3.420 \(\int \cos (x) (5 \cos ^2(x)+\sin ^2(x))^{5/2} \, dx\)

Optimal. Leaf size=69 \[ \frac {625}{32} \sin ^{-1}\left (\frac {2 \sin (x)}{\sqrt {5}}\right )+\frac {1}{6} \sin (x) \left (5-4 \sin ^2(x)\right )^{5/2}+\frac {25}{24} \sin (x) \left (5-4 \sin ^2(x)\right )^{3/2}+\frac {125}{16} \sin (x) \sqrt {5-4 \sin ^2(x)} \]

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Rubi [A]  time = 0.05, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4356, 195, 216} \[ \frac {1}{6} \sin (x) \left (5-4 \sin ^2(x)\right )^{5/2}+\frac {25}{24} \sin (x) \left (5-4 \sin ^2(x)\right )^{3/2}+\frac {125}{16} \sin (x) \sqrt {5-4 \sin ^2(x)}+\frac {625}{32} \sin ^{-1}\left (\frac {2 \sin (x)}{\sqrt {5}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(625*ArcSin[(2*Sin[x])/Sqrt[5]])/32 + (125*Sin[x]*Sqrt[5 - 4*Sin[x]^2])/16 + (25*Sin[x]*(5 - 4*Sin[x]^2)^(3/2)
)/24 + (Sin[x]*(5 - 4*Sin[x]^2)^(5/2))/6

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 4356

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]] /; FreeQ[{a, b, c}, x] && (EqQ[F, Cos] || EqQ[F, cos])

Rubi steps

\begin {align*} \int \cos (x) \left (5 \cos ^2(x)+\sin ^2(x)\right )^{5/2} \, dx &=\operatorname {Subst}\left (\int \left (5-4 x^2\right )^{5/2} \, dx,x,\sin (x)\right )\\ &=\frac {1}{6} \sin (x) \left (5-4 \sin ^2(x)\right )^{5/2}+\frac {25}{6} \operatorname {Subst}\left (\int \left (5-4 x^2\right )^{3/2} \, dx,x,\sin (x)\right )\\ &=\frac {25}{24} \sin (x) \left (5-4 \sin ^2(x)\right )^{3/2}+\frac {1}{6} \sin (x) \left (5-4 \sin ^2(x)\right )^{5/2}+\frac {125}{8} \operatorname {Subst}\left (\int \sqrt {5-4 x^2} \, dx,x,\sin (x)\right )\\ &=\frac {125}{16} \sin (x) \sqrt {5-4 \sin ^2(x)}+\frac {25}{24} \sin (x) \left (5-4 \sin ^2(x)\right )^{3/2}+\frac {1}{6} \sin (x) \left (5-4 \sin ^2(x)\right )^{5/2}+\frac {625}{16} \operatorname {Subst}\left (\int \frac {1}{\sqrt {5-4 x^2}} \, dx,x,\sin (x)\right )\\ &=\frac {625}{32} \sin ^{-1}\left (\frac {2 \sin (x)}{\sqrt {5}}\right )+\frac {125}{16} \sin (x) \sqrt {5-4 \sin ^2(x)}+\frac {25}{24} \sin (x) \left (5-4 \sin ^2(x)\right )^{3/2}+\frac {1}{6} \sin (x) \left (5-4 \sin ^2(x)\right )^{5/2}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 48, normalized size = 0.70 \[ \frac {1}{96} \left (1875 \sin ^{-1}\left (\frac {2 \sin (x)}{\sqrt {5}}\right )+2 (515 \sin (x)+90 \sin (3 x)+8 \sin (5 x)) \sqrt {2 \cos (2 x)+3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1875*ArcSin[(2*Sin[x])/Sqrt[5]] + 2*Sqrt[3 + 2*Cos[2*x]]*(515*Sin[x] + 90*Sin[3*x] + 8*Sin[5*x]))/96

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

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[Cos[x]*(5*Cos[x]^2 + Sin[x]^2)^(5/2),x]

[Out]

