\(\int \frac {\sin (2 x)}{\cos ^4(x)+\sin ^4(x)} \, dx\) [375]

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

Optimal result

Integrand size = 16, antiderivative size = 7 \[ \int \frac {\sin (2 x)}{\cos ^4(x)+\sin ^4(x)} \, dx=-\arctan (\cos (2 x)) \]

[Out]

-arctan(cos(2*x))

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 7, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {12, 1121, 631, 210} \[ \int \frac {\sin (2 x)}{\cos ^4(x)+\sin ^4(x)} \, dx=-\arctan (\cos (2 x)) \]

[In]

Int[Sin[2*x]/(Cos[x]^4 + Sin[x]^4),x]

[Out]

-ArcTan[Cos[2*x]]

Rule 12

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

Rule 210

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

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
 FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 1121

Int[(x_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(a + b*x + c*x^2)^p, x],
 x, x^2], x] /; FreeQ[{a, b, c, p}, x]

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 7, normalized size of antiderivative = 1.00 \[ \int \frac {\sin (2 x)}{\cos ^4(x)+\sin ^4(x)} \, dx=-\arctan (\cos (2 x)) \]

[In]

Integrate[Sin[2*x]/(Cos[x]^4 + Sin[x]^4),x]

[Out]

-ArcTan[Cos[2*x]]

Maple [A] (verified)

Time = 12.97 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.71

method result size
derivativedivides \(-\arctan \left (2 \left (\cos ^{2}\left (x \right )\right )-1\right )\) \(12\)
default \(-\arctan \left (2 \left (\cos ^{2}\left (x \right )\right )-1\right )\) \(12\)
risch \(-\frac {i \ln \left ({\mathrm e}^{4 i x}+2 i {\mathrm e}^{2 i x}+1\right )}{2}+\frac {i \ln \left ({\mathrm e}^{4 i x}-2 i {\mathrm e}^{2 i x}+1\right )}{2}\) \(40\)

[In]

int(sin(2*x)/(cos(x)^4+sin(x)^4),x,method=_RETURNVERBOSE)

[Out]

-arctan(2*cos(x)^2-1)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.57 \[ \int \frac {\sin (2 x)}{\cos ^4(x)+\sin ^4(x)} \, dx=-\arctan \left (2 \, \cos \left (x\right )^{2} - 1\right ) \]

[In]

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

[Out]

-arctan(2*cos(x)^2 - 1)

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin (2 x)}{\cos ^4(x)+\sin ^4(x)} \, dx=\text {Timed out} \]

[In]

integrate(sin(2*x)/(cos(x)**4+sin(x)**4),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.29 \[ \int \frac {\sin (2 x)}{\cos ^4(x)+\sin ^4(x)} \, dx=\arctan \left (2 \, \sin \left (x\right )^{2} - 1\right ) \]

[In]

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

[Out]

arctan(2*sin(x)^2 - 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.29 \[ \int \frac {\sin (2 x)}{\cos ^4(x)+\sin ^4(x)} \, dx=\arctan \left (2 \, \sin \left (x\right )^{2} - 1\right ) \]

[In]

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

[Out]

arctan(2*sin(x)^2 - 1)

Mupad [B] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 5, normalized size of antiderivative = 0.71 \[ \int \frac {\sin (2 x)}{\cos ^4(x)+\sin ^4(x)} \, dx=\mathrm {atan}\left ({\mathrm {tan}\left (x\right )}^2\right ) \]

[In]

int(sin(2*x)/(cos(x)^4 + sin(x)^4),x)

[Out]

atan(tan(x)^2)