3.275 \(\int \frac{\tan (x)}{\sqrt{a+a \tan ^2(x)}} \, dx\)

Optimal. Leaf size=12 \[ -\frac{1}{\sqrt{a \sec ^2(x)}} \]

[Out]

-(1/Sqrt[a*Sec[x]^2])

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Rubi [A]  time = 0.0464586, antiderivative size = 12, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3657, 4124, 32} \[ -\frac{1}{\sqrt{a \sec ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[x]/Sqrt[a + a*Tan[x]^2],x]

[Out]

-(1/Sqrt[a*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 4124

Int[((b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> Dist[b/(2*f), Subst[In
t[(-1 + x)^((m - 1)/2)*(b*x)^(p - 1), x], x, Sec[e + f*x]^2], x] /; FreeQ[{b, e, f, p}, x] &&  !IntegerQ[p] &&
IntegerQ[(m - 1)/2]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0128336, size = 12, normalized size = 1. \[ -\frac{1}{\sqrt{a \sec ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[x]/Sqrt[a + a*Tan[x]^2],x]

[Out]

-(1/Sqrt[a*Sec[x]^2])

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Maple [A]  time = 0.023, size = 13, normalized size = 1.1 \begin{align*} -{\frac{1}{\sqrt{a+a \left ( \tan \left ( x \right ) \right ) ^{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(a+a*tan(x)^2)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (x\right )}{\sqrt{a \tan \left (x\right )^{2} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(tan(x)/sqrt(a*tan(x)^2 + a), x)

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Fricas [A]  time = 1.37088, size = 34, normalized size = 2.83 \begin{align*} -\frac{1}{\sqrt{a \tan \left (x\right )^{2} + a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/sqrt(a*tan(x)^2 + a)

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Sympy [A]  time = 0.609833, size = 14, normalized size = 1.17 \begin{align*} - \frac{1}{\sqrt{a \tan ^{2}{\left (x \right )} + a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)/(a+a*tan(x)**2)**(1/2),x)

[Out]

-1/sqrt(a*tan(x)**2 + a)

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Giac [A]  time = 1.08739, size = 16, normalized size = 1.33 \begin{align*} -\frac{1}{\sqrt{a \tan \left (x\right )^{2} + a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/sqrt(a*tan(x)^2 + a)