3.679 \(\int \frac{\sqrt{\sec (c+d x)}}{\sqrt{2-3 \sec (c+d x)}} \, dx\)

Optimal. Leaf size=54 \[ \frac{2 \sqrt{3-2 \cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),-4\right )}{d \sqrt{2-3 \sec (c+d x)}} \]

[Out]

(2*Sqrt[3 - 2*Cos[c + d*x]]*EllipticF[(c + d*x)/2, -4]*Sqrt[Sec[c + d*x]])/(d*Sqrt[2 - 3*Sec[c + d*x]])

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Rubi [A]  time = 0.0692775, antiderivative size = 54, 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 = {3858, 2663, 2661} \[ \frac{2 \sqrt{3-2 \cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |-4\right )}{d \sqrt{2-3 \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Sec[c + d*x]]/Sqrt[2 - 3*Sec[c + d*x]],x]

[Out]

(2*Sqrt[3 - 2*Cos[c + d*x]]*EllipticF[(c + d*x)/2, -4]*Sqrt[Sec[c + d*x]])/(d*Sqrt[2 - 3*Sec[c + d*x]])

Rule 3858

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(Sqrt[d*
Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]])/Sqrt[a + b*Csc[e + f*x]], Int[1/Sqrt[b + a*Sin[e + f*x]], x], x] /; Fr
eeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rubi steps

\begin{align*} \int \frac{\sqrt{\sec (c+d x)}}{\sqrt{2-3 \sec (c+d x)}} \, dx &=\frac{\left (\sqrt{-3+2 \cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{-3+2 \cos (c+d x)}} \, dx}{\sqrt{2-3 \sec (c+d x)}}\\ &=\frac{\left (\sqrt{3-2 \cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{3-2 \cos (c+d x)}} \, dx}{\sqrt{2-3 \sec (c+d x)}}\\ &=\frac{2 \sqrt{3-2 \cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |-4\right ) \sqrt{\sec (c+d x)}}{d \sqrt{2-3 \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.0352445, size = 54, normalized size = 1. \[ \frac{2 \sqrt{3-2 \cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),-4\right )}{d \sqrt{2-3 \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Sec[c + d*x]]/Sqrt[2 - 3*Sec[c + d*x]],x]

[Out]

(2*Sqrt[3 - 2*Cos[c + d*x]]*EllipticF[(c + d*x)/2, -4]*Sqrt[Sec[c + d*x]])/(d*Sqrt[2 - 3*Sec[c + d*x]])

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Maple [A]  time = 0.226, size = 144, normalized size = 2.7 \begin{align*}{\frac{-{\frac{i}{5}}\sqrt{5}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}\sqrt{2}}{d \left ( 2\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}-5\,\cos \left ( dx+c \right ) +3 \right ) }\sqrt{ \left ( \cos \left ( dx+c \right ) \right ) ^{-1}}\sqrt{{\frac{-3+2\,\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }}}{\it EllipticF} \left ({\frac{i\sqrt{5} \left ( -1+\cos \left ( dx+c \right ) \right ) }{\sin \left ( dx+c \right ) }},{\frac{\sqrt{5}}{5}} \right ) \sqrt{-2\,{\frac{-3+2\,\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^(1/2)/(2-3*sec(d*x+c))^(1/2),x)

[Out]

-1/5*I/d*5^(1/2)*cos(d*x+c)*sin(d*x+c)^2*(1/cos(d*x+c))^(1/2)*((-3+2*cos(d*x+c))/cos(d*x+c))^(1/2)*EllipticF(I
*5^(1/2)*(-1+cos(d*x+c))/sin(d*x+c),1/5*5^(1/2))*2^(1/2)*(-2*(-3+2*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d
*x+c)+1))^(1/2)/(2*cos(d*x+c)^2-5*cos(d*x+c)+3)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\sec \left (d x + c\right )}}{\sqrt{-3 \, \sec \left (d x + c\right ) + 2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(1/2)/(2-3*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(sec(d*x + c))/sqrt(-3*sec(d*x + c) + 2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{-3 \, \sec \left (d x + c\right ) + 2} \sqrt{\sec \left (d x + c\right )}}{3 \, \sec \left (d x + c\right ) - 2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(1/2)/(2-3*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-3*sec(d*x + c) + 2)*sqrt(sec(d*x + c))/(3*sec(d*x + c) - 2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\sec{\left (c + d x \right )}}}{\sqrt{2 - 3 \sec{\left (c + d x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**(1/2)/(2-3*sec(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(sec(c + d*x))/sqrt(2 - 3*sec(c + d*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\sec \left (d x + c\right )}}{\sqrt{-3 \, \sec \left (d x + c\right ) + 2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(1/2)/(2-3*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(sec(d*x + c))/sqrt(-3*sec(d*x + c) + 2), x)