3.92 \(\int \frac{\sec (e+f x)}{(a+a \sec (e+f x)) (c-c \sec (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=156 \[ -\frac{15 \tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{32 \sqrt{2} a c^{5/2} f}-\frac{15 \tan (e+f x)}{32 a c f (c-c \sec (e+f x))^{3/2}}-\frac{5 \tan (e+f x)}{8 a f (c-c \sec (e+f x))^{5/2}}+\frac{\tan (e+f x)}{f (a \sec (e+f x)+a) (c-c \sec (e+f x))^{5/2}} \]

[Out]

(-15*ArcTan[(Sqrt[c]*Tan[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sec[e + f*x]])])/(32*Sqrt[2]*a*c^(5/2)*f) - (5*Tan[e +
f*x])/(8*a*f*(c - c*Sec[e + f*x])^(5/2)) + Tan[e + f*x]/(f*(a + a*Sec[e + f*x])*(c - c*Sec[e + f*x])^(5/2)) -
(15*Tan[e + f*x])/(32*a*c*f*(c - c*Sec[e + f*x])^(3/2))

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Rubi [A]  time = 0.255264, antiderivative size = 156, 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} \[ -\frac{15 \tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{32 \sqrt{2} a c^{5/2} f}-\frac{15 \tan (e+f x)}{32 a c f (c-c \sec (e+f x))^{3/2}}-\frac{5 \tan (e+f x)}{8 a f (c-c \sec (e+f x))^{5/2}}+\frac{\tan (e+f x)}{f (a \sec (e+f x)+a) (c-c \sec (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]/((a + a*Sec[e + f*x])*(c - c*Sec[e + f*x])^(5/2)),x]

[Out]

(-15*ArcTan[(Sqrt[c]*Tan[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sec[e + f*x]])])/(32*Sqrt[2]*a*c^(5/2)*f) - (5*Tan[e +
f*x])/(8*a*f*(c - c*Sec[e + f*x])^(5/2)) + Tan[e + f*x]/(f*(a + a*Sec[e + f*x])*(c - c*Sec[e + f*x])^(5/2)) -
(15*Tan[e + f*x])/(32*a*c*f*(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))^{5/2}} \, dx &=\frac{\tan (e+f x)}{f (a+a \sec (e+f x)) (c-c \sec (e+f x))^{5/2}}+\frac{5 \int \frac{\sec (e+f x)}{(c-c \sec (e+f x))^{5/2}} \, dx}{2 a}\\ &=-\frac{5 \tan (e+f x)}{8 a f (c-c \sec (e+f x))^{5/2}}+\frac{\tan (e+f x)}{f (a+a \sec (e+f x)) (c-c \sec (e+f x))^{5/2}}+\frac{15 \int \frac{\sec (e+f x)}{(c-c \sec (e+f x))^{3/2}} \, dx}{16 a c}\\ &=-\frac{5 \tan (e+f x)}{8 a f (c-c \sec (e+f x))^{5/2}}+\frac{\tan (e+f x)}{f (a+a \sec (e+f x)) (c-c \sec (e+f x))^{5/2}}-\frac{15 \tan (e+f x)}{32 a c f (c-c \sec (e+f x))^{3/2}}+\frac{15 \int \frac{\sec (e+f x)}{\sqrt{c-c \sec (e+f x)}} \, dx}{64 a c^2}\\ &=-\frac{5 \tan (e+f x)}{8 a f (c-c \sec (e+f x))^{5/2}}+\frac{\tan (e+f x)}{f (a+a \sec (e+f x)) (c-c \sec (e+f x))^{5/2}}-\frac{15 \tan (e+f x)}{32 a c f (c-c \sec (e+f x))^{3/2}}-\frac{15 \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 )}{32 a c^2 f}\\ &=-\frac{15 \tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{32 \sqrt{2} a c^{5/2} f}-\frac{5 \tan (e+f x)}{8 a f (c-c \sec (e+f x))^{5/2}}+\frac{\tan (e+f x)}{f (a+a \sec (e+f x)) (c-c \sec (e+f x))^{5/2}}-\frac{15 \tan (e+f x)}{32 a c f (c-c \sec (e+f x))^{3/2}}\\ \end{align*}

Mathematica [C]  time = 3.01863, size = 306, normalized size = 1.96 \[ \frac{e^{-\frac{1}{2} i (e+f x)} \sin \left (\frac{1}{2} (e+f x)\right ) \cos \left (\frac{1}{2} (e+f x)\right ) \sec ^{\frac{7}{2}}(e+f x) \left (-\frac{1}{32} e^{-\frac{5}{2} i (e+f x)} \left (40 e^{i (e+f x)}-51 e^{2 i (e+f x)}+80 e^{3 i (e+f x)}-51 e^{4 i (e+f x)}+40 e^{5 i (e+f x)}+3 e^{6 i (e+f x)}+3\right ) \sqrt{\sec (e+f x)}-\frac{15}{2} \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt{1+e^{2 i (e+f x)}} \sin (e+f x) \sin ^3\left (\frac{1}{2} (e+f x)\right ) \tanh ^{-1}\left (\frac{1+e^{i (e+f x)}}{\sqrt{2} \sqrt{1+e^{2 i (e+f x)}}}\right )\right )}{4 a c^2 f (\sec (e+f x)-1)^2 (\sec (e+f x)+1) \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]/((a + a*Sec[e + f*x])*(c - c*Sec[e + f*x])^(5/2)),x]

