3.3.15 \(\int a \cos (5+3 x) \sin ^2(5+3 x) \, dx\) [215]

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

[Out]

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

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Rubi [A]
time = 0.01, 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, 2644, 30} \begin {gather*} \frac {1}{9} a \sin ^3(3 x+5) \end {gather*}

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 2644

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 \text {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 = 0.00, size = 13, normalized size = 1.00 \begin {gather*} \frac {1}{9} a \sin ^3(5+3 x) \end {gather*}

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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Maple [A]
time = 0.04, size = 12, normalized size = 0.92

method	result	size
derivativedivides	\(\frac {a \left (\sin ^{3}\left (5+3 x \right )\right )}{9}\)	\(12\)
default	\(\frac {a \left (\sin ^{3}\left (5+3 x \right )\right )}{9}\)	\(12\)
risch	\(\frac {a \sin \left (5+3 x \right )}{12}-\frac {a \sin \left (15+9 x \right )}{36}\)	\(20\)
norman	\(\frac {8 a \left (\tan ^{3}\left (\frac {5}{2}+\frac {3 x}{2}\right )\right )}{9 \left (1+\tan ^{2}\left (\frac {5}{2}+\frac {3 x}{2}\right )\right )^{3}}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 3.12, size = 11, normalized size = 0.85 \begin {gather*} \frac {1}{9} \, a \sin \left (3 \, x + 5\right )^{3} \end {gather*}

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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Fricas [A]
time = 0.62, size = 22, normalized size = 1.69 \begin {gather*} -\frac {1}{9} \, {\left (a \cos \left (3 \, x + 5\right )^{2} - a\right )} \sin \left (3 \, x + 5\right ) \end {gather*}

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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Sympy [A]
time = 0.08, size = 10, normalized size = 0.77 \begin {gather*} \frac {a \sin ^{3}{\left (3 x + 5 \right )}}{9} \end {gather*}

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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Giac [A]
time = 0.45, size = 11, normalized size = 0.85 \begin {gather*} \frac {1}{9} \, a \sin \left (3 \, x + 5\right )^{3} \end {gather*}

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

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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