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

Optimal. Leaf size=127 \[ \frac {2 \sqrt {5} \sqrt {2 \sec (c+d x)+3} E\left (\frac {1}{2} (c+d x)|\frac {6}{5}\right )}{3 d \sqrt {3 \cos (c+d x)+2} \sqrt {\sec (c+d x)}}-\frac {4 \sqrt {3 \cos (c+d x)+2} \sqrt {\sec (c+d x)} F\left (\frac {1}{2} (c+d x)|\frac {6}{5}\right )}{3 \sqrt {5} d \sqrt {2 \sec (c+d x)+3}} \]

[Out]

-4/15*(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),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)+2/3*(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),1/5*30^(1/2))*5^(1/2)*(3+2*sec(d*x+c))^(1/2)/d/(2+3*cos(d*x+c))^(1/2)/sec(d*x+c
)^(1/2)

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Rubi [A]  time = 0.17, antiderivative size = 127, 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} \[ \frac {2 \sqrt {5} \sqrt {2 \sec (c+d x)+3} E\left (\frac {1}{2} (c+d x)|\frac {6}{5}\right )}{3 d \sqrt {3 \cos (c+d x)+2} \sqrt {\sec (c+d x)}}-\frac {4 \sqrt {3 \cos (c+d x)+2} \sqrt {\sec (c+d x)} F\left (\frac {1}{2} (c+d x)|\frac {6}{5}\right )}{3 \sqrt {5} d \sqrt {2 \sec (c+d x)+3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*Sqrt[2 + 3*Cos[c + d*x]]*EllipticF[(c + d*x)/2, 6/5]*Sqrt[Sec[c + d*x]])/(3*Sqrt[5]*d*Sqrt[3 + 2*Sec[c + d
*x]]) + (2*Sqrt[5]*EllipticE[(c + d*x)/2, 6/5]*Sqrt[3 + 2*Sec[c + d*x]])/(3*d*Sqrt[2 + 3*Cos[c + d*x]]*Sqrt[Se
c[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 {3+2 \sec (c+d x)}} \, dx &=\frac {1}{3} \int \frac {\sqrt {3+2 \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx-\frac {2}{3} \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {3+2 \sec (c+d x)}} \, dx\\ &=-\frac {\left (2 \sqrt {2+3 \cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {2+3 \cos (c+d x)}} \, dx}{3 \sqrt {3+2 \sec (c+d x)}}+\frac {\sqrt {3+2 \sec (c+d x)} \int \sqrt {2+3 \cos (c+d x)} \, dx}{3 \sqrt {2+3 \cos (c+d x)} \sqrt {\sec (c+d x)}}\\ &=-\frac {4 \sqrt {2+3 \cos (c+d x)} F\left (\frac {1}{2} (c+d x)|\frac {6}{5}\right ) \sqrt {\sec (c+d x)}}{3 \sqrt {5} d \sqrt {3+2 \sec (c+d x)}}+\frac {2 \sqrt {5} E\left (\frac {1}{2} (c+d x)|\frac {6}{5}\right ) \sqrt {3+2 \sec (c+d x)}}{3 d \sqrt {2+3 \cos (c+d x)} \sqrt {\sec (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 81, normalized size = 0.64 \[ \frac {2 \sqrt {3 \cos (c+d x)+2} \sqrt {\sec (c+d x)} \left (5 E\left (\frac {1}{2} (c+d x)|\frac {6}{5}\right )-2 F\left (\frac {1}{2} (c+d x)|\frac {6}{5}\right )\right )}{3 \sqrt {5} d \sqrt {2 \sec (c+d x)+3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {2 \, \sec \left (d x + c\right ) + 3} \sqrt {\sec \left (d x + c\right )}}{2 \, \sec \left (d x + c\right )^{2} + 3 \, \sec \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 1.80, size = 409, normalized size = 3.22 \[ \frac {\left (3 \sin \left (d x +c \right ) \cos \left (d x +c \right ) \EllipticF \left (\frac {\sqrt {5}\, \left (-1+\cos \left (d x +c \right )\right )}{5 \sin \left (d x +c \right )}, i \sqrt {5}\right ) \sqrt {5}\, \sqrt {10}\, \sqrt {\frac {2+3 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {2}-\sin \left (d x +c \right ) \cos \left (d x +c \right ) \EllipticE \left (\frac {\sqrt {5}\, \left (-1+\cos \left (d x +c \right )\right )}{5 \sin \left (d x +c \right )}, i \sqrt {5}\right ) \sqrt {5}\, \sqrt {10}\, \sqrt {\frac {2+3 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {2}+3 \sqrt {5}\, \EllipticF \left (\frac {\sqrt {5}\, \left (-1+\cos \left (d x +c \right )\right )}{5 \sin \left (d x +c \right )}, i \sqrt {5}\right ) \sqrt {10}\, \sqrt {\frac {2+3 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}\, \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-\sqrt {5}\, \EllipticE \left (\frac {\sqrt {5}\, \left (-1+\cos \left (d x +c \right )\right )}{5 \sin \left (d x +c \right )}, i \sqrt {5}\right ) \sqrt {10}\, \sqrt {\frac {2+3 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}\, \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-30 \left (\cos ^{2}\left (d x +c \right )\right )+10 \cos \left (d x +c \right )+20\right ) \sqrt {\frac {2+3 \cos \left (d x +c \right )}{\cos \left (d x +c \right )}}}{15 d \sqrt {\frac {1}{\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \left (2+3 \cos \left (d x +c \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15/d*(3*sin(d*x+c)*cos(d*x+c)*EllipticF(1/5*5^(1/2)*(-1+cos(d*x+c))/sin(d*x+c),I*5^(1/2))*5^(1/2)*10^(1/2)*(
(2+3*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*2^(1/2)-sin(d*x+c)*cos(d*x+c)*EllipticE(1/5*5^
(1/2)*(-1+cos(d*x+c))/sin(d*x+c),I*5^(1/2))*5^(1/2)*10^(1/2)*((2+3*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(1/(1+cos
(d*x+c)))^(1/2)*2^(1/2)+3*5^(1/2)*EllipticF(1/5*5^(1/2)*(-1+cos(d*x+c))/sin(d*x+c),I*5^(1/2))*10^(1/2)*((2+3*c
os(d*x+c))/(1+cos(d*x+c)))^(1/2)*2^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-5^(1/2)*EllipticE(1/5*5^(1/2)*(-1
+cos(d*x+c))/sin(d*x+c),I*5^(1/2))*10^(1/2)*((2+3*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*2^(1/2)*(1/(1+cos(d*x+c)))
^(1/2)*sin(d*x+c)-30*cos(d*x+c)^2+10*cos(d*x+c)+20)*((2+3*cos(d*x+c))/cos(d*x+c))^(1/2)/(1/cos(d*x+c))^(1/2)/s
in(d*x+c)/(2+3*cos(d*x+c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/((2/cos(c + d*x) + 3)^(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 \[ \int \frac {1}{\sqrt {2 \sec {\left (c + d x \right )} + 3} \sqrt {\sec {\left (c + d x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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