\(\int \frac {\sin (8 x)}{9+\sin ^4(4 x)} \, dx\) [762]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F(-1)]
   Maxima [F]
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 15, antiderivative size = 15 \[ \int \frac {\sin (8 x)}{9+\sin ^4(4 x)} \, dx=\frac {1}{12} \arctan \left (\frac {1}{3} \sin ^2(4 x)\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {12, 281, 209} \[ \int \frac {\sin (8 x)}{9+\sin ^4(4 x)} \, dx=\frac {1}{12} \arctan \left (\frac {1}{3} \sin ^2(4 x)\right ) \]

[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 209

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

Rule 281

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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \text {Subst}\left (\int \frac {2 x}{9+x^4} \, dx,x,\sin (4 x)\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {x}{9+x^4} \, dx,x,\sin (4 x)\right ) \\ & = \frac {1}{4} \text {Subst}\left (\int \frac {1}{9+x^2} \, dx,x,\sin ^2(4 x)\right ) \\ & = \frac {1}{12} \arctan \left (\frac {1}{3} \sin ^2(4 x)\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {\sin (8 x)}{9+\sin ^4(4 x)} \, dx=\frac {1}{12} \arctan \left (\frac {1}{3} \sin ^2(4 x)\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 26.90 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80

method result size
derivativedivides \(\frac {\arctan \left (\frac {\sin \left (4 x \right )^{2}}{3}\right )}{12}\) \(12\)
default \(\frac {\arctan \left (\frac {\sin \left (4 x \right )^{2}}{3}\right )}{12}\) \(12\)
risch \(\frac {i \ln \left ({\mathrm e}^{16 i x}+\left (-2-12 i\right ) {\mathrm e}^{8 i x}+1\right )}{24}-\frac {i \ln \left ({\mathrm e}^{16 i x}+\left (-2+12 i\right ) {\mathrm e}^{8 i x}+1\right )}{24}\) \(42\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \frac {\sin (8 x)}{9+\sin ^4(4 x)} \, dx=-\frac {1}{12} \, \arctan \left (\frac {1}{3} \, \cos \left (4 \, x\right )^{2} - \frac {1}{3}\right ) \]

[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)

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin (8 x)}{9+\sin ^4(4 x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {\sin (8 x)}{9+\sin ^4(4 x)} \, dx=\int { \frac {\sin \left (8 \, x\right )}{\sin \left (4 \, x\right )^{4} + 9} \,d x } \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int \frac {\sin (8 x)}{9+\sin ^4(4 x)} \, dx=\frac {1}{12} \, \arctan \left (\frac {1}{3} \, \sin \left (4 \, x\right )^{2}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 26.98 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \frac {\sin (8 x)}{9+\sin ^4(4 x)} \, dx=\frac {\mathrm {atan}\left (\frac {10\,{\mathrm {tan}\left (4\,x\right )}^2}{3}+3\right )}{12} \]

[In]

int(sin(8*x)/(sin(4*x)^4 + 9),x)

[Out]

atan((10*tan(4*x)^2)/3 + 3)/12