3.187 \(\int (d \cos (a+b x))^{3/2} \sin (a+b x) \, dx\)

Optimal. Leaf size=22 \[ -\frac{2 (d \cos (a+b x))^{5/2}}{5 b d} \]

[Out]

(-2*(d*Cos[a + b*x])^(5/2))/(5*b*d)

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Rubi [A]  time = 0.0252584, antiderivative size = 22, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2565, 30} \[ -\frac{2 (d \cos (a+b x))^{5/2}}{5 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(d*Cos[a + b*x])^(5/2))/(5*b*d)

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int (d \cos (a+b x))^{3/2} \sin (a+b x) \, dx &=-\frac{\operatorname{Subst}\left (\int x^{3/2} \, dx,x,d \cos (a+b x)\right )}{b d}\\ &=-\frac{2 (d \cos (a+b x))^{5/2}}{5 b d}\\ \end{align*}

Mathematica [A]  time = 0.0325464, size = 22, normalized size = 1. \[ -\frac{2 (d \cos (a+b x))^{5/2}}{5 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(d*Cos[a + b*x])^(5/2))/(5*b*d)

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Maple [A]  time = 0.007, size = 19, normalized size = 0.9 \begin{align*} -{\frac{2}{5\,bd} \left ( d\cos \left ( bx+a \right ) \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*(d*cos(b*x+a))^(5/2)/b/d

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Maxima [A]  time = 0.974904, size = 24, normalized size = 1.09 \begin{align*} -\frac{2 \, \left (d \cos \left (b x + a\right )\right )^{\frac{5}{2}}}{5 \, b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*(d*cos(b*x + a))^(5/2)/(b*d)

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Fricas [A]  time = 1.95647, size = 62, normalized size = 2.82 \begin{align*} -\frac{2 \, \sqrt{d \cos \left (b x + a\right )} d \cos \left (b x + a\right )^{2}}{5 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*sqrt(d*cos(b*x + a))*d*cos(b*x + a)^2/b

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Sympy [A]  time = 63.4996, size = 34, normalized size = 1.55 \begin{align*} \begin{cases} - \frac{2 d^{\frac{3}{2}} \cos ^{\frac{5}{2}}{\left (a + b x \right )}}{5 b} & \text{for}\: b \neq 0 \\x \left (d \cos{\left (a \right )}\right )^{\frac{3}{2}} \sin{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-2*d**(3/2)*cos(a + b*x)**(5/2)/(5*b), Ne(b, 0)), (x*(d*cos(a))**(3/2)*sin(a), True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \cos \left (b x + a\right )\right )^{\frac{3}{2}} \sin \left (b x + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*cos(b*x + a))^(3/2)*sin(b*x + a), x)