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

Optimal. Leaf size=123 \[ -\frac {3 \sqrt {2 \cos (c+d x)+3} \sqrt {\sec (c+d x)} F\left (\frac {1}{2} (c+d x)|\frac {4}{5}\right )}{\sqrt {5} d \sqrt {-3 \sec (c+d x)-2}}-\frac {\sqrt {5} \sqrt {-3 \sec (c+d x)-2} E\left (\frac {1}{2} (c+d x)|\frac {4}{5}\right )}{d \sqrt {2 \cos (c+d x)+3} \sqrt {\sec (c+d x)}} \]

[Out]

-(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),2/5*5^(1/2))*5^(1/2)*(-2-3*sec(d
*x+c))^(1/2)/d/(3+2*cos(d*x+c))^(1/2)/sec(d*x+c)^(1/2)-3/5*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ell
ipticF(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 = 0.20, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3862, 3856, 2655, 2653, 3858, 2663, 2661} \[ -\frac {3 \sqrt {2 \cos (c+d x)+3} \sqrt {\sec (c+d x)} F\left (\frac {1}{2} (c+d x)|\frac {4}{5}\right )}{\sqrt {5} d \sqrt {-3 \sec (c+d x)-2}}-\frac {\sqrt {5} \sqrt {-3 \sec (c+d x)-2} E\left (\frac {1}{2} (c+d x)|\frac {4}{5}\right )}{d \sqrt {2 \cos (c+d x)+3} \sqrt {\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[5]*EllipticE[(c + d*x)/2, 4/5]*Sqrt[-2 - 3*Sec[c + d*x]])/(d*Sqrt[3 + 2*Cos[c + d*x]]*Sqrt[Sec[c + d*x
]])) - (3*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 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 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[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]

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 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 {-2-3 \sec (c+d x)} \sqrt {\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 {\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 {\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 {\left (\sqrt {5} \sqrt {-2-3 \sec (c+d x)}\right ) \int \sqrt {\frac {3}{5}+\frac {2}{5} \cos (c+d x)} \, dx}{2 \sqrt {3+2 \cos (c+d x)} \sqrt {\sec (c+d x)}}-\frac {\left (3 \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}{2 \sqrt {5} \sqrt {-2-3 \sec (c+d x)}}\\ &=-\frac {\sqrt {5} E\left (\frac {1}{2} (c+d x)|\frac {4}{5}\right ) \sqrt {-2-3 \sec (c+d x)}}{d \sqrt {3+2 \cos (c+d x)} \sqrt {\sec (c+d x)}}-\frac {3 \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 = 0.10, size = 78, normalized size = 0.63 \[ \frac {\sqrt {2 \cos (c+d x)+3} \sqrt {\sec (c+d x)} \left (5 E\left (\frac {1}{2} (c+d x)|\frac {4}{5}\right )-3 F\left (\frac {1}{2} (c+d x)|\frac {4}{5}\right )\right )}{\sqrt {5} d \sqrt {-3 \sec (c+d x)-2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[3 + 2*Cos[c + d*x]]*(5*EllipticE[(c + d*x)/2, 4/5] - 3*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 = 0.44, 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/(-2-3*sec(d*x+c))^(1/2)/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/(-2-3*sec(d*x+c))^(1/2)/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 [C]  time = 1.86, size = 390, normalized size = 3.17 \[ -\frac {\left (2 i \cos \left (d x +c \right ) \sqrt {2}\, \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {10}\, \sqrt {\frac {3+2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, \frac {\sqrt {5}}{5}\right )-5 i \cos \left (d x +c \right ) \sqrt {2}\, \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {10}\, \sqrt {\frac {3+2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, \frac {\sqrt {5}}{5}\right )+2 i \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, \frac {\sqrt {5}}{5}\right ) \sqrt {2}\, \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {10}\, \sqrt {\frac {3+2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-5 i \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, \frac {\sqrt {5}}{5}\right ) \sqrt {2}\, \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {10}\, \sqrt {\frac {3+2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-20 \left (\cos ^{2}\left (d x +c \right )\right )-10 \cos \left (d x +c \right )+30\right ) \sqrt {-\frac {3+2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )}}}{10 d \sqrt {\frac {1}{\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \left (3+2 \cos \left (d x +c \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/10/d*(2*I*cos(d*x+c)*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)*sin(
d*x+c)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),1/5*5^(1/2))-5*I*cos(d*x+c)*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)*sin(d*x+c)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),1/5*5^(1/2))+2
*I*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),1/5*5^(1/2))*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)*sin(d*x+c)-5*I*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),1/5*5^(1/2))*2^(1/2)*(1/(1+co
s(d*x+c)))^(1/2)*10^(1/2)*((3+2*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-20*cos(d*x+c)^2-10*cos(d*x+c)+30)
*(-(3+2*cos(d*x+c))/cos(d*x+c))^(1/2)/(1/cos(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/(-2-3*sec(d*x+c))^(1/2)/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/(-2-3*sec(d*x+c))**(1/2)/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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