3.450 \(\int \frac {\cos (x) \sin ^8(x)}{(2-5 \sin ^3(x))^{4/3}} \, dx\)

Optimal. Leaf size=49 \[ -\frac {1}{625} \left (2-5 \sin ^3(x)\right )^{5/3}+\frac {2}{125} \left (2-5 \sin ^3(x)\right )^{2/3}+\frac {4}{125 \sqrt [3]{2-5 \sin ^3(x)}} \]

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Rubi [A]  time = 0.11, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {4334, 266, 43} \[ -\frac {1}{625} \left (2-5 \sin ^3(x)\right )^{5/3}+\frac {2}{125} \left (2-5 \sin ^3(x)\right )^{2/3}+\frac {4}{125 \sqrt [3]{2-5 \sin ^3(x)}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[x]*Sin[x]^8)/(2 - 5*Sin[x]^3)^(4/3),x]

[Out]

4/(125*(2 - 5*Sin[x]^3)^(1/3)) + (2*(2 - 5*Sin[x]^3)^(2/3))/125 - (2 - 5*Sin[x]^3)^(5/3)/625

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 4334

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

Rubi steps

\begin {align*} \int \frac {\cos (x) \sin ^8(x)}{\left (2-5 \sin ^3(x)\right )^{4/3}} \, dx &=\operatorname {Subst}\left (\int \frac {x^8}{\left (2-5 x^3\right )^{4/3}} \, dx,x,\sin (x)\right )\\ &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x^2}{(2-5 x)^{4/3}} \, dx,x,\sin ^3(x)\right )\\ &=\frac {1}{3} \operatorname {Subst}\left (\int \left (\frac {4}{25 (2-5 x)^{4/3}}-\frac {4}{25 \sqrt [3]{2-5 x}}+\frac {1}{25} (2-5 x)^{2/3}\right ) \, dx,x,\sin ^3(x)\right )\\ &=\frac {4}{125 \sqrt [3]{2-5 \sin ^3(x)}}+\frac {2}{125} \left (2-5 \sin ^3(x)\right )^{2/3}-\frac {1}{625} \left (2-5 \sin ^3(x)\right )^{5/3}\\ \end {align*}

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Mathematica [A]  time = 0.50, size = 30, normalized size = 0.61 \[ \frac {-25 \sin ^6(x)-30 \sin ^3(x)+36}{625 \sqrt [3]{2-5 \sin ^3(x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[x]*Sin[x]^8)/(2 - 5*Sin[x]^3)^(4/3),x]

[Out]

(36 - 30*Sin[x]^3 - 25*Sin[x]^6)/(625*(2 - 5*Sin[x]^3)^(1/3))

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos (x) \sin ^8(x)}{\left (2-5 \sin ^3(x)\right )^{4/3}} \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[(Cos[x]*Sin[x]^8)/(2 - 5*Sin[x]^3)^(4/3),x]

[Out]

Could not integrate

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fricas [A]  time = 1.27, size = 46, normalized size = 0.94 \[ \frac {25 \, \cos \relax (x)^{6} - 75 \, \cos \relax (x)^{4} + 75 \, \cos \relax (x)^{2} + 30 \, {\left (\cos \relax (x)^{2} - 1\right )} \sin \relax (x) + 11}{625 \, {\left (5 \, {\left (\cos \relax (x)^{2} - 1\right )} \sin \relax (x) + 2\right )}^{\frac {1}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)*sin(x)^9/(2-5*sin(x)^3)^(4/3),x, algorithm="fricas")

[Out]

1/625*(25*cos(x)^6 - 75*cos(x)^4 + 75*cos(x)^2 + 30*(cos(x)^2 - 1)*sin(x) + 11)/(5*(cos(x)^2 - 1)*sin(x) + 2)^
(1/3)

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giac [A]  time = 0.91, size = 37, normalized size = 0.76 \[ -\frac {1}{625} \, {\left (-5 \, \sin \relax (x)^{3} + 2\right )}^{\frac {5}{3}} + \frac {2}{125} \, {\left (-5 \, \sin \relax (x)^{3} + 2\right )}^{\frac {2}{3}} + \frac {4}{125 \, {\left (-5 \, \sin \relax (x)^{3} + 2\right )}^{\frac {1}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)*sin(x)^9/(2-5*sin(x)^3)^(4/3),x, algorithm="giac")

[Out]

-1/625*(-5*sin(x)^3 + 2)^(5/3) + 2/125*(-5*sin(x)^3 + 2)^(2/3) + 4/125/(-5*sin(x)^3 + 2)^(1/3)

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maple [F]  time = 0.96, size = 0, normalized size = 0.00 \[\int \frac {\cot \relax (x ) \left (\sin ^{9}\relax (x )\right )}{\left (2-5 \left (\sin ^{3}\relax (x )\right )\right )^{\frac {4}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(x)*sin(x)^9/(2-5*sin(x)^3)^(4/3),x)

[Out]

int(cot(x)*sin(x)^9/(2-5*sin(x)^3)^(4/3),x)

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maxima [A]  time = 0.43, size = 37, normalized size = 0.76 \[ -\frac {1}{625} \, {\left (-5 \, \sin \relax (x)^{3} + 2\right )}^{\frac {5}{3}} + \frac {2}{125} \, {\left (-5 \, \sin \relax (x)^{3} + 2\right )}^{\frac {2}{3}} + \frac {4}{125 \, {\left (-5 \, \sin \relax (x)^{3} + 2\right )}^{\frac {1}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)*sin(x)^9/(2-5*sin(x)^3)^(4/3),x, algorithm="maxima")

[Out]

-1/625*(-5*sin(x)^3 + 2)^(5/3) + 2/125*(-5*sin(x)^3 + 2)^(2/3) + 4/125/(-5*sin(x)^3 + 2)^(1/3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\mathrm {cot}\relax (x)\,{\sin \relax (x)}^9}{{\left (2-5\,{\sin \relax (x)}^3\right )}^{4/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cot(x)*sin(x)^9)/(2 - 5*sin(x)^3)^(4/3),x)

[Out]

int((cot(x)*sin(x)^9)/(2 - 5*sin(x)^3)^(4/3), x)

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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(cot(x)*sin(x)**9/(2-5*sin(x)**3)**(4/3),x)

[Out]

Timed out

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