\(\int \frac {\tan (e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx\) [470]

Optimal result

Integrand size = 24, antiderivative size = 18 \[ \int \frac {\tan (e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\frac {1}{f \sqrt {a \cos ^2(e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3255, 3284, 16, 32} \[ \int \frac {\tan (e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\frac {1}{f \sqrt {a \cos ^2(e+f x)}} \]

[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 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]

Rule 3255

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 3284

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
]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\tan (e+f x)}{\sqrt {a \cos ^2(e+f x)}} \, dx \\ & = -\frac {\text {Subst}\left (\int \frac {1}{x \sqrt {a x}} \, dx,x,\cos ^2(e+f x)\right )}{2 f} \\ & = -\frac {a \text {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] (verified)

Time = 0.03 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {\tan (e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\frac {1}{f \sqrt {a \cos ^2(e+f x)}} \]

[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])

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11

method	result	size
derivativedivides	\(\frac {1}{\sqrt {a -a \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\)	\(20\)
default	\(\frac {1}{\sqrt {a -a \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\)	\(20\)
risch	\(\frac {2}{\sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, f}\)	\(32\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.50 \[ \int \frac {\tan (e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\frac {\sqrt {a \cos \left (f x + e\right )^{2}}}{a f \cos \left (f x + e\right )^{2}} \]

[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)

Sympy [F]

\[ \int \frac {\tan (e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\int \frac {\tan {\left (e + f x \right )}}{\sqrt {- a \left (\sin {\left (e + f x \right )} - 1\right ) \left (\sin {\left (e + f x \right )} + 1\right )}}\, dx \]

[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)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (16) = 32\).

Time = 0.31 (sec) , antiderivative size = 65, normalized size of antiderivative = 3.61 \[ \int \frac {\tan (e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\frac {\frac {\sqrt {-a \sin \left (f x + e\right )^{2} + a}}{a \sin \left (f x + e\right ) + a} - \frac {\sqrt {-a \sin \left (f x + e\right )^{2} + a}}{a \sin \left (f x + e\right ) - a}}{2 \, f} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 39 vs. \(2 (16) = 32\).

Time = 0.42 (sec) , antiderivative size = 39, normalized size of antiderivative = 2.17 \[ \int \frac {\tan (e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\frac {2}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} \sqrt {a} f \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right )} \]

[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))

Mupad [B] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 61, normalized size of antiderivative = 3.39 \[ \int \frac {\tan (e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\frac {2\,\sqrt {2}\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )\,\sqrt {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )}}{a\,f\,\left (4\,\cos \left (2\,e+2\,f\,x\right )+\cos \left (4\,e+4\,f\,x\right )+3\right )} \]

[In]

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

[Out]

(2*2^(1/2)*(cos(2*e + 2*f*x) + 1)*(a*(cos(2*e + 2*f*x) + 1))^(1/2))/(a*f*(4*cos(2*e + 2*f*x) + cos(4*e + 4*f*x
) + 3))