3.3.55 \(\int \frac {\sec (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx\) [255]

Optimal. Leaf size=47 \[ \frac {F\left (\left .e-\frac {\pi }{4}+f x\right |2\right ) \sec (e+f x) \sqrt {\sin (2 e+2 f x)}}{f \sqrt {d \tan (e+f x)}} \]

[Out]

-(sin(e+1/4*Pi+f*x)^2)^(1/2)/sin(e+1/4*Pi+f*x)*EllipticF(cos(e+1/4*Pi+f*x),2^(1/2))*sec(f*x+e)*sin(2*f*x+2*e)^
(1/2)/f/(d*tan(f*x+e))^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2694, 2653, 2720} \begin {gather*} \frac {\sqrt {\sin (2 e+2 f x)} \sec (e+f x) F\left (\left .e+f x-\frac {\pi }{4}\right |2\right )}{f \sqrt {d \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]/Sqrt[d*Tan[e + f*x]],x]

[Out]

(EllipticF[e - Pi/4 + f*x, 2]*Sec[e + f*x]*Sqrt[Sin[2*e + 2*f*x]])/(f*Sqrt[d*Tan[e + f*x]])

Rule 2653

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt[Sin[2*
e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b*Cos[e + f*x]]), Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b,
e, f}, x]

Rule 2694

Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[e + f*x]]/(Sqrt[Co
s[e + f*x]]*Sqrt[b*Tan[e + f*x]]), Int[1/(Sqrt[Cos[e + f*x]]*Sqrt[Sin[e + f*x]]), x], x] /; FreeQ[{b, e, f}, x
]

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]

Rubi steps

\begin {align*} \int \frac {\sec (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx &=\frac {\sqrt {\sin (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)} \sqrt {\sin (e+f x)}} \, dx}{\sqrt {\cos (e+f x)} \sqrt {d \tan (e+f x)}}\\ &=\frac {\left (\sec (e+f x) \sqrt {\sin (2 e+2 f x)}\right ) \int \frac {1}{\sqrt {\sin (2 e+2 f x)}} \, dx}{\sqrt {d \tan (e+f x)}}\\ &=\frac {F\left (\left .e-\frac {\pi }{4}+f x\right |2\right ) \sec (e+f x) \sqrt {\sin (2 e+2 f x)}}{f \sqrt {d \tan (e+f x)}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.15, size = 77, normalized size = 1.64 \begin {gather*} -\frac {2 \sqrt [4]{-1} F\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-1} \sqrt {\tan (e+f x)}\right )\right |-1\right ) \sec ^3(e+f x) \sqrt {\tan (e+f x)}}{f \sqrt {d \tan (e+f x)} \left (1+\tan ^2(e+f x)\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]/Sqrt[d*Tan[e + f*x]],x]

[Out]

(-2*(-1)^(1/4)*EllipticF[I*ArcSinh[(-1)^(1/4)*Sqrt[Tan[e + f*x]]], -1]*Sec[e + f*x]^3*Sqrt[Tan[e + f*x]])/(f*S
qrt[d*Tan[e + f*x]]*(1 + Tan[e + f*x]^2)^(3/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(166\) vs. \(2(69)=138\).
time = 0.26, size = 167, normalized size = 3.55

method	result	size
default	\(-\frac {\EllipticF \left (\sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {\cos \left (f x +e \right )-1}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \left (\cos \left (f x +e \right )-1\right ) \left (\cos \left (f x +e \right )+1\right )^{2} \sqrt {2}}{f \sin \left (f x +e \right )^{2} \cos \left (f x +e \right ) \sqrt {\frac {d \sin \left (f x +e \right )}{\cos \left (f x +e \right )}}}\)	\(167\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)/(d*tan(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/f*EllipticF((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))*((cos(f*x+e)-1)/sin(f*x+e))^(1/2)*(
(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*(cos(f*x+e)-1)/sin
(f*x+e)^2/cos(f*x+e)*(cos(f*x+e)+1)^2/(d*sin(f*x+e)/cos(f*x+e))^(1/2)*2^(1/2)

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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(f*x+e)/(d*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(sec(f*x + e)/sqrt(d*tan(f*x + e)), x)

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Fricas [C] Result contains complex when optimal does not.
time = 0.11, size = 59, normalized size = 1.26 \begin {gather*} -\frac {\sqrt {i \, d} {\rm ellipticF}\left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right ), -1\right ) + \sqrt {-i \, d} {\rm ellipticF}\left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right ), -1\right )}{d f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)/(d*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

-(sqrt(I*d)*ellipticF(cos(f*x + e) + I*sin(f*x + e), -1) + sqrt(-I*d)*ellipticF(cos(f*x + e) - I*sin(f*x + e),
-1))/(d*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)/(d*tan(f*x+e))**(1/2),x)

[Out]

Integral(sec(e + f*x)/sqrt(d*tan(e + f*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(f*x+e)/(d*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate(sec(f*x + e)/sqrt(d*tan(f*x + e)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{\cos \left (e+f\,x\right )\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(e + f*x)*(d*tan(e + f*x))^(1/2)),x)

[Out]

int(1/(cos(e + f*x)*(d*tan(e + f*x))^(1/2)), x)

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