\(\int \cos ^3(x) \sin ^{\frac {3}{2}}(x) \, dx\) [254]

Optimal result

Integrand size = 11, antiderivative size = 21 \[ \int \cos ^3(x) \sin ^{\frac {3}{2}}(x) \, dx=\frac {2}{5} \sin ^{\frac {5}{2}}(x)-\frac {2}{9} \sin ^{\frac {9}{2}}(x) \]

[Out]

2/5*sin(x)^(5/2)-2/9*sin(x)^(9/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2644, 14} \[ \int \cos ^3(x) \sin ^{\frac {3}{2}}(x) \, dx=\frac {2}{5} \sin ^{\frac {5}{2}}(x)-\frac {2}{9} \sin ^{\frac {9}{2}}(x) \]

[In]

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

[Out]

(2*Sin[x]^(5/2))/5 - (2*Sin[x]^(9/2))/9

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]]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \cos ^3(x) \sin ^{\frac {3}{2}}(x) \, dx=\frac {1}{45} (13+5 \cos (2 x)) \sin ^{\frac {5}{2}}(x) \]

[In]

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

[Out]

((13 + 5*Cos[2*x])*Sin[x]^(5/2))/45

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67

method	result	size
derivativedivides	\(\frac {2 \left (\sin ^{\frac {5}{2}}\left (x \right )\right )}{5}-\frac {2 \left (\sin ^{\frac {9}{2}}\left (x \right )\right )}{9}\)	\(14\)
default	\(\frac {2 \left (\sin ^{\frac {5}{2}}\left (x \right )\right )}{5}-\frac {2 \left (\sin ^{\frac {9}{2}}\left (x \right )\right )}{9}\)	\(14\)

[In]

int(cos(x)^3*sin(x)^(3/2),x,method=_RETURNVERBOSE)

[Out]

2/5*sin(x)^(5/2)-2/9*sin(x)^(9/2)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \cos ^3(x) \sin ^{\frac {3}{2}}(x) \, dx=-\frac {2}{45} \, {\left (5 \, \cos \left (x\right )^{4} - \cos \left (x\right )^{2} - 4\right )} \sqrt {\sin \left (x\right )} \]

[In]

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

[Out]

-2/45*(5*cos(x)^4 - cos(x)^2 - 4)*sqrt(sin(x))

Sympy [A] (verification not implemented)

Time = 3.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \cos ^3(x) \sin ^{\frac {3}{2}}(x) \, dx=\frac {8 \sin ^{\frac {9}{2}}{\left (x \right )}}{45} + \frac {2 \sin ^{\frac {5}{2}}{\left (x \right )} \cos ^{2}{\left (x \right )}}{5} \]

[In]

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

[Out]

8*sin(x)**(9/2)/45 + 2*sin(x)**(5/2)*cos(x)**2/5

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \cos ^3(x) \sin ^{\frac {3}{2}}(x) \, dx=-\frac {2}{9} \, \sin \left (x\right )^{\frac {9}{2}} + \frac {2}{5} \, \sin \left (x\right )^{\frac {5}{2}} \]

[In]

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

[Out]

-2/9*sin(x)^(9/2) + 2/5*sin(x)^(5/2)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \cos ^3(x) \sin ^{\frac {3}{2}}(x) \, dx=-\frac {2}{9} \, \sin \left (x\right )^{\frac {9}{2}} + \frac {2}{5} \, \sin \left (x\right )^{\frac {5}{2}} \]

[In]

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

[Out]

-2/9*sin(x)^(9/2) + 2/5*sin(x)^(5/2)

Mupad [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19 \[ \int \cos ^3(x) \sin ^{\frac {3}{2}}(x) \, dx=-\frac {{\cos \left (x\right )}^4\,{\sin \left (x\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},2;\ 3;\ {\cos \left (x\right )}^2\right )}{4\,{\left ({\sin \left (x\right )}^2\right )}^{5/4}} \]

[In]

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

[Out]

-(cos(x)^4*sin(x)^(5/2)*hypergeom([-1/4, 2], 3, cos(x)^2))/(4*(sin(x)^2)^(5/4))