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

Optimal. Leaf size=95 \[ -\frac{i c^4 \tan ^2(e+f x)}{2 a f}+\frac{5 c^4 \tan (e+f x)}{a f}+\frac{8 i c^4}{f (a+i a \tan (e+f x))}-\frac{12 i c^4 \log (\cos (e+f x))}{a f}-\frac{12 c^4 x}{a} \]

[Out]

(-12*c^4*x)/a - ((12*I)*c^4*Log[Cos[e + f*x]])/(a*f) + (5*c^4*Tan[e + f*x])/(a*f) - ((I/2)*c^4*Tan[e + f*x]^2)
/(a*f) + ((8*I)*c^4)/(f*(a + I*a*Tan[e + f*x]))

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Rubi [A]  time = 0.129874, antiderivative size = 95, 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} \[ -\frac{i c^4 \tan ^2(e+f x)}{2 a f}+\frac{5 c^4 \tan (e+f x)}{a f}+\frac{8 i c^4}{f (a+i a \tan (e+f x))}-\frac{12 i c^4 \log (\cos (e+f x))}{a f}-\frac{12 c^4 x}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-12*c^4*x)/a - ((12*I)*c^4*Log[Cos[e + f*x]])/(a*f) + (5*c^4*Tan[e + f*x])/(a*f) - ((I/2)*c^4*Tan[e + f*x]^2)
/(a*f) + ((8*I)*c^4)/(f*(a + I*a*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)} \, dx &=\left (a^4 c^4\right ) \int \frac{\sec ^8(e+f x)}{(a+i a \tan (e+f x))^5} \, dx\\ &=-\frac{\left (i c^4\right ) \operatorname{Subst}\left (\int \frac{(a-x)^3}{(a+x)^2} \, dx,x,i a \tan (e+f x)\right )}{a^3 f}\\ &=-\frac{\left (i c^4\right ) \operatorname{Subst}\left (\int \left (5 a-x+\frac{8 a^3}{(a+x)^2}-\frac{12 a^2}{a+x}\right ) \, dx,x,i a \tan (e+f x)\right )}{a^3 f}\\ &=-\frac{12 c^4 x}{a}-\frac{12 i c^4 \log (\cos (e+f x))}{a f}+\frac{5 c^4 \tan (e+f x)}{a f}-\frac{i c^4 \tan ^2(e+f x)}{2 a f}+\frac{8 i c^4}{f (a+i a \tan (e+f x))}\\ \end{align*}

Mathematica [B]  time = 2.67013, size = 194, normalized size = 2.04 \[ \frac{c^4 \cos (e) \sec (e+f x) (\cos (f x)+i \sin (f x)) \left (-24 f x \tan ^2(e)-24 i (\tan (e)-i) \tan ^{-1}(\tan (f x))+24 f x \sec ^2(e)-i \sec ^2(e+f x)-8 i \tan (e) \sin (2 f x)-12 i \log \left (\cos ^2(e+f x)\right )+8 (\tan (e)+i) \cos (2 f x)+\tan (e) \sec ^2(e+f x)+10 \sec (e) \sin (f x) \sec (e+f x)+12 \tan (e) \log \left (\cos ^2(e+f x)\right )+10 i \tan (e) \sec (e) \sin (f x) \sec (e+f x)-24 f x+8 \sin (2 f x)\right )}{2 f (a+i a \tan (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^4*Cos[e]*Sec[e + f*x]*(Cos[f*x] + I*Sin[f*x])*(-24*f*x - (12*I)*Log[Cos[e + f*x]^2] + 24*f*x*Sec[e]^2 - I*S
ec[e + f*x]^2 + 10*Sec[e]*Sec[e + f*x]*Sin[f*x] + 8*Sin[2*f*x] + 12*Log[Cos[e + f*x]^2]*Tan[e] + Sec[e + f*x]^
2*Tan[e] + (10*I)*Sec[e]*Sec[e + f*x]*Sin[f*x]*Tan[e] - (8*I)*Sin[2*f*x]*Tan[e] - 24*f*x*Tan[e]^2 - (24*I)*Arc
Tan[Tan[f*x]]*(-I + Tan[e]) + 8*Cos[2*f*x]*(I + Tan[e])))/(2*f*(a + I*a*Tan[e + f*x]))

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Maple [A]  time = 0.027, size = 83, normalized size = 0.9 \begin{align*} 5\,{\frac{{c}^{4}\tan \left ( fx+e \right ) }{af}}-{\frac{{\frac{i}{2}}{c}^{4} \left ( \tan \left ( fx+e \right ) \right ) ^{2}}{af}}+{\frac{12\,i{c}^{4}\ln \left ( \tan \left ( fx+e \right ) -i \right ) }{af}}+8\,{\frac{{c}^{4}}{af \left ( \tan \left ( fx+e \right ) -i \right ) }} \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)),x)

