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

Optimal. Leaf size=14 \[ \tanh ^{-1}\left (\frac {\tan (x)}{\sqrt {\tan ^2(x)-4}}\right ) \]

[Out]

arctanh(tan(x)/(-4+tan(x)^2)^(1/2))

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Rubi [A]  time = 0.04, antiderivative size = 14, 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 = {3675, 217, 206} \[ \tanh ^{-1}\left (\frac {\tan (x)}{\sqrt {\tan ^2(x)-4}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]^2/Sqrt[-4 + Tan[x]^2],x]

[Out]

ArcTanh[Tan[x]/Sqrt[-4 + Tan[x]^2]]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

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

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])

Rubi steps

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

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Mathematica [B]  time = 0.05, size = 46, normalized size = 3.29 \[ \frac {\sqrt {5 \cos (2 x)+3} \sec (x) \tan ^{-1}\left (\frac {\sin (x)}{\sqrt {4-5 \sin ^2(x)}}\right )}{\sqrt {2} \sqrt {\tan ^2(x)-4}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]^2/Sqrt[-4 + Tan[x]^2],x]

[Out]

(ArcTan[Sin[x]/Sqrt[4 - 5*Sin[x]^2]]*Sqrt[3 + 5*Cos[2*x]]*Sec[x])/(Sqrt[2]*Sqrt[-4 + Tan[x]^2])

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fricas [B]  time = 1.00, size = 67, normalized size = 4.79 \[ \frac {1}{4} \, \log \left (\frac {1}{2} \, \sqrt {-\frac {5 \, \cos \relax (x)^{2} - 1}{\cos \relax (x)^{2}}} \cos \relax (x) \sin \relax (x) - \frac {3}{2} \, \cos \relax (x)^{2} + \frac {1}{2}\right ) - \frac {1}{4} \, \log \left (-\frac {1}{2} \, \sqrt {-\frac {5 \, \cos \relax (x)^{2} - 1}{\cos \relax (x)^{2}}} \cos \relax (x) \sin \relax (x) - \frac {3}{2} \, \cos \relax (x)^{2} + \frac {1}{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*log(1/2*sqrt(-(5*cos(x)^2 - 1)/cos(x)^2)*cos(x)*sin(x) - 3/2*cos(x)^2 + 1/2) - 1/4*log(-1/2*sqrt(-(5*cos(x
)^2 - 1)/cos(x)^2)*cos(x)*sin(x) - 3/2*cos(x)^2 + 1/2)

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giac [A]  time = 0.18, size = 17, normalized size = 1.21 \[ -\log \left ({\left | \sqrt {\tan \relax (x)^{2} - 4} - \tan \relax (x) \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(abs(sqrt(tan(x)^2 - 4) - tan(x)))

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maple [C]  time = 0.72, size = 171, normalized size = 12.21 \[ -\frac {\sqrt {-\frac {2 \left (\cos \relax (x ) \sqrt {5}-5 \cos \relax (x )-\sqrt {5}+1\right )}{1+\cos \relax (x )}}\, \sqrt {2}\, \sqrt {\frac {\cos \relax (x ) \sqrt {5}-\sqrt {5}+5 \cos \relax (x )-1}{1+\cos \relax (x )}}\, \left (\EllipticF \left (\frac {\left (-1+\cos \relax (x )\right ) \left (\sqrt {5}-1\right )}{2 \sin \relax (x )}, \frac {3}{2}+\frac {\sqrt {5}}{2}\right )-2 \EllipticPi \left (\frac {\sqrt {\frac {3}{2}-\frac {\sqrt {5}}{2}}\, \left (-1+\cos \relax (x )\right )}{\sin \relax (x )}, -\frac {2}{\sqrt {5}-3}, \frac {\sqrt {\frac {3}{2}+\frac {\sqrt {5}}{2}}}{\sqrt {\frac {3}{2}-\frac {\sqrt {5}}{2}}}\right )\right ) \left (\sin ^{2}\relax (x )\right )}{4 \sqrt {-\frac {5 \left (\cos ^{2}\relax (x )\right )-1}{\cos \relax (x )^{2}}}\, \cos \relax (x ) \left (-1+\cos \relax (x )\right ) \sqrt {\frac {3}{2}-\frac {\sqrt {5}}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(-2*(cos(x)*5^(1/2)-5*cos(x)-5^(1/2)+1)/(1+cos(x)))^(1/2)*2^(1/2)*((cos(x)*5^(1/2)-5^(1/2)+5*cos(x)-1)/(1
+cos(x)))^(1/2)*(EllipticF(1/2*(-1+cos(x))*(5^(1/2)-1)/sin(x),3/2+1/2*5^(1/2))-2*EllipticPi((3/2-1/2*5^(1/2))^
(1/2)*(-1+cos(x))/sin(x),-2/(5^(1/2)-3),(3/2+1/2*5^(1/2))^(1/2)/(3/2-1/2*5^(1/2))^(1/2)))*sin(x)^2/(-(5*cos(x)
^2-1)/cos(x)^2)^(1/2)/cos(x)/(-1+cos(x))/(3/2-1/2*5^(1/2))^(1/2)

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maxima [A]  time = 0.31, size = 16, normalized size = 1.14 \[ \log \left (2 \, \sqrt {\tan \relax (x)^{2} - 4} + 2 \, \tan \relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(2*sqrt(tan(x)^2 - 4) + 2*tan(x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.07 \[ \int \frac {1}{{\cos \relax (x)}^2\,\sqrt {{\mathrm {tan}\relax (x)}^2-4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{2}{\relax (x )}}{\sqrt {\left (\tan {\relax (x )} - 2\right ) \left (\tan {\relax (x )} + 2\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sec(x)**2/sqrt((tan(x) - 2)*(tan(x) + 2)), x)

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