3.424 \(\int \frac{\sin (5 x)}{(5 \cos ^2(x)+9 \sin ^2(x))^{5/2}} \, dx\)

Optimal. Leaf size=48 \[ \frac{295 \cos (x)}{243 \sqrt{9-4 \cos ^2(x)}}-\frac{55 \cos (x)}{27 \left (9-4 \cos ^2(x)\right )^{3/2}}-\frac{1}{2} \sin ^{-1}\left (\frac{2 \cos (x)}{3}\right ) \]

[Out]

-ArcSin[(2*Cos[x])/3]/2 - (55*Cos[x])/(27*(9 - 4*Cos[x]^2)^(3/2)) + (295*Cos[x])/(243*Sqrt[9 - 4*Cos[x]^2])

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Rubi [A]  time = 0.0722855, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {1157, 385, 216} \[ \frac{295 \cos (x)}{243 \sqrt{9-4 \cos ^2(x)}}-\frac{55 \cos (x)}{27 \left (9-4 \cos ^2(x)\right )^{3/2}}-\frac{1}{2} \sin ^{-1}\left (\frac{2 \cos (x)}{3}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sin[5*x]/(5*Cos[x]^2 + 9*Sin[x]^2)^(5/2),x]

[Out]

-ArcSin[(2*Cos[x])/3]/2 - (55*Cos[x])/(27*(9 - 4*Cos[x]^2)^(3/2)) + (295*Cos[x])/(243*Sqrt[9 - 4*Cos[x]^2])

Rule 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

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 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]

Rubi steps

\begin{align*} \int \frac{\sin (5 x)}{\left (5 \cos ^2(x)+9 \sin ^2(x)\right )^{5/2}} \, dx &=-\operatorname{Subst}\left (\int \frac{1-12 x^2+16 x^4}{\left (9-4 x^2\right )^{5/2}} \, dx,x,\cos (x)\right )\\ &=-\frac{55 \cos (x)}{27 \left (9-4 \cos ^2(x)\right )^{3/2}}+\frac{1}{27} \operatorname{Subst}\left (\int \frac{52+108 x^2}{\left (9-4 x^2\right )^{3/2}} \, dx,x,\cos (x)\right )\\ &=-\frac{55 \cos (x)}{27 \left (9-4 \cos ^2(x)\right )^{3/2}}+\frac{295 \cos (x)}{243 \sqrt{9-4 \cos ^2(x)}}-\operatorname{Subst}\left (\int \frac{1}{\sqrt{9-4 x^2}} \, dx,x,\cos (x)\right )\\ &=-\frac{1}{2} \sin ^{-1}\left (\frac{2 \cos (x)}{3}\right )-\frac{55 \cos (x)}{27 \left (9-4 \cos ^2(x)\right )^{3/2}}+\frac{295 \cos (x)}{243 \sqrt{9-4 \cos ^2(x)}}\\ \end{align*}

Mathematica [C]  time = 0.303501, size = 63, normalized size = 1.31 \[ \frac{2550 \cos (x)-590 \cos (3 x)+243 i (7-2 \cos (2 x))^{3/2} \log \left (\sqrt{7-2 \cos (2 x)}+2 i \cos (x)\right )}{486 (7-2 \cos (2 x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[5*x]/(5*Cos[x]^2 + 9*Sin[x]^2)^(5/2),x]

[Out]

(2550*Cos[x] - 590*Cos[3*x] + (243*I)*(7 - 2*Cos[2*x])^(3/2)*Log[(2*I)*Cos[x] + Sqrt[7 - 2*Cos[2*x]]])/(486*(7
- 2*Cos[2*x])^(3/2))

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Maple [A]  time = 0.074, size = 53, normalized size = 1.1 \begin{align*} -{\frac{4\, \left ( \cos \left ( x \right ) \right ) ^{3}}{3} \left ( 9-4\, \left ( \cos \left ( x \right ) \right ) ^{2} \right ) ^{-{\frac{3}{2}}}}+{\frac{214\,\cos \left ( x \right ) }{243}{\frac{1}{\sqrt{9-4\, \left ( \cos \left ( x \right ) \right ) ^{2}}}}}-{\frac{1}{2}\arcsin \left ({\frac{2\,\cos \left ( x \right ) }{3}} \right ) }+{\frac{26\,\cos \left ( x \right ) }{27} \left ( 9-4\, \left ( \cos \left ( x \right ) \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(5*x)/(5*cos(x)^2+9*sin(x)^2)^(5/2),x)

[Out]

