3.1.12 \(\int \sqrt {\sec (a+b x)} \, dx\) [12]

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

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3856, 2720} \begin {gather*} \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} \end {gather*}

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 2720

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

Rule 3856

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]

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*}

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Mathematica [A]
time = 0.02, size = 36, normalized size = 1.00 \begin {gather*} \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 {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(132\) vs. \(2(58)=116\).
time = 1.58, size = 133, normalized size = 3.69

method	result	size
default	\(-\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}\, \sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+1}\, \EllipticF \left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right )}{\sqrt {-2 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )}\, \sin \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1}\, b}\)	\(133\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.81, size = 51, normalized size = 1.42 \begin {gather*} \frac {-i \, \sqrt {2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + i \, \sqrt {2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-I*sqrt(2)*weierstrassPInverse(-4, 0, cos(b*x + a) + I*sin(b*x + a)) + I*sqrt(2)*weierstrassPInverse(-4, 0, c
os(b*x + a) - I*sin(b*x + a)))/b

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {\sec {\left (a + b x \right )}}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [B]
time = 0.13, size = 33, normalized size = 0.92 \begin {gather*} \frac {2\,\sqrt {\cos \left (a+b\,x\right )}\,\sqrt {\frac {1}{\cos \left (a+b\,x\right )}}\,\mathrm {F}\left (\frac {a}{2}+\frac {b\,x}{2}\middle |2\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/cos(a + b*x))^(1/2),x)

[Out]

(2*cos(a + b*x)^(1/2)*(1/cos(a + b*x))^(1/2)*ellipticF(a/2 + (b*x)/2, 2))/b

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