3.484 \(\int \frac{1}{\sec ^2(x)+\tan ^2(x)} \, dx\)

Optimal. Leaf size=36 \[ \sqrt{2} x-x+\sqrt{2} \tan ^{-1}\left (\frac{\sin (x) \cos (x)}{\sin ^2(x)+\sqrt{2}+1}\right ) \]

[Out]

-x + Sqrt[2]*x + Sqrt[2]*ArcTan[(Cos[x]*Sin[x])/(1 + Sqrt[2] + Sin[x]^2)]

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Rubi [A]  time = 0.0277595, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {1093, 203} \[ \sqrt{2} x-x+\sqrt{2} \tan ^{-1}\left (\frac{\sin (x) \cos (x)}{\sin ^2(x)+\sqrt{2}+1}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(Sec[x]^2 + Tan[x]^2)^(-1),x]

[Out]

-x + Sqrt[2]*x + Sqrt[2]*ArcTan[(Cos[x]*Sin[x])/(1 + Sqrt[2] + Sin[x]^2)]

Rule 1093

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

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

Mathematica [A]  time = 0.0465269, size = 19, normalized size = 0.53 \[ \sqrt{2} \tan ^{-1}\left (\sqrt{2} \tan (x)\right )-x \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[x]^2 + Tan[x]^2)^(-1),x]

[Out]

-x + Sqrt[2]*ArcTan[Sqrt[2]*Tan[x]]

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Maple [A]  time = 0.046, size = 16, normalized size = 0.4 \begin{align*} \sqrt{2}\arctan \left ( \tan \left ( x \right ) \sqrt{2} \right ) -x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sec(x)^2+tan(x)^2),x)

[Out]

2^(1/2)*arctan(tan(x)*2^(1/2))-x

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Maxima [A]  time = 1.46873, size = 20, normalized size = 0.56 \begin{align*} \sqrt{2} \arctan \left (\sqrt{2} \tan \left (x\right )\right ) - x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sec(x)^2+tan(x)^2),x, algorithm="maxima")

[Out]

sqrt(2)*arctan(sqrt(2)*tan(x)) - x

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Fricas [A]  time = 1.77284, size = 107, normalized size = 2.97 \begin{align*} -\frac{1}{2} \, \sqrt{2} \arctan \left (\frac{3 \, \sqrt{2} \cos \left (x\right )^{2} - 2 \, \sqrt{2}}{4 \, \cos \left (x\right ) \sin \left (x\right )}\right ) - x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sec(x)^2+tan(x)^2),x, algorithm="fricas")

[Out]

-1/2*sqrt(2)*arctan(1/4*(3*sqrt(2)*cos(x)^2 - 2*sqrt(2))/(cos(x)*sin(x))) - x

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\tan ^{2}{\left (x \right )} + \sec ^{2}{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sec(x)**2+tan(x)**2),x)

[Out]

Integral(1/(tan(x)**2 + sec(x)**2), x)

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Giac [A]  time = 1.11276, size = 20, normalized size = 0.56 \begin{align*} \sqrt{2} \arctan \left (\sqrt{2} \tan \left (x\right )\right ) - x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sec(x)^2+tan(x)^2),x, algorithm="giac")

[Out]

sqrt(2)*arctan(sqrt(2)*tan(x)) - x