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

Optimal. Leaf size=71 \[ \frac{a^3 \tan (e+f x)}{c f}-\frac{4 i a^3}{f (c-i c \tan (e+f x))}+\frac{4 i a^3 \log (\cos (e+f x))}{c f}-\frac{4 a^3 x}{c} \]

[Out]

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

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Rubi [A]  time = 0.122143, antiderivative size = 71, 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} \[ \frac{a^3 \tan (e+f x)}{c f}-\frac{4 i a^3}{f (c-i c \tan (e+f x))}+\frac{4 i a^3 \log (\cos (e+f x))}{c f}-\frac{4 a^3 x}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [B]  time = 2.36298, size = 214, normalized size = 3.01 \[ \frac{a^3 \sec (e) (\tan (e+f x)-i) \left (-2 f x \sin (e+2 f x)+2 i \sin (e+2 f x)-2 f x \sin (3 e+2 f x)+i \sin (3 e+2 f x)-2 i f x \cos (3 e+2 f x)+\cos (3 e+2 f x)-\cos (3 e+2 f x) \log \left (\cos ^2(e+f x)\right )+\cos (e) \left (-2 \log \left (\cos ^2(e+f x)\right )-4 i f x+3\right )+\cos (e+2 f x) \left (-\log \left (\cos ^2(e+f x)\right )-2 i f x\right )+i \sin (e+2 f x) \log \left (\cos ^2(e+f x)\right )+i \sin (3 e+2 f x) \log \left (\cos ^2(e+f x)\right )-i \sin (e)\right )}{2 c f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.026, size = 62, normalized size = 0.9 \begin{align*}{\frac{{a}^{3}\tan \left ( fx+e \right ) }{cf}}+4\,{\frac{{a}^{3}}{cf \left ( \tan \left ( fx+e \right ) +i \right ) }}-{\frac{4\,i{a}^{3}\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))^3/(c-I*c*tan(f*x+e)),x)

[Out]

a^3*tan(f*x+e)/c/f+4/f*a^3/c/(tan(f*x+e)+I)-4*I/f*a^3/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))^3/(c-I*c*tan(f*x+e)),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.36956, size = 236, normalized size = 3.32 \begin{align*} \frac{-2 i \, a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} - 2 i \, a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i \, a^{3} +{\left (4 i \, a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 4 i \, a^{3}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{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))^3/(c-I*c*tan(f*x+e)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.82855, size = 92, normalized size = 1.3 \begin{align*} \frac{4 a^{3} \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{4 i a^{3} \log{\left (e^{2 i f x} + e^{- 2 i e} \right )}}{c f} + \frac{2 i a^{3} e^{- 2 i e}}{c f \left (e^{2 i f x} + e^{- 2 i e}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.52526, size = 250, normalized size = 3.52 \begin{align*} \frac{2 \,{\left (-\frac{4 i \, a^{3} \log \left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + i\right )}{c} + \frac{2 i \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{c} + \frac{2 i \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{c} + \frac{-2 i \, a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 2 i \, a^{3}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right )} c} + \frac{6 i \, a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 16 \, a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 6 i \, a^{3}}{c{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + i\right )}^{2}}\right )}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(-4*I*a^3*log(tan(1/2*f*x + 1/2*e) + I)/c + 2*I*a^3*log(abs(tan(1/2*f*x + 1/2*e) + 1))/c + 2*I*a^3*log(abs(t
an(1/2*f*x + 1/2*e) - 1))/c + (-2*I*a^3*tan(1/2*f*x + 1/2*e)^2 - a^3*tan(1/2*f*x + 1/2*e) + 2*I*a^3)/((tan(1/2
*f*x + 1/2*e)^2 - 1)*c) + (6*I*a^3*tan(1/2*f*x + 1/2*e)^2 - 16*a^3*tan(1/2*f*x + 1/2*e) - 6*I*a^3)/(c*(tan(1/2
*f*x + 1/2*e) + I)^2))/f