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

Optimal. Leaf size=62 \[ \frac{i a^2 \tan (c+d x)}{d}-\frac{2 a^2 \log (\cos (c+d x))}{d}-2 i a^2 x+\frac{(a+i a \tan (c+d x))^2}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.0424361, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {3527, 3477, 3475} \[ \frac{i a^2 \tan (c+d x)}{d}-\frac{2 a^2 \log (\cos (c+d x))}{d}-2 i a^2 x+\frac{(a+i a \tan (c+d x))^2}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3527

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] + Dist[(b*c + a*d)/b, Int[(a + b*Tan[e + f*x])^m, x], x] /; FreeQ[{a, b, c,
d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] &&  !LtQ[m, 0]

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

Mathematica [A]  time = 0.154691, size = 51, normalized size = 0.82 \[ \frac{a^2 \left (-\tan ^2(c+d x)-4 i \tan ^{-1}(\tan (c+d x))+4 i \tan (c+d x)-4 \log (\cos (c+d x))\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.004, size = 67, normalized size = 1.1 \begin{align*}{\frac{2\,i{a}^{2}\tan \left ( dx+c \right ) }{d}}-{\frac{{a}^{2} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+{\frac{{a}^{2}\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{d}}-{\frac{2\,i{a}^{2}\arctan \left ( \tan \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 2.07361, size = 74, normalized size = 1.19 \begin{align*} -\frac{a^{2} \tan \left (d x + c\right )^{2} + 4 i \,{\left (d x + c\right )} a^{2} - 2 \, a^{2} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 4 i \, a^{2} \tan \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.17084, size = 250, normalized size = 4.03 \begin{align*} -\frac{2 \,{\left (3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 \, a^{2} +{\left (a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )\right )}}{d e^{\left (4 i \, d x + 4 i \, c\right )} + 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(tan(d*x+c)*(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.90786, size = 95, normalized size = 1.53 \begin{align*} - \frac{2 a^{2} \log{\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac{- \frac{6 a^{2} e^{- 2 i c} e^{2 i d x}}{d} - \frac{4 a^{2} e^{- 4 i c}}{d}}{e^{4 i d x} + 2 e^{- 2 i c} e^{2 i d x} + e^{- 4 i c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a**2*log(exp(2*I*d*x) + exp(-2*I*c))/d + (-6*a**2*exp(-2*I*c)*exp(2*I*d*x)/d - 4*a**2*exp(-4*I*c)/d)/(exp(4
*I*d*x) + 2*exp(-2*I*c)*exp(2*I*d*x) + exp(-4*I*c))

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Giac [B]  time = 1.32071, size = 157, normalized size = 2.53 \begin{align*} -\frac{2 \,{\left (a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 2 \, 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 ) + 3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 2 \, a^{2}\right )}}{d e^{\left (4 i \, d x + 4 i \, c\right )} + 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(tan(d*x+c)*(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

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