3.97 \(\int \frac{\sec (e+f x)}{(a+a \sec (e+f x))^2 \sqrt{c-c \sec (e+f x)}} \, dx\)

Optimal. Leaf size=138 \[ -\frac{\tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{2 \sqrt{2} a^2 \sqrt{c} f}+\frac{\tan (e+f x)}{2 f \left (a^2 \sec (e+f x)+a^2\right ) \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x)}{3 f (a \sec (e+f x)+a)^2 \sqrt{c-c \sec (e+f x)}} \]

[Out]

-ArcTan[(Sqrt[c]*Tan[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sec[e + f*x]])]/(2*Sqrt[2]*a^2*Sqrt[c]*f) + Tan[e + f*x]/(3
*f*(a + a*Sec[e + f*x])^2*Sqrt[c - c*Sec[e + f*x]]) + Tan[e + f*x]/(2*f*(a^2 + a^2*Sec[e + f*x])*Sqrt[c - c*Se
c[e + f*x]])

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Rubi [A]  time = 0.271098, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.088, Rules used = {3960, 3795, 203} \[ -\frac{\tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{2 \sqrt{2} a^2 \sqrt{c} f}+\frac{\tan (e+f x)}{2 f \left (a^2 \sec (e+f x)+a^2\right ) \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x)}{3 f (a \sec (e+f x)+a)^2 \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]/((a + a*Sec[e + f*x])^2*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]])]/(2*Sqrt[2]*a^2*Sqrt[c]*f) + Tan[e + f*x]/(3
*f*(a + a*Sec[e + f*x])^2*Sqrt[c - c*Sec[e + f*x]]) + Tan[e + f*x]/(2*f*(a^2 + a^2*Sec[e + f*x])*Sqrt[c - c*Se
c[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))^2 \sqrt{c-c \sec (e+f x)}} \, dx &=\frac{\tan (e+f x)}{3 f (a+a \sec (e+f x))^2 \sqrt{c-c \sec (e+f x)}}+\frac{\int \frac{\sec (e+f x)}{(a+a \sec (e+f x)) \sqrt{c-c \sec (e+f x)}} \, dx}{2 a}\\ &=\frac{\tan (e+f x)}{3 f (a+a \sec (e+f x))^2 \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x)}{2 f \left (a^2+a^2 \sec (e+f x)\right ) \sqrt{c-c \sec (e+f x)}}+\frac{\int \frac{\sec (e+f x)}{\sqrt{c-c \sec (e+f x)}} \, dx}{4 a^2}\\ &=\frac{\tan (e+f x)}{3 f (a+a \sec (e+f x))^2 \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x)}{2 f \left (a^2+a^2 \sec (e+f x)\right ) \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 )}{2 a^2 f}\\ &=-\frac{\tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{2 \sqrt{2} a^2 \sqrt{c} f}+\frac{\tan (e+f x)}{3 f (a+a \sec (e+f x))^2 \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x)}{2 f \left (a^2+a^2 \sec (e+f x)\right ) \sqrt{c-c \sec (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 2.14287, size = 259, normalized size = 1.88 \[ \frac{2 e^{-\frac{1}{2} i (e+f x)} \sin \left (\frac{1}{2} (e+f x)\right ) \cos \left (\frac{1}{2} (e+f x)\right ) \sec ^{\frac{5}{2}}(e+f x) \left (\frac{1}{8} e^{-\frac{3}{2} i (e+f x)} \left (6 e^{i (e+f x)}+10 e^{2 i (e+f x)}+6 e^{3 i (e+f x)}+5 e^{4 i (e+f x)}+5\right ) \sqrt{\sec (e+f x)}-3 \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt{1+e^{2 i (e+f x)}} \cos ^3\left (\frac{1}{2} (e+f x)\right ) \tanh ^{-1}\left (\frac{1+e^{i (e+f x)}}{\sqrt{2} \sqrt{1+e^{2 i (e+f x)}}}\right )\right )}{3 a^2 f (\sec (e+f x)+1)^2 \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]/((a + a*Sec[e + f*x])^2*Sqrt[c - c*Sec[e + f*x]]),x]

[Out]

