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

Optimal. Leaf size=117 \[ -\frac{4 \sqrt{2} a^2 \tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{\sqrt{c} f}-\frac{2 a^2 \tan (e+f x) \sqrt{c-c \sec (e+f x)}}{3 c f}+\frac{16 a^2 \tan (e+f x)}{3 f \sqrt{c-c \sec (e+f x)}} \]

[Out]

(-4*Sqrt[2]*a^2*ArcTan[(Sqrt[c]*Tan[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sec[e + f*x]])])/(Sqrt[c]*f) + (16*a^2*Tan[e
+ f*x])/(3*f*Sqrt[c - c*Sec[e + f*x]]) - (2*a^2*Sqrt[c - c*Sec[e + f*x]]*Tan[e + f*x])/(3*c*f)

________________________________________________________________________________________

Rubi [A]  time = 0.213263, antiderivative size = 123, normalized size of antiderivative = 1.05, 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 = {3956, 3795, 203} \[ -\frac{4 \sqrt{2} a^2 \tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{\sqrt{c} f}+\frac{4 a^2 \tan (e+f x)}{f \sqrt{c-c \sec (e+f x)}}+\frac{2 \tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{3 f \sqrt{c-c \sec (e+f x)}} \]

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]

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

Mathematica [C]  time = 0.889227, size = 173, normalized size = 1.48 \[ \frac{4 a^2 e^{-\frac{1}{2} i (e+f x)} \sin \left (\frac{1}{2} (e+f x)\right ) \sec (e+f x) \left (\cos \left (\frac{1}{2} (e+f x)\right )+i \sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right ) (\sec (e+f x)+7)-3 \sqrt{2} e^{-\frac{1}{2} i (e+f x)} \sqrt{1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\frac{1+e^{i (e+f x)}}{\sqrt{2} \sqrt{1+e^{2 i (e+f x)}}}\right )\right )}{3 f \sqrt{c-c \sec (e+f x)}} \]

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]

(4*a^2*Sec[e + f*x]*((-3*Sqrt[2]*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))])])/E^((I/2)*(e + f*x)) + Cos[(e + f*x)/2]*(7 + Sec[e + f*x]))*(Cos[(e + f*x)/2] + I*Sin[
(e + f*x)/2])*Sin[(e + f*x)/2])/(3*E^((I/2)*(e + f*x))*f*Sqrt[c - c*Sec[e + f*x]])

________________________________________________________________________________________

Maple [A]  time = 0.25, size = 145, normalized size = 1.2 \begin{align*}{\frac{2\,{a}^{2}\sin \left ( fx+e \right ) }{3\,f \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \left ( 3\,\cos \left ( fx+e \right ) \arctan \left ({\frac{1}{\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}}} \right ) \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{3/2}+3\,\arctan \left ({\frac{1}{\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}}} \right ) \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{3/2}+7\,\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))^2/(c-c*sec(f*x+e))^(1/2),x)

[Out]

2/3*a^2/f*(3*cos(f*x+e)*arctan(1/(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2))*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(3/2)+3*
arctan(1/(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2))*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(3/2)+7*cos(f*x+e)+1)*sin(f*x+e)
/cos(f*x+e)^2/(c*(-1+cos(f*x+e))/cos(f*x+e))^(1/2)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 0.596823, size = 851, normalized size = 7.27 \begin{align*} \left [\frac{2 \,{\left (3 \, \sqrt{2} a^{2} c \sqrt{-\frac{1}{c}} \cos \left (f x + e\right ) \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 ) -{\left (7 \, a^{2} \cos \left (f x + e\right )^{2} + 8 \, a^{2} \cos \left (f x + e\right ) + a^{2}\right )} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}\right )}}{3 \, c f \cos \left (f x + e\right ) \sin \left (f x + e\right )}, \frac{2 \,{\left (6 \, \sqrt{2} a^{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 ) \cos \left (f x + e\right ) \sin \left (f x + e\right ) -{\left (7 \, a^{2} \cos \left (f x + e\right )^{2} + 8 \, a^{2} \cos \left (f x + e\right ) + a^{2}\right )} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}\right )}}{3 \, c f \cos \left (f x + e\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]

[2/3*(3*sqrt(2)*a^2*c*sqrt(-1/c)*cos(f*x + e)*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) - (7*a^2*cos(f*x + e)^2 + 8*a^2*cos(f*x + e) + a^2)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)))/(c*f*co
s(f*x + e)*sin(f*x + e)), 2/3*(6*sqrt(2)*a^2*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)))*cos(f*x + e)*sin(f*x + e) - (7*a^2*cos(f*x + e)^2 + 8*a^2*cos(f*x + e) + a^
2)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)))/(c*f*cos(f*x + e)*sin(f*x + e))]

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{2} \left (\int \frac{\sec{\left (e + f x \right )}}{\sqrt{- c \sec{\left (e + f x \right )} + c}}\, dx + \int \frac{2 \sec ^{2}{\left (e + f x \right )}}{\sqrt{- c \sec{\left (e + f x \right )} + c}}\, dx + \int \frac{\sec ^{3}{\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))**2/(c-c*sec(f*x+e))**(1/2),x)

[Out]

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

________________________________________________________________________________________

Giac [C]  time = 1.95031, size = 267, normalized size = 2.28 \begin{align*} -\frac{2 \,{\left (2 \, a^{2} c^{2}{\left (\frac{3 \, \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{5}{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}{\left (3 \, c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 4 \, c\right )}}{{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{\frac{3}{2}} c^{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 )}\right )} + \frac{{\left (6 i \, \sqrt{2} a^{2} \sqrt{-c} \arctan \left (-i\right ) - 8 \, \sqrt{2} a^{2} \sqrt{-c}\right )} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}{c}\right )}}{3 \, 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]

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