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

Optimal. Leaf size=5 \[ \text {csch}^{-1}(2 \cos (x)) \]

[Out]

arccsch(2*cos(x))

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Rubi [A]  time = 0.04, antiderivative size = 5, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4339, 335, 215} \[ \text {csch}^{-1}(2 \cos (x)) \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcCsch[2*Cos[x]]

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 335

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

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

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Mathematica [B]  time = 0.03, size = 38, normalized size = 7.60 \[ \frac {\sqrt {2 \cos (2 x)+3} \sec (x) \tanh ^{-1}\left (\sqrt {4 \cos ^2(x)+1}\right )}{\sqrt {\sec ^2(x)+4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(ArcTanh[Sqrt[1 + 4*Cos[x]^2]]*Sqrt[3 + 2*Cos[2*x]]*Sec[x])/Sqrt[4 + Sec[x]^2]

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fricas [B]  time = 0.95, size = 27, normalized size = 5.40 \[ \log \left (-\frac {\sqrt {\frac {4 \, \cos \relax (x)^{2} + 1}{\cos \relax (x)^{2}}} \cos \relax (x) + 1}{\cos \relax (x)}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)*tan(x)/(4+sec(x)^2)^(1/2),x, algorithm="fricas")

[Out]

log(-(sqrt((4*cos(x)^2 + 1)/cos(x)^2)*cos(x) + 1)/cos(x))

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giac [B]  time = 0.15, size = 36, normalized size = 7.20 \[ \frac {\log \left (\sqrt {4 \, \cos \relax (x)^{2} + 1} + 1\right ) - \log \left (\sqrt {4 \, \cos \relax (x)^{2} + 1} - 1\right )}{2 \, \mathrm {sgn}\left (\cos \relax (x)\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)*tan(x)/(4+sec(x)^2)^(1/2),x, algorithm="giac")

[Out]

1/2*(log(sqrt(4*cos(x)^2 + 1) + 1) - log(sqrt(4*cos(x)^2 + 1) - 1))/sgn(cos(x))

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maple [A]  time = 0.10, size = 6, normalized size = 1.20 \[ \arcsinh \left (\frac {\sec \relax (x )}{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x)*tan(x)/(4+sec(x)^2)^(1/2),x)

[Out]

arcsinh(1/2*sec(x))

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maxima [B]  time = 0.31, size = 33, normalized size = 6.60 \[ \frac {1}{2} \, \log \left (\sqrt {\frac {1}{\cos \relax (x)^{2}} + 4} \cos \relax (x) + 1\right ) - \frac {1}{2} \, \log \left (\sqrt {\frac {1}{\cos \relax (x)^{2}} + 4} \cos \relax (x) - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)*tan(x)/(4+sec(x)^2)^(1/2),x, algorithm="maxima")

[Out]

1/2*log(sqrt(1/cos(x)^2 + 4)*cos(x) + 1) - 1/2*log(sqrt(1/cos(x)^2 + 4)*cos(x) - 1)

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mupad [B]  time = 3.10, size = 7, normalized size = 1.40 \[ \mathrm {asinh}\left (\frac {1}{2\,\cos \relax (x)}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)/(cos(x)*(1/cos(x)^2 + 4)^(1/2)),x)

[Out]

asinh(1/(2*cos(x)))

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sympy [A]  time = 0.73, size = 5, normalized size = 1.00 \[ \operatorname {asinh}{\left (\frac {\sec {\relax (x )}}{2} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)*tan(x)/(4+sec(x)**2)**(1/2),x)

[Out]

asinh(sec(x)/2)

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