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

Optimal. Leaf size=25 \[ -\frac{i a^3}{d (a-i a \tan (c+d x))} \]

[Out]

((-I)*a^3)/(d*(a - I*a*Tan[c + d*x]))

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Rubi [A]  time = 0.0372104, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {3487, 32} \[ -\frac{i a^3}{d (a-i a \tan (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*a^3)/(d*(a - I*a*Tan[c + d*x]))

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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

\begin{align*} \int \cos ^2(c+d x) (a+i a \tan (c+d x))^2 \, dx &=-\frac{\left (i a^3\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x)^2} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=-\frac{i a^3}{d (a-i a \tan (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.0535852, size = 31, normalized size = 1.24 \[ -\frac{i a^2 (\cos (c+d x)+i \sin (c+d x))^2}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.047, size = 73, normalized size = 2.9 \begin{align*}{\frac{1}{d} \left ( -{a}^{2} \left ( -{\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) -i{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}+{a}^{2} \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-a^2*(-1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)-I*a^2*cos(d*x+c)^2+a^2*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x
+1/2*c))

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Maxima [A]  time = 1.64713, size = 43, normalized size = 1.72 \begin{align*} \frac{a^{2} \tan \left (d x + c\right ) - i \, a^{2}}{{\left (\tan \left (d x + c\right )^{2} + 1\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.35485, size = 46, normalized size = 1.84 \begin{align*} -\frac{i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*I*a^2*e^(2*I*d*x + 2*I*c)/d

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Sympy [A]  time = 0.298465, size = 37, normalized size = 1.48 \begin{align*} \begin{cases} - \frac{i a^{2} e^{2 i c} e^{2 i d x}}{2 d} & \text{for}\: 2 d \neq 0 \\a^{2} x e^{2 i c} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-I*a**2*exp(2*I*c)*exp(2*I*d*x)/(2*d), Ne(2*d, 0)), (a**2*x*exp(2*I*c), True))

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Giac [A]  time = 1.19159, size = 23, normalized size = 0.92 \begin{align*} -\frac{i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*I*a^2*e^(2*I*d*x + 2*I*c)/d