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*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]))

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Rubi [A]  time = 0.132704, antiderivative size = 101, normalized size of antiderivative = 1., 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 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]))

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 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])

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 = 2.37058, 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)

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Maple [A]  time = 0.025, size = 86, normalized size = 0.9 \begin{align*} -{\frac{{c}^{4}\tan \left ( fx+e \right ) }{{a}^{2}f}}-{\frac{6\,i{c}^{4}\ln \left ( \tan \left ( fx+e \right ) -i \right ) }{{a}^{2}f}}-12\,{\frac{{c}^{4}}{{a}^{2}f \left ( \tan \left ( fx+e \right ) -i \right ) }}-{\frac{4\,i{c}^{4}}{{a}^{2}f \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}} \end{align*}

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

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

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Fricas [A]  time = 1.29552, size = 356, normalized size = 3.52 \begin{align*} \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 )}} \end{align*}

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))

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Sympy [A]  time = 5.04238, size = 158, normalized size = 1.56 \begin{align*} \frac{6 i c^{4} \log{\left (e^{2 i f x} + e^{- 2 i e} \right )}}{a^{2} f} - \frac{2 i c^{4} e^{- 2 i e}}{a^{2} f \left (e^{2 i f x} + e^{- 2 i e}\right )} + \frac{\left (\begin{cases} 12 c^{4} x e^{4 i e} - \frac{4 i c^{4} e^{2 i e} e^{- 2 i f x}}{f} + \frac{i c^{4} e^{- 4 i f x}}{f} & \text{for}\: f \neq 0 \\x \left (12 c^{4} e^{4 i e} - 8 c^{4} e^{2 i e} + 4 c^{4}\right ) & \text{otherwise} \end{cases}\right ) e^{- 4 i e}}{a^{2}} \end{align*}

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]

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

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Giac [B]  time = 1.5997, size = 296, normalized size = 2.93 \begin{align*} -\frac{\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 ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{a^{2}} - \frac{6 i \, c^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\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} \end{align*}

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]

-(12*I*c^4*log(tan(1/2*f*x + 1/2*e) - I)/a^2 - 6*I*c^4*log(abs(tan(1/2*f*x + 1/2*e) + 1))/a^2 - 6*I*c^4*log(ab
s(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