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

Optimal. Leaf size=47 \[ \frac{\sqrt{\sin (2 e+2 f x)} \sec (e+f x) F\left (\left .e+f x-\frac{\pi }{4}\right |2\right )}{f \sqrt{d \tan (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.0587222, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2614, 2573, 2641} \[ \frac{\sqrt{\sin (2 e+2 f x)} \sec (e+f x) F\left (\left .e+f x-\frac{\pi }{4}\right |2\right )}{f \sqrt{d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2614

Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[e + f*x]]/(Sqrt[Co
s[e + f*x]]*Sqrt[b*Tan[e + f*x]]), Int[1/(Sqrt[Cos[e + f*x]]*Sqrt[Sin[e + f*x]]), x], x] /; FreeQ[{b, e, f}, x
]

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

Mathematica [C]  time = 0.124286, size = 77, normalized size = 1.64 \[ -\frac{2 \sqrt [4]{-1} \sqrt{\tan (e+f x)} \sec ^3(e+f x) F\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-1} \sqrt{\tan (e+f x)}\right )\right |-1\right )}{f \left (\tan ^2(e+f x)+1\right )^{3/2} \sqrt{d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(-1)^(1/4)*EllipticF[I*ArcSinh[(-1)^(1/4)*Sqrt[Tan[e + f*x]]], -1]*Sec[e + f*x]^3*Sqrt[Tan[e + f*x]])/(f*S
qrt[d*Tan[e + f*x]]*(1 + Tan[e + f*x]^2)^(3/2))

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Maple [B]  time = 0.122, size = 165, normalized size = 3.5 \begin{align*} -{\frac{\sqrt{2} \left ( \cos \left ( fx+e \right ) -1 \right ) \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}{f \left ( \sin \left ( fx+e \right ) \right ) ^{2}\cos \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 ) \sqrt{{\frac{\cos \left ( fx+e \right ) -1}{\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{\cos \left ( fx+e \right ) -1+\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}}{\frac{1}{\sqrt{{\frac{d\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/f*2^(1/2)*EllipticF(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))*((cos(f*x+e)-1)/sin(f*x+e))^(
1/2)*((cos(f*x+e)-1+sin(f*x+e))/sin(f*x+e))^(1/2)*((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*(cos(f*x+e)-1)/
sin(f*x+e)^2/cos(f*x+e)*(cos(f*x+e)+1)^2/(d*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{\sec \left (f x + e\right )}{\sqrt{d \tan \left (f x + e\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sec(f*x + e)/sqrt(d*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 \tan \left (f x + e\right )} \sec \left (f x + e\right )}{d \tan \left (f x + e\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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