3.910 \(\int \frac{(c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^2} \, dx\)

Optimal. Leaf size=83 \[ -\frac{4 i c^3}{f \left (a^2+i a^2 \tan (e+f x)\right )}+\frac{i c^3 \log (\cos (e+f x))}{a^2 f}+\frac{c^3 x}{a^2}+\frac{2 i c^3}{f (a+i a \tan (e+f x))^2} \]

[Out]

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

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Rubi [A]  time = 0.121946, antiderivative size = 83, 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} \[ -\frac{4 i c^3}{f \left (a^2+i a^2 \tan (e+f x)\right )}+\frac{i c^3 \log (\cos (e+f x))}{a^2 f}+\frac{c^3 x}{a^2}+\frac{2 i c^3}{f (a+i a \tan (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

Int[(c - I*c*Tan[e + f*x])^3/(a + I*a*Tan[e + f*x])^2,x]

[Out]

(c^3*x)/a^2 + (I*c^3*Log[Cos[e + f*x]])/(a^2*f) + ((2*I)*c^3)/(f*(a + I*a*Tan[e + f*x])^2) - ((4*I)*c^3)/(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))^3}{(a+i a \tan (e+f x))^2} \, dx &=\left (a^3 c^3\right ) \int \frac{\sec ^6(e+f x)}{(a+i a \tan (e+f x))^5} \, dx\\ &=-\frac{\left (i c^3\right ) \operatorname{Subst}\left (\int \frac{(a-x)^2}{(a+x)^3} \, dx,x,i a \tan (e+f x)\right )}{a^2 f}\\ &=-\frac{\left (i c^3\right ) \operatorname{Subst}\left (\int \left (\frac{4 a^2}{(a+x)^3}-\frac{4 a}{(a+x)^2}+\frac{1}{a+x}\right ) \, dx,x,i a \tan (e+f x)\right )}{a^2 f}\\ &=\frac{c^3 x}{a^2}+\frac{i c^3 \log (\cos (e+f x))}{a^2 f}+\frac{2 i c^3}{f (a+i a \tan (e+f x))^2}-\frac{4 i c^3}{f \left (a^2+i a^2 \tan (e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 1.8576, size = 115, normalized size = 1.39 \[ -\frac{c^3 \sec ^2(e+f x) \left (\sin (2 (e+f x))+i \cos (2 (e+f x)) \left (\log \left (\cos ^2(e+f x)\right )+1\right )-\sin (2 (e+f x)) \log \left (\cos ^2(e+f x)\right )+2 \tan ^{-1}(\tan (f x)) (\cos (2 (e+f x))+i \sin (2 (e+f x)))-2 i\right )}{2 a^2 f (\tan (e+f x)-i)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(c - I*c*Tan[e + f*x])^3/(a + I*a*Tan[e + f*x])^2,x]

[Out]

-(c^3*Sec[e + f*x]^2*(-2*I + I*Cos[2*(e + f*x)]*(1 + Log[Cos[e + f*x]^2]) + 2*ArcTan[Tan[f*x]]*(Cos[2*(e + f*x
)] + I*Sin[2*(e + f*x)]) + Sin[2*(e + f*x)] - Log[Cos[e + f*x]^2]*Sin[2*(e + f*x)]))/(2*a^2*f*(-I + Tan[e + f*
x])^2)

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Maple [A]  time = 0.03, size = 69, normalized size = 0.8 \begin{align*}{\frac{-2\,i{c}^{3}}{f{a}^{2} \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}-{\frac{i{c}^{3}\ln \left ( \tan \left ( fx+e \right ) -i \right ) }{f{a}^{2}}}-4\,{\frac{{c}^{3}}{f{a}^{2} \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))^2,x)

[Out]

-2*I/f*c^3/a^2/(tan(f*x+e)-I)^2-I/f*c^3/a^2*ln(tan(f*x+e)-I)-4/f*c^3/a^2/(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))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.45054, size = 220, normalized size = 2.65 \begin{align*} \frac{{\left (4 \, c^{3} f x e^{\left (4 i \, f x + 4 i \, e\right )} + 2 i \, c^{3} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - 2 i \, c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c^{3}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{2 \, a^{2} 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))^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 4.15936, size = 121, normalized size = 1.46 \begin{align*} \frac{i c^{3} \log{\left (e^{2 i f x} + e^{- 2 i e} \right )}}{a^{2} f} + \frac{\left (\begin{cases} 2 c^{3} x e^{4 i e} - \frac{i c^{3} e^{2 i e} e^{- 2 i f x}}{f} + \frac{i c^{3} e^{- 4 i f x}}{2 f} & \text{for}\: f \neq 0 \\x \left (2 c^{3} e^{4 i e} - 2 c^{3} e^{2 i e} + 2 c^{3}\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))**3/(a+I*a*tan(f*x+e))**2,x)

[Out]

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

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Giac [B]  time = 1.58941, size = 217, normalized size = 2.61 \begin{align*} -\frac{\frac{12 i \, c^{3} \log \left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - i\right )}{a^{2}} - \frac{6 i \, c^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{a^{2}} - \frac{6 i \, c^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{a^{2}} + \frac{-25 i \, c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 100 \, c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 198 i \, c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 100 \, c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 25 i \, c^{3}}{a^{2}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - i\right )}^{4}}}{6 \, 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))^2,x, algorithm="giac")

[Out]

-1/6*(12*I*c^3*log(tan(1/2*f*x + 1/2*e) - I)/a^2 - 6*I*c^3*log(abs(tan(1/2*f*x + 1/2*e) + 1))/a^2 - 6*I*c^3*lo
g(abs(tan(1/2*f*x + 1/2*e) - 1))/a^2 + (-25*I*c^3*tan(1/2*f*x + 1/2*e)^4 - 100*c^3*tan(1/2*f*x + 1/2*e)^3 + 19
8*I*c^3*tan(1/2*f*x + 1/2*e)^2 + 100*c^3*tan(1/2*f*x + 1/2*e) - 25*I*c^3)/(a^2*(tan(1/2*f*x + 1/2*e) - I)^4))/
f