-4/3*cos(x)^3/(9-4*cos(x)^2)^(3/2)+214/243*cos(x)/(9-4*cos(x)^2)^(1/2)-1/2*arcsin(2/3*cos(x))+26/27*cos(x)/(9-
4*cos(x)^2)^(3/2)

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Maxima [A]  time = 1.48914, size = 93, normalized size = 1.94 \begin{align*} -2 \,{\left (\frac{2 \, \cos \left (x\right )^{2}}{{\left (-4 \, \cos \left (x\right )^{2} + 9\right )}^{\frac{3}{2}}} - \frac{3}{{\left (-4 \, \cos \left (x\right )^{2} + 9\right )}^{\frac{3}{2}}}\right )} \cos \left (x\right ) + \frac{52 \, \cos \left (x\right )}{243 \, \sqrt{-4 \, \cos \left (x\right )^{2} + 9}} + \frac{26 \, \cos \left (x\right )}{27 \,{\left (-4 \, \cos \left (x\right )^{2} + 9\right )}^{\frac{3}{2}}} - \frac{1}{2} \, \arcsin \left (\frac{2}{3} \, \cos \left (x\right )\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(5*x)/(5*cos(x)^2+9*sin(x)^2)^(5/2),x, algorithm="maxima")

[Out]

-2*(2*cos(x)^2/(-4*cos(x)^2 + 9)^(3/2) - 3/(-4*cos(x)^2 + 9)^(3/2))*cos(x) + 52/243*cos(x)/sqrt(-4*cos(x)^2 +
9) + 26/27*cos(x)/(-4*cos(x)^2 + 9)^(3/2) - 1/2*arcsin(2/3*cos(x))

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Fricas [B]  time = 3.00686, size = 413, normalized size = 8.6 \begin{align*} \frac{243 \,{\left (16 \, \cos \left (x\right )^{4} - 72 \, \cos \left (x\right )^{2} + 81\right )} \arctan \left (-\frac{81 \, \cos \left (x\right ) \sin \left (x\right ) - 4 \,{\left (8 \, \cos \left (x\right )^{3} - 9 \, \cos \left (x\right )\right )} \sqrt{-4 \, \cos \left (x\right )^{2} + 9}}{64 \, \cos \left (x\right )^{4} - 225 \, \cos \left (x\right )^{2} + 81}\right ) - 243 \,{\left (16 \, \cos \left (x\right )^{4} - 72 \, \cos \left (x\right )^{2} + 81\right )} \arctan \left (\frac{\sin \left (x\right )}{\cos \left (x\right )}\right ) - 80 \,{\left (59 \, \cos \left (x\right )^{3} - 108 \, \cos \left (x\right )\right )} \sqrt{-4 \, \cos \left (x\right )^{2} + 9}}{972 \,{\left (16 \, \cos \left (x\right )^{4} - 72 \, \cos \left (x\right )^{2} + 81\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(5*x)/(5*cos(x)^2+9*sin(x)^2)^(5/2),x, algorithm="fricas")

[Out]

1/972*(243*(16*cos(x)^4 - 72*cos(x)^2 + 81)*arctan(-(81*cos(x)*sin(x) - 4*(8*cos(x)^3 - 9*cos(x))*sqrt(-4*cos(
x)^2 + 9))/(64*cos(x)^4 - 225*cos(x)^2 + 81)) - 243*(16*cos(x)^4 - 72*cos(x)^2 + 81)*arctan(sin(x)/cos(x)) - 8
0*(59*cos(x)^3 - 108*cos(x))*sqrt(-4*cos(x)^2 + 9))/(16*cos(x)^4 - 72*cos(x)^2 + 81)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(5*x)/(5*cos(x)**2+9*sin(x)**2)**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.11154, size = 54, normalized size = 1.12 \begin{align*} -\frac{20 \,{\left (59 \, \cos \left (x\right )^{2} - 108\right )} \sqrt{-4 \, \cos \left (x\right )^{2} + 9} \cos \left (x\right )}{243 \,{\left (4 \, \cos \left (x\right )^{2} - 9\right )}^{2}} - \frac{1}{2} \, \arcsin \left (\frac{2}{3} \, \cos \left (x\right )\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(5*x)/(5*cos(x)^2+9*sin(x)^2)^(5/2),x, algorithm="giac")

[Out]

-20/243*(59*cos(x)^2 - 108)*sqrt(-4*cos(x)^2 + 9)*cos(x)/(4*cos(x)^2 - 9)^2 - 1/2*arcsin(2/3*cos(x))