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

Optimal. Leaf size=47 \[ -\frac{\tan (e+f x) \tanh ^{-1}(\cos (e+f x))}{f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}} \]

[Out]

-((ArcTanh[Cos[e + f*x]]*Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]))

________________________________________________________________________________________

Rubi [A]  time = 0.13497, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.056, Rules used = {3959, 3770} \[ -\frac{\tan (e+f x) \tanh ^{-1}(\cos (e+f x))}{f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((ArcTanh[Cos[e + f*x]]*Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]))

Rule 3959

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

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{\sec (e+f x)}{\sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}} \, dx &=\frac{\tan (e+f x) \int \csc (e+f x) \, dx}{\sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ &=-\frac{\tanh ^{-1}(\cos (e+f x)) \tan (e+f x)}{f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.792874, size = 94, normalized size = 2. \[ \frac{4 i \left (-1+e^{i (e+f x)}\right ) \cos ^2\left (\frac{1}{2} (e+f x)\right ) \sec (e+f x) \tanh ^{-1}\left (e^{i (e+f x)}\right )}{f \left (1+e^{i (e+f x)}\right ) \sqrt{a (\sec (e+f x)+1)} \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*I)*(-1 + E^(I*(e + f*x)))*ArcTanh[E^(I*(e + f*x))]*Cos[(e + f*x)/2]^2*Sec[e + f*x])/((1 + E^(I*(e + f*x)))
*f*Sqrt[a*(1 + Sec[e + f*x])]*Sqrt[c - c*Sec[e + f*x]])

________________________________________________________________________________________

Maple [A]  time = 0.267, size = 85, normalized size = 1.8 \begin{align*} -{\frac{\cos \left ( fx+e \right ) }{af\sin \left ( fx+e \right ) c}\sqrt{{\frac{a \left ( 1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }}}\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) \sqrt{{\frac{c \left ( -1+\cos \left ( fx+e \right ) \right ) }{\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-c*sec(f*x+e))^(1/2),x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.83463, size = 59, normalized size = 1.26 \begin{align*} -\frac{\arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right ) + 1\right ) - \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right ) - 1\right )}{\sqrt{a} \sqrt{c} f} \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-c*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

-(arctan2(sin(f*x + e), cos(f*x + e) + 1) - arctan2(sin(f*x + e), cos(f*x + e) - 1))/(sqrt(a)*sqrt(c)*f)

________________________________________________________________________________________

Fricas [A]  time = 0.607107, size = 487, normalized size = 10.36 \begin{align*} \left [-\frac{\sqrt{-a c} \log \left (-\frac{4 \,{\left (2 \, \sqrt{-a c} \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 ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )^{2} +{\left (a c \cos \left (f x + e\right )^{2} + a c\right )} \sin \left (f x + e\right )\right )}}{{\left (\cos \left (f x + e\right )^{2} - 1\right )} \sin \left (f x + e\right )}\right )}{2 \, a c f}, \frac{\sqrt{a c} \arctan \left (\frac{\sqrt{a c} \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 ) - c}{\cos \left (f x + e\right )}}}{a c \sin \left (f x + e\right )}\right )}{a c 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-c*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

[-1/2*sqrt(-a*c)*log(-4*(2*sqrt(-a*c)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*
x + e))*cos(f*x + e)^2 + (a*c*cos(f*x + e)^2 + a*c)*sin(f*x + e))/((cos(f*x + e)^2 - 1)*sin(f*x + e)))/(a*c*f)
, sqrt(a*c)*arctan(sqrt(a*c)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))/(
a*c*sin(f*x + e)))/(a*c*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 \left (\sec{\left (e + f x \right )} - 1\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-c*sec(f*x+e))**(1/2),x)

[Out]

Integral(sec(e + f*x)/(sqrt(a*(sec(e + f*x) + 1))*sqrt(-c*(sec(e + f*x) - 1))), 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-c*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Timed out