3.670 \(\int \frac {1}{\sqrt {\sec (c+d x)} \sqrt {-2+3 \sec (c+d x)}} \, dx\)

Optimal. Leaf size=109 \[ \frac {3 \sqrt {3-2 \cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |-4\right )}{d \sqrt {3 \sec (c+d x)-2}}-\frac {\sqrt {3 \sec (c+d x)-2} E\left (\left .\frac {1}{2} (c+d x)\right |-4\right )}{d \sqrt {3-2 \cos (c+d x)} \sqrt {\sec (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3862, 3856, 2653, 3858, 2661} \[ \frac {3 \sqrt {3-2 \cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |-4\right )}{d \sqrt {3 \sec (c+d x)-2}}-\frac {\sqrt {3 \sec (c+d x)-2} E\left (\left .\frac {1}{2} (c+d x)\right |-4\right )}{d \sqrt {3-2 \cos (c+d x)} \sqrt {\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[Sec[c + d*x]]*Sqrt[-2 + 3*Sec[c + d*x]]),x]

[Out]

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

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, 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]

Rule 3856

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

Rule 3862

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

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 68, normalized size = 0.62 \[ -\frac {\sqrt {3-2 \cos (c+d x)} \sqrt {\sec (c+d x)} \left (E\left (\left .\frac {1}{2} (c+d x)\right |-4\right )-3 F\left (\left .\frac {1}{2} (c+d x)\right |-4\right )\right )}{d \sqrt {3 \sec (c+d x)-2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[Sec[c + d*x]]*Sqrt[-2 + 3*Sec[c + d*x]]),x]

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/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 - 2*sec(d*x + c)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {3 \, \sec \left (d x + c\right ) - 2} \sqrt {\sec \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 1.88, size = 374, normalized size = 3.43 \[ \frac {\left (-2 i \cos \left (d x +c \right ) \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, \sqrt {5}\right ) \sqrt {2}\, \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {-\frac {2 \left (-3+2 \cos \left (d x +c \right )\right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-i \cos \left (d x +c \right ) \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, \sqrt {5}\right ) \sqrt {2}\, \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {-\frac {2 \left (-3+2 \cos \left (d x +c \right )\right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-2 i \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, \sqrt {5}\right ) \sqrt {2}\, \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {-\frac {2 \left (-3+2 \cos \left (d x +c \right )\right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-i \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, \sqrt {5}\right ) \sqrt {2}\, \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {-\frac {2 \left (-3+2 \cos \left (d x +c \right )\right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+4 \left (\cos ^{2}\left (d x +c \right )\right )-10 \cos \left (d x +c \right )+6\right ) \sqrt {-\frac {-3+2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )}}}{2 d \sqrt {\frac {1}{\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \left (-3+2 \cos \left (d x +c \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d*(-2*I*cos(d*x+c)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),5^(1/2))*2^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*(-2*(-
3+2*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-I*cos(d*x+c)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),5^(1/2))*
2^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*(-2*(-3+2*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-2*I*EllipticF(I*(-1+co
s(d*x+c))/sin(d*x+c),5^(1/2))*2^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*(-2*(-3+2*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*sin
(d*x+c)-I*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),5^(1/2))*2^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*(-2*(-3+2*cos(d*x+c
))/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+4*cos(d*x+c)^2-10*cos(d*x+c)+6)*(-(-3+2*cos(d*x+c))/cos(d*x+c))^(1/2)/(1/c
os(d*x+c))^(1/2)/sin(d*x+c)/(-3+2*cos(d*x+c))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {3 \, \sec \left (d x + c\right ) - 2} \sqrt {\sec \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\sqrt {\frac {3}{\cos \left (c+d\,x\right )}-2}\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {3 \sec {\left (c + d x \right )} - 2} \sqrt {\sec {\left (c + d x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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