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

Optimal. Leaf size=25 \[ -\frac {3}{64} \cos ^{\frac {8}{3}}(2 x)-\frac {3}{40} \cos ^{\frac {5}{3}}(2 x) \]

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4357, 266, 43} \[ -\frac {3}{64} \cos ^{\frac {8}{3}}(2 x)-\frac {3}{40} \cos ^{\frac {5}{3}}(2 x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Cos[2*x]^(5/3))/40 - (3*Cos[2*x]^(8/3))/64

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 4357

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{d = FreeFactors[Cos[c*(a + b*x)], x]}, -Dist[d/(
b*c), Subst[Int[SubstFor[1, Cos[c*(a + b*x)]/d, u, x], x], x, Cos[c*(a + b*x)]/d], x] /; FunctionOfQ[Cos[c*(a
+ b*x)]/d, u, x]] /; FreeQ[{a, b, c}, x] && (EqQ[F, Sin] || EqQ[F, sin])

Rubi steps

\begin {align*} \int \cos ^3(x) \cos ^{\frac {2}{3}}(2 x) \sin (x) \, dx &=-\operatorname {Subst}\left (\int x^3 \left (-1+2 x^2\right )^{2/3} \, dx,x,\cos (x)\right )\\ &=-\left (\frac {1}{2} \operatorname {Subst}\left (\int x (-1+2 x)^{2/3} \, dx,x,\cos ^2(x)\right )\right )\\ &=-\left (\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {1}{2} (-1+2 x)^{2/3}+\frac {1}{2} (-1+2 x)^{5/3}\right ) \, dx,x,\cos ^2(x)\right )\right )\\ &=-\frac {3}{40} \left (-1+2 \cos ^2(x)\right )^{5/3}-\frac {3}{64} \left (-1+2 \cos ^2(x)\right )^{8/3}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.32, size = 140, normalized size = 5.60 \[ -\frac {3}{40} \cos ^{\frac {5}{3}}(2 x)-\frac {3 e^{-6 i x} \sqrt [3]{1+e^{4 i x}} \left (2 e^{4 i x} \, _2F_1\left (-\frac {1}{3},\frac {1}{3};\frac {2}{3};-e^{4 i x}\right )+e^{8 i x} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {5}{3};-e^{4 i x}\right )+\left (1+e^{4 i x}\right )^{2/3} \left (1+e^{8 i x}\right )\right )}{256\ 2^{2/3} \sqrt [3]{e^{-2 i x}+e^{2 i x}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Cos[2*x]^(5/3))/40 - (3*(1 + E^((4*I)*x))^(1/3)*((1 + E^((4*I)*x))^(2/3)*(1 + E^((8*I)*x)) + 2*E^((4*I)*x)
*Hypergeometric2F1[-1/3, 1/3, 2/3, -E^((4*I)*x)] + E^((8*I)*x)*Hypergeometric2F1[1/3, 2/3, 5/3, -E^((4*I)*x)])
)/(256*2^(2/3)*E^((6*I)*x)*(E^((-2*I)*x) + E^((2*I)*x))^(1/3))

________________________________________________________________________________________

IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \cos ^3(x) \cos ^{\frac {2}{3}}(2 x) \sin (x) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

Could not integrate

________________________________________________________________________________________

fricas [A]  time = 1.71, size = 26, normalized size = 1.04 \[ -\frac {3}{320} \, {\left (20 \, \cos \relax (x)^{4} - 4 \, \cos \relax (x)^{2} - 3\right )} {\left (2 \, \cos \relax (x)^{2} - 1\right )}^{\frac {2}{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^4*cos(2*x)^(2/3)*tan(x),x, algorithm="fricas")

[Out]

-3/320*(20*cos(x)^4 - 4*cos(x)^2 - 3)*(2*cos(x)^2 - 1)^(2/3)

________________________________________________________________________________________

giac [A]  time = 0.64, size = 25, normalized size = 1.00 \[ -\frac {3}{64} \, {\left (2 \, \cos \relax (x)^{2} - 1\right )}^{\frac {8}{3}} - \frac {3}{40} \, {\left (2 \, \cos \relax (x)^{2} - 1\right )}^{\frac {5}{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^4*cos(2*x)^(2/3)*tan(x),x, algorithm="giac")

[Out]

-3/64*(2*cos(x)^2 - 1)^(8/3) - 3/40*(2*cos(x)^2 - 1)^(5/3)

________________________________________________________________________________________

maple [F]  time = 0.26, size = 0, normalized size = 0.00 \[\int \left (\cos ^{4}\relax (x )\right ) \left (\cos ^{\frac {2}{3}}\left (2 x \right )\right ) \tan \relax (x )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^4*cos(2*x)^(2/3)*tan(x),x)

[Out]

int(cos(x)^4*cos(2*x)^(2/3)*tan(x),x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \cos \left (2 \, x\right )^{\frac {2}{3}} \cos \relax (x)^{4} \tan \relax (x)\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^4*cos(2*x)^(2/3)*tan(x),x, algorithm="maxima")

[Out]

integrate(cos(2*x)^(2/3)*cos(x)^4*tan(x), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \[ \int {\cos \left (2\,x\right )}^{2/3}\,{\cos \relax (x)}^4\,\mathrm {tan}\relax (x) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(2*x)^(2/3)*cos(x)^4*tan(x),x)

[Out]

int(cos(2*x)^(2/3)*cos(x)^4*tan(x), x)

________________________________________________________________________________________

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(cos(x)**4*cos(2*x)**(2/3)*tan(x),x)

[Out]

Timed out

________________________________________________________________________________________