3.292 \(\int \frac {1}{\sqrt {-1-\tan ^2(x)}} \, dx\)

Optimal. Leaf size=13 \[ \frac {\tan (x)}{\sqrt {-\sec ^2(x)}} \]

[Out]

tan(x)/(-sec(x)^2)^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3657, 4122, 191} \[ \frac {\tan (x)}{\sqrt {-\sec ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[-1 - Tan[x]^2],x]

[Out]

Tan[x]/Sqrt[-Sec[x]^2]

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 3657

Int[(u_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(a*sec[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a, b]

Rule 4122

Int[((b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(b*ff)
/f, Subst[Int[(b + b*ff^2*x^2)^(p - 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{b, e, f, p}, x] &&  !IntegerQ[p
]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 13, normalized size = 1.00 \[ \frac {\tan (x)}{\sqrt {-\sec ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[-1 - Tan[x]^2],x]

[Out]

Tan[x]/Sqrt[-Sec[x]^2]

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fricas [C]  time = 0.39, size = 12, normalized size = 0.92 \[ -\frac {1}{2} \, {\left (e^{\left (2 i \, x\right )} - 1\right )} e^{\left (-i \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(e^(2*I*x) - 1)*e^(-I*x)

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giac [C]  time = 0.35, size = 12, normalized size = 0.92 \[ -\frac {i \, \tan \relax (x)}{\sqrt {\tan \relax (x)^{2} + 1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*tan(x)/sqrt(tan(x)^2 + 1)

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maple [A]  time = 0.24, size = 14, normalized size = 1.08 \[ \frac {\tan \relax (x )}{\sqrt {-1-\left (\tan ^{2}\relax (x )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-1-tan(x)^2)^(1/2),x)

[Out]

tan(x)/(-1-tan(x)^2)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {-\tan \relax (x)^{2} - 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt(-tan(x)^2 - 1), x)

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mupad [B]  time = 11.98, size = 13, normalized size = 1.00 \[ -\frac {\sqrt {2}\,\sin \left (2\,x\right )\,1{}\mathrm {i}}{2\,\sqrt {2\,{\cos \relax (x)}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(- tan(x)^2 - 1)^(1/2),x)

[Out]

-(2^(1/2)*sin(2*x)*1i)/(2*(2*cos(x)^2)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {- \tan ^{2}{\relax (x )} - 1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-1-tan(x)**2)**(1/2),x)

[Out]

Integral(1/sqrt(-tan(x)**2 - 1), x)

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