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

Optimal. Leaf size=169 \[ -\frac{5 \tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{8 \sqrt{2} a^2 c^{3/2} f}-\frac{5 \tan (e+f x)}{8 a^2 f (c-c \sec (e+f x))^{3/2}}+\frac{5 \tan (e+f x)}{6 f \left (a^2 \sec (e+f x)+a^2\right ) (c-c \sec (e+f x))^{3/2}}+\frac{\tan (e+f x)}{3 f (a \sec (e+f x)+a)^2 (c-c \sec (e+f x))^{3/2}} \]

[Out]

(-5*ArcTan[(Sqrt[c]*Tan[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sec[e + f*x]])])/(8*Sqrt[2]*a^2*c^(3/2)*f) - (5*Tan[e +
f*x])/(8*a^2*f*(c - c*Sec[e + f*x])^(3/2)) + Tan[e + f*x]/(3*f*(a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^(3/
2)) + (5*Tan[e + f*x])/(6*f*(a^2 + a^2*Sec[e + f*x])*(c - c*Sec[e + f*x])^(3/2))

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Rubi [A]  time = 0.343459, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {3960, 3796, 3795, 203} \[ -\frac{5 \tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{8 \sqrt{2} a^2 c^{3/2} f}-\frac{5 \tan (e+f x)}{8 a^2 f (c-c \sec (e+f x))^{3/2}}+\frac{5 \tan (e+f x)}{6 f \left (a^2 \sec (e+f x)+a^2\right ) (c-c \sec (e+f x))^{3/2}}+\frac{\tan (e+f x)}{3 f (a \sec (e+f x)+a)^2 (c-c \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*ArcTan[(Sqrt[c]*Tan[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sec[e + f*x]])])/(8*Sqrt[2]*a^2*c^(3/2)*f) - (5*Tan[e +
f*x])/(8*a^2*f*(c - c*Sec[e + f*x])^(3/2)) + Tan[e + f*x]/(3*f*(a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^(3/
2)) + (5*Tan[e + f*x])/(6*f*(a^2 + a^2*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))^2 (c-c \sec (e+f x))^{3/2}} \, dx &=\frac{\tan (e+f x)}{3 f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2}}+\frac{5 \int \frac{\sec (e+f x)}{(a+a \sec (e+f x)) (c-c \sec (e+f x))^{3/2}} \, dx}{6 a}\\ &=\frac{\tan (e+f x)}{3 f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2}}+\frac{5 \tan (e+f x)}{6 f \left (a^2+a^2 \sec (e+f x)\right ) (c-c \sec (e+f x))^{3/2}}+\frac{5 \int \frac{\sec (e+f x)}{(c-c \sec (e+f x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{5 \tan (e+f x)}{8 a^2 f (c-c \sec (e+f x))^{3/2}}+\frac{\tan (e+f x)}{3 f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2}}+\frac{5 \tan (e+f x)}{6 f \left (a^2+a^2 \sec (e+f x)\right ) (c-c \sec (e+f x))^{3/2}}+\frac{5 \int \frac{\sec (e+f x)}{\sqrt{c-c \sec (e+f x)}} \, dx}{16 a^2 c}\\ &=-\frac{5 \tan (e+f x)}{8 a^2 f (c-c \sec (e+f x))^{3/2}}+\frac{\tan (e+f x)}{3 f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2}}+\frac{5 \tan (e+f x)}{6 f \left (a^2+a^2 \sec (e+f x)\right ) (c-c \sec (e+f x))^{3/2}}-\frac{5 \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 )}{8 a^2 c f}\\ &=-\frac{5 \tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{8 \sqrt{2} a^2 c^{3/2} f}-\frac{5 \tan (e+f x)}{8 a^2 f (c-c \sec (e+f x))^{3/2}}+\frac{\tan (e+f x)}{3 f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2}}+\frac{5 \tan (e+f x)}{6 f \left (a^2+a^2 \sec (e+f x)\right ) (c-c \sec (e+f x))^{3/2}}\\ \end{align*}

Mathematica [C]  time = 1.39913, size = 365, normalized size = 2.16 \[ \frac{-\frac{15 i \sqrt{2} \left (-1+e^{i (e+f x)}\right )^3 \left (1+e^{i (e+f x)}\right )^4 \tanh ^{-1}\left (\frac{1+e^{i (e+f x)}}{\sqrt{2} \sqrt{1+e^{2 i (e+f x)}}}\right )}{\left (1+e^{2 i (e+f x)}\right )^{7/2}}-416 \sin \left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right ) \sin ^8(e+f x) \csc \left (\frac{1}{2} (e+f x)\right ) \csc ^4(2 (e+f x))-40 \tan ^3(e+f x) \sec (e+f x)+416 \cos \left (\frac{e}{2}\right ) \cos \left (\frac{f x}{2}\right ) \sin ^3\left (\frac{1}{2} (e+f x)\right ) \cos ^4\left (\frac{1}{2} (e+f x)\right ) \sec ^4(e+f x)+32 \sin ^3\left (\frac{1}{2} (e+f x)\right ) \cos \left (\frac{1}{2} (e+f x)\right ) \sec ^4(e+f x)+48 \cot \left (\frac{e}{2}\right ) \sin ^2\left (\frac{1}{2} (e+f x)\right ) \cos ^4\left (\frac{1}{2} (e+f x)\right ) \sec ^4(e+f x)-48 \csc \left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right ) \sin \left (\frac{1}{2} (e+f x)\right ) \cos ^4\left (\frac{1}{2} (e+f x)\right ) \sec ^4(e+f x)}{48 a^2 c f (\sec (e+f x)-1) (\sec (e+f x)+1)^2 \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-15*I)*Sqrt[2]*(-1 + E^(I*(e + f*x)))^3*(1 + E^(I*(e + f*x)))^4*ArcTanh[(1 + E^(I*(e + f*x)))/(Sqrt[2]*Sqrt
[1 + E^((2*I)*(e + f*x))])])/(1 + E^((2*I)*(e + f*x)))^(7/2) - 48*Cos[(e + f*x)/2]^4*Csc[e/2]*Sec[e + f*x]^4*S
in[(f*x)/2]*Sin[(e + f*x)/2] + 48*Cos[(e + f*x)/2]^4*Cot[e/2]*Sec[e + f*x]^4*Sin[(e + f*x)/2]^2 + 32*Cos[(e +
f*x)/2]*Sec[e + f*x]^4*Sin[(e + f*x)/2]^3 + 416*Cos[e/2]*Cos[(f*x)/2]*Cos[(e + f*x)/2]^4*Sec[e + f*x]^4*Sin[(e
+ f*x)/2]^3 - 416*Csc[(e + f*x)/2]*Csc[2*(e + f*x)]^4*Sin[e/2]*Sin[(f*x)/2]*Sin[e + f*x]^8 - 40*Sec[e + f*x]*
Tan[e + f*x]^3)/(48*a^2*c*f*(-1 + Sec[e + f*x])*(1 + Sec[e + f*x])^2*Sqrt[c - c*Sec[e + f*x]])