(2*Cos[(e + f*x)/2]*(-3*Sqrt[E^(I*(e + f*x))/(1 + E^((2*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))])]*Cos[(e + f*x)/2]^3 + ((5 + 6*E^(I*(e + f*x)) +
10*E^((2*I)*(e + f*x)) + 6*E^((3*I)*(e + f*x)) + 5*E^((4*I)*(e + f*x)))*Sqrt[Sec[e + f*x]])/(8*E^(((3*I)/2)*(e
+ f*x))))*Sec[e + f*x]^(5/2)*Sin[(e + f*x)/2])/(3*a^2*E^((I/2)*(e + f*x))*f*(1 + Sec[e + f*x])^2*Sqrt[c - c*S
ec[e + f*x]])

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Maple [A]  time = 0.219, size = 131, normalized size = 1. \begin{align*}{\frac{-1+\cos \left ( fx+e \right ) }{6\,f{a}^{2}\sin \left ( fx+e \right ) } \left ( \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{{\frac{3}{2}}}-3\,\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}-3\,\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))^2/(c-c*sec(f*x+e))^(1/2),x)

[Out]

1/6/a^2/f*(-1+cos(f*x+e))*((-2*cos(f*x+e)/(1+cos(f*x+e)))^(3/2)-3*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)-3*arcta
n(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+c
os(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 )}^{2} \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))^2/(c-c*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.617205, size = 864, normalized size = 6.26 \begin{align*} \left [-\frac{3 \, \sqrt{2} \sqrt{-c}{\left (\cos \left (f x + e\right ) + 1\right )} \log \left (\frac{2 \, \sqrt{2}{\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt{-c} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} +{\left (3 \, c \cos \left (f x + e\right ) + c\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 \,{\left (5 \, \cos \left (f x + e\right )^{2} + 3 \, \cos \left (f x + e\right )\right )} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{24 \,{\left (a^{2} c f \cos \left (f x + e\right ) + a^{2} c f\right )} \sin \left (f x + e\right )}, \frac{3 \, \sqrt{2} \sqrt{c}{\left (\cos \left (f x + e\right ) + 1\right )} \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 \,{\left (5 \, \cos \left (f x + e\right )^{2} + 3 \, \cos \left (f x + e\right )\right )} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{12 \,{\left (a^{2} c f \cos \left (f x + e\right ) + a^{2} c f\right )} \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))^2/(c-c*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

[-1/24*(3*sqrt(2)*sqrt(-c)*(cos(f*x + e) + 1)*log((2*sqrt(2)*(cos(f*x + e)^2 + cos(f*x + e))*sqrt(-c)*sqrt((c*
cos(f*x + e) - c)/cos(f*x + e)) + (3*c*cos(f*x + e) + c)*sin(f*x + e))/((cos(f*x + e) - 1)*sin(f*x + e)))*sin(
f*x + e) + 4*(5*cos(f*x + e)^2 + 3*cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)))/((a^2*c*f*cos(f*x +
e) + a^2*c*f)*sin(f*x + e)), 1/12*(3*sqrt(2)*sqrt(c)*(cos(f*x + e) + 1)*arctan(sqrt(2)*sqrt((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*(5*cos(f*x + e)^2 + 3*cos(f*x + e))*sqr
t((c*cos(f*x + e) - c)/cos(f*x + e)))/((a^2*c*f*cos(f*x + e) + a^2*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 ^{2}{\left (e + f x \right )} + 2 \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^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Giac [C]  time = 1.57959, size = 224, normalized size = 1.62 \begin{align*} -\frac{\frac{\sqrt{2}{\left (3 i \, \sqrt{-c} \arctan \left (-i\right ) - 4 \, \sqrt{-c}\right )} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}{a^{2} c} + \frac{\sqrt{2}{\left (\frac{3 \, \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{{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{\frac{3}{2}} c^{4} - 3 \, \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c} c^{5}}{c^{6}}\right )}}{a^{2} \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 )}}{12 \, 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))^2/(c-c*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

-1/12*(sqrt(2)*(3*I*sqrt(-c)*arctan(-I) - 4*sqrt(-c))*sgn(tan(1/2*f*x + 1/2*e))/(a^2*c) + sqrt(2)*(3*arctan(sq
rt(c*tan(1/2*f*x + 1/2*e)^2 - c)/sqrt(c))/sqrt(c) + ((c*tan(1/2*f*x + 1/2*e)^2 - c)^(3/2)*c^4 - 3*sqrt(c*tan(1
/2*f*x + 1/2*e)^2 - c)*c^5)/c^6)/(a^2*sgn(tan(1/2*f*x + 1/2*e)^2 - 1)*sgn(tan(1/2*f*x + 1/2*e))))/f