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

Optimal. Leaf size=7 \[ -\tan ^{-1}(\cos (2 x)) \]

[Out]

-ArcTan[Cos[2*x]]

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Rubi [A]  time = 0.0391762, antiderivative size = 7, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {12, 1107, 617, 204} \[ -\tan ^{-1}(\cos (2 x)) \]

Antiderivative was successfully verified.

[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 1107

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]

Rule 617

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 204

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

Rubi steps

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

Mathematica [A]  time = 0.0310197, size = 7, normalized size = 1. \[ -\tan ^{-1}(\cos (2 x)) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcTan[Cos[2*x]]

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Maple [A]  time = 0.035, size = 12, normalized size = 1.7 \begin{align*} -\arctan \left ( 2\, \left ( \cos \left ( x \right ) \right ) ^{2}-1 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.42329, size = 12, normalized size = 1.71 \begin{align*} \arctan \left (2 \, \sin \left (x\right )^{2} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan(2*sin(x)^2 - 1)

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Fricas [A]  time = 2.27802, size = 34, normalized size = 4.86 \begin{align*} -\arctan \left (2 \, \cos \left (x\right )^{2} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arctan(2*cos(x)^2 - 1)

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

[Out]

Timed out

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Giac [A]  time = 1.07174, size = 12, normalized size = 1.71 \begin{align*} \arctan \left (2 \, \sin \left (x\right )^{2} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan(2*sin(x)^2 - 1)