3.318 \(\int \frac{\sqrt{d \sec (e+f x)}}{\sqrt{b \tan (e+f x)}} \, dx\)

Optimal. Leaf size=55 \[ \frac{2 \sqrt{\sin (e+f x)} F\left (\left .\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{d \sec (e+f x)}}{f \sqrt{b \tan (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.061367, antiderivative size = 55, 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 = {2616, 2642, 2641} \[ \frac{2 \sqrt{\sin (e+f x)} F\left (\left .\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{d \sec (e+f x)}}{f \sqrt{b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[d*Sec[e + f*x]]/Sqrt[b*Tan[e + f*x]],x]

[Out]

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

Rule 2616

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

Rule 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, 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{d \sec (e+f x)}}{\sqrt{b \tan (e+f x)}} \, dx &=\frac{\left (\sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}\right ) \int \frac{1}{\sqrt{b \sin (e+f x)}} \, dx}{\sqrt{b \tan (e+f x)}}\\ &=\frac{\left (\sqrt{d \sec (e+f x)} \sqrt{\sin (e+f x)}\right ) \int \frac{1}{\sqrt{\sin (e+f x)}} \, dx}{\sqrt{b \tan (e+f x)}}\\ &=\frac{2 F\left (\left .\frac{1}{2} \left (e-\frac{\pi }{2}+f x\right )\right |2\right ) \sqrt{d \sec (e+f x)} \sqrt{\sin (e+f x)}}{f \sqrt{b \tan (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.444065, size = 89, normalized size = 1.62 \[ \frac{2 \sqrt{b \tan (e+f x)} \sqrt{d \sec (e+f x)} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )}{b f \sqrt{\sec (e+f x)} \sqrt{\cos ^2\left (\frac{1}{2} (e+f x)\right ) \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[d*Sec[e + f*x]]/Sqrt[b*Tan[e + f*x]],x]

[Out]

(2*Hypergeometric2F1[1/4, 1/2, 5/4, -Tan[(e + f*x)/2]^2]*Sqrt[d*Sec[e + f*x]]*Sqrt[b*Tan[e + f*x]])/(b*f*Sqrt[
Sec[e + f*x]]*Sqrt[Cos[(e + f*x)/2]^2*Sec[e + f*x]])

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Maple [C]  time = 0.207, size = 175, normalized size = 3.2 \begin{align*}{\frac{-i \left ( \sin \left ( fx+e \right ) \right ) ^{2}\sqrt{2}}{f \left ( \cos \left ( fx+e \right ) -1 \right ) }\sqrt{{\frac{i\cos \left ( fx+e \right ) -i+\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}}\sqrt{-{\frac{i\cos \left ( fx+e \right ) -i-\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}}{\it EllipticF} \left ( \sqrt{{\frac{i\cos \left ( fx+e \right ) -i+\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}},{\frac{\sqrt{2}}{2}} \right ) \sqrt{{\frac{-i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }}}\sqrt{{\frac{d}{\cos \left ( fx+e \right ) }}}{\frac{1}{\sqrt{{\frac{b\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sec(f*x+e))^(1/2)/(b*tan(f*x+e))^(1/2),x)

[Out]

-I/f*EllipticF(((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))*(-I*(cos(f*x+e)-1)/sin(f*x+e))^(1/2
)*(-(I*cos(f*x+e)-I-sin(f*x+e))/sin(f*x+e))^(1/2)*((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2)*(d/cos(f*x+e)
)^(1/2)*sin(f*x+e)^2*2^(1/2)/(cos(f*x+e)-1)/(b*sin(f*x+e)/cos(f*x+e))^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(d*sec(f*x + e))/sqrt(b*tan(f*x + e)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(d*sec(f*x + e))*sqrt(b*tan(f*x + e))/(b*tan(f*x + e)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))**(1/2)/(b*tan(f*x+e))**(1/2),x)

[Out]

Integral(sqrt(d*sec(e + f*x))/sqrt(b*tan(e + f*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(d*sec(f*x + e))/sqrt(b*tan(f*x + e)), x)