3.188 \(\int \sqrt {d \cos (a+b x)} \sin (a+b x) \, dx\)

Optimal. Leaf size=22 \[ -\frac {2 (d \cos (a+b x))^{3/2}}{3 b d} \]

[Out]

-2/3*(d*cos(b*x+a))^(3/2)/b/d

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Rubi [A]  time = 0.02, antiderivative size = 22, normalized size of antiderivative = 1.00, 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} \[ -\frac {2 (d \cos (a+b x))^{3/2}}{3 b d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[d*Cos[a + b*x]]*Sin[a + b*x],x]

[Out]

(-2*(d*Cos[a + b*x])^(3/2))/(3*b*d)

Rule 30

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

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])

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 22, normalized size = 1.00 \[ -\frac {2 (d \cos (a+b x))^{3/2}}{3 b d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[d*Cos[a + b*x]]*Sin[a + b*x],x]

[Out]

(-2*(d*Cos[a + b*x])^(3/2))/(3*b*d)

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fricas [A]  time = 0.44, size = 21, normalized size = 0.95 \[ -\frac {2 \, \sqrt {d \cos \left (b x + a\right )} \cos \left (b x + a\right )}{3 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(b*x+a))^(1/2)*sin(b*x+a),x, algorithm="fricas")

[Out]

-2/3*sqrt(d*cos(b*x + a))*cos(b*x + a)/b

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giac [A]  time = 1.34, size = 18, normalized size = 0.82 \[ -\frac {2 \, \left (d \cos \left (b x + a\right )\right )^{\frac {3}{2}}}{3 \, b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(b*x+a))^(1/2)*sin(b*x+a),x, algorithm="giac")

[Out]

-2/3*(d*cos(b*x + a))^(3/2)/(b*d)

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maple [A]  time = 0.01, size = 19, normalized size = 0.86 \[ -\frac {2 \left (d \cos \left (b x +a \right )\right )^{\frac {3}{2}}}{3 b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*cos(b*x+a))^(1/2)*sin(b*x+a),x)

[Out]

-2/3*(d*cos(b*x+a))^(3/2)/b/d

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maxima [A]  time = 0.46, size = 18, normalized size = 0.82 \[ -\frac {2 \, \left (d \cos \left (b x + a\right )\right )^{\frac {3}{2}}}{3 \, b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(b*x+a))^(1/2)*sin(b*x+a),x, algorithm="maxima")

[Out]

-2/3*(d*cos(b*x + a))^(3/2)/(b*d)

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mupad [B]  time = 0.42, size = 18, normalized size = 0.82 \[ -\frac {2\,{\left (d\,\cos \left (a+b\,x\right )\right )}^{3/2}}{3\,b\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)*(d*cos(a + b*x))^(1/2),x)

[Out]

-(2*(d*cos(a + b*x))^(3/2))/(3*b*d)

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sympy [A]  time = 1.67, size = 34, normalized size = 1.55 \[ \begin {cases} - \frac {2 \sqrt {d} \cos ^{\frac {3}{2}}{\left (a + b x \right )}}{3 b} & \text {for}\: b \neq 0 \\x \sqrt {d \cos {\relax (a )}} \sin {\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(b*x+a))**(1/2)*sin(b*x+a),x)

[Out]

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

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