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

Optimal. Leaf size=61 \[ \frac {2 \sqrt {2 \cos (c+d x)+3} \sqrt {\sec (c+d x)} F\left (\frac {1}{2} (c+d x)|\frac {4}{5}\right )}{\sqrt {5} d \sqrt {-3 \sec (c+d x)-2}} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {3858, 2663, 2661} \[ \frac {2 \sqrt {2 \cos (c+d x)+3} \sqrt {\sec (c+d x)} F\left (\frac {1}{2} (c+d x)|\frac {4}{5}\right )}{\sqrt {5} d \sqrt {-3 \sec (c+d x)-2}} \]

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, 4/5]*Sqrt[Sec[c + d*x]])/(Sqrt[5]*d*Sqrt[-2 - 3*Sec[c + d*x
]])

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]

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 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 {-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 {\frac {3}{5}+\frac {2}{5} \cos (c+d x)}} \, dx}{\sqrt {5} \sqrt {-2-3 \sec (c+d x)}}\\ &=\frac {2 \sqrt {3+2 \cos (c+d x)} F\left (\frac {1}{2} (c+d x)|\frac {4}{5}\right ) \sqrt {\sec (c+d x)}}{\sqrt {5} d \sqrt {-2-3 \sec (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 61, normalized size = 1.00 \[ \frac {2 \sqrt {2 \cos (c+d x)+3} \sqrt {\sec (c+d x)} F\left (\frac {1}{2} (c+d x)|\frac {4}{5}\right )}{\sqrt {5} d \sqrt {-3 \sec (c+d x)-2}} \]

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, 4/5]*Sqrt[Sec[c + d*x]])/(Sqrt[5]*d*Sqrt[-2 - 3*Sec[c + d*x
]])

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-3 \, \sec \left (d x + c\right ) - 2} \sqrt {\sec \left (d x + c\right )}}{3 \, \sec \left (d x + c\right ) + 2}, x\right ) \]

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

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)

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maple [C]  time = 1.65, size = 139, normalized size = 2.28 \[ \frac {i \sqrt {2}\, \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {10}\, \sqrt {\frac {3+2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )}}\, \sqrt {-\frac {3+2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, \frac {\sqrt {5}}{5}\right ) \left (\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{5 d \left (2 \left (\cos ^{2}\left (d x +c \right )\right )+\cos \left (d x +c \right )-3\right )} \]

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*2^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*10^(1/2)*((3+2*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(1/cos(d*x+c))^(1/2)
*(-(3+2*cos(d*x+c))/cos(d*x+c))^(1/2)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),1/5*5^(1/2))*sin(d*x+c)^2*cos(d*x
+c)/(2*cos(d*x+c)^2+cos(d*x+c)-3)

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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(-3*sec(c + d*x) - 2), x)

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