3.74 \(\int \sqrt {\cos (x)} \sin ^3(x) \, dx\)

Optimal. Leaf size=21 \[ \frac {2}{7} \cos ^{\frac {7}{2}}(x)-\frac {2}{3} \cos ^{\frac {3}{2}}(x) \]

[Out]

-2/3*cos(x)^(3/2)+2/7*cos(x)^(7/2)

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Rubi [A]  time = 0.02, antiderivative size = 21, normalized size of antiderivative = 1.00, 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 = {2565, 14} \[ \frac {2}{7} \cos ^{\frac {7}{2}}(x)-\frac {2}{3} \cos ^{\frac {3}{2}}(x) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Cos[x]]*Sin[x]^3,x]

[Out]

(-2*Cos[x]^(3/2))/3 + (2*Cos[x]^(7/2))/7

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

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Mathematica [A]  time = 0.06, size = 34, normalized size = 1.62 \[ \frac {(3 \cos (2 x)-11) \cos ^2(x)+8 \sqrt [4]{\cos ^2(x)}}{21 \sqrt {\cos (x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Cos[x]]*Sin[x]^3,x]

[Out]

(8*(Cos[x]^2)^(1/4) + Cos[x]^2*(-11 + 3*Cos[2*x]))/(21*Sqrt[Cos[x]])

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fricas [A]  time = 0.43, size = 17, normalized size = 0.81 \[ \frac {2}{21} \, {\left (3 \, \cos \relax (x)^{3} - 7 \, \cos \relax (x)\right )} \sqrt {\cos \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^3*cos(x)^(1/2),x, algorithm="fricas")

[Out]

2/21*(3*cos(x)^3 - 7*cos(x))*sqrt(cos(x))

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giac [A]  time = 0.90, size = 13, normalized size = 0.62 \[ \frac {2}{7} \, \cos \relax (x)^{\frac {7}{2}} - \frac {2}{3} \, \cos \relax (x)^{\frac {3}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^3*cos(x)^(1/2),x, algorithm="giac")

[Out]

2/7*cos(x)^(7/2) - 2/3*cos(x)^(3/2)

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maple [B]  time = 0.06, size = 39, normalized size = 1.86 \[ -\frac {8 \sqrt {-2 \left (\sin ^{2}\left (\frac {x}{2}\right )\right )+1}\, \left (6 \left (\sin ^{6}\left (\frac {x}{2}\right )\right )-9 \left (\sin ^{4}\left (\frac {x}{2}\right )\right )+\sin ^{2}\left (\frac {x}{2}\right )+1\right )}{21} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^3*cos(x)^(1/2),x)

[Out]

-8/21*(-2*sin(1/2*x)^2+1)^(1/2)*(6*sin(1/2*x)^6-9*sin(1/2*x)^4+sin(1/2*x)^2+1)

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maxima [A]  time = 0.44, size = 13, normalized size = 0.62 \[ \frac {2}{7} \, \cos \relax (x)^{\frac {7}{2}} - \frac {2}{3} \, \cos \relax (x)^{\frac {3}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^3*cos(x)^(1/2),x, algorithm="maxima")

[Out]

2/7*cos(x)^(7/2) - 2/3*cos(x)^(3/2)

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mupad [B]  time = 0.09, size = 13, normalized size = 0.62 \[ {\cos \relax (x)}^{3/2}\,\left (\frac {2\,{\cos \relax (x)}^2}{7}-\frac {2}{3}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^(1/2)*sin(x)^3,x)

[Out]

cos(x)^(3/2)*((2*cos(x)^2)/7 - 2/3)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)**3*cos(x)**(1/2),x)

[Out]

Timed out

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