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

Optimal. Leaf size=19 \[ -\frac{\sqrt{a \cos ^2(e+f x)}}{f} \]

[Out]

-(Sqrt[a*Cos[e + f*x]^2]/f)

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Rubi [A]  time = 0.0629348, antiderivative size = 19, 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} \[ -\frac{\sqrt{a \cos ^2(e+f x)}}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[a*Cos[e + f*x]^2]/f)

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

Mathematica [A]  time = 0.0397946, size = 19, normalized size = 1. \[ -\frac{\sqrt{a \cos ^2(e+f x)}}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[a*Cos[e + f*x]^2]/f)

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Maple [A]  time = 0.131, size = 21, normalized size = 1.1 \begin{align*} -{\frac{1}{f}\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((a-a*sin(f*x+e)^2)^(1/2)*tan(f*x+e),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((a-a*sin(f*x+e)^2)^(1/2)*tan(f*x+e),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.6049, size = 36, normalized size = 1.89 \begin{align*} -\frac{\sqrt{a \cos \left (f x + e\right )^{2}}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(a*cos(f*x + e)^2)/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.1338, size = 53, normalized size = 2.79 \begin{align*} \frac{2 \, \sqrt{a} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 1\right )}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1\right )} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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