3.250 \(\int (a \cos (c+d x)+i a \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=31 \[ -\frac{i (a \cos (c+d x)+i a \sin (c+d x))^3}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.0162244, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.045, Rules used = {3071} \[ -\frac{i (a \cos (c+d x)+i a \sin (c+d x))^3}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3071

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(a*Cos[c + d*x]
+ b*Sin[c + d*x])^n)/(b*d*n), x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 + b^2, 0]

Rubi steps

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

Mathematica [A]  time = 0.0821072, size = 31, normalized size = 1. \[ -\frac{i (a \cos (c+d x)+i a \sin (c+d x))^3}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.058, size = 76, normalized size = 2.5 \begin{align*}{\frac{1}{d} \left ({\frac{i}{3}}{a}^{3} \left ( 2+ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) -{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{3}-i{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{{a}^{3} \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((a*cos(d*x+c)+I*a*sin(d*x+c))^3,x)

[Out]

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

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Maxima [B]  time = 1.01103, size = 112, normalized size = 3.61 \begin{align*} -\frac{i \, a^{3} \cos \left (d x + c\right )^{3}}{d} - \frac{a^{3} \sin \left (d x + c\right )^{3}}{d} - \frac{i \,{\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} a^{3}}{3 \, d} - \frac{{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{3}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.03724, size = 46, normalized size = 1.48 \begin{align*} -\frac{i \, a^{3} e^{\left (3 i \, d x + 3 i \, c\right )}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*I*a^3*e^(3*I*d*x + 3*I*c)/d

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Sympy [A]  time = 0.166956, size = 37, normalized size = 1.19 \begin{align*} \begin{cases} - \frac{i a^{3} e^{3 i c} e^{3 i d x}}{3 d} & \text{for}\: 3 d \neq 0 \\a^{3} x e^{3 i c} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.10807, size = 70, normalized size = 2.26 \begin{align*} -\frac{i \, a^{3} e^{\left (3 i \, d x + 3 i \, c\right )}}{6 \, d} - \frac{i \, a^{3} e^{\left (-3 i \, d x - 3 i \, c\right )}}{6 \, d} + \frac{a^{3} \sin \left (3 \, d x + 3 \, c\right )}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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