3.1073 \(\int (a+i a \tan (e+f x)) (c+d \tan (e+f x))^2 \, dx\)

Optimal. Leaf size=78 \[ \frac{i a (c+d \tan (e+f x))^2}{2 f}+\frac{a d (d+i c) \tan (e+f x)}{f}-\frac{i a (c-i d)^2 \log (\cos (e+f x))}{f}+a x (c-i d)^2 \]

[Out]

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

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Rubi [A]  time = 0.0921949, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {3528, 3525, 3475} \[ \frac{i a (c+d \tan (e+f x))^2}{2 f}+\frac{a d (d+i c) \tan (e+f x)}{f}-\frac{i a (c-i d)^2 \log (\cos (e+f x))}{f}+a x (c-i d)^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3528

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

Rule 3525

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(a*c - b
*d)*x, x] + (Dist[b*c + a*d, Int[Tan[e + f*x], x], x] + Simp[(b*d*Tan[e + f*x])/f, x]) /; FreeQ[{a, b, c, d, e
, f}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 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 (a+i a \tan (e+f x)) (c+d \tan (e+f x))^2 \, dx &=\frac{i a (c+d \tan (e+f x))^2}{2 f}+\int (c+d \tan (e+f x)) (a (c-i d)+a (i c+d) \tan (e+f x)) \, dx\\ &=a (c-i d)^2 x+\frac{a d (i c+d) \tan (e+f x)}{f}+\frac{i a (c+d \tan (e+f x))^2}{2 f}+\left (i a (c-i d)^2\right ) \int \tan (e+f x) \, dx\\ &=a (c-i d)^2 x-\frac{i a (c-i d)^2 \log (\cos (e+f x))}{f}+\frac{a d (i c+d) \tan (e+f x)}{f}+\frac{i a (c+d \tan (e+f x))^2}{2 f}\\ \end{align*}

Mathematica [B]  time = 1.78658, size = 175, normalized size = 2.24 \[ \frac{(\cos (f x)-i \sin (f x)) (a+i a \tan (e+f x)) \left (2 d (2 c-i d) (\tan (e)+i) \sin (f x)+4 f x (c-i d)^2 (\cos (e)-i \sin (e)) \cos (e+f x)-i (c-i d)^2 (\cos (e)-i \sin (e)) \cos (e+f x) \log \left (\cos ^2(e+f x)\right )-2 (c-i d)^2 (\cos (e)-i \sin (e)) \cos (e+f x) \tan ^{-1}(\tan (2 e+f x))+d^2 (\sin (e)+i \cos (e)) \sec (e+f x)\right )}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.006, size = 156, normalized size = 2. \begin{align*}{\frac{{\frac{i}{2}}a{d}^{2} \left ( \tan \left ( fx+e \right ) \right ) ^{2}}{f}}+{\frac{2\,iacd\tan \left ( fx+e \right ) }{f}}+{\frac{a\tan \left ( fx+e \right ){d}^{2}}{f}}-{\frac{{\frac{i}{2}}a\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ){d}^{2}}{f}}+{\frac{{\frac{i}{2}}a\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ){c}^{2}}{f}}+{\frac{a\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) cd}{f}}-{\frac{2\,ia\arctan \left ( \tan \left ( fx+e \right ) \right ) cd}{f}}+{\frac{a\arctan \left ( \tan \left ( fx+e \right ) \right ){c}^{2}}{f}}-{\frac{a\arctan \left ( \tan \left ( fx+e \right ) \right ){d}^{2}}{f}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.49244, size = 127, normalized size = 1.63 \begin{align*} -\frac{-i \, a d^{2} \tan \left (f x + e\right )^{2} - 2 \,{\left (a c^{2} - 2 i \, a c d - a d^{2}\right )}{\left (f x + e\right )} +{\left (-i \, a c^{2} - 2 \, a c d + i \, a d^{2}\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right ) + 2 \,{\left (-2 i \, a c d - a d^{2}\right )} \tan \left (f x + e\right )}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.58174, size = 394, normalized size = 5.05 \begin{align*} -\frac{4 \, a c d - 2 i \, a d^{2} + 4 \,{\left (a c d - i \, a d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )} -{\left (-i \, a c^{2} - 2 \, a c d + i \, a d^{2} +{\left (-i \, a c^{2} - 2 \, a c d + i \, a d^{2}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (-2 i \, a c^{2} - 4 \, a c d + 2 i \, a d^{2}\right )} 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 )}{f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 2.88525, size = 128, normalized size = 1.64 \begin{align*} \frac{a \left (- i c^{2} - 2 c d + i d^{2}\right ) \log{\left (e^{2 i f x} + e^{- 2 i e} \right )}}{f} + \frac{- \frac{\left (4 a c d - 4 i a d^{2}\right ) e^{- 2 i e} e^{2 i f x}}{f} - \frac{\left (4 a c d - 2 i a d^{2}\right ) e^{- 4 i e}}{f}}{e^{4 i f x} + 2 e^{- 2 i e} e^{2 i f x} + e^{- 4 i e}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.58246, size = 406, normalized size = 5.21 \begin{align*} \frac{-i \, a c^{2} 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 \, a c d e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) + i \, a d^{2} 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 \, a 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 ) - 4 \, a c d e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) + 2 i \, a 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 ) - 4 \, a c d e^{\left (2 i \, f x + 2 i \, e\right )} + 4 i \, a d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a c^{2} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - 2 \, a c d \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) + i \, a d^{2} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - 4 \, a c d + 2 i \, a d^{2}}{f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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