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

Optimal. Leaf size=39 \[ \frac {2 \sin ^{-1}\left (\sqrt {\frac {5}{2}} \sin (x)\right )}{5 \sqrt {5}}+\frac {\sin (x)}{10 \sqrt {2-5 \sin ^2(x)}} \]

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Rubi [A]  time = 0.07, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {4356, 385, 216} \[ \frac {2 \sin ^{-1}\left (\sqrt {\frac {5}{2}} \sin (x)\right )}{5 \sqrt {5}}+\frac {\sin (x)}{10 \sqrt {2-5 \sin ^2(x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*ArcSin[Sqrt[5/2]*Sin[x]])/(5*Sqrt[5]) + Sin[x]/(10*Sqrt[2 - 5*Sin[x]^2])

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 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

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 \frac {\cos (x) \cos (2 x)}{\left (2-5 \sin ^2(x)\right )^{3/2}} \, dx &=\operatorname {Subst}\left (\int \frac {1-2 x^2}{\left (2-5 x^2\right )^{3/2}} \, dx,x,\sin (x)\right )\\ &=\frac {\sin (x)}{10 \sqrt {2-5 \sin ^2(x)}}+\frac {2}{5} \operatorname {Subst}\left (\int \frac {1}{\sqrt {2-5 x^2}} \, dx,x,\sin (x)\right )\\ &=\frac {2 \sin ^{-1}\left (\sqrt {\frac {5}{2}} \sin (x)\right )}{5 \sqrt {5}}+\frac {\sin (x)}{10 \sqrt {2-5 \sin ^2(x)}}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 39, normalized size = 1.00 \[ \frac {1}{50} \left (4 \sqrt {5} \sin ^{-1}\left (\sqrt {\frac {5}{2}} \sin (x)\right )+\frac {5 \sin (x)}{\sqrt {2-5 \sin ^2(x)}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[5]*ArcSin[Sqrt[5/2]*Sin[x]] + (5*Sin[x])/Sqrt[2 - 5*Sin[x]^2])/50

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

Verification is Not applicable to the result.

[In]

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

[Out]

Could not integrate

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fricas [B]  time = 0.81, size = 100, normalized size = 2.56 \[ -\frac {{\left (5 \, \sqrt {5} \cos \relax (x)^{2} - 3 \, \sqrt {5}\right )} \arctan \left (\frac {{\left (50 \, \sqrt {5} \cos \relax (x)^{4} - 80 \, \sqrt {5} \cos \relax (x)^{2} + 31 \, \sqrt {5}\right )} \sqrt {5 \, \cos \relax (x)^{2} - 3}}{10 \, {\left (25 \, \cos \relax (x)^{4} - 35 \, \cos \relax (x)^{2} + 12\right )} \sin \relax (x)}\right ) - 5 \, \sqrt {5 \, \cos \relax (x)^{2} - 3} \sin \relax (x)}{50 \, {\left (5 \, \cos \relax (x)^{2} - 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/50*((5*sqrt(5)*cos(x)^2 - 3*sqrt(5))*arctan(1/10*(50*sqrt(5)*cos(x)^4 - 80*sqrt(5)*cos(x)^2 + 31*sqrt(5))*s
qrt(5*cos(x)^2 - 3)/((25*cos(x)^4 - 35*cos(x)^2 + 12)*sin(x))) - 5*sqrt(5*cos(x)^2 - 3)*sin(x))/(5*cos(x)^2 -
3)

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giac [A]  time = 0.70, size = 38, normalized size = 0.97 \[ \frac {2}{25} \, \sqrt {5} \arcsin \left (\frac {1}{2} \, \sqrt {10} \sin \relax (x)\right ) - \frac {\sqrt {-5 \, \sin \relax (x)^{2} + 2} \sin \relax (x)}{10 \, {\left (5 \, \sin \relax (x)^{2} - 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/25*sqrt(5)*arcsin(1/2*sqrt(10)*sin(x)) - 1/10*sqrt(-5*sin(x)^2 + 2)*sin(x)/(5*sin(x)^2 - 2)

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maple [B]  time = 0.28, size = 58, normalized size = 1.49

method	result	size
default	\(\frac {20 \arcsin \left (\frac {\sin \relax (x ) \sqrt {10}}{2}\right ) \sqrt {5}\, \left (\cos ^{2}\relax (x )\right )+5 \sin \relax (x ) \sqrt {5 \left (\cos ^{2}\relax (x )\right )-3}-12 \arcsin \left (\frac {\sin \relax (x ) \sqrt {10}}{2}\right ) \sqrt {5}}{250 \left (\cos ^{2}\relax (x )\right )-150}\)	\(58\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)*cos(2*x)/(2-5*sin(x)^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/50/(5*cos(x)^2-3)*(20*arcsin(1/2*sin(x)*10^(1/2))*5^(1/2)*cos(x)^2+5*sin(x)*(5*cos(x)^2-3)^(1/2)-12*arcsin(1
/2*sin(x)*10^(1/2))*5^(1/2))

