\(\int \frac {\sec (2 x) \tan (2 x)}{(1+\sec (2 x))^{3/2}} \, dx\) [732]

Optimal result

Integrand size = 19, antiderivative size = 12 \[ \int \frac {\sec (2 x) \tan (2 x)}{(1+\sec (2 x))^{3/2}} \, dx=-\frac {1}{\sqrt {1+\sec (2 x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {4424, 267} \[ \int \frac {\sec (2 x) \tan (2 x)}{(1+\sec (2 x))^{3/2}} \, dx=-\frac {1}{\sqrt {\sec (2 x)+1}} \]

[In]

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

[Out]

-(1/Sqrt[1 + Sec[2*x]])

Rule 267

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]

Rule 4424

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*} \text {integral}& = -\left (\frac {1}{2} \text {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] (verified)

Time = 0.10 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {\sec (2 x) \tan (2 x)}{(1+\sec (2 x))^{3/2}} \, dx=-\frac {1}{\sqrt {1+\sec (2 x)}} \]

[In]

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

[Out]

-(1/Sqrt[1 + Sec[2*x]])

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.92

method	result	size
derivativedivides	\(-\frac {1}{\sqrt {1+\sec \left (2 x \right )}}\)	\(11\)
default	\(-\frac {1}{\sqrt {1+\sec \left (2 x \right )}}\)	\(11\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 29 vs. \(2 (10) = 20\).

Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 2.42 \[ \int \frac {\sec (2 x) \tan (2 x)}{(1+\sec (2 x))^{3/2}} \, dx=-\frac {\sqrt {\frac {\cos \left (2 \, x\right ) + 1}{\cos \left (2 \, x\right )}} \cos \left (2 \, x\right )}{\cos \left (2 \, x\right ) + 1} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {\sec (2 x) \tan (2 x)}{(1+\sec (2 x))^{3/2}} \, dx=- \frac {1}{\sqrt {\sec {\left (2 x \right )} + 1}} \]

[In]

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

[Out]

-1/sqrt(sec(2*x) + 1)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \frac {\sec (2 x) \tan (2 x)}{(1+\sec (2 x))^{3/2}} \, dx=-\frac {1}{\sqrt {\sec \left (2 \, x\right ) + 1}} \]

[In]

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

[Out]

-1/sqrt(sec(2*x) + 1)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 31 vs. \(2 (10) = 20\).

Time = 0.29 (sec) , antiderivative size = 31, normalized size of antiderivative = 2.58 \[ \int \frac {\sec (2 x) \tan (2 x)}{(1+\sec (2 x))^{3/2}} \, dx=\frac {1}{{\left (\sqrt {\cos \left (2 \, x\right )^{2} + \cos \left (2 \, x\right )} - \cos \left (2 \, x\right ) - 1\right )} \mathrm {sgn}\left (\cos \left (2 \, x\right )\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 25.97 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.50 \[ \int \frac {\sec (2 x) \tan (2 x)}{(1+\sec (2 x))^{3/2}} \, dx=-\frac {1}{\sqrt {\cos \left (2\,x\right )+1}\,\sqrt {\frac {1}{\cos \left (2\,x\right )}}} \]

[In]

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

[Out]

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