3.692 \(\int \frac{\sec ^2(x)}{9+\tan ^2(x)} \, dx\)

Optimal. Leaf size=27 \[ \frac{x}{3}-\frac{1}{3} \tan ^{-1}\left (\frac{2 \sin (x) \cos (x)}{2 \cos ^2(x)+1}\right ) \]

[Out]

x/3 - ArcTan[(2*Cos[x]*Sin[x])/(1 + 2*Cos[x]^2)]/3

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Rubi [A]  time = 0.0292568, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3675, 203} \[ \frac{x}{3}-\frac{1}{3} \tan ^{-1}\left (\frac{2 \sin (x) \cos (x)}{2 \cos ^2(x)+1}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]^2/(9 + Tan[x]^2),x]

[Out]

x/3 - ArcTan[(2*Cos[x]*Sin[x])/(1 + 2*Cos[x]^2)]/3

Rule 3675

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/(c^(m - 1)*f), Subst[Int[(c^2 + ff^2*x^2)^(m/2 - 1)*(a + b*(ff*x)
^n)^p, x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2] && (IntegersQ[n, p
] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

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

Mathematica [A]  time = 0.0239948, size = 9, normalized size = 0.33 \[ -\frac{1}{3} \tan ^{-1}(3 \cot (x)) \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]^2/(9 + Tan[x]^2),x]

[Out]

-ArcTan[3*Cot[x]]/3

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Maple [A]  time = 0.031, size = 8, normalized size = 0.3 \begin{align*}{\frac{1}{3}\arctan \left ({\frac{\tan \left ( x \right ) }{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x)^2/(9+tan(x)^2),x)

[Out]

1/3*arctan(1/3*tan(x))

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Maxima [A]  time = 1.44673, size = 9, normalized size = 0.33 \begin{align*} \frac{1}{3} \, \arctan \left (\frac{1}{3} \, \tan \left (x\right )\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^2/(9+tan(x)^2),x, algorithm="maxima")

[Out]

1/3*arctan(1/3*tan(x))

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Fricas [A]  time = 2.13205, size = 70, normalized size = 2.59 \begin{align*} -\frac{1}{6} \, \arctan \left (\frac{10 \, \cos \left (x\right )^{2} - 1}{6 \, \cos \left (x\right ) \sin \left (x\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^2/(9+tan(x)^2),x, algorithm="fricas")

[Out]

-1/6*arctan(1/6*(10*cos(x)^2 - 1)/(cos(x)*sin(x)))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{2}{\left (x \right )}}{\tan ^{2}{\left (x \right )} + 9}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)**2/(9+tan(x)**2),x)

[Out]

Integral(sec(x)**2/(tan(x)**2 + 9), x)

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Giac [A]  time = 1.08678, size = 9, normalized size = 0.33 \begin{align*} \frac{1}{3} \, \arctan \left (\frac{1}{3} \, \tan \left (x\right )\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^2/(9+tan(x)^2),x, algorithm="giac")

[Out]

1/3*arctan(1/3*tan(x))