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

Optimal. Leaf size=89 \[ \frac{\tan (e+f x)}{f (a \sec (e+f x)+a) \sqrt{c-c \sec (e+f x)}}-\frac{\tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{\sqrt{2} a \sqrt{c} f} \]

[Out]

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

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Rubi [A]  time = 0.157163, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.088, Rules used = {3960, 3795, 203} \[ \frac{\tan (e+f x)}{f (a \sec (e+f x)+a) \sqrt{c-c \sec (e+f x)}}-\frac{\tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{\sqrt{2} a \sqrt{c} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3960

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

Rule 3795

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2/f, Subst[Int[1/(2
*a + x^2), x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x]]], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^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)}{(a+a \sec (e+f x)) \sqrt{c-c \sec (e+f x)}} \, dx &=\frac{\tan (e+f x)}{f (a+a \sec (e+f x)) \sqrt{c-c \sec (e+f x)}}+\frac{\int \frac{\sec (e+f x)}{\sqrt{c-c \sec (e+f x)}} \, dx}{2 a}\\ &=\frac{\tan (e+f x)}{f (a+a \sec (e+f x)) \sqrt{c-c \sec (e+f x)}}-\frac{\operatorname{Subst}\left (\int \frac{1}{2 c+x^2} \, dx,x,\frac{c \tan (e+f x)}{\sqrt{c-c \sec (e+f x)}}\right )}{a f}\\ &=-\frac{\tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{\sqrt{2} a \sqrt{c} f}+\frac{\tan (e+f x)}{f (a+a \sec (e+f x)) \sqrt{c-c \sec (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.495587, size = 155, normalized size = 1.74 \[ -\frac{i \left (-1+e^{2 i (e+f x)}\right ) \left (2 \left (1+e^{2 i (e+f x)}\right )-\sqrt{2} \left (1+e^{i (e+f x)}\right ) \sqrt{1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\frac{1+e^{i (e+f x)}}{\sqrt{2} \sqrt{1+e^{2 i (e+f x)}}}\right )\right )}{2 a f \left (1+e^{2 i (e+f x)}\right )^2 (\sec (e+f x)+1) \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.194, size = 107, normalized size = 1.2 \begin{align*} -{\frac{-1+\cos \left ( fx+e \right ) }{fa\sin \left ( fx+e \right ) } \left ( \sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}+\arctan \left ({\frac{1}{\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}}} \right ) \right ){\frac{1}{\sqrt{{\frac{c \left ( -1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }}}}}{\frac{1}{\sqrt{-2\,{\frac{\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))/(c-c*sec(f*x+e))^(1/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (f x + e\right )}{{\left (a \sec \left (f x + e\right ) + a\right )} \sqrt{-c \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))/(c-c*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.588019, size = 675, normalized size = 7.58 \begin{align*} \left [\frac{\sqrt{2} c \sqrt{-\frac{1}{c}} \log \left (-\frac{2 \, \sqrt{2}{\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \sqrt{-\frac{1}{c}} -{\left (3 \, \cos \left (f x + e\right ) + 1\right )} \sin \left (f x + e\right )}{{\left (\cos \left (f x + e\right ) - 1\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 4 \, \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{4 \, a c f \sin \left (f x + e\right )}, \frac{\sqrt{2} \sqrt{c} \arctan \left (\frac{\sqrt{2} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt{c} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 2 \, \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{2 \, a c f \sin \left (f x + e\right )}\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))/(c-c*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

[1/4*(sqrt(2)*c*sqrt(-1/c)*log(-(2*sqrt(2)*(cos(f*x + e)^2 + cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x +
e))*sqrt(-1/c) - (3*cos(f*x + e) + 1)*sin(f*x + e))/((cos(f*x + e) - 1)*sin(f*x + e)))*sin(f*x + e) - 4*sqrt(
(c*cos(f*x + e) - c)/cos(f*x + e))*cos(f*x + e))/(a*c*f*sin(f*x + e)), 1/2*(sqrt(2)*sqrt(c)*arctan(sqrt(2)*sqr
t((c*cos(f*x + e) - c)/cos(f*x + e))*cos(f*x + e)/(sqrt(c)*sin(f*x + e)))*sin(f*x + e) - 2*sqrt((c*cos(f*x + e
) - c)/cos(f*x + e))*cos(f*x + e))/(a*c*f*sin(f*x + e))]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [C]  time = 1.40476, size = 184, normalized size = 2.07 \begin{align*} -\frac{\frac{\sqrt{2}{\left (i \, \sqrt{-c} \arctan \left (-i\right ) - \sqrt{-c}\right )} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}{a c} + \frac{\sqrt{2}{\left (\frac{\arctan \left (\frac{\sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c}}{\sqrt{c}}\right )}{\sqrt{c}} - \frac{\sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c}}{c}\right )}}{a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right ) \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}}{2 \, 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))/(c-c*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

-1/2*(sqrt(2)*(I*sqrt(-c)*arctan(-I) - sqrt(-c))*sgn(tan(1/2*f*x + 1/2*e))/(a*c) + sqrt(2)*(arctan(sqrt(c*tan(
1/2*f*x + 1/2*e)^2 - c)/sqrt(c))/sqrt(c) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)/c)/(a*sgn(tan(1/2*f*x + 1/2*e)^2
- 1)*sgn(tan(1/2*f*x + 1/2*e))))/f