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

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

[Out]

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

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Rubi [A]  time = 0.121027, 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} \[ \frac{c^3 \tan (e+f x)}{a f}+\frac{4 i c^3}{f (a+i a \tan (e+f x))}-\frac{4 i c^3 \log (\cos (e+f x))}{a f}-\frac{4 c^3 x}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [B]  time = 1.76348, size = 234, normalized size = 3.3 \[ \frac{i c^3 \sec ^2(e+f x) \left (-2 \sin (e+2 f x)-\sin (3 e+2 f x)-i \cos (3 e+2 f x)+i \cos (3 e+2 f x) \log \left (\cos ^2(e+f x)\right )+i \cos (e+2 f x) \log \left (\cos ^2(e+f x)\right )+i \cos (e) \left (2 \log \left (\cos ^2(e+f x)\right )-3\right )-\sin (e+2 f x) \log \left (\cos ^2(e+f x)\right )-\sin (3 e+2 f x) \log \left (\cos ^2(e+f x)\right )+8 \cos (e) \tan ^{-1}(\tan (f x)) \cos (e+f x) (\cos (e+f x)+i \sin (e+f x))+\sin (e)\right )}{2 a f \left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \left (\sin \left (\frac{e}{2}\right )+\cos \left (\frac{e}{2}\right )\right ) (\tan (e+f x)-i)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/2)*c^3*Sec[e + f*x]^2*((-I)*Cos[3*e + 2*f*x] + I*Cos[e + 2*f*x]*Log[Cos[e + f*x]^2] + I*Cos[3*e + 2*f*x]*L
og[Cos[e + f*x]^2] + I*Cos[e]*(-3 + 2*Log[Cos[e + f*x]^2]) + Sin[e] + 8*ArcTan[Tan[f*x]]*Cos[e]*Cos[e + f*x]*(
Cos[e + f*x] + I*Sin[e + f*x]) - 2*Sin[e + 2*f*x] - Log[Cos[e + f*x]^2]*Sin[e + 2*f*x] - Sin[3*e + 2*f*x] - Lo
g[Cos[e + f*x]^2]*Sin[3*e + 2*f*x]))/(a*f*(Cos[e/2] - Sin[e/2])*(Cos[e/2] + Sin[e/2])*(-I + Tan[e + f*x]))

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Maple [A]  time = 0.024, size = 62, normalized size = 0.9 \begin{align*}{\frac{{c}^{3}\tan \left ( fx+e \right ) }{af}}+{\frac{4\,i{c}^{3}\ln \left ( \tan \left ( fx+e \right ) -i \right ) }{af}}+4\,{\frac{{c}^{3}}{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))^3/(a+I*a*tan(f*x+e)),x)

[Out]

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.34351, size = 313, normalized size = 4.41 \begin{align*} -\frac{8 \, c^{3} f x e^{\left (4 i \, f x + 4 i \, e\right )} - 2 i \, c^{3} +{\left (8 \, c^{3} f x - 4 i \, c^{3}\right )} e^{\left (2 i \, f x + 2 i \, e\right )} -{\left (-4 i \, c^{3} e^{\left (4 i \, f x + 4 i \, e\right )} - 4 i \, c^{3} 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 (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))^3/(a+I*a*tan(f*x+e)),x, algorithm="fricas")

[Out]

-(8*c^3*f*x*e^(4*I*f*x + 4*I*e) - 2*I*c^3 + (8*c^3*f*x - 4*I*c^3)*e^(2*I*f*x + 2*I*e) - (-4*I*c^3*e^(4*I*f*x +
4*I*e) - 4*I*c^3*e^(2*I*f*x + 2*I*e))*log(e^(2*I*f*x + 2*I*e) + 1))/(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 = 3.16319, size = 119, normalized size = 1.68 \begin{align*} - \frac{4 i c^{3} \log{\left (e^{2 i f x} + e^{- 2 i e} \right )}}{a f} + \frac{2 i c^{3} e^{- 2 i e}}{a f \left (e^{2 i f x} + e^{- 2 i e}\right )} - \frac{\left (\begin{cases} 8 c^{3} x e^{2 i e} - \frac{2 i c^{3} e^{- 2 i f x}}{f} & \text{for}\: f \neq 0 \\x \left (8 c^{3} e^{2 i e} - 4 c^{3}\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))**3/(a+I*a*tan(f*x+e)),x)

[Out]

-4*I*c**3*log(exp(2*I*f*x) + exp(-2*I*e))/(a*f) + 2*I*c**3*exp(-2*I*e)/(a*f*(exp(2*I*f*x) + exp(-2*I*e))) - Pi
ecewise((8*c**3*x*exp(2*I*e) - 2*I*c**3*exp(-2*I*f*x)/f, Ne(f, 0)), (x*(8*c**3*exp(2*I*e) - 4*c**3), True))*ex
p(-2*I*e)/a

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Giac [B]  time = 1.49415, size = 250, normalized size = 3.52 \begin{align*} \frac{2 \,{\left (\frac{4 i \, c^{3} \log \left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - i\right )}{a} - \frac{2 i \, c^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{a} - \frac{2 i \, c^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{a} + \frac{2 i \, c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 2 i \, c^{3}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right )} a} + \frac{-6 i \, c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 16 \, c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 6 i \, c^{3}}{a{\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((c-I*c*tan(f*x+e))^3/(a+I*a*tan(f*x+e)),x, algorithm="giac")

[Out]

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