\(\int \sqrt {1+3 \cos ^2(x)} \sin (2 x) \, dx\) [55]

Optimal result

Integrand size = 17, antiderivative size = 16 \[ \int \sqrt {1+3 \cos ^2(x)} \sin (2 x) \, dx=-\frac {2}{9} \left (4-3 \sin ^2(x)\right )^{3/2} \]

[Out]

-2/9*(4-3*sin(x)^2)^(3/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {12, 267} \[ \int \sqrt {1+3 \cos ^2(x)} \sin (2 x) \, dx=-\frac {2}{9} \left (4-3 \sin ^2(x)\right )^{3/2} \]

[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 267

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*} \text {integral}& = \text {Subst}\left (\int 2 x \sqrt {4-3 x^2} \, dx,x,\sin (x)\right ) \\ & = 2 \text {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] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \sqrt {1+3 \cos ^2(x)} \sin (2 x) \, dx=-\frac {2}{9} \left (4-3 \sin ^2(x)\right )^{3/2} \]

[In]

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

[Out]

(-2*(4 - 3*Sin[x]^2)^(3/2))/9

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81

method	result	size
derivativedivides	\(-\frac {2 {\left (1+3 \left (\cos ^{2}\left (x \right )\right )\right )}^{\frac {3}{2}}}{9}\)	\(13\)
default	\(-\frac {2 {\left (1+3 \left (\cos ^{2}\left (x \right )\right )\right )}^{\frac {3}{2}}}{9}\)	\(13\)

[In]

int(sin(2*x)*(1+3*cos(x)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \sqrt {1+3 \cos ^2(x)} \sin (2 x) \, dx=-\frac {2}{9} \, {\left (3 \, \cos \left (x\right )^{2} + 1\right )}^{\frac {3}{2}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.94 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \sqrt {1+3 \cos ^2(x)} \sin (2 x) \, dx=- \frac {2 \left (3 \cos ^{2}{\left (x \right )} + 1\right )^{\frac {3}{2}}}{9} \]

[In]

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

[Out]

-2*(3*cos(x)**2 + 1)**(3/2)/9

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \sqrt {1+3 \cos ^2(x)} \sin (2 x) \, dx=-\frac {2}{9} \, {\left (3 \, \cos \left (x\right )^{2} + 1\right )}^{\frac {3}{2}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \sqrt {1+3 \cos ^2(x)} \sin (2 x) \, dx=-\frac {2}{9} \, {\left (3 \, \cos \left (x\right )^{2} + 1\right )}^{\frac {3}{2}} \]

[In]

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

[Out]

-2/9*(3*cos(x)^2 + 1)^(3/2)

Mupad [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \sqrt {1+3 \cos ^2(x)} \sin (2 x) \, dx=-\frac {2\,{\left (3\,{\cos \left (x\right )}^2+1\right )}^{3/2}}{9} \]

[In]

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

[Out]

-(2*(3*cos(x)^2 + 1)^(3/2))/9