3.1074 \(\int \frac{(c+d \tan (e+f x))^2}{a+i a \tan (e+f x)} \, dx\)

Optimal. Leaf size=75 \[ \frac{x \left (c^2-2 i c d+d^2\right )}{2 a}+\frac{i (c+i d)^2}{2 f (a+i a \tan (e+f x))}+\frac{i d^2 \log (\cos (e+f x))}{a f} \]

[Out]

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

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Rubi [A]  time = 0.0825378, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {3540, 3475} \[ \frac{x \left (c^2-2 i c d+d^2\right )}{2 a}+\frac{i (c+i d)^2}{2 f (a+i a \tan (e+f x))}+\frac{i d^2 \log (\cos (e+f x))}{a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3540

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> -Simp[
(b*(a*c + b*d)^2*(a + b*Tan[e + f*x])^m)/(2*a^3*f*m), x] + Dist[1/(2*a^2), Int[(a + b*Tan[e + f*x])^(m + 1)*Si
mp[a*c^2 - 2*b*c*d + a*d^2 - 2*b*d^2*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0] && LeQ[m, -1] && EqQ[a^2 + b^2, 0]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{(c+d \tan (e+f x))^2}{a+i a \tan (e+f x)} \, dx &=\frac{i (c+i d)^2}{2 f (a+i a \tan (e+f x))}+\frac{\int \left (a \left (c^2-2 i c d+d^2\right )-2 i a d^2 \tan (e+f x)\right ) \, dx}{2 a^2}\\ &=\frac{\left (c^2-2 i c d+d^2\right ) x}{2 a}+\frac{i (c+i d)^2}{2 f (a+i a \tan (e+f x))}-\frac{\left (i d^2\right ) \int \tan (e+f x) \, dx}{a}\\ &=\frac{\left (c^2-2 i c d+d^2\right ) x}{2 a}+\frac{i d^2 \log (\cos (e+f x))}{a f}+\frac{i (c+i d)^2}{2 f (a+i a \tan (e+f x))}\\ \end{align*}

Mathematica [B]  time = 1.27354, size = 155, normalized size = 2.07 \[ \frac{\tan (e+f x) \left (c^2 (2 f x-i)+2 c (d-2 i d f x)+2 i d^2 \log \left (\cos ^2(e+f x)\right )+d^2 (-2 f x+i)\right )-2 i c^2 f x+c^2-4 c d f x+2 i c d+4 d^2 \tan ^{-1}(\tan (f x)) (\tan (e+f x)-i)+2 d^2 \log \left (\cos ^2(e+f x)\right )+2 i d^2 f x-d^2}{4 a f (\tan (e+f x)-i)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.026, size = 196, normalized size = 2.6 \begin{align*} -{\frac{\ln \left ( \tan \left ( fx+e \right ) -i \right ) cd}{2\,af}}-{\frac{{\frac{i}{4}}{c}^{2}\ln \left ( \tan \left ( fx+e \right ) -i \right ) }{af}}-{\frac{{\frac{3\,i}{4}}\ln \left ( \tan \left ( fx+e \right ) -i \right ){d}^{2}}{af}}+{\frac{icd}{af \left ( \tan \left ( fx+e \right ) -i \right ) }}+{\frac{{c}^{2}}{2\,af \left ( \tan \left ( fx+e \right ) -i \right ) }}-{\frac{{d}^{2}}{2\,af \left ( \tan \left ( fx+e \right ) -i \right ) }}+{\frac{\ln \left ( \tan \left ( fx+e \right ) +i \right ) cd}{2\,af}}+{\frac{{\frac{i}{4}}\ln \left ( \tan \left ( fx+e \right ) +i \right ){c}^{2}}{af}}-{\frac{{\frac{i}{4}}\ln \left ( \tan \left ( fx+e \right ) +i \right ){d}^{2}}{af}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c+d*tan(f*x+e))^2/(a+I*a*tan(f*x+e)),x)

[Out]

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

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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+d*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 = 1.68124, size = 225, normalized size = 3. \begin{align*} \frac{{\left (2 \,{\left (c^{2} - 2 i \, c d + 3 \, d^{2}\right )} f x e^{\left (2 i \, f x + 2 i \, e\right )} + 4 i \, d^{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} - 2 \, c d - i \, d^{2}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 2.03764, size = 175, normalized size = 2.33 \begin{align*} \frac{i d^{2} \log{\left (e^{2 i f x} + e^{- 2 i e} \right )}}{a f} + \frac{\left (\begin{cases} c^{2} x e^{2 i e} + \frac{i c^{2} e^{- 2 i f x}}{2 f} - 2 i c d x e^{2 i e} - \frac{c d e^{- 2 i f x}}{f} + 3 d^{2} x e^{2 i e} - \frac{i d^{2} e^{- 2 i f x}}{2 f} & \text{for}\: f \neq 0 \\x \left (c^{2} e^{2 i e} + c^{2} - 2 i c d e^{2 i e} + 2 i c d + 3 d^{2} e^{2 i e} - d^{2}\right ) & \text{otherwise} \end{cases}\right ) e^{- 2 i e}}{2 a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))**2/(a+I*a*tan(f*x+e)),x)

[Out]

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

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Giac [B]  time = 1.55325, size = 177, normalized size = 2.36 \begin{align*} -\frac{\frac{{\left (i \, c^{2} + 2 \, c d + 3 i \, d^{2}\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{a} + \frac{{\left (-i \, c^{2} - 2 \, c d + i \, d^{2}\right )} \log \left (i \, \tan \left (f x + e\right ) - 1\right )}{a} + \frac{-i \, c^{2} \tan \left (f x + e\right ) - 2 \, c d \tan \left (f x + e\right ) - 3 i \, d^{2} \tan \left (f x + e\right ) - 3 \, c^{2} - 2 i \, c d - d^{2}}{a{\left (\tan \left (f x + e\right ) - i\right )}}}{4 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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