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

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

[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]])

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Rubi [A]  time = 0.171297, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, 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 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]

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 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 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 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]

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*}

Mathematica [A]  time = 0.139241, size = 81, normalized size = 0.64 \[ \frac{2 \sqrt{3 \cos (c+d x)+2} \sqrt{\sec (c+d x)} \left (5 E\left (\frac{1}{2} (c+d x)|\frac{6}{5}\right )-2 \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{6}{5}\right )\right )}{3 \sqrt{5} d \sqrt{2 \sec (c+d x)+3}} \]

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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Maple [C]  time = 0.461, size = 409, normalized size = 3.2 \begin{align*}{\frac{1}{15\,d\sin \left ( dx+c \right ) \left ( 2+3\,\cos \left ( dx+c \right ) \right ) } \left ( 3\,\sqrt{5}\sin \left ( dx+c \right ) \cos \left ( dx+c \right ){\it EllipticF} \left ( 1/5\,{\frac{ \left ( -1+\cos \left ( dx+c \right ) \right ) \sqrt{5}}{\sin \left ( dx+c \right ) }},i\sqrt{5} \right ) \sqrt{2}\sqrt{10}\sqrt{{\frac{2+3\,\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}-\sqrt{5}\sin \left ( dx+c \right ) \cos \left ( dx+c \right ){\it EllipticE} \left ({\frac{ \left ( -1+\cos \left ( dx+c \right ) \right ) \sqrt{5}}{5\,\sin \left ( dx+c \right ) }},i\sqrt{5} \right ) \sqrt{2}\sqrt{10}\sqrt{{\frac{2+3\,\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}+3\,\sqrt{5}{\it EllipticF} \left ( 1/5\,{\frac{ \left ( -1+\cos \left ( dx+c \right ) \right ) \sqrt{5}}{\sin \left ( dx+c \right ) }},i\sqrt{5} \right ) \sqrt{10}\sqrt{{\frac{2+3\,\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\sqrt{2}\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sin \left ( dx+c \right ) -\sqrt{5}{\it EllipticE} \left ({\frac{ \left ( -1+\cos \left ( dx+c \right ) \right ) \sqrt{5}}{5\,\sin \left ( dx+c \right ) }},i\sqrt{5} \right ) \sqrt{10}\sqrt{{\frac{2+3\,\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\sqrt{2}\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sin \left ( dx+c \right ) -30\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}+10\,\cos \left ( dx+c \right ) +20 \right ) \sqrt{{\frac{2+3\,\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }}}{\frac{1}{\sqrt{ \left ( \cos \left ( dx+c \right ) \right ) ^{-1}}}}} \end{align*}

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*5^(1/2)*sin(d*x+c)*cos(d*x+c)*EllipticF(1/5*(-1+cos(d*x+c))*5^(1/2)/sin(d*x+c),I*5^(1/2))*2^(1/2)*10
^(1/2)*((2+3*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)-5^(1/2)*sin(d*x+c)*cos(d*x+c)*Elliptic
E(1/5*(-1+cos(d*x+c))*5^(1/2)/sin(d*x+c),I*5^(1/2))*2^(1/2)*10^(1/2)*((2+3*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(
1/(cos(d*x+c)+1))^(1/2)+3*5^(1/2)*EllipticF(1/5*(-1+cos(d*x+c))*5^(1/2)/sin(d*x+c),I*5^(1/2))*10^(1/2)*((2+3*c
os(d*x+c))/(cos(d*x+c)+1))^(1/2)*2^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-5^(1/2)*EllipticE(1/5*(-1+cos(d*x
+c))*5^(1/2)/sin(d*x+c),I*5^(1/2))*10^(1/2)*((2+3*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*2^(1/2)*(1/(cos(d*x+c)+1))
^(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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{2 \, \sec \left (d x + c\right ) + 3} \sqrt{\sec \left (d x + c\right )}}\,{d x} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{2 \sec{\left (c + d x \right )} + 3} \sqrt{\sec{\left (c + d x \right )}}}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{2 \, \sec \left (d x + c\right ) + 3} \sqrt{\sec \left (d x + c\right )}}\,{d x} \end{align*}

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)