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

Optimal. Leaf size=62 \[ \frac {2 \sqrt {2-3 \cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\frac {1}{2} (c+d x+\pi )|\frac {6}{5}\right )}{\sqrt {5} d \sqrt {2 \sec (c+d x)-3}} \]

[Out]

-2/5*(sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)*EllipticF(cos(1/2*d*x+1/2*c),1/5*30^(1/2))*(2-3*cos(d*x+c
))^(1/2)*sec(d*x+c)^(1/2)/d*5^(1/2)/(-3+2*sec(d*x+c))^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {3858, 2662} \[ \frac {2 \sqrt {2-3 \cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\frac {1}{2} (c+d x+\pi )|\frac {6}{5}\right )}{\sqrt {5} d \sqrt {2 \sec (c+d x)-3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[2 - 3*Cos[c + d*x]]*EllipticF[(c + Pi + d*x)/2, 6/5]*Sqrt[Sec[c + d*x]])/(Sqrt[5]*d*Sqrt[-3 + 2*Sec[c
+ d*x]])

Rule 2662

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]

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]

Rubi steps

\begin {align*} \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {-3+2 \sec (c+d x)}} \, dx &=\frac {\left (\sqrt {2-3 \cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {2-3 \cos (c+d x)}} \, dx}{\sqrt {-3+2 \sec (c+d x)}}\\ &=\frac {2 \sqrt {2-3 \cos (c+d x)} F\left (\frac {1}{2} (c+\pi +d x)|\frac {6}{5}\right ) \sqrt {\sec (c+d x)}}{\sqrt {5} d \sqrt {-3+2 \sec (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 54, normalized size = 0.87 \[ \frac {2 \sqrt {3 \cos (c+d x)-2} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |6\right )}{d \sqrt {2 \sec (c+d x)-3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[-2 + 3*Cos[c + d*x]]*EllipticF[(c + d*x)/2, 6]*Sqrt[Sec[c + d*x]])/(d*Sqrt[-3 + 2*Sec[c + d*x]])

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {\sec \left (d x + c\right )}}{\sqrt {2 \, \sec \left (d x + c\right ) - 3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\sec \left (d x + c\right )}}{\sqrt {2 \, \sec \left (d x + c\right ) - 3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 1.61, size = 136, normalized size = 2.19 \[ \frac {2 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {-2+3 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i \sqrt {5}\right ) \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )}}\, \sqrt {-\frac {-2+3 \cos \left (d x +c \right )}{\cos \left (d x +c \right )}}}{d \left (3 \left (\cos ^{2}\left (d x +c \right )\right )-5 \cos \left (d x +c \right )+2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\sec \left (d x + c\right )}}{\sqrt {2 \, \sec \left (d x + c\right ) - 3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}}{\sqrt {\frac {2}{\cos \left (c+d\,x\right )}-3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\sec {\left (c + d x \right )}}}{\sqrt {2 \sec {\left (c + d x \right )} - 3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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