\(\int (a^2-4 \cos ^2(x))^{3/4} \sin (2 x) \, dx\) [64]

Optimal result

Integrand size = 19, antiderivative size = 18 \[ \int \left (a^2-4 \cos ^2(x)\right )^{3/4} \sin (2 x) \, dx=\frac {1}{7} \left (a^2-4 \cos ^2(x)\right )^{7/4} \]

[Out]

1/7*(a^2-4*cos(x)^2)^(7/4)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {12, 267} \[ \int \left (a^2-4 \cos ^2(x)\right )^{3/4} \sin (2 x) \, dx=\frac {1}{7} \left (a^2+4 \sin ^2(x)-4\right )^{7/4} \]

[In]

Int[(a^2 - 4*Cos[x]^2)^(3/4)*Sin[2*x],x]

[Out]

(-4 + a^2 + 4*Sin[x]^2)^(7/4)/7

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 \left (-4+a^2+4 x^2\right )^{3/4} \, dx,x,\sin (x)\right ) \\ & = 2 \text {Subst}\left (\int x \left (-4+a^2+4 x^2\right )^{3/4} \, dx,x,\sin (x)\right ) \\ & = \frac {1}{7} \left (-4+a^2+4 \sin ^2(x)\right )^{7/4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \left (a^2-4 \cos ^2(x)\right )^{3/4} \sin (2 x) \, dx=\frac {1}{7} \left (-4+a^2+4 \sin ^2(x)\right )^{7/4} \]

[In]

Integrate[(a^2 - 4*Cos[x]^2)^(3/4)*Sin[2*x],x]

[Out]

(-4 + a^2 + 4*Sin[x]^2)^(7/4)/7

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(\frac {\left (a^{2}-4 \left (\cos ^{2}\left (x \right )\right )\right )^{\frac {7}{4}}}{7}\)	\(15\)
default	\(\frac {\left (a^{2}-4 \left (\cos ^{2}\left (x \right )\right )\right )^{\frac {7}{4}}}{7}\)	\(15\)

[In]

int((a^2-4*cos(x)^2)^(3/4)*sin(2*x),x,method=_RETURNVERBOSE)

[Out]

1/7*(a^2-4*cos(x)^2)^(7/4)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \left (a^2-4 \cos ^2(x)\right )^{3/4} \sin (2 x) \, dx=\frac {1}{7} \, {\left (a^{2} - 4 \, \cos \left (x\right )^{2}\right )}^{\frac {7}{4}} \]

[In]

integrate((a^2-4*cos(x)^2)^(3/4)*sin(2*x),x, algorithm="fricas")

[Out]

1/7*(a^2 - 4*cos(x)^2)^(7/4)

Sympy [F(-1)]

Timed out. \[ \int \left (a^2-4 \cos ^2(x)\right )^{3/4} \sin (2 x) \, dx=\text {Timed out} \]

[In]

integrate((a**2-4*cos(x)**2)**(3/4)*sin(2*x),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \left (a^2-4 \cos ^2(x)\right )^{3/4} \sin (2 x) \, dx=\frac {1}{7} \, {\left (a^{2} - 4 \, \cos \left (x\right )^{2}\right )}^{\frac {7}{4}} \]

[In]

integrate((a^2-4*cos(x)^2)^(3/4)*sin(2*x),x, algorithm="maxima")

[Out]

1/7*(a^2 - 4*cos(x)^2)^(7/4)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \left (a^2-4 \cos ^2(x)\right )^{3/4} \sin (2 x) \, dx=\frac {1}{7} \, {\left (a^{2} - 4 \, \cos \left (x\right )^{2}\right )}^{\frac {7}{4}} \]

[In]

integrate((a^2-4*cos(x)^2)^(3/4)*sin(2*x),x, algorithm="giac")

[Out]

1/7*(a^2 - 4*cos(x)^2)^(7/4)

Mupad [B] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \left (a^2-4 \cos ^2(x)\right )^{3/4} \sin (2 x) \, dx=\frac {{\left (a^2-4\,{\cos \left (x\right )}^2\right )}^{7/4}}{7} \]

[In]

int(sin(2*x)*(a^2 - 4*cos(x)^2)^(3/4),x)

[Out]

(a^2 - 4*cos(x)^2)^(7/4)/7