3.236 \(\int \frac{\sec (e+f x)}{\sqrt{a+a \sec (e+f x)} \sqrt{c+d \sec (e+f x)}} \, dx\)

Optimal. Leaf size=78 \[ \frac{\sqrt{2} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{c-d} \tan (e+f x)}{\sqrt{2} \sqrt{a \sec (e+f x)+a} \sqrt{c+d \sec (e+f x)}}\right )}{\sqrt{a} f \sqrt{c-d}} \]

[Out]

(Sqrt[2]*ArcTan[(Sqrt[a]*Sqrt[c - d]*Tan[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c + d*Sec[e + f*x]])
])/(Sqrt[a]*Sqrt[c - d]*f)

________________________________________________________________________________________

Rubi [A]  time = 0.157262, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.057, Rules used = {3983, 203} \[ \frac{\sqrt{2} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{c-d} \tan (e+f x)}{\sqrt{2} \sqrt{a \sec (e+f x)+a} \sqrt{c+d \sec (e+f x)}}\right )}{\sqrt{a} f \sqrt{c-d}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]/(Sqrt[a + a*Sec[e + f*x]]*Sqrt[c + d*Sec[e + f*x]]),x]

[Out]

(Sqrt[2]*ArcTan[(Sqrt[a]*Sqrt[c - d]*Tan[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c + d*Sec[e + f*x]])
])/(Sqrt[a]*Sqrt[c - d]*f)

Rule 3983

Int[csc[(e_.) + (f_.)*(x_)]/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.) + (
c_)]), x_Symbol] :> Dist[(-2*a)/(b*f), Subst[Int[1/(2 + (a*c - b*d)*x^2), x], x, Cot[e + f*x]/(Sqrt[a + b*Csc[
e + f*x]]*Sqrt[c + d*Csc[e + f*x]])], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2
, 0] && NeQ[c^2 - d^2, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sec (e+f x)}{\sqrt{a+a \sec (e+f x)} \sqrt{c+d \sec (e+f x)}} \, dx &=-\frac{2 \operatorname{Subst}\left (\int \frac{1}{2+(a c-a d) x^2} \, dx,x,-\frac{\tan (e+f x)}{\sqrt{a+a \sec (e+f x)} \sqrt{c+d \sec (e+f x)}}\right )}{f}\\ &=\frac{\sqrt{2} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{c-d} \tan (e+f x)}{\sqrt{2} \sqrt{a+a \sec (e+f x)} \sqrt{c+d \sec (e+f x)}}\right )}{\sqrt{a} \sqrt{c-d} f}\\ \end{align*}

Mathematica [A]  time = 0.217764, size = 107, normalized size = 1.37 \[ \frac{2 \cos \left (\frac{1}{2} (e+f x)\right ) \sec (e+f x) \sqrt{c \cos (e+f x)+d} \tan ^{-1}\left (\frac{\sqrt{c-d} \sin \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c \cos (e+f x)+d}}\right )}{f \sqrt{c-d} \sqrt{a (\sec (e+f x)+1)} \sqrt{c+d \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]/(Sqrt[a + a*Sec[e + f*x]]*Sqrt[c + d*Sec[e + f*x]]),x]

[Out]

(2*ArcTan[(Sqrt[c - d]*Sin[(e + f*x)/2])/Sqrt[d + c*Cos[e + f*x]]]*Cos[(e + f*x)/2]*Sqrt[d + c*Cos[e + f*x]]*S
ec[e + f*x])/(Sqrt[c - d]*f*Sqrt[a*(1 + Sec[e + f*x])]*Sqrt[c + d*Sec[e + f*x]])

________________________________________________________________________________________

Maple [B]  time = 0.331, size = 170, normalized size = 2.2 \begin{align*} 2\,{\frac{\cos \left ( fx+e \right ) \left ( -1+\cos \left ( fx+e \right ) \right ) }{af\sqrt{c-d} \left ( \sin \left ( fx+e \right ) \right ) ^{2}}\sqrt{{\frac{a \left ( 1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }}}\sqrt{{\frac{d+c\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }}}\ln \left ( -{\frac{1}{\sin \left ( fx+e \right ) } \left ( \sqrt{c-d}\cos \left ( fx+e \right ) -\sqrt{-2\,{\frac{d+c\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}\sin \left ( fx+e \right ) -\sqrt{c-d} \right ) } \right ){\frac{1}{\sqrt{-2\,{\frac{d+c\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)/(a+a*sec(f*x+e))^(1/2)/(c+d*sec(f*x+e))^(1/2),x)

[Out]

2/f/a/(c-d)^(1/2)*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/2)*cos(f*x+e)*(-1+cos(f*x+e))*((d+c*cos(f*x+e))/cos(f*x+e
))^(1/2)*ln(-((c-d)^(1/2)*cos(f*x+e)-(-2*(d+c*cos(f*x+e))/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)-(c-d)^(1/2))/sin(f*
x+e))/sin(f*x+e)^2/(-2*(d+c*cos(f*x+e))/(1+cos(f*x+e)))^(1/2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 0.600526, size = 625, normalized size = 8.01 \begin{align*} \left [\frac{\sqrt{2} \sqrt{-\frac{1}{a c - a d}} \log \left (-\frac{2 \, \sqrt{2}{\left (c - d\right )} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt{\frac{c \cos \left (f x + e\right ) + d}{\cos \left (f x + e\right )}} \sqrt{-\frac{1}{a c - a d}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) -{\left (3 \, c - d\right )} \cos \left (f x + e\right )^{2} - 2 \,{\left (c + d\right )} \cos \left (f x + e\right ) + c - 3 \, d}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right )}{2 \, f}, -\frac{\sqrt{2} \arctan \left (\frac{\sqrt{2} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt{\frac{c \cos \left (f x + e\right ) + d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt{a c - a d} \sin \left (f x + e\right )}\right )}{\sqrt{a c - a d} f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*sqrt(2)*sqrt(-1/(a*c - a*d))*log(-(2*sqrt(2)*(c - d)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(
f*x + e) + d)/cos(f*x + e))*sqrt(-1/(a*c - a*d))*cos(f*x + e)*sin(f*x + e) - (3*c - d)*cos(f*x + e)^2 - 2*(c +
d)*cos(f*x + e) + c - 3*d)/(cos(f*x + e)^2 + 2*cos(f*x + e) + 1))/f, -sqrt(2)*arctan(sqrt(2)*sqrt((a*cos(f*x
+ e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) + d)/cos(f*x + e))*cos(f*x + e)/(sqrt(a*c - a*d)*sin(f*x + e)))/(
sqrt(a*c - a*d)*f)]

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sec(e + f*x)/(sqrt(a*(sec(e + f*x) + 1))*sqrt(c + d*sec(e + f*x))), x)

________________________________________________________________________________________

Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out