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

Optimal. Leaf size=55 \[ \frac{2 i c^2}{f (a+i a \tan (e+f x))}-\frac{i c^2 \log (\cos (e+f x))}{a f}-\frac{c^2 x}{a} \]

[Out]

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

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Rubi [A]  time = 0.114738, antiderivative size = 55, 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} \[ \frac{2 i c^2}{f (a+i a \tan (e+f x))}-\frac{i c^2 \log (\cos (e+f x))}{a f}-\frac{c^2 x}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 1.19416, size = 74, normalized size = 1.35 \[ -\frac{c^2 \left (2 \tan ^{-1}(\tan (f x)) (\tan (e+f x)-i)+\log \left (\cos ^2(e+f x)\right )+i \tan (e+f x) \left (\log \left (\cos ^2(e+f x)\right )+2\right )-2\right )}{2 a f (\tan (e+f x)-i)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.026, size = 46, normalized size = 0.8 \begin{align*}{\frac{i{c}^{2}\ln \left ( \tan \left ( fx+e \right ) -i \right ) }{fa}}+2\,{\frac{{c}^{2}}{fa \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))^2/(a+I*a*tan(f*x+e)),x)

[Out]

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.23292, size = 170, normalized size = 3.09 \begin{align*} -\frac{{\left (2 \, c^{2} f x e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - i \, c^{2}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))^2/(a+I*a*tan(f*x+e)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 2.77431, size = 83, normalized size = 1.51 \begin{align*} - \frac{i c^{2} \log{\left (e^{2 i f x} + e^{- 2 i e} \right )}}{a f} - \frac{\left (\begin{cases} 2 c^{2} x e^{2 i e} - \frac{i c^{2} e^{- 2 i f x}}{f} & \text{for}\: f \neq 0 \\x \left (2 c^{2} e^{2 i e} - 2 c^{2}\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))**2/(a+I*a*tan(f*x+e)),x)

[Out]

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

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Giac [B]  time = 1.68204, size = 171, normalized size = 3.11 \begin{align*} -\frac{-\frac{2 i \, c^{2} \log \left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - i\right )}{a} + \frac{i \, c^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{a} + \frac{i \, c^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{a} + \frac{3 i \, c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 10 \, c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 3 i \, c^{2}}{a{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - i\right )}^{2}}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))^2/(a+I*a*tan(f*x+e)),x, algorithm="giac")

[Out]

-(-2*I*c^2*log(tan(1/2*f*x + 1/2*e) - I)/a + I*c^2*log(abs(tan(1/2*f*x + 1/2*e) + 1))/a + I*c^2*log(abs(tan(1/
2*f*x + 1/2*e) - 1))/a + (3*I*c^2*tan(1/2*f*x + 1/2*e)^2 + 10*c^2*tan(1/2*f*x + 1/2*e) - 3*I*c^2)/(a*(tan(1/2*
f*x + 1/2*e) - I)^2))/f