Could not integrate

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fricas [A]  time = 1.49, size = 88, normalized size = 1.28 \[ \frac {1}{48} \, {\left (128 \, \cos \relax (x)^{4} + 264 \, \cos \relax (x)^{2} + 433\right )} \sqrt {4 \, \cos \relax (x)^{2} + 1} \sin \relax (x) + \frac {625}{64} \, \arctan \left (\frac {4 \, {\left (8 \, \cos \relax (x)^{2} - 3\right )} \sqrt {4 \, \cos \relax (x)^{2} + 1} \sin \relax (x) - 25 \, \cos \relax (x) \sin \relax (x)}{64 \, \cos \relax (x)^{4} - 23 \, \cos \relax (x)^{2} - 16}\right ) + \frac {625}{64} \, \arctan \left (\frac {\sin \relax (x)}{\cos \relax (x)}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(128*cos(x)^4 + 264*cos(x)^2 + 433)*sqrt(4*cos(x)^2 + 1)*sin(x) + 625/64*arctan((4*(8*cos(x)^2 - 3)*sqrt(
4*cos(x)^2 + 1)*sin(x) - 25*cos(x)*sin(x))/(64*cos(x)^4 - 23*cos(x)^2 - 16)) + 625/64*arctan(sin(x)/cos(x))

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giac [A]  time = 0.68, size = 41, normalized size = 0.59 \[ \frac {1}{48} \, {\left (8 \, {\left (16 \, \sin \relax (x)^{2} - 65\right )} \sin \relax (x)^{2} + 825\right )} \sqrt {-4 \, \sin \relax (x)^{2} + 5} \sin \relax (x) + \frac {625}{32} \, \arcsin \left (\frac {2}{5} \, \sqrt {5} \sin \relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(8*(16*sin(x)^2 - 65)*sin(x)^2 + 825)*sqrt(-4*sin(x)^2 + 5)*sin(x) + 625/32*arcsin(2/5*sqrt(5)*sin(x))

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maple [A]  time = 0.23, size = 103, normalized size = 1.49

method	result	size
default	\(\frac {\sqrt {\left (4 \left (\cos ^{2}\relax (x )\right )+1\right ) \left (\sin ^{2}\relax (x )\right )}\, \left (512 \sqrt {-4 \left (\sin ^{4}\relax (x )\right )+5 \left (\sin ^{2}\relax (x )\right )}\, \left (\sin ^{4}\relax (x )\right )-2080 \sqrt {-4 \left (\sin ^{4}\relax (x )\right )+5 \left (\sin ^{2}\relax (x )\right )}\, \left (\sin ^{2}\relax (x )\right )+3300 \sqrt {-4 \left (\sin ^{4}\relax (x )\right )+5 \left (\sin ^{2}\relax (x )\right )}+1875 \arcsin \left (-1+\frac {8 \left (\sin ^{2}\relax (x )\right )}{5}\right )\right )}{192 \sin \relax (x ) \sqrt {4 \left (\cos ^{2}\relax (x )\right )+1}}\)	\(103\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)*(5*cos(x)^2+sin(x)^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/192*((4*cos(x)^2+1)*sin(x)^2)^(1/2)*(512*(-4*sin(x)^4+5*sin(x)^2)^(1/2)*sin(x)^4-2080*(-4*sin(x)^4+5*sin(x)^
2)^(1/2)*sin(x)^2+3300*(-4*sin(x)^4+5*sin(x)^2)^(1/2)+1875*arcsin(-1+8/5*sin(x)^2))/sin(x)/(4*cos(x)^2+1)^(1/2
)

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maxima [A]  time = 0.98, size = 53, normalized size = 0.77 \[ \frac {1}{6} \, {\left (-4 \, \sin \relax (x)^{2} + 5\right )}^{\frac {5}{2}} \sin \relax (x) + \frac {25}{24} \, {\left (-4 \, \sin \relax (x)^{2} + 5\right )}^{\frac {3}{2}} \sin \relax (x) + \frac {125}{16} \, \sqrt {-4 \, \sin \relax (x)^{2} + 5} \sin \relax (x) + \frac {625}{32} \, \arcsin \left (\frac {2}{5} \, \sqrt {5} \sin \relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(-4*sin(x)^2 + 5)^(5/2)*sin(x) + 25/24*(-4*sin(x)^2 + 5)^(3/2)*sin(x) + 125/16*sqrt(-4*sin(x)^2 + 5)*sin(x
) + 625/32*arcsin(2/5*sqrt(5)*sin(x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \cos \relax (x)\,{\left (5\,{\cos \relax (x)}^2+{\sin \relax (x)}^2\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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