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

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(d*x+c))

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Rubi [A]
time = 0.04, antiderivative size = 25, normalized size of antiderivative = 1.00, 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 = {3568, 32} \begin {gather*} -\frac {i a^3}{d (a-i a \tan (c+d x))} \end {gather*}

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 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*} \int \cos ^2(c+d x) (a+i a \tan (c+d x))^2 \, dx &=-\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*}

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Mathematica [A]
time = 0.08, size = 31, normalized size = 1.24 \begin {gather*} -\frac {i a^2 (\cos (c+d x)+i \sin (c+d x))^2}{2 d} \end {gather*}

Antiderivative was successfully verified.

[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

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (23 ) = 46\).
time = 0.24, size = 73, normalized size = 2.92

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	\(\frac {-a^{2} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-i a^{2} \left (\cos ^{2}\left (d x +c \right )\right )+a^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(73\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.49, size = 32, normalized size = 1.28 \begin {gather*} \frac {a^{2} \tan \left (d x + c\right ) - i \, a^{2}}{{\left (\tan \left (d x + c\right )^{2} + 1\right )} d} \end {gather*}

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 = 0.36, size = 17, normalized size = 0.68 \begin {gather*} -\frac {i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )}}{2 \, d} \end {gather*}

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 = 0.10, size = 36, normalized size = 1.44 \begin {gather*} \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} \end {gather*}

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(d, 0)), (a**2*x*exp(2*I*c), True))

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Giac [A]
time = 0.56, size = 17, normalized size = 0.68 \begin {gather*} -\frac {i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )}}{2 \, d} \end {gather*}

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

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Mupad [B]
time = 3.28, size = 18, normalized size = 0.72 \begin {gather*} \frac {a^2}{d\,\left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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