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

Optimal. Leaf size=18 \[ \frac {1}{7} \left (a^2-4 \cos ^2(x)\right )^{7/4} \]

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Rubi [A]  time = 0.05, antiderivative size = 19, normalized size of antiderivative = 1.06, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {12, 261} \[ \frac {1}{7} \left (a^2+4 \sin ^2(x)-4\right )^{7/4} \]

Antiderivative was successfully verified.

[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 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 \left (a^2-4 \cos ^2(x)\right )^{3/4} \sin (2 x) \, dx &=\operatorname {Subst}\left (\int 2 x \left (-4+a^2+4 x^2\right )^{3/4} \, dx,x,\sin (x)\right )\\ &=2 \operatorname {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*}

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Mathematica [A]  time = 0.02, size = 19, normalized size = 1.06 \[ \frac {1}{7} \left (a^2+4 \sin ^2(x)-4\right )^{7/4} \]

Antiderivative was successfully verified.

[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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a^2-4 \cos ^2(x)\right )^{3/4} \sin (2 x) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

Could not integrate

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fricas [A]  time = 1.06, size = 14, normalized size = 0.78 \[ \frac {1}{7} \, {\left (a^{2} - 4 \, \cos \relax (x)^{2}\right )}^{\frac {7}{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 1.00, size = 14, normalized size = 0.78 \[ \frac {1}{7} \, {\left (a^{2} - 4 \, \cos \relax (x)^{2}\right )}^{\frac {7}{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.06, size = 15, normalized size = 0.83

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.43, size = 14, normalized size = 0.78 \[ \frac {1}{7} \, {\left (a^{2} - 4 \, \cos \relax (x)^{2}\right )}^{\frac {7}{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.31, size = 14, normalized size = 0.78 \[ \frac {{\left (a^2-4\,{\cos \relax (x)}^2\right )}^{7/4}}{7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2 - 4*cos(x)^2)^(7/4)/7

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

[Out]

Timed out

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