3.762 \(\int \frac{\sin (8 x)}{9+\sin ^4(4 x)} \, dx\)

Optimal. Leaf size=15 \[ \frac{1}{12} \tan ^{-1}\left (\frac{1}{3} \sin ^2(4 x)\right ) \]

[Out]

ArcTan[Sin[4*x]^2/3]/12

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Rubi [A]  time = 0.0290622, antiderivative size = 15, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {12, 275, 203} \[ \frac{1}{12} \tan ^{-1}\left (\frac{1}{3} \sin ^2(4 x)\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sin[8*x]/(9 + Sin[4*x]^4),x]

[Out]

ArcTan[Sin[4*x]^2/3]/12

Rule 12

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

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sin (8 x)}{9+\sin ^4(4 x)} \, dx &=\frac{1}{4} \operatorname{Subst}\left (\int \frac{2 x}{9+x^4} \, dx,x,\sin (4 x)\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x}{9+x^4} \, dx,x,\sin (4 x)\right )\\ &=\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{9+x^2} \, dx,x,\sin ^2(4 x)\right )\\ &=\frac{1}{12} \tan ^{-1}\left (\frac{1}{3} \sin ^2(4 x)\right )\\ \end{align*}

Mathematica [A]  time = 0.0162902, size = 15, normalized size = 1. \[ \frac{1}{12} \tan ^{-1}\left (\frac{1}{3} \sin ^2(4 x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[8*x]/(9 + Sin[4*x]^4),x]

[Out]

ArcTan[Sin[4*x]^2/3]/12

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Maple [A]  time = 0.05, size = 12, normalized size = 0.8 \begin{align*}{\frac{1}{12}\arctan \left ({\frac{ \left ( \sin \left ( 4\,x \right ) \right ) ^{2}}{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(8*x)/(9+sin(4*x)^4),x)

[Out]

1/12*arctan(1/3*sin(4*x)^2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (8 \, x\right )}{\sin \left (4 \, x\right )^{4} + 9}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(8*x)/(9+sin(4*x)^4),x, algorithm="maxima")

[Out]

integrate(sin(8*x)/(sin(4*x)^4 + 9), x)

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Fricas [A]  time = 2.14511, size = 49, normalized size = 3.27 \begin{align*} -\frac{1}{12} \, \arctan \left (\frac{1}{3} \, \cos \left (4 \, x\right )^{2} - \frac{1}{3}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(8*x)/(9+sin(4*x)^4),x, algorithm="fricas")

[Out]

-1/12*arctan(1/3*cos(4*x)^2 - 1/3)

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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(8*x)/(9+sin(4*x)**4),x)

[Out]

Timed out

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Giac [A]  time = 1.11829, size = 20, normalized size = 1.33 \begin{align*} \frac{1}{12} \, \arctan \left (\frac{3}{\cos \left (4 \, x\right )^{2} - 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(8*x)/(9+sin(4*x)^4),x, algorithm="giac")

[Out]

1/12*arctan(3/(cos(4*x)^2 - 1))