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

Optimal. Leaf size=122 \[ -\frac{64 c^3 \tan (e+f x) (a \sec (e+f x)+a)}{105 f \sqrt{c-c \sec (e+f x)}}-\frac{16 c^2 \tan (e+f x) (a \sec (e+f x)+a) \sqrt{c-c \sec (e+f x)}}{35 f}-\frac{2 c \tan (e+f x) (a \sec (e+f x)+a) (c-c \sec (e+f x))^{3/2}}{7 f} \]

[Out]

(-64*c^3*(a + a*Sec[e + f*x])*Tan[e + f*x])/(105*f*Sqrt[c - c*Sec[e + f*x]]) - (16*c^2*(a + a*Sec[e + f*x])*Sq
rt[c - c*Sec[e + f*x]]*Tan[e + f*x])/(35*f) - (2*c*(a + a*Sec[e + f*x])*(c - c*Sec[e + f*x])^(3/2)*Tan[e + f*x
])/(7*f)

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Rubi [A]  time = 0.200355, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.062, Rules used = {3955, 3953} \[ -\frac{64 c^3 \tan (e+f x) (a \sec (e+f x)+a)}{105 f \sqrt{c-c \sec (e+f x)}}-\frac{16 c^2 \tan (e+f x) (a \sec (e+f x)+a) \sqrt{c-c \sec (e+f x)}}{35 f}-\frac{2 c \tan (e+f x) (a \sec (e+f x)+a) (c-c \sec (e+f x))^{3/2}}{7 f} \]

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]

(-64*c^3*(a + a*Sec[e + f*x])*Tan[e + f*x])/(105*f*Sqrt[c - c*Sec[e + f*x]]) - (16*c^2*(a + a*Sec[e + f*x])*Sq
rt[c - c*Sec[e + f*x]]*Tan[e + f*x])/(35*f) - (2*c*(a + a*Sec[e + f*x])*(c - c*Sec[e + f*x])^(3/2)*Tan[e + f*x
])/(7*f)

Rule 3955

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

Rule 3953

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

Rubi steps

\begin{align*} \int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x))^{5/2} \, dx &=-\frac{2 c (a+a \sec (e+f x)) (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{7 f}+\frac{1}{7} (8 c) \int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x))^{3/2} \, dx\\ &=-\frac{16 c^2 (a+a \sec (e+f x)) \sqrt{c-c \sec (e+f x)} \tan (e+f x)}{35 f}-\frac{2 c (a+a \sec (e+f x)) (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{7 f}+\frac{1}{35} \left (32 c^2\right ) \int \sec (e+f x) (a+a \sec (e+f x)) \sqrt{c-c \sec (e+f x)} \, dx\\ &=-\frac{64 c^3 (a+a \sec (e+f x)) \tan (e+f x)}{105 f \sqrt{c-c \sec (e+f x)}}-\frac{16 c^2 (a+a \sec (e+f x)) \sqrt{c-c \sec (e+f x)} \tan (e+f x)}{35 f}-\frac{2 c (a+a \sec (e+f x)) (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{7 f}\\ \end{align*}

Mathematica [A]  time = 0.479334, size = 76, normalized size = 0.62 \[ \frac{2 a c^2 \cos ^2\left (\frac{1}{2} (e+f x)\right ) (-108 \cos (e+f x)+71 \cos (2 (e+f x))+101) \cot \left (\frac{1}{2} (e+f x)\right ) \sec ^3(e+f x) \sqrt{c-c \sec (e+f x)}}{105 f} \]

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]

(2*a*c^2*Cos[(e + f*x)/2]^2*(101 - 108*Cos[e + f*x] + 71*Cos[2*(e + f*x)])*Cot[(e + f*x)/2]*Sec[e + f*x]^3*Sqr
t[c - c*Sec[e + f*x]])/(105*f)

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Maple [A]  time = 0.18, size = 73, normalized size = 0.6 \begin{align*}{\frac{2\,a \left ( \sin \left ( fx+e \right ) \right ) ^{3} \left ( 71\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}-54\,\cos \left ( fx+e \right ) +15 \right ) }{105\,f \left ( -1+\cos \left ( fx+e \right ) \right ) ^{4}\cos \left ( fx+e \right ) } \left ({\frac{c \left ( -1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{{\frac{5}{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))^(5/2),x)

[Out]

2/105*a/f*(c*(-1+cos(f*x+e))/cos(f*x+e))^(5/2)*sin(f*x+e)^3*(71*cos(f*x+e)^2-54*cos(f*x+e)+15)/(-1+cos(f*x+e))
^4/cos(f*x+e)

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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]

Timed out

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Fricas [A]  time = 0.469479, size = 259, normalized size = 2.12 \begin{align*} \frac{2 \,{\left (71 \, a c^{2} \cos \left (f x + e\right )^{4} + 88 \, a c^{2} \cos \left (f x + e\right )^{3} - 22 \, a c^{2} \cos \left (f x + e\right )^{2} - 24 \, a c^{2} \cos \left (f x + e\right ) + 15 \, a c^{2}\right )} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{105 \, f \cos \left (f x + e\right )^{3} \sin \left (f x + e\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]

2/105*(71*a*c^2*cos(f*x + e)^4 + 88*a*c^2*cos(f*x + e)^3 - 22*a*c^2*cos(f*x + e)^2 - 24*a*c^2*cos(f*x + e) + 1
5*a*c^2)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))/(f*cos(f*x + e)^3*sin(f*x + e))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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]

Timed out

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Giac [A]  time = 1.83178, size = 113, normalized size = 0.93 \begin{align*} \frac{16 \, \sqrt{2}{\left (35 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{2} c^{3} + 42 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )} c^{4} + 15 \, c^{5}\right )} a c}{105 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{\frac{7}{2}} 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))^(5/2),x, algorithm="giac")

[Out]

16/105*sqrt(2)*(35*(c*tan(1/2*f*x + 1/2*e)^2 - c)^2*c^3 + 42*(c*tan(1/2*f*x + 1/2*e)^2 - c)*c^4 + 15*c^5)*a*c/
((c*tan(1/2*f*x + 1/2*e)^2 - c)^(7/2)*f)