[Out]

5*c^4*tan(f*x+e)/a/f-1/2*I*c^4*tan(f*x+e)^2/a/f+12*I/f*c^4/a*ln(tan(f*x+e)-I)+8/f*c^4/a/(tan(f*x+e)-I)

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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)),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.43827, size = 462, normalized size = 4.86 \begin{align*} -\frac{24 \, c^{4} f x e^{\left (6 i \, f x + 6 i \, e\right )} - 4 i \, c^{4} +{\left (48 \, c^{4} f x - 12 i \, c^{4}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (24 \, c^{4} f x - 18 i \, c^{4}\right )} e^{\left (2 i \, f x + 2 i \, e\right )} -{\left (-12 i \, c^{4} e^{\left (6 i \, f x + 6 i \, e\right )} - 24 i \, c^{4} e^{\left (4 i \, f x + 4 i \, e\right )} - 12 i \, c^{4} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{a f e^{\left (6 i \, f x + 6 i \, e\right )} + 2 \, a f e^{\left (4 i \, f x + 4 i \, e\right )} + a f e^{\left (2 i \, f x + 2 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)),x, algorithm="fricas")

[Out]

-(24*c^4*f*x*e^(6*I*f*x + 6*I*e) - 4*I*c^4 + (48*c^4*f*x - 12*I*c^4)*e^(4*I*f*x + 4*I*e) + (24*c^4*f*x - 18*I*
c^4)*e^(2*I*f*x + 2*I*e) - (-12*I*c^4*e^(6*I*f*x + 6*I*e) - 24*I*c^4*e^(4*I*f*x + 4*I*e) - 12*I*c^4*e^(2*I*f*x
+ 2*I*e))*log(e^(2*I*f*x + 2*I*e) + 1))/(a*f*e^(6*I*f*x + 6*I*e) + 2*a*f*e^(4*I*f*x + 4*I*e) + a*f*e^(2*I*f*x
+ 2*I*e))

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Sympy [A]  time = 3.91423, size = 162, normalized size = 1.71 \begin{align*} \frac{\frac{8 i c^{4} e^{- 2 i e} e^{2 i f x}}{a f} + \frac{10 i c^{4} e^{- 4 i e}}{a f}}{e^{4 i f x} + 2 e^{- 2 i e} e^{2 i f x} + e^{- 4 i e}} - \frac{12 i c^{4} \log{\left (e^{2 i f x} + e^{- 2 i e} \right )}}{a f} - \frac{\left (\begin{cases} 24 c^{4} x e^{2 i e} - \frac{4 i c^{4} e^{- 2 i f x}}{f} & \text{for}\: f \neq 0 \\x \left (24 c^{4} e^{2 i e} - 8 c^{4}\right ) & \text{otherwise} \end{cases}\right ) e^{- 2 i e}}{a} \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)),x)

[Out]

(8*I*c**4*exp(-2*I*e)*exp(2*I*f*x)/(a*f) + 10*I*c**4*exp(-4*I*e)/(a*f))/(exp(4*I*f*x) + 2*exp(-2*I*e)*exp(2*I*
f*x) + exp(-4*I*e)) - 12*I*c**4*log(exp(2*I*f*x) + exp(-2*I*e))/(a*f) - Piecewise((24*c**4*x*exp(2*I*e) - 4*I*
c**4*exp(-2*I*f*x)/f, Ne(f, 0)), (x*(24*c**4*exp(2*I*e) - 8*c**4), True))*exp(-2*I*e)/a

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Giac [B]  time = 1.56524, size = 273, normalized size = 2.87 \begin{align*} \frac{2 \,{\left (\frac{12 i \, c^{4} \log \left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - i\right )}{a} - \frac{6 i \, c^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{a} - \frac{6 i \, c^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{a} - \frac{13 \, c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 9 i \, c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 24 \, c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 9 i \, c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 13 \, c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - i \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + i\right )}^{2} a}\right )}}{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)),x, algorithm="giac")

[Out]

2*(12*I*c^4*log(tan(1/2*f*x + 1/2*e) - I)/a - 6*I*c^4*log(abs(tan(1/2*f*x + 1/2*e) + 1))/a - 6*I*c^4*log(abs(t
an(1/2*f*x + 1/2*e) - 1))/a - (13*c^4*tan(1/2*f*x + 1/2*e)^5 - 9*I*c^4*tan(1/2*f*x + 1/2*e)^4 - 24*c^4*tan(1/2
*f*x + 1/2*e)^3 + 9*I*c^4*tan(1/2*f*x + 1/2*e)^2 + 13*c^4*tan(1/2*f*x + 1/2*e))/((tan(1/2*f*x + 1/2*e)^3 - I*t
an(1/2*f*x + 1/2*e)^2 - tan(1/2*f*x + 1/2*e) + I)^2*a))/f