3.732 \(\int \frac{\sec (2 x) \tan (2 x)}{(1+\sec (2 x))^{3/2}} \, dx\)

Optimal. Leaf size=12 \[ -\frac{1}{\sqrt{\sec (2 x)+1}} \]

[Out]

-(1/Sqrt[1 + Sec[2*x]])

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Rubi [A]  time = 0.0545603, antiderivative size = 12, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {4339, 261} \[ -\frac{1}{\sqrt{\sec (2 x)+1}} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[2*x]*Tan[2*x])/(1 + Sec[2*x])^(3/2),x]

[Out]

-(1/Sqrt[1 + Sec[2*x]])

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])

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps

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

Mathematica [A]  time = 0.0767221, size = 20, normalized size = 1.67 \[ -\frac{2 \cos ^2(x) \sec (2 x)}{(\sec (2 x)+1)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[2*x]*Tan[2*x])/(1 + Sec[2*x])^(3/2),x]

[Out]

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

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Maple [A]  time = 0.023, size = 11, normalized size = 0.9 \begin{align*} -{\frac{1}{\sqrt{1+\sec \left ( 2\,x \right ) }}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(2*x)*tan(2*x)/(1+sec(2*x))^(3/2),x)

[Out]

-1/(1+sec(2*x))^(1/2)

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Maxima [A]  time = 0.974531, size = 14, normalized size = 1.17 \begin{align*} -\frac{1}{\sqrt{\sec \left (2 \, x\right ) + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(2*x)*tan(2*x)/(1+sec(2*x))^(3/2),x, algorithm="maxima")

[Out]

-1/sqrt(sec(2*x) + 1)

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Fricas [B]  time = 2.39472, size = 76, normalized size = 6.33 \begin{align*} -\frac{\sqrt{\frac{\cos \left (2 \, x\right ) + 1}{\cos \left (2 \, x\right )}} \cos \left (2 \, x\right )}{\cos \left (2 \, x\right ) + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(2*x)*tan(2*x)/(1+sec(2*x))^(3/2),x, algorithm="fricas")

[Out]

-sqrt((cos(2*x) + 1)/cos(2*x))*cos(2*x)/(cos(2*x) + 1)

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Sympy [A]  time = 1.28341, size = 12, normalized size = 1. \begin{align*} - \frac{1}{\sqrt{\sec{\left (2 x \right )} + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(2*x)*tan(2*x)/(1+sec(2*x))**(3/2),x)

[Out]

-1/sqrt(sec(2*x) + 1)

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Giac [B]  time = 1.46652, size = 32, normalized size = 2.67 \begin{align*} \frac{\sqrt{2} \sqrt{-\tan \left (x\right )^{2} + 1}}{2 \, \mathrm{sgn}\left (\tan \left (x\right )^{2} - 1\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(2*x)*tan(2*x)/(1+sec(2*x))^(3/2),x, algorithm="giac")

[Out]

1/2*sqrt(2)*sqrt(-tan(x)^2 + 1)/sgn(tan(x)^2 - 1)