3.726 \(\int \frac {\sec (x) \tan (x)}{9+4 \sec ^2(x)} \, dx\)

Optimal. Leaf size=11 \[ -\frac {1}{6} \tan ^{-1}\left (\frac {3 \cos (x)}{2}\right ) \]

[Out]

-1/6*arctan(3/2*cos(x))

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Rubi [A]  time = 0.03, antiderivative size = 11, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {4339, 203} \[ -\frac {1}{6} \tan ^{-1}\left (\frac {3 \cos (x)}{2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(Sec[x]*Tan[x])/(9 + 4*Sec[x]^2),x]

[Out]

-ArcTan[(3*Cos[x])/2]/6

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])

Rule 4339

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{d = FreeFactors[Cos[c*(a + b*x)], x]}, -Dist[(b*
c)^(-1), Subst[Int[SubstFor[1/x, Cos[c*(a + b*x)]/d, u, x], x], x, Cos[c*(a + b*x)]/d], x] /; FunctionOfQ[Cos[
c*(a + b*x)]/d, u, x, True]] /; FreeQ[{a, b, c}, x] && (EqQ[F, Tan] || EqQ[F, tan])

Rubi steps

\begin {align*} \int \frac {\sec (x) \tan (x)}{9+4 \sec ^2(x)} \, dx &=-\operatorname {Subst}\left (\int \frac {1}{4+9 x^2} \, dx,x,\cos (x)\right )\\ &=-\frac {1}{6} \tan ^{-1}\left (\frac {3 \cos (x)}{2}\right )\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 11, normalized size = 1.00 \[ -\frac {1}{6} \tan ^{-1}\left (\frac {3 \cos (x)}{2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[x]*Tan[x])/(9 + 4*Sec[x]^2),x]

[Out]

-1/6*ArcTan[(3*Cos[x])/2]

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fricas [A]  time = 0.97, size = 7, normalized size = 0.64 \[ -\frac {1}{6} \, \arctan \left (\frac {3}{2} \, \cos \relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)*tan(x)/(9+4*sec(x)^2),x, algorithm="fricas")

[Out]

-1/6*arctan(3/2*cos(x))

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giac [A]  time = 0.13, size = 7, normalized size = 0.64 \[ -\frac {1}{6} \, \arctan \left (\frac {3}{2} \, \cos \relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)*tan(x)/(9+4*sec(x)^2),x, algorithm="giac")

[Out]

-1/6*arctan(3/2*cos(x))

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maple [A]  time = 0.06, size = 8, normalized size = 0.73 \[ \frac {\arctan \left (\frac {2 \sec \relax (x )}{3}\right )}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x)*tan(x)/(9+4*sec(x)^2),x)

[Out]

1/6*arctan(2/3*sec(x))

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maxima [A]  time = 0.44, size = 7, normalized size = 0.64 \[ -\frac {1}{6} \, \arctan \left (\frac {3}{2} \, \cos \relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)*tan(x)/(9+4*sec(x)^2),x, algorithm="maxima")

[Out]

-1/6*arctan(3/2*cos(x))

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mupad [B]  time = 3.04, size = 13, normalized size = 1.18 \[ \frac {\mathrm {atan}\left (\frac {13\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{12}-\frac {5}{12}\right )}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)/(cos(x)*(4/cos(x)^2 + 9)),x)

[Out]

atan((13*tan(x/2)^2)/12 - 5/12)/6

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sympy [A]  time = 0.23, size = 8, normalized size = 0.73 \[ \frac {\operatorname {atan}{\left (\frac {2 \sec {\relax (x )}}{3} \right )}}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)*tan(x)/(9+4*sec(x)**2),x)

[Out]

atan(2*sec(x)/3)/6

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