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

Optimal. Leaf size=77 \[ \frac{2 a \tan (e+f x)}{f \sqrt{c-c \sec (e+f x)}}-\frac{2 \sqrt{2} a \tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{\sqrt{c} f} \]

[Out]

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

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Rubi [A]  time = 0.106849, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.094, Rules used = {3956, 3795, 203} \[ \frac{2 a \tan (e+f x)}{f \sqrt{c-c \sec (e+f x)}}-\frac{2 \sqrt{2} a \tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{\sqrt{c} f} \]

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]

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

Rule 3956

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

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{2 a \tan (e+f x)}{f \sqrt{c-c \sec (e+f x)}}+(2 a) \int \frac{\sec (e+f x)}{\sqrt{c-c \sec (e+f x)}} \, dx\\ &=\frac{2 a \tan (e+f x)}{f \sqrt{c-c \sec (e+f x)}}-\frac{(4 a) \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 )}{f}\\ &=-\frac{2 \sqrt{2} a \tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{\sqrt{c} f}+\frac{2 a \tan (e+f x)}{f \sqrt{c-c \sec (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.616451, size = 132, normalized size = 1.71 \[ -\frac{i \sqrt{2} a \left (-1+e^{i (e+f x)}\right ) \left (\sqrt{2} \left (1+e^{i (e+f x)}\right )-2 \sqrt{1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\frac{1+e^{i (e+f x)}}{\sqrt{2} \sqrt{1+e^{2 i (e+f x)}}}\right )\right )}{f \left (1+e^{2 i (e+f x)}\right ) \sqrt{c-c \sec (e+f x)}} \]

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

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Maple [A]  time = 0.239, size = 85, normalized size = 1.1 \begin{align*} -2\,{\frac{a\sin \left ( fx+e \right ) }{f\cos \left ( fx+e \right ) } \left ( \arctan \left ({\frac{1}{\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}}} \right ) \sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}-1 \right ){\frac{1}{\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))/(c-c*sec(f*x+e))^(1/2),x)

[Out]

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

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

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Fricas [A]  time = 0.58336, size = 686, normalized size = 8.91 \begin{align*} \left [\frac{\sqrt{2} a 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 ) - 2 \,{\left (a \cos \left (f x + e\right ) + a\right )} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{c f \sin \left (f x + e\right )}, \frac{2 \,{\left (\sqrt{2} a \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 ) -{\left (a \cos \left (f x + e\right ) + a\right )} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}\right )}}{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]

[(sqrt(2)*a*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) - 2*(a*cos(
f*x + e) + a)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)))/(c*f*sin(f*x + e)), 2*(sqrt(2)*a*sqrt(c)*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) - (a*cos(f*x + e) +
a)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)))/(c*f*sin(f*x + e))]

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

[Out]

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

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Giac [C]  time = 1.82451, size = 228, normalized size = 2.96 \begin{align*} -\frac{2 \,{\left (a c{\left (\frac{\sqrt{2} \arctan \left (\frac{\sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c}}{\sqrt{c}}\right )}{c^{\frac{3}{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 )} + \frac{\sqrt{2}}{\sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c} c \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 )}\right )} - \frac{{\left (-i \, \sqrt{2} a \sqrt{-c} \arctan \left (-i\right ) + \sqrt{2} a \sqrt{-c}\right )} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}{c}\right )}}{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]

-2*(a*c*(sqrt(2)*arctan(sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)/sqrt(c))/(c^(3/2)*sgn(tan(1/2*f*x + 1/2*e)^2 - 1)*s
gn(tan(1/2*f*x + 1/2*e))) + sqrt(2)/(sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)*c*sgn(tan(1/2*f*x + 1/2*e)^2 - 1)*sgn(
tan(1/2*f*x + 1/2*e)))) - (-I*sqrt(2)*a*sqrt(-c)*arctan(-I) + sqrt(2)*a*sqrt(-c))*sgn(tan(1/2*f*x + 1/2*e))/c)
/f