3.215 \(\int a \cos (5+3 x) \sin ^2(5+3 x) \, dx\)

Optimal. Leaf size=13 \[ \frac {1}{9} a \sin ^3(3 x+5) \]

[Out]

1/9*a*sin(5+3*x)^3

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Rubi [A]  time = 0.02, antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {12, 2564, 30} \[ \frac {1}{9} a \sin ^3(3 x+5) \]

Antiderivative was successfully verified.

[In]

Int[a*Cos[5 + 3*x]*Sin[5 + 3*x]^2,x]

[Out]

(a*Sin[5 + 3*x]^3)/9

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 30

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

Rule 2564

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

Rubi steps

\begin {align*} \int a \cos (5+3 x) \sin ^2(5+3 x) \, dx &=a \int \cos (5+3 x) \sin ^2(5+3 x) \, dx\\ &=\frac {1}{3} a \operatorname {Subst}\left (\int x^2 \, dx,x,\sin (5+3 x)\right )\\ &=\frac {1}{9} a \sin ^3(5+3 x)\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 13, normalized size = 1.00 \[ \frac {1}{9} a \sin ^3(3 x+5) \]

Antiderivative was successfully verified.

[In]

Integrate[a*Cos[5 + 3*x]*Sin[5 + 3*x]^2,x]

[Out]

(a*Sin[5 + 3*x]^3)/9

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fricas [A]  time = 0.42, size = 22, normalized size = 1.69 \[ -\frac {1}{9} \, {\left (a \cos \left (3 \, x + 5\right )^{2} - a\right )} \sin \left (3 \, x + 5\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(a*cos(5+3*x)*sin(5+3*x)^2,x, algorithm="fricas")

[Out]

-1/9*(a*cos(3*x + 5)^2 - a)*sin(3*x + 5)

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giac [A]  time = 1.04, size = 11, normalized size = 0.85 \[ \frac {1}{9} \, a \sin \left (3 \, x + 5\right )^{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(a*cos(5+3*x)*sin(5+3*x)^2,x, algorithm="giac")

[Out]

1/9*a*sin(3*x + 5)^3

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maple [A]  time = 0.01, size = 12, normalized size = 0.92 \[ \frac {a \left (\sin ^{3}\left (3 x +5\right )\right )}{9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(a*cos(5+3*x)*sin(5+3*x)^2,x)

[Out]

1/9*a*sin(5+3*x)^3

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maxima [A]  time = 0.42, size = 11, normalized size = 0.85 \[ \frac {1}{9} \, a \sin \left (3 \, x + 5\right )^{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(a*cos(5+3*x)*sin(5+3*x)^2,x, algorithm="maxima")

[Out]

1/9*a*sin(3*x + 5)^3

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mupad [B]  time = 0.09, size = 11, normalized size = 0.85 \[ \frac {a\,{\sin \left (3\,x+5\right )}^3}{9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(a*cos(3*x + 5)*sin(3*x + 5)^2,x)

[Out]

(a*sin(3*x + 5)^3)/9

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sympy [A]  time = 0.32, size = 10, normalized size = 0.77 \[ \frac {a \sin ^{3}{\left (3 x + 5 \right )}}{9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(a*cos(5+3*x)*sin(5+3*x)**2,x)

[Out]

a*sin(3*x + 5)**3/9

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