3.91 \(\int \frac{\sec (e+f x)}{(a+a \sec (e+f x)) (c-c \sec (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=122 \[ -\frac{3 \tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{4 \sqrt{2} a c^{3/2} f}-\frac{3 \tan (e+f x)}{4 a f (c-c \sec (e+f x))^{3/2}}+\frac{\tan (e+f x)}{f (a \sec (e+f x)+a) (c-c \sec (e+f x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.210511, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {3960, 3796, 3795, 203} \[ -\frac{3 \tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{4 \sqrt{2} a c^{3/2} f}-\frac{3 \tan (e+f x)}{4 a f (c-c \sec (e+f x))^{3/2}}+\frac{\tan (e+f x)}{f (a \sec (e+f x)+a) (c-c \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]/((a + a*Sec[e + f*x])*(c - c*Sec[e + f*x])^(3/2)),x]

[Out]

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

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 3796

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

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

Mathematica [C]  time = 1.43111, size = 183, normalized size = 1.5 \[ \frac{e^{-\frac{1}{2} i (e+f x)} \csc (e+f x) \left (\cos \left (\frac{1}{2} (e+f x)\right )+i \sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (-8 (\cos (e+f x)-3)+\frac{6 \sqrt{2} e^{-i (e+f x)} \left (-1+e^{i (e+f x)}\right )^2 \left (1+e^{i (e+f x)}\right ) \tanh ^{-1}\left (\frac{1+e^{i (e+f x)}}{\sqrt{2} \sqrt{1+e^{2 i (e+f x)}}}\right )}{\sqrt{1+e^{2 i (e+f x)}}}\right )}{32 a c f \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]/((a + a*Sec[e + f*x])*(c - c*Sec[e + f*x])^(3/2)),x]

[Out]

(((6*Sqrt[2]*(-1 + E^(I*(e + f*x)))^2*(1 + E^(I*(e + f*x)))*ArcTanh[(1 + E^(I*(e + f*x)))/(Sqrt[2]*Sqrt[1 + E^
((2*I)*(e + f*x))])])/(E^(I*(e + f*x))*Sqrt[1 + E^((2*I)*(e + f*x))]) - 8*(-3 + Cos[e + f*x]))*Csc[e + f*x]*(C
os[(e + f*x)/2] + I*Sin[(e + f*x)/2]))/(32*a*c*E^((I/2)*(e + f*x))*f*Sqrt[c - c*Sec[e + f*x]])

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Maple [B]  time = 0.21, size = 266, normalized size = 2.2 \begin{align*}{\frac{ \left ( -1+\cos \left ( fx+e \right ) \right ) ^{2}}{2\,fa \left ( \sin \left ( fx+e \right ) \right ) ^{3}} \left ( \cos \left ( fx+e \right ) \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{{\frac{5}{2}}}+ \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{{\frac{5}{2}}}+\cos \left ( fx+e \right ) \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{{\frac{3}{2}}}- \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 ) }}}\cos \left ( fx+e \right ) -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 ) +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 ) \left ({\frac{c \left ( -1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{3}{2}}} \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{3}{2}}}} \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))^(3/2),x)

[Out]

1/2/a/f*(-1+cos(f*x+e))^2*(cos(f*x+e)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(5/2)+(-2*cos(f*x+e)/(1+cos(f*x+e)))^(5/2
)+cos(f*x+e)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(3/2)-(-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)*cos(f*x+e)-3*cos(f*x+e)*arctan(1/(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2))+3*(-2*cos(f*x+e)/(1+cos
(f*x+e)))^(1/2)+3*arctan(1/(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)))/(c*(-1+cos(f*x+e))/cos(f*x+e))^(3/2)/sin(f*x
+e)^3/(-2*cos(f*x+e)/(1+cos(f*x+e)))^(3/2)

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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 )}{\left (-c \sec \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}\,{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))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.617447, size = 857, normalized size = 7.02 \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 (\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 )}}}{16 \,{\left (a c^{2} f \cos \left (f x + e\right ) - a c^{2} 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 (\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 )}}}{8 \,{\left (a c^{2} f \cos \left (f x + e\right ) - a c^{2} 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))/(c-c*sec(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

[-1/16*(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*(cos(f*x + e)^2 - 3*cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)))/((a*c^2*f*cos(f*x + e)
- a*c^2*f)*sin(f*x + e)), 1/8*(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*(cos(f*x + e)^2 - 3*cos(f*x + e))*sqrt((c*
cos(f*x + e) - c)/cos(f*x + e)))/((a*c^2*f*cos(f*x + e) - a*c^2*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 )}}{- c \sqrt{- c \sec{\left (e + f x \right )} + c} \sec ^{2}{\left (e + f x \right )} + c \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))**(3/2),x)

[Out]

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

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Giac [A]  time = 1.60222, size = 177, normalized size = 1.45 \begin{align*} -\frac{\sqrt{2}{\left (3 \, \sqrt{c} \arctan \left (\frac{\sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c}}{\sqrt{c}}\right ) - 2 \, \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c} - \frac{\sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c}}{\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2}}\right )}}{8 \, a c^{2} f \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 )} \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))^(3/2),x, algorithm="giac")

[Out]

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