(a^2*f)

________________________________________________________________________________________

sympy [A]  time = 0.59, size = 201, normalized size = 1.99 \[ \frac {2 i c^{4}}{- a^{2} f e^{2 i e} e^{2 i f x} - a^{2} f} + \begin {cases} \frac {\left (- 4 i a^{2} c^{4} f e^{4 i e} e^{- 2 i f x} + i a^{2} c^{4} f e^{2 i e} e^{- 4 i f x}\right ) e^{- 6 i e}}{a^{4} f^{2}} & \text {for}\: a^{4} f^{2} e^{6 i e} \neq 0 \\x \left (- \frac {12 c^{4}}{a^{2}} + \frac {\left (12 c^{4} e^{4 i e} - 8 c^{4} e^{2 i e} + 4 c^{4}\right ) e^{- 4 i e}}{a^{2}}\right ) & \text {otherwise} \end {cases} + \frac {12 c^{4} x}{a^{2}} + \frac {6 i c^{4} \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{a^{2} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))**4/(a+I*a*tan(f*x+e))**2,x)

[Out]

2*I*c**4/(-a**2*f*exp(2*I*e)*exp(2*I*f*x) - a**2*f) + Piecewise(((-4*I*a**2*c**4*f*exp(4*I*e)*exp(-2*I*f*x) +
I*a**2*c**4*f*exp(2*I*e)*exp(-4*I*f*x))*exp(-6*I*e)/(a**4*f**2), Ne(a**4*f**2*exp(6*I*e), 0)), (x*(-12*c**4/a*
*2 + (12*c**4*exp(4*I*e) - 8*c**4*exp(2*I*e) + 4*c**4)*exp(-4*I*e)/a**2), True)) + 12*c**4*x/a**2 + 6*I*c**4*l
og(exp(2*I*f*x) + exp(-2*I*e))/(a**2*f)

________________________________________________________________________________________