3.12 \(\int \sqrt{\sec (a+b x)} \, dx\)

Optimal. Leaf size=36 \[ \frac{2 \sqrt{\cos (a+b x)} \sqrt{\sec (a+b x)} \text{EllipticF}\left (\frac{1}{2} (a+b x),2\right )}{b} \]

[Out]

(2*Sqrt[Cos[a + b*x]]*EllipticF[(a + b*x)/2, 2]*Sqrt[Sec[a + b*x]])/b

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Rubi [A]  time = 0.0167242, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3771, 2641} \[ \frac{2 \sqrt{\cos (a+b x)} \sqrt{\sec (a+b x)} F\left (\left .\frac{1}{2} (a+b x)\right |2\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Sec[a + b*x]],x]

[Out]

(2*Sqrt[Cos[a + b*x]]*EllipticF[(a + b*x)/2, 2]*Sqrt[Sec[a + b*x]])/b

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin{align*} \int \sqrt{\sec (a+b x)} \, dx &=\left (\sqrt{\cos (a+b x)} \sqrt{\sec (a+b x)}\right ) \int \frac{1}{\sqrt{\cos (a+b x)}} \, dx\\ &=\frac{2 \sqrt{\cos (a+b x)} F\left (\left .\frac{1}{2} (a+b x)\right |2\right ) \sqrt{\sec (a+b x)}}{b}\\ \end{align*}

Mathematica [A]  time = 0.028913, size = 36, normalized size = 1. \[ \frac{2 \sqrt{\cos (a+b x)} \sqrt{\sec (a+b x)} \text{EllipticF}\left (\frac{1}{2} (a+b x),2\right )}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Sec[a + b*x]],x]

[Out]

(2*Sqrt[Cos[a + b*x]]*EllipticF[(a + b*x)/2, 2]*Sqrt[Sec[a + b*x]])/b

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Maple [B]  time = 0.883, size = 133, normalized size = 3.7 \begin{align*} -2\,{\frac{\sqrt{ \left ( 2\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}}\sqrt{ \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}+1}{\it EllipticF} \left ( \cos \left ( 1/2\,bx+a/2 \right ) ,\sqrt{2} \right ) }{\sqrt{-2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{4}+ \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}}\sin \left ( 1/2\,bx+a/2 \right ) \sqrt{2\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1}b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^(1/2),x)

[Out]

-2*((2*cos(1/2*b*x+1/2*a)^2-1)*sin(1/2*b*x+1/2*a)^2)^(1/2)*(sin(1/2*b*x+1/2*a)^2)^(1/2)*(-2*cos(1/2*b*x+1/2*a)
^2+1)^(1/2)/(-2*sin(1/2*b*x+1/2*a)^4+sin(1/2*b*x+1/2*a)^2)^(1/2)*EllipticF(cos(1/2*b*x+1/2*a),2^(1/2))/sin(1/2
*b*x+1/2*a)/(2*cos(1/2*b*x+1/2*a)^2-1)^(1/2)/b

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{\sec \left (b x + a\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(sec(b*x + a)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{\sec \left (b x + a\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(sec(b*x + a)), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{\sec{\left (a + b x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)**(1/2),x)

[Out]

Integral(sqrt(sec(a + b*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(sec(b*x + a)), x)