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

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Rubi [A]  time = 0.0198659, antiderivative size = 13, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, 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 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
]

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]

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*}

Mathematica [A]  time = 0.0070587, size = 13, normalized size = 1. \[ \frac{\tan (x)}{\sqrt{-\sec ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[-1 - Tan[x]^2],x]

[Out]

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

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Maple [A]  time = 0.029, size = 14, normalized size = 1.1 \begin{align*}{\tan \left ( x \right ){\frac{1}{\sqrt{-1- \left ( \tan \left ( x \right ) \right ) ^{2}}}}} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{-\tan \left (x\right )^{2} - 1}}\,{d x} \end{align*}

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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Fricas [C]  time = 1.71258, size = 42, normalized size = 3.23 \begin{align*} -\frac{1}{2} \,{\left (e^{\left (2 i \, x\right )} - 1\right )} e^{\left (-i \, x\right )} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{- \tan ^{2}{\left (x \right )} - 1}}\, dx \end{align*}

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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Giac [C]  time = 1.08866, size = 16, normalized size = 1.23 \begin{align*} -\frac{i \, \tan \left (x\right )}{\sqrt{\tan \left (x\right )^{2} + 1}} \end{align*}

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)