3.55 \(\int \sqrt{1+3 \cos ^2(x)} \sin (2 x) \, dx\)

Optimal. Leaf size=16 \[ -\frac{2}{9} \left (4-3 \sin ^2(x)\right )^{3/2} \]

[Out]

(-2*(4 - 3*Sin[x]^2)^(3/2))/9

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Rubi [A]  time = 0.0305212, antiderivative size = 16, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {12, 261} \[ -\frac{2}{9} \left (4-3 \sin ^2(x)\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 + 3*Cos[x]^2]*Sin[2*x],x]

[Out]

(-2*(4 - 3*Sin[x]^2)^(3/2))/9

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 261

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

Rubi steps

\begin{align*} \int \sqrt{1+3 \cos ^2(x)} \sin (2 x) \, dx &=\operatorname{Subst}\left (\int 2 x \sqrt{4-3 x^2} \, dx,x,\sin (x)\right )\\ &=2 \operatorname{Subst}\left (\int x \sqrt{4-3 x^2} \, dx,x,\sin (x)\right )\\ &=-\frac{2}{9} \left (4-3 \sin ^2(x)\right )^{3/2}\\ \end{align*}

Mathematica [A]  time = 0.0111561, size = 16, normalized size = 1. \[ -\frac{2}{9} \left (4-3 \sin ^2(x)\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 + 3*Cos[x]^2]*Sin[2*x],x]

[Out]

(-2*(4 - 3*Sin[x]^2)^(3/2))/9

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Maple [A]  time = 0.018, size = 13, normalized size = 0.8 \begin{align*} -{\frac{2}{9} \left ( 1+3\, \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(2*x)*(1+3*cos(x)^2)^(1/2),x)

[Out]

-2/9*(1+3*cos(x)^2)^(3/2)

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Maxima [A]  time = 0.949571, size = 16, normalized size = 1. \begin{align*} -\frac{2}{9} \,{\left (3 \, \cos \left (x\right )^{2} + 1\right )}^{\frac{3}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(2*x)*(1+3*cos(x)^2)^(1/2),x, algorithm="maxima")

[Out]

-2/9*(3*cos(x)^2 + 1)^(3/2)

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Fricas [A]  time = 0.5205, size = 39, normalized size = 2.44 \begin{align*} -\frac{2}{9} \,{\left (3 \, \cos \left (x\right )^{2} + 1\right )}^{\frac{3}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(2*x)*(1+3*cos(x)^2)^(1/2),x, algorithm="fricas")

[Out]

-2/9*(3*cos(x)^2 + 1)^(3/2)

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Sympy [A]  time = 2.32179, size = 15, normalized size = 0.94 \begin{align*} - \frac{2 \left (3 \cos ^{2}{\left (x \right )} + 1\right )^{\frac{3}{2}}}{9} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(2*x)*(1+3*cos(x)**2)**(1/2),x)

[Out]

-2*(3*cos(x)**2 + 1)**(3/2)/9

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Giac [B]  time = 1.11583, size = 248, normalized size = 15.5 \begin{align*} -\frac{16 \,{\left ({\left (\tan \left (\frac{1}{2} \, x\right )^{2} - \sqrt{\tan \left (\frac{1}{2} \, x\right )^{4} - \tan \left (\frac{1}{2} \, x\right )^{2} + 1}\right )}^{5} -{\left (\tan \left (\frac{1}{2} \, x\right )^{2} - \sqrt{\tan \left (\frac{1}{2} \, x\right )^{4} - \tan \left (\frac{1}{2} \, x\right )^{2} + 1}\right )}^{3} - 2 \,{\left (\tan \left (\frac{1}{2} \, x\right )^{2} - \sqrt{\tan \left (\frac{1}{2} \, x\right )^{4} - \tan \left (\frac{1}{2} \, x\right )^{2} + 1}\right )}^{2} + 3 \, \tan \left (\frac{1}{2} \, x\right )^{2} - 3 \, \sqrt{\tan \left (\frac{1}{2} \, x\right )^{4} - \tan \left (\frac{1}{2} \, x\right )^{2} + 1} - 1\right )}}{{\left ({\left (\tan \left (\frac{1}{2} \, x\right )^{2} - \sqrt{\tan \left (\frac{1}{2} \, x\right )^{4} - \tan \left (\frac{1}{2} \, x\right )^{2} + 1}\right )}^{2} + 2 \, \tan \left (\frac{1}{2} \, x\right )^{2} - 2 \, \sqrt{\tan \left (\frac{1}{2} \, x\right )^{4} - \tan \left (\frac{1}{2} \, x\right )^{2} + 1} - 2\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(2*x)*(1+3*cos(x)^2)^(1/2),x, algorithm="giac")

[Out]

-16*((tan(1/2*x)^2 - sqrt(tan(1/2*x)^4 - tan(1/2*x)^2 + 1))^5 - (tan(1/2*x)^2 - sqrt(tan(1/2*x)^4 - tan(1/2*x)
^2 + 1))^3 - 2*(tan(1/2*x)^2 - sqrt(tan(1/2*x)^4 - tan(1/2*x)^2 + 1))^2 + 3*tan(1/2*x)^2 - 3*sqrt(tan(1/2*x)^4
- tan(1/2*x)^2 + 1) - 1)/((tan(1/2*x)^2 - sqrt(tan(1/2*x)^4 - tan(1/2*x)^2 + 1))^2 + 2*tan(1/2*x)^2 - 2*sqrt(
tan(1/2*x)^4 - tan(1/2*x)^2 + 1) - 2)^3