3.258 \(\int (a \cos (c+d x)+i a \sin (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=33 \[ -\frac {2 i (a \cos (c+d x)+i a \sin (c+d x))^{3/2}}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {3071} \[ -\frac {2 i (a \cos (c+d x)+i a \sin (c+d x))^{3/2}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-2*I)/3)*(a*Cos[c + d*x] + I*a*Sin[c + d*x])^(3/2))/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))^{3/2} \, dx &=-\frac {2 i (a \cos (c+d x)+i a \sin (c+d x))^{3/2}}{3 d}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 32, normalized size = 0.97 \[ -\frac {2 i (a (\cos (c+d x)+i \sin (c+d x)))^{3/2}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.08, size = 17, normalized size = 0.52 \[ -\frac {2 i \, a^{\frac {3}{2}} e^{\left (\frac {3}{2} i \, d x + \frac {3}{2} i \, c\right )}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.46, size = 17, normalized size = 0.52 \[ -\frac {2 i \, a^{\frac {3}{2}} e^{\left (\frac {3}{2} i \, d x + \frac {3}{2} i \, c\right )}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.22, size = 28, normalized size = 0.85 \[ -\frac {2 i \left (a \cos \left (d x +c \right )+i a \sin \left (d x +c \right )\right )^{\frac {3}{2}}}{3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.90, size = 51, normalized size = 1.55 \[ -\frac {2 i \, a^{\frac {3}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + i\right )}^{\frac {3}{2}}}{3 \, d {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + i\right )}^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.38, size = 33, normalized size = 1.00 \[ -\frac {a\,{\mathrm {e}}^{c\,1{}\mathrm {i}}\,{\mathrm {e}}^{d\,x\,1{}\mathrm {i}}\,\sqrt {a\,{\mathrm {e}}^{c\,1{}\mathrm {i}}\,{\mathrm {e}}^{d\,x\,1{}\mathrm {i}}}\,2{}\mathrm {i}}{3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*cos(c + d*x) + a*sin(c + d*x)*1i)^(3/2),x)

[Out]

-(a*exp(c*1i)*exp(d*x*1i)*(a*exp(c*1i)*exp(d*x*1i))^(1/2)*2i)/(3*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (i a \sin {\left (c + d x \right )} + a \cos {\left (c + d x \right )}\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((I*a*sin(c + d*x) + a*cos(c + d*x))**(3/2), x)

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