3.104 \(\int \frac{\sec (e+f x)}{(a+a \sec (e+f x))^3 \sqrt{c-c \sec (e+f x)}} \, dx\)

Optimal. Leaf size=181 \[ -\frac{\tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{4 \sqrt{2} a^3 \sqrt{c} f}+\frac{\tan (e+f x)}{4 f \left (a^3 \sec (e+f x)+a^3\right ) \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x)}{6 a f (a \sec (e+f x)+a)^2 \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x)}{5 f (a \sec (e+f x)+a)^3 \sqrt{c-c \sec (e+f x)}} \]

[Out]

-ArcTan[(Sqrt[c]*Tan[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sec[e + f*x]])]/(4*Sqrt[2]*a^3*Sqrt[c]*f) + Tan[e + f*x]/(5
*f*(a + a*Sec[e + f*x])^3*Sqrt[c - c*Sec[e + f*x]]) + Tan[e + f*x]/(6*a*f*(a + a*Sec[e + f*x])^2*Sqrt[c - c*Se
c[e + f*x]]) + Tan[e + f*x]/(4*f*(a^3 + a^3*Sec[e + f*x])*Sqrt[c - c*Sec[e + f*x]])

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Rubi [A]  time = 0.394111, antiderivative size = 181, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.088, Rules used = {3960, 3795, 203} \[ -\frac{\tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{4 \sqrt{2} a^3 \sqrt{c} f}+\frac{\tan (e+f x)}{4 f \left (a^3 \sec (e+f x)+a^3\right ) \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x)}{6 a f (a \sec (e+f x)+a)^2 \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x)}{5 f (a \sec (e+f x)+a)^3 \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]/((a + a*Sec[e + f*x])^3*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]])]/(4*Sqrt[2]*a^3*Sqrt[c]*f) + Tan[e + f*x]/(5
*f*(a + a*Sec[e + f*x])^3*Sqrt[c - c*Sec[e + f*x]]) + Tan[e + f*x]/(6*a*f*(a + a*Sec[e + f*x])^2*Sqrt[c - c*Se
c[e + f*x]]) + Tan[e + f*x]/(4*f*(a^3 + a^3*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))^3 \sqrt{c-c \sec (e+f x)}} \, dx &=\frac{\tan (e+f x)}{5 f (a+a \sec (e+f x))^3 \sqrt{c-c \sec (e+f x)}}+\frac{\int \frac{\sec (e+f x)}{(a+a \sec (e+f x))^2 \sqrt{c-c \sec (e+f x)}} \, dx}{2 a}\\ &=\frac{\tan (e+f x)}{5 f (a+a \sec (e+f x))^3 \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x)}{6 a 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}{4 a^2}\\ &=\frac{\tan (e+f x)}{5 f (a+a \sec (e+f x))^3 \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x)}{6 a f (a+a \sec (e+f x))^2 \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x)}{4 f \left (a^3+a^3 \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}{8 a^3}\\ &=\frac{\tan (e+f x)}{5 f (a+a \sec (e+f x))^3 \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x)}{6 a f (a+a \sec (e+f x))^2 \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x)}{4 f \left (a^3+a^3 \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 )}{4 a^3 f}\\ &=-\frac{\tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{4 \sqrt{2} a^3 \sqrt{c} f}+\frac{\tan (e+f x)}{5 f (a+a \sec (e+f x))^3 \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x)}{6 a f (a+a \sec (e+f x))^2 \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x)}{4 f \left (a^3+a^3 \sec (e+f x)\right ) \sqrt{c-c \sec (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 1.58037, size = 225, normalized size = 1.24 \[ \frac{2 e^{-\frac{1}{2} i (e+f x)} \sin \left (\frac{1}{2} (e+f x)\right ) \cos \left (\frac{1}{2} (e+f x)\right ) \sec ^{\frac{7}{2}}(e+f x) \left (\frac{e^{\frac{1}{2} i (e+f x)} (80 \cos (e+f x)+37 \cos (2 (e+f x))+67)}{8 \sqrt{\sec (e+f x)}}-15 \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt{1+e^{2 i (e+f x)}} \cos ^5\left (\frac{1}{2} (e+f x)\right ) \tanh ^{-1}\left (\frac{1+e^{i (e+f x)}}{\sqrt{2} \sqrt{1+e^{2 i (e+f x)}}}\right )\right )}{15 a^3 f (\sec (e+f x)+1)^3 \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]/((a + a*Sec[e + f*x])^3*Sqrt[c - c*Sec[e + f*x]]),x]

[Out]

(2*Cos[(e + f*x)/2]*(-15*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]^5 + (E^((I/2)*(e + f*x))*(67
+ 80*Cos[e + f*x] + 37*Cos[2*(e + f*x)]))/(8*Sqrt[Sec[e + f*x]]))*Sec[e + f*x]^(7/2)*Sin[(e + f*x)/2])/(15*a^3
*E^((I/2)*(e + f*x))*f*(1 + Sec[e + f*x])^3*Sqrt[c - c*Sec[e + f*x]])