[Out]

(Cos[(e + f*x)/2]*Sec[e + f*x]^(7/2)*Sin[(e + f*x)/2]*(-((3 + 40*E^(I*(e + f*x)) - 51*E^((2*I)*(e + f*x)) + 80
*E^((3*I)*(e + f*x)) - 51*E^((4*I)*(e + f*x)) + 40*E^((5*I)*(e + f*x)) + 3*E^((6*I)*(e + f*x)))*Sqrt[Sec[e + f
*x]])/(32*E^(((5*I)/2)*(e + f*x))) - (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))])]*Sin[(e + f*x)/2]^3*Sin[e + f*x
])/2))/(4*a*c^2*E^((I/2)*(e + f*x))*f*(-1 + Sec[e + f*x])^2*(1 + Sec[e + f*x])*Sqrt[c - c*Sec[e + f*x]])

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Maple [B]  time = 0.22, size = 471, normalized size = 3. \begin{align*} \text{result too large to display} \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))^(5/2),x)

[Out]

-1/8/a/f*(-1+cos(f*x+e))^3*(5*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(7/2)*cos(f*x+e)^2+4*(-2*cos(f*x+e)/(1+cos(f*x+e)
))^(7/2)*cos(f*x+e)-(-2*cos(f*x+e)/(1+cos(f*x+e)))^(7/2)+3*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(5/2)*cos(f*x+e)^2-6
*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*(-2*cos
(f*x+e)/(1+cos(f*x+e)))^(3/2)+10*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*cos(f*x+e)^2*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)+15*cos(f*x+e)^2*arctan(1/(-2*cos(f*x+e)/(1+cos(
f*x+e)))^(1/2))-30*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*cos(f*x+e)-30*cos(f*x+e)*arctan(1/(-2*cos(f*x+e)/(1+co
s(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))^(5/2)/sin(f*x+e)^5/(-2*cos(f*x+e)/(1+cos(f*x+e)))^(5/2)

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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{5}{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))^(5/2),x, algorithm="maxima")

[Out]

integrate(sec(f*x + e)/((a*sec(f*x + e) + a)*(-c*sec(f*x + e) + c)^(5/2)), x)

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Fricas [A]  time = 0.664488, size = 1049, normalized size = 6.72 \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 (3 \, \cos \left (f x + e\right )^{3} + 20 \, \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 )}}}{128 \,{\left (a c^{3} f \cos \left (f x + e\right )^{2} - 2 \, a c^{3} f \cos \left (f x + e\right ) + a c^{3} 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 (3 \, \cos \left (f x + e\right )^{3} + 20 \, \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 )}}}{64 \,{\left (a c^{3} f \cos \left (f x + e\right )^{2} - 2 \, a c^{3} f \cos \left (f x + e\right ) + a c^{3} 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))^(5/2),x, algorithm="fricas")

[Out]

[-1/128*(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*(3*cos(f*x + e)^3 + 20*cos(f*x + e)^2 - 15*cos(f*x + e))*sqrt((c*cos(f*x + e
) - c)/cos(f*x + e)))/((a*c^3*f*cos(f*x + e)^2 - 2*a*c^3*f*cos(f*x + e) + a*c^3*f)*sin(f*x + e)), 1/64*(15*sqr
t(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))*cos(
f*x + e)/(sqrt(c)*sin(f*x + e)))*sin(f*x + e) + 2*(3*cos(f*x + e)^3 + 20*cos(f*x + e)^2 - 15*cos(f*x + e))*sqr
t((c*cos(f*x + e) - c)/cos(f*x + e)))/((a*c^3*f*cos(f*x + e)^2 - 2*a*c^3*f*cos(f*x + e) + a*c^3*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^{2} \sqrt{- c \sec{\left (e + f x \right )} + c} \sec ^{3}{\left (e + f x \right )} - c^{2} \sqrt{- c \sec{\left (e + f x \right )} + c} \sec ^{2}{\left (e + f x \right )} - c^{2} \sqrt{- c \sec{\left (e + f x \right )} + c} \sec{\left (e + f x \right )} + c^{2} \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))**(5/2),x)

[Out]

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

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Giac [A]  time = 1.58549, size = 220, normalized size = 1.41 \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 ) - 8 \, \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c} - \frac{9 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{\frac{3}{2}} c + 7 \, \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c} c^{2}}{c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4}}\right )}}{64 \, a c^{3} 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))^(5/2),x, algorithm="giac")

[Out]

-1/64*sqrt(2)*(15*sqrt(c)*arctan(sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)/sqrt(c)) - 8*sqrt(c*tan(1/2*f*x + 1/2*e)^2
- c) - (9*(c*tan(1/2*f*x + 1/2*e)^2 - c)^(3/2)*c + 7*sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)*c^2)/(c^2*tan(1/2*f*x
+ 1/2*e)^4))/(a*c^3*f*sgn(tan(1/2*f*x + 1/2*e)^2 - 1)*sgn(tan(1/2*f*x + 1/2*e)))