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Maple [B]  time = 0.233, size = 320, normalized size = 1.9 \begin{align*} -{\frac{ \left ( -1+\cos \left ( fx+e \right ) \right ) ^{2}}{12\,f{a}^{2} \left ( \sin \left ( fx+e \right ) \right ) ^{3}} \left ( 3\, \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{7/2}\cos \left ( fx+e \right ) +3\, \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{7/2}+3\,\cos \left ( fx+e \right ) \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{5/2}-3\, \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{5/2}-5\,\cos \left ( fx+e \right ) \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{3/2}+5\, \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{3/2}+15\,\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}\cos \left ( fx+e \right ) +15\,\cos \left ( fx+e \right ) \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 ) }}}-15\,\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))^2/(c-c*sec(f*x+e))^(3/2),x)

[Out]

-1/12/a^2/f*(-1+cos(f*x+e))^2*(3*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(7/2)*cos(f*x+e)+3*(-2*cos(f*x+e)/(1+cos(f*x+e
)))^(7/2)+3*cos(f*x+e)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(5/2)-3*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(5/2)-5*cos(f*x+e
)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(3/2)+5*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(3/2)+15*(-2*cos(f*x+e)/(1+cos(f*x+e))
)^(1/2)*cos(f*x+e)+15*cos(f*x+e)*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)-15*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 )}^{2}{\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))^2/(c-c*sec(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.660829, size = 948, normalized size = 5.61 \begin{align*} \left [-\frac{15 \, \sqrt{2}{\left (\cos \left (f x + e\right )^{2} - 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 (13 \, \cos \left (f x + e\right )^{3} - 10 \, \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 )}}}{96 \,{\left (a^{2} c^{2} f \cos \left (f x + e\right )^{2} - a^{2} c^{2} f\right )} \sin \left (f x + e\right )}, \frac{15 \, \sqrt{2}{\left (\cos \left (f x + e\right )^{2} - 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 (13 \, \cos \left (f x + e\right )^{3} - 10 \, \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 )}}}{48 \,{\left (a^{2} c^{2} f \cos \left (f x + e\right )^{2} - a^{2} 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))^2/(c-c*sec(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

[-1/96*(15*sqrt(2)*(cos(f*x + e)^2 - 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)))*s
in(f*x + e) + 4*(13*cos(f*x + e)^3 - 10*cos(f*x + e)^2 - 15*cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x +
e)))/((a^2*c^2*f*cos(f*x + e)^2 - a^2*c^2*f)*sin(f*x + e)), 1/48*(15*sqrt(2)*(cos(f*x + e)^2 - 1)*sqrt(c)*arct
an(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*(13*c
os(f*x + e)^3 - 10*cos(f*x + e)^2 - 15*cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)))/((a^2*c^2*f*cos(
f*x + e)^2 - a^2*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 ^{3}{\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} \sec{\left (e + f x \right )} + c \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))**(3/2),x)

[Out]

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

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Giac [A]  time = 1.51185, size = 221, normalized size = 1.31 \begin{align*} -\frac{\sqrt{2}{\left (15 \, \sqrt{c} \arctan \left (\frac{\sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c}}{\sqrt{c}}\right ) - \frac{3 \, \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}} + \frac{2 \,{\left ({\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{\frac{3}{2}} c^{2} - 6 \, \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c} c^{3}\right )}}{c^{3}}\right )}}{48 \, a^{2} 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))^2/(c-c*sec(f*x+e))^(3/2),x, algorithm="giac")

[Out]

-1/48*sqrt(2)*(15*sqrt(c)*arctan(sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)/sqrt(c)) - 3*sqrt(c*tan(1/2*f*x + 1/2*e)^2
- c)/tan(1/2*f*x + 1/2*e)^2 + 2*((c*tan(1/2*f*x + 1/2*e)^2 - c)^(3/2)*c^2 - 6*sqrt(c*tan(1/2*f*x + 1/2*e)^2 -
c)*c^3)/c^3)/(a^2*c^2*f*sgn(tan(1/2*f*x + 1/2*e)^2 - 1)*sgn(tan(1/2*f*x + 1/2*e)))