3.470 \(\int \frac{\tan (e+f x)}{\sqrt{a-a \sin ^2(e+f x)}} \, dx\)

Optimal. Leaf size=18 \[ \frac{1}{f \sqrt{a \cos ^2(e+f x)}} \]

[Out]

1/(f*Sqrt[a*Cos[e + f*x]^2])

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Rubi [A]  time = 0.0645136, antiderivative size = 18, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3176, 3205, 16, 32} \[ \frac{1}{f \sqrt{a \cos ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[e + f*x]/Sqrt[a - a*Sin[e + f*x]^2],x]

[Out]

1/(f*Sqrt[a*Cos[e + f*x]^2])

Rule 3176

Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(a*cos[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0]

Rule 3205

Int[((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFact
ors[Sin[e + f*x]^2, x]}, Dist[ff^((m + 1)/2)/(2*f), Subst[Int[(x^((m - 1)/2)*(b*ff^(n/2)*x^(n/2))^p)/(1 - ff*x
)^((m + 1)/2), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{b, e, f, p}, x] && IntegerQ[(m - 1)/2] && IntegerQ[n/2
]

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

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

Mathematica [A]  time = 0.0265371, size = 18, normalized size = 1. \[ \frac{1}{f \sqrt{a \cos ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[e + f*x]/Sqrt[a - a*Sin[e + f*x]^2],x]

[Out]

1/(f*Sqrt[a*Cos[e + f*x]^2])

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Maple [A]  time = 0.148, size = 20, normalized size = 1.1 \begin{align*}{\frac{1}{f}{\frac{1}{\sqrt{a-a \left ( \sin \left ( fx+e \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(f*x+e)/(a-a*sin(f*x+e)^2)^(1/2),x)

[Out]

1/f/(a-a*sin(f*x+e)^2)^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.5705, size = 61, normalized size = 3.39 \begin{align*} \frac{\sqrt{a \cos \left (f x + e\right )^{2}}}{a f \cos \left (f x + e\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(a*cos(f*x + e)^2)/(a*f*cos(f*x + e)^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)/(a-a*sin(f*x+e)**2)**(1/2),x)

[Out]

Integral(tan(e + f*x)/sqrt(-a*(sin(e + f*x) - 1)*(sin(e + f*x) + 1)), x)

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Giac [B]  time = 1.21751, size = 55, normalized size = 3.06 \begin{align*} \frac{2}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right )} \sqrt{a} f \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 1\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/((tan(1/2*f*x + 1/2*e)^2 - 1)*sqrt(a)*f*sgn(tan(1/2*f*x + 1/2*e)^4 - 1))