\(\int \cos ^2(c+d x) (a+i a \tan (c+d x))^2 \, dx\) [24]

Optimal result

Integrand size = 24, antiderivative size = 25 \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {i a^3}{d (a-i a \tan (c+d x))} \]

[Out]

-I*a^3/d/(a-I*a*tan(d*x+c))

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3568, 32} \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {i a^3}{d (a-i a \tan (c+d x))} \]

[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 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]

Rule 3568

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (i a^3\right ) \text {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] (verified)

Time = 0.13 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {i a^2 (\cos (c+d x)+i \sin (c+d x))^2}{2 d} \]

[In]

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

[Out]

((-1/2*I)*a^2*(Cos[c + d*x] + I*Sin[c + d*x])^2)/d

Maple [A] (verified)

Time = 1.35 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76

method	result	size
risch	\(-\frac {i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{2 d}\)	\(19\)
derivativedivides	\(\frac {-a^{2} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-i a^{2} \left (\cos ^{2}\left (d x +c \right )\right )+a^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(73\)
default	\(\frac {-a^{2} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-i a^{2} \left (\cos ^{2}\left (d x +c \right )\right )+a^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(73\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )}}{2 \, d} \]

[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

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44 \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^2 \, dx=\begin {cases} - \frac {i a^{2} e^{2 i c} e^{2 i d x}}{2 d} & \text {for}\: d \neq 0 \\a^{2} x e^{2 i c} & \text {otherwise} \end {cases} \]

[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(d, 0)), (a**2*x*exp(2*I*c), True))

Maxima [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {a^{2} \tan \left (d x + c\right ) - i \, a^{2}}{{\left (\tan \left (d x + c\right )^{2} + 1\right )} d} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.50 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )}}{2 \, d} \]

[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

Mupad [B] (verification not implemented)

Time = 3.82 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {a^2}{d\,\left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )} \]

[In]

int(cos(c + d*x)^2*(a + a*tan(c + d*x)*1i)^2,x)

[Out]

a^2/(d*(tan(c + d*x) + 1i))