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

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

[Out]

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

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Rubi [A]  time = 0.133885, 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} \[ \frac{i a^4 \tan ^2(e+f x)}{2 c f}+\frac{5 a^4 \tan (e+f x)}{c f}-\frac{8 i a^4}{f (c-i c \tan (e+f x))}+\frac{12 i a^4 \log (\cos (e+f x))}{c f}-\frac{12 a^4 x}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [B]  time = 2.8567, size = 376, normalized size = 3.96 \[ -\frac{a^4 \sec (e) \sec ^2(e+f x) (\cos (e+5 f x)+i \sin (e+5 f x)) \left (-6 i f x \sin (2 e+f x)+2 \sin (2 e+f x)-6 i f x \sin (2 e+3 f x)-7 \sin (2 e+3 f x)-6 i f x \sin (4 e+3 f x)-2 \sin (4 e+3 f x)+6 f x \cos (2 e+3 f x)-3 i \cos (2 e+3 f x)+6 f x \cos (4 e+3 f x)+2 i \cos (4 e+3 f x)-3 i \cos (2 e+3 f x) \log \left (\cos ^2(e+f x)\right )+\cos (f x) \left (-9 i \log \left (\cos ^2(e+f x)\right )+18 f x+5 i\right )+\cos (2 e+f x) \left (-9 i \log \left (\cos ^2(e+f x)\right )+18 f x+10 i\right )-3 i \cos (4 e+3 f x) \log \left (\cos ^2(e+f x)\right )-3 \sin (f x) \log \left (\cos ^2(e+f x)\right )-3 \sin (2 e+f x) \log \left (\cos ^2(e+f x)\right )-3 \sin (2 e+3 f x) \log \left (\cos ^2(e+f x)\right )-3 \sin (4 e+3 f x) \log \left (\cos ^2(e+f x)\right )-6 i f x \sin (f x)-13 \sin (f x)\right )}{4 c f (\cos (f x)+i \sin (f x))^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^4*Sec[e]*Sec[e + f*x]^2*((-3*I)*Cos[2*e + 3*f*x] + 6*f*x*Cos[2*e + 3*f*x] + (2*I)*Cos[4*e + 3*f*x] + 6*f*x
*Cos[4*e + 3*f*x] + Cos[f*x]*(5*I + 18*f*x - (9*I)*Log[Cos[e + f*x]^2]) + Cos[2*e + f*x]*(10*I + 18*f*x - (9*I
)*Log[Cos[e + f*x]^2]) - (3*I)*Cos[2*e + 3*f*x]*Log[Cos[e + f*x]^2] - (3*I)*Cos[4*e + 3*f*x]*Log[Cos[e + f*x]^
2] - 13*Sin[f*x] - (6*I)*f*x*Sin[f*x] - 3*Log[Cos[e + f*x]^2]*Sin[f*x] + 2*Sin[2*e + f*x] - (6*I)*f*x*Sin[2*e
+ f*x] - 3*Log[Cos[e + f*x]^2]*Sin[2*e + f*x] - 7*Sin[2*e + 3*f*x] - (6*I)*f*x*Sin[2*e + 3*f*x] - 3*Log[Cos[e
+ f*x]^2]*Sin[2*e + 3*f*x] - 2*Sin[4*e + 3*f*x] - (6*I)*f*x*Sin[4*e + 3*f*x] - 3*Log[Cos[e + f*x]^2]*Sin[4*e +
3*f*x])*(Cos[e + 5*f*x] + I*Sin[e + 5*f*x]))/(4*c*f*(Cos[f*x] + I*Sin[f*x])^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.46219, size = 360, normalized size = 3.79 \begin{align*} \frac{-4 i \, a^{4} e^{\left (6 i \, f x + 6 i \, e\right )} - 8 i \, a^{4} e^{\left (4 i \, f x + 4 i \, e\right )} + 8 i \, a^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + 10 i \, a^{4} +{\left (12 i \, a^{4} e^{\left (4 i \, f x + 4 i \, e\right )} + 24 i \, a^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + 12 i \, a^{4}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{c f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, c f e^{\left (2 i \, f x + 2 i \, e\right )} + c f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-4*I*a^4*e^(6*I*f*x + 6*I*e) - 8*I*a^4*e^(4*I*f*x + 4*I*e) + 8*I*a^4*e^(2*I*f*x + 2*I*e) + 10*I*a^4 + (12*I*a
^4*e^(4*I*f*x + 4*I*e) + 24*I*a^4*e^(2*I*f*x + 2*I*e) + 12*I*a^4)*log(e^(2*I*f*x + 2*I*e) + 1))/(c*f*e^(4*I*f*
x + 4*I*e) + 2*c*f*e^(2*I*f*x + 2*I*e) + c*f)

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Sympy [A]  time = 2.89058, size = 134, normalized size = 1.41 \begin{align*} \frac{8 a^{4} \left (\begin{cases} - \frac{i e^{2 i f x}}{2 f} & \text{for}\: f \neq 0 \\x & \text{otherwise} \end{cases}\right ) e^{2 i e}}{c} + \frac{12 i a^{4} \log{\left (e^{2 i f x} + e^{- 2 i e} \right )}}{c f} + \frac{\frac{12 i a^{4} e^{- 2 i e} e^{2 i f x}}{c f} + \frac{10 i a^{4} e^{- 4 i e}}{c f}}{e^{4 i f x} + 2 e^{- 2 i e} e^{2 i f x} + e^{- 4 i e}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.66352, size = 273, normalized size = 2.87 \begin{align*} \frac{2 \,{\left (-\frac{12 i \, a^{4} \log \left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + i\right )}{c} + \frac{6 i \, a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{c} + \frac{6 i \, a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{c} - \frac{13 \, a^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 9 i \, a^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 24 \, a^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 9 i \, a^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 13 \, a^{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} c}\right )}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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