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Maple [A]  time = 0.244, size = 155, normalized size = 0.9 \begin{align*} -{\frac{-1+\cos \left ( fx+e \right ) }{60\,f{a}^{3}\sin \left ( fx+e \right ) } \left ( 3\, \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{5/2}-5\, \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{3/2}+15\,\arctan \left ({\frac{1}{\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}}} \right ) +15\,\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \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))^3/(c-c*sec(f*x+e))^(1/2),x)

[Out]

-1/60/a^3/f*(-1+cos(f*x+e))*(3*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(5/2)-5*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(3/2)+15*
arctan(1/(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2))+15*(-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 )}^{3} \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))^3/(c-c*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(sec(f*x + e)/((a*sec(f*x + e) + a)^3*sqrt(-c*sec(f*x + e) + c)), x)

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Fricas [A]  time = 0.659831, size = 1053, normalized size = 5.82 \begin{align*} \left [-\frac{15 \, \sqrt{2}{\left (\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1\right )} \sqrt{-c} \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 (37 \, \cos \left (f x + e\right )^{3} + 40 \, \cos \left (f x + e\right )^{2} + 15 \, \cos \left (f x + e\right )\right )} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{240 \,{\left (a^{3} c f \cos \left (f x + e\right )^{2} + 2 \, a^{3} c f \cos \left (f x + e\right ) + a^{3} c f\right )} \sin \left (f x + e\right )}, \frac{15 \, \sqrt{2}{\left (\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1\right )} \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 \,{\left (37 \, \cos \left (f x + e\right )^{3} + 40 \, \cos \left (f x + e\right )^{2} + 15 \, \cos \left (f x + e\right )\right )} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{120 \,{\left (a^{3} c f \cos \left (f x + e\right )^{2} + 2 \, a^{3} c f \cos \left (f x + e\right ) + a^{3} 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))^3/(c-c*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

[-1/240*(15*sqrt(2)*(cos(f*x + e)^2 + 2*cos(f*x + e) + 1)*sqrt(-c)*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*(37*cos(f*x + e)^3 + 40*cos(f*x + e)^2 + 15*cos(f*x + e))*sqrt((c*cos(f*x +
e) - c)/cos(f*x + e)))/((a^3*c*f*cos(f*x + e)^2 + 2*a^3*c*f*cos(f*x + e) + a^3*c*f)*sin(f*x + e)), 1/120*(15*s
qrt(2)*(cos(f*x + e)^2 + 2*cos(f*x + e) + 1)*sqrt(c)*arctan(sqrt(2)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))*co
s(f*x + e)/(sqrt(c)*sin(f*x + e)))*sin(f*x + e) - 2*(37*cos(f*x + e)^3 + 40*cos(f*x + e)^2 + 15*cos(f*x + e))*
sqrt((c*cos(f*x + e) - c)/cos(f*x + e)))/((a^3*c*f*cos(f*x + e)^2 + 2*a^3*c*f*cos(f*x + e) + a^3*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 ^{3}{\left (e + f x \right )} + 3 \sqrt{- c \sec{\left (e + f x \right )} + c} \sec ^{2}{\left (e + f x \right )} + 3 \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^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)/(a+a*sec(f*x+e))**3/(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)**3 + 3*sqrt(-c*sec(e + f*x) + c)*sec(e + f*x)**2
+ 3*sqrt(-c*sec(e + f*x) + c)*sec(e + f*x) + sqrt(-c*sec(e + f*x) + c)), x)/a**3

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Giac [C]  time = 1.61584, size = 261, normalized size = 1.44 \begin{align*} -\frac{\frac{\sqrt{2}{\left (15 i \, \sqrt{-c} \arctan \left (-i\right ) - 23 \, \sqrt{-c}\right )} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}{a^{3} c} + \frac{\sqrt{2}{\left (\frac{15 \, \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{3 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{\frac{5}{2}} c^{12} - 5 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{\frac{3}{2}} c^{13} + 15 \, \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c} c^{14}}{c^{15}}\right )}}{a^{3} \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 )}}{120 \, 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))^3/(c-c*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

-1/120*(sqrt(2)*(15*I*sqrt(-c)*arctan(-I) - 23*sqrt(-c))*sgn(tan(1/2*f*x + 1/2*e))/(a^3*c) + sqrt(2)*(15*arcta
n(sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)/sqrt(c))/sqrt(c) - (3*(c*tan(1/2*f*x + 1/2*e)^2 - c)^(5/2)*c^12 - 5*(c*ta
n(1/2*f*x + 1/2*e)^2 - c)^(3/2)*c^13 + 15*sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)*c^14)/c^15)/(a^3*sgn(tan(1/2*f*x
+ 1/2*e)^2 - 1)*sgn(tan(1/2*f*x + 1/2*e))))/f