3.23 \(\int (a+i a \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=38 \[ -\frac{a^2 \tan (c+d x)}{d}-\frac{2 i a^2 \log (\cos (c+d x))}{d}+2 a^2 x \]

[Out]

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

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Rubi [A]  time = 0.0169765, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {3477, 3475} \[ -\frac{a^2 \tan (c+d x)}{d}-\frac{2 i a^2 \log (\cos (c+d x))}{d}+2 a^2 x \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^2,x]

[Out]

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

Rule 3477

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^2, x_Symbol] :> Simp[(a^2 - b^2)*x, x] + (Dist[2*a*b, Int[Tan[c + d
*x], x], x] + Simp[(b^2*Tan[c + d*x])/d, x]) /; FreeQ[{a, b, c, d}, x]

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

Mathematica [B]  time = 0.632605, size = 100, normalized size = 2.63 \[ -\frac{a^2 \sec (c) \sec (c+d x) \left (-4 d x \cos (2 c+d x)+\cos (d x) \left (-4 d x+i \log \left (\cos ^2(c+d x)\right )\right )+i \cos (2 c+d x) \log \left (\cos ^2(c+d x)\right )+4 \cos (c) \cos (c+d x) \tan ^{-1}(\tan (3 c+d x))+2 \sin (d x)\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])^2,x]

[Out]

-(a^2*Sec[c]*Sec[c + d*x]*(4*ArcTan[Tan[3*c + d*x]]*Cos[c]*Cos[c + d*x] - 4*d*x*Cos[2*c + d*x] + Cos[d*x]*(-4*
d*x + I*Log[Cos[c + d*x]^2]) + I*Cos[2*c + d*x]*Log[Cos[c + d*x]^2] + 2*Sin[d*x]))/(2*d)

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Maple [A]  time = 0.005, size = 51, normalized size = 1.3 \begin{align*}{\frac{i{a}^{2}\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{d}}+2\,{\frac{{a}^{2}\arctan \left ( \tan \left ( dx+c \right ) \right ) }{d}}-{\frac{{a}^{2}\tan \left ( dx+c \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.66464, size = 55, normalized size = 1.45 \begin{align*} a^{2} x + \frac{{\left (d x + c - \tan \left (d x + c\right )\right )} a^{2}}{d} + \frac{2 i \, a^{2} \log \left (\sec \left (d x + c\right )\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*x + (d*x + c - tan(d*x + c))*a^2/d + 2*I*a^2*log(sec(d*x + c))/d

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Fricas [A]  time = 1.06008, size = 151, normalized size = 3.97 \begin{align*} \frac{-2 i \, a^{2} +{\left (-2 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 2 i \, a^{2}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}{d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.937538, size = 60, normalized size = 1.58 \begin{align*} - \frac{2 i a^{2} \log{\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} - \frac{2 i a^{2} e^{- 2 i c}}{d \left (e^{2 i d x} + e^{- 2 i c}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.13516, size = 88, normalized size = 2.32 \begin{align*} \frac{-2 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 2 i \, a^{2} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 2 i \, a^{2}}{d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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