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maxima [B]  time = 1.32, size = 716, normalized size = 18.36 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/50*(5*cos(1/2*arctan2(5*sin(4*x) - 2*sin(2*x), 5*cos(4*x) - 2*cos(2*x) + 5))*sin(2*x) - 5*(cos(2*x) - 1)*sin
(1/2*arctan2(5*sin(4*x) - 2*sin(2*x), 5*cos(4*x) - 2*cos(2*x) + 5)) + 2*(-10*(2*cos(2*x) - 5)*cos(4*x) + 25*co
s(4*x)^2 + 4*cos(2*x)^2 + 25*sin(4*x)^2 - 20*sin(4*x)*sin(2*x) + 4*sin(2*x)^2 - 20*cos(2*x) + 25)^(1/4)*(sqrt(
5)*arctan2(1/12*sqrt(6)*(sqrt(6)*(25/36)^(1/4)*(25*cos(2*x)^4 + 25*sin(2*x)^4 - 20*cos(2*x)^3 + 2*(25*cos(2*x)
^2 - 10*cos(2*x) - 23)*sin(2*x)^2 + 54*cos(2*x)^2 - 20*cos(2*x) + 25)^(1/4)*sin(1/2*arctan2(5/12*(5*cos(2*x) -
1)*sin(2*x), 25/24*cos(2*x)^2 - 25/24*sin(2*x)^2 - 5/12*cos(2*x) + 25/24)) + 5*sin(2*x)), 5/12*sqrt(6)*cos(2*
x) + 1/2*(25/36)^(1/4)*(25*cos(2*x)^4 + 25*sin(2*x)^4 - 20*cos(2*x)^3 + 2*(25*cos(2*x)^2 - 10*cos(2*x) - 23)*s
in(2*x)^2 + 54*cos(2*x)^2 - 20*cos(2*x) + 25)^(1/4)*cos(1/2*arctan2(5/12*(5*cos(2*x) - 1)*sin(2*x), 25/24*cos(
2*x)^2 - 25/24*sin(2*x)^2 - 5/12*cos(2*x) + 25/24)) - 1/12*sqrt(6)) + sqrt(5)*arctan2(1/12*sqrt(6)*(sqrt(6)*(1
/36)^(1/4)*(cos(2*x)^4 + sin(2*x)^4 - 20*cos(2*x)^3 + 2*(cos(2*x)^2 - 10*cos(2*x) + 1)*sin(2*x)^2 + 198*cos(2*
x)^2 - 980*cos(2*x) + 2401)^(1/4)*sin(1/2*arctan2(1/12*(cos(2*x) - 5)*sin(2*x), 1/24*cos(2*x)^2 - 1/24*sin(2*x
)^2 - 5/12*cos(2*x) + 49/24)) + sin(2*x)), 1/12*sqrt(6)*cos(2*x) + 1/2*(1/36)^(1/4)*(cos(2*x)^4 + sin(2*x)^4 -
20*cos(2*x)^3 + 2*(cos(2*x)^2 - 10*cos(2*x) + 1)*sin(2*x)^2 + 198*cos(2*x)^2 - 980*cos(2*x) + 2401)^(1/4)*cos
(1/2*arctan2(1/12*(cos(2*x) - 5)*sin(2*x), 1/24*cos(2*x)^2 - 1/24*sin(2*x)^2 - 5/12*cos(2*x) + 49/24)) - 5/12*
sqrt(6))))/(-10*(2*cos(2*x) - 5)*cos(4*x) + 25*cos(4*x)^2 + 4*cos(2*x)^2 + 25*sin(4*x)^2 - 20*sin(4*x)*sin(2*x
) + 4*sin(2*x)^2 - 20*cos(2*x) + 25)^(1/4)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {\cos \left (2\,x\right )\,\cos \relax (x)}{{\left (2-5\,{\sin \relax (x)}^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(2*x)*cos(x))/(2 - 5*sin(x)^2)^(3/2),x)

[Out]

int((cos(2*x)*cos(x))/(2 - 5*sin(x)^2)^(3/2), 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)*cos(2*x)/(2-5*sin(x)**2)**(3/2),x)

[Out]

Timed out

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