3.32 \(\int \cos ^3(c+d x) (a+i a \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=51 \[ \frac{a^2 \sin (c+d x)}{3 d}-\frac{2 i \cos ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.0419488, antiderivative size = 51, 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 = {3496, 2637} \[ \frac{a^2 \sin (c+d x)}{3 d}-\frac{2 i \cos ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*(a + I*a*Tan[c + d*x])^2,x]

[Out]

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

Rule 3496

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(2*b*(
d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1))/(f*m), x] - Dist[(b^2*(m + 2*n - 2))/(d^2*m), Int[(d*Sec[e + f
*x])^(m + 2)*(a + b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n,
1] && ((IGtQ[n/2, 0] && ILtQ[m - 1/2, 0]) || (EqQ[n, 2] && LtQ[m, 0]) || (LeQ[m, -1] && GtQ[m + n, 0]) || (ILt
Q[m, 0] && LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) && IntegerQ[2*m]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [A]  time = 0.178426, size = 50, normalized size = 0.98 \[ \frac{a^2 (2 \cos (c+d x)-i \sin (c+d x)) (\sin (2 (c+d x))-i \cos (2 (c+d x)))}{3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*(a + I*a*Tan[c + d*x])^2,x]

[Out]

(a^2*(2*Cos[c + d*x] - I*Sin[c + d*x])*((-I)*Cos[2*(c + d*x)] + Sin[2*(c + d*x)]))/(3*d)

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Maple [A]  time = 0.052, size = 54, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ( -{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3}}-{\frac{2\,i}{3}}{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{{a}^{2} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^2,x)

[Out]

1/d*(-1/3*a^2*sin(d*x+c)^3-2/3*I*a^2*cos(d*x+c)^3+1/3*a^2*(2+cos(d*x+c)^2)*sin(d*x+c))

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Maxima [A]  time = 1.06649, size = 70, normalized size = 1.37 \begin{align*} -\frac{2 i \, a^{2} \cos \left (d x + c\right )^{3} + a^{2} \sin \left (d x + c\right )^{3} +{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{2}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/3*(2*I*a^2*cos(d*x + c)^3 + a^2*sin(d*x + c)^3 + (sin(d*x + c)^3 - 3*sin(d*x + c))*a^2)/d

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Fricas [A]  time = 1.19644, size = 84, normalized size = 1.65 \begin{align*} \frac{-i \, a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} - 3 i \, a^{2} e^{\left (i \, d x + i \, c\right )}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/6*(-I*a^2*e^(3*I*d*x + 3*I*c) - 3*I*a^2*e^(I*d*x + I*c))/d

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Sympy [A]  time = 0.43183, size = 76, normalized size = 1.49 \begin{align*} \begin{cases} \frac{- 2 i a^{2} d e^{3 i c} e^{3 i d x} - 6 i a^{2} d e^{i c} e^{i d x}}{12 d^{2}} & \text{for}\: 12 d^{2} \neq 0 \\x \left (\frac{a^{2} e^{3 i c}}{2} + \frac{a^{2} e^{i c}}{2}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3*(a+I*a*tan(d*x+c))**2,x)

[Out]

Piecewise(((-2*I*a**2*d*exp(3*I*c)*exp(3*I*d*x) - 6*I*a**2*d*exp(I*c)*exp(I*d*x))/(12*d**2), Ne(12*d**2, 0)),
(x*(a**2*exp(3*I*c)/2 + a**2*exp(I*c)/2), True))

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Giac [B]  time = 1.25377, size = 717, normalized size = 14.06 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/96*(24*a^2*e^(4*I*d*x + 2*I*c)*log(I*e^(I*d*x + I*c) + 1) + 48*a^2*e^(2*I*d*x)*log(I*e^(I*d*x + I*c) + 1) +
24*a^2*e^(-2*I*c)*log(I*e^(I*d*x + I*c) + 1) + 27*a^2*e^(4*I*d*x + 2*I*c)*log(I*e^(I*d*x + I*c) - 1) + 54*a^2
*e^(2*I*d*x)*log(I*e^(I*d*x + I*c) - 1) + 27*a^2*e^(-2*I*c)*log(I*e^(I*d*x + I*c) - 1) - 24*a^2*e^(4*I*d*x + 2
*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 48*a^2*e^(2*I*d*x)*log(-I*e^(I*d*x + I*c) + 1) - 24*a^2*e^(-2*I*c)*log(-I*
e^(I*d*x + I*c) + 1) - 27*a^2*e^(4*I*d*x + 2*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 54*a^2*e^(2*I*d*x)*log(-I*e^(I
*d*x + I*c) - 1) - 27*a^2*e^(-2*I*c)*log(-I*e^(I*d*x + I*c) - 1) + 3*a^2*e^(4*I*d*x + 2*I*c)*log(I*e^(I*d*x) +
e^(-I*c)) + 6*a^2*e^(2*I*d*x)*log(I*e^(I*d*x) + e^(-I*c)) + 3*a^2*e^(-2*I*c)*log(I*e^(I*d*x) + e^(-I*c)) - 3*
a^2*e^(4*I*d*x + 2*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) - 6*a^2*e^(2*I*d*x)*log(-I*e^(I*d*x) + e^(-I*c)) - 3*a^2*
e^(-2*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) + 16*I*a^2*e^(7*I*d*x + 5*I*c) + 80*I*a^2*e^(5*I*d*x + 3*I*c) + 112*I*
a^2*e^(3*I*d*x + I*c) + 48*I*a^2*e^(I*d*x - I*c))/(d*e^(4*I*d*x + 2*I*c) + 2*d*e^(2*I*d*x) + d*e^(-2*I*c))