3.458 \(\int \frac{\sqrt{b \sec (e+f x)}}{\sqrt{a \sin (e+f x)}} \, dx\)

Optimal. Leaf size=53 \[ \frac{\sqrt{\sin (2 e+2 f x)} F\left (\left .e+f x-\frac{\pi }{4}\right |2\right ) \sqrt{b \sec (e+f x)}}{f \sqrt{a \sin (e+f x)}} \]

[Out]

(EllipticF[e - Pi/4 + f*x, 2]*Sqrt[b*Sec[e + f*x]]*Sqrt[Sin[2*e + 2*f*x]])/(f*Sqrt[a*Sin[e + f*x]])

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Rubi [A]  time = 0.0990154, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {2585, 2573, 2641} \[ \frac{\sqrt{\sin (2 e+2 f x)} F\left (\left .e+f x-\frac{\pi }{4}\right |2\right ) \sqrt{b \sec (e+f x)}}{f \sqrt{a \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*Sec[e + f*x]]/Sqrt[a*Sin[e + f*x]],x]

[Out]

(EllipticF[e - Pi/4 + f*x, 2]*Sqrt[b*Sec[e + f*x]]*Sqrt[Sin[2*e + 2*f*x]])/(f*Sqrt[a*Sin[e + f*x]])

Rule 2585

Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[(b*Cos[e + f*
x])^n*(b*Sec[e + f*x])^n, Int[(a*Sin[e + f*x])^m/(b*Cos[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, m, n}, x] &&
IntegerQ[m - 1/2] && IntegerQ[n - 1/2]

Rule 2573

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt[Sin[2*
e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b*Cos[e + f*x]]), Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b,
e, f}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{\sqrt{b \sec (e+f x)}}{\sqrt{a \sin (e+f x)}} \, dx &=\left (\sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}\right ) \int \frac{1}{\sqrt{b \cos (e+f x)} \sqrt{a \sin (e+f x)}} \, dx\\ &=\frac{\left (\sqrt{b \sec (e+f x)} \sqrt{\sin (2 e+2 f x)}\right ) \int \frac{1}{\sqrt{\sin (2 e+2 f x)}} \, dx}{\sqrt{a \sin (e+f x)}}\\ &=\frac{F\left (\left .e-\frac{\pi }{4}+f x\right |2\right ) \sqrt{b \sec (e+f x)} \sqrt{\sin (2 e+2 f x)}}{f \sqrt{a \sin (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.371877, size = 66, normalized size = 1.25 \[ \frac{\left (-\tan ^2(e+f x)\right )^{3/4} \cot (e+f x) \sqrt{b \sec (e+f x)} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{3}{2};\sec ^2(e+f x)\right )}{f \sqrt{a \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*Sec[e + f*x]]/Sqrt[a*Sin[e + f*x]],x]

[Out]

(Cot[e + f*x]*Hypergeometric2F1[1/2, 3/4, 3/2, Sec[e + f*x]^2]*Sqrt[b*Sec[e + f*x]]*(-Tan[e + f*x]^2)^(3/4))/(
f*Sqrt[a*Sin[e + f*x]])

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Maple [B]  time = 0.13, size = 153, normalized size = 2.9 \begin{align*} -{\frac{\sqrt{2} \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{f \left ( -1+\cos \left ( fx+e \right ) \right ) }\sqrt{{\frac{b}{\cos \left ( fx+e \right ) }}}\sqrt{{\frac{1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}}\sqrt{{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}}\sqrt{{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}}{\it EllipticF} \left ( \sqrt{{\frac{1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}},{\frac{\sqrt{2}}{2}} \right ){\frac{1}{\sqrt{a\sin \left ( fx+e \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*sec(f*x+e))^(1/2)/(a*sin(f*x+e))^(1/2),x)

[Out]

-1/f*2^(1/2)*(b/cos(f*x+e))^(1/2)*sin(f*x+e)^2*((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+si
n(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticF(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e)
)^(1/2),1/2*2^(1/2))/(a*sin(f*x+e))^(1/2)/(-1+cos(f*x+e))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{b \sec \left (f x + e\right )}}{\sqrt{a \sin \left (f x + e\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*sec(f*x + e))/sqrt(a*sin(f*x + e)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b \sec \left (f x + e\right )} \sqrt{a \sin \left (f x + e\right )}}{a \sin \left (f x + e\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*sec(f*x + e))*sqrt(a*sin(f*x + e))/(a*sin(f*x + e)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))**(1/2)/(a*sin(f*x+e))**(1/2),x)

[Out]

Integral(sqrt(b*sec(e + f*x))/sqrt(a*sin(e + f*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{b \sec \left (f x + e\right )}}{\sqrt{a \sin \left (f x + e\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*sec(f*x + e))/sqrt(a*sin(f*x + e)), x)