3.255 \(\int \cos ^3(x) \sin ^{\frac{5}{2}}(x) \, dx\)

Optimal. Leaf size=21 \[ \frac{2}{7} \sin ^{\frac{7}{2}}(x)-\frac{2}{11} \sin ^{\frac{11}{2}}(x) \]

[Out]

(2*Sin[x]^(7/2))/7 - (2*Sin[x]^(11/2))/11

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Rubi [A]  time = 0.0249217, antiderivative size = 21, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {2564, 14} \[ \frac{2}{7} \sin ^{\frac{7}{2}}(x)-\frac{2}{11} \sin ^{\frac{11}{2}}(x) \]

Antiderivative was successfully verified.

[In]

Int[Cos[x]^3*Sin[x]^(5/2),x]

[Out]

(2*Sin[x]^(7/2))/7 - (2*Sin[x]^(11/2))/11

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])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int \cos ^3(x) \sin ^{\frac{5}{2}}(x) \, dx &=\operatorname{Subst}\left (\int x^{5/2} \left (1-x^2\right ) \, dx,x,\sin (x)\right )\\ &=\operatorname{Subst}\left (\int \left (x^{5/2}-x^{9/2}\right ) \, dx,x,\sin (x)\right )\\ &=\frac{2}{7} \sin ^{\frac{7}{2}}(x)-\frac{2}{11} \sin ^{\frac{11}{2}}(x)\\ \end{align*}

Mathematica [A]  time = 0.0115088, size = 18, normalized size = 0.86 \[ \frac{1}{77} \sin ^{\frac{7}{2}}(x) (7 \cos (2 x)+15) \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]^3*Sin[x]^(5/2),x]

[Out]

((15 + 7*Cos[2*x])*Sin[x]^(7/2))/77

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Maple [A]  time = 0.036, size = 14, normalized size = 0.7 \begin{align*}{\frac{2}{7} \left ( \sin \left ( x \right ) \right ) ^{{\frac{7}{2}}}}-{\frac{2}{11} \left ( \sin \left ( x \right ) \right ) ^{{\frac{11}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*sin(x)^(7/2)-2/11*sin(x)^(11/2)

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Maxima [A]  time = 0.963436, size = 18, normalized size = 0.86 \begin{align*} -\frac{2}{11} \, \sin \left (x\right )^{\frac{11}{2}} + \frac{2}{7} \, \sin \left (x\right )^{\frac{7}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/11*sin(x)^(11/2) + 2/7*sin(x)^(7/2)

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Fricas [A]  time = 2.34417, size = 68, normalized size = 3.24 \begin{align*} -\frac{2}{77} \,{\left (7 \, \cos \left (x\right )^{4} - 3 \, \cos \left (x\right )^{2} - 4\right )} \sin \left (x\right )^{\frac{3}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/77*(7*cos(x)^4 - 3*cos(x)^2 - 4)*sin(x)^(3/2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)**3*sin(x)**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.09783, size = 18, normalized size = 0.86 \begin{align*} -\frac{2}{11} \, \sin \left (x\right )^{\frac{11}{2}} + \frac{2}{7} \, \sin \left (x\right )^{\frac{7}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/11*sin(x)^(11/2) + 2/7*sin(x)^(7/2)