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

Optimal. Leaf size=85 \[ -\frac{8 c^2 \tan (e+f x) (a \sec (e+f x)+a)^3}{63 f \sqrt{c-c \sec (e+f x)}}-\frac{2 c \tan (e+f x) (a \sec (e+f x)+a)^3 \sqrt{c-c \sec (e+f x)}}{9 f} \]

[Out]

(-8*c^2*(a + a*Sec[e + f*x])^3*Tan[e + f*x])/(63*f*Sqrt[c - c*Sec[e + f*x]]) - (2*c*(a + a*Sec[e + f*x])^3*Sqr
t[c - c*Sec[e + f*x]]*Tan[e + f*x])/(9*f)

________________________________________________________________________________________

Rubi [A]  time = 0.208853, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.059, Rules used = {3955, 3953} \[ -\frac{8 c^2 \tan (e+f x) (a \sec (e+f x)+a)^3}{63 f \sqrt{c-c \sec (e+f x)}}-\frac{2 c \tan (e+f x) (a \sec (e+f x)+a)^3 \sqrt{c-c \sec (e+f x)}}{9 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*c^2*(a + a*Sec[e + f*x])^3*Tan[e + f*x])/(63*f*Sqrt[c - c*Sec[e + f*x]]) - (2*c*(a + a*Sec[e + f*x])^3*Sqr
t[c - c*Sec[e + f*x]]*Tan[e + f*x])/(9*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))^3 (c-c \sec (e+f x))^{3/2} \, dx &=-\frac{2 c (a+a \sec (e+f x))^3 \sqrt{c-c \sec (e+f x)} \tan (e+f x)}{9 f}+\frac{1}{9} (4 c) \int \sec (e+f x) (a+a \sec (e+f x))^3 \sqrt{c-c \sec (e+f x)} \, dx\\ &=-\frac{8 c^2 (a+a \sec (e+f x))^3 \tan (e+f x)}{63 f \sqrt{c-c \sec (e+f x)}}-\frac{2 c (a+a \sec (e+f x))^3 \sqrt{c-c \sec (e+f x)} \tan (e+f x)}{9 f}\\ \end{align*}

Mathematica [A]  time = 1.02977, size = 66, normalized size = 0.78 \[ \frac{16 a^3 c \cos ^6\left (\frac{1}{2} (e+f x)\right ) (11 \cos (e+f x)-7) \cot \left (\frac{1}{2} (e+f x)\right ) \sec ^4(e+f x) \sqrt{c-c \sec (e+f x)}}{63 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(16*a^3*c*Cos[(e + f*x)/2]^6*(-7 + 11*Cos[e + f*x])*Cot[(e + f*x)/2]*Sec[e + f*x]^4*Sqrt[c - c*Sec[e + f*x]])/
(63*f)

________________________________________________________________________________________

Maple [A]  time = 0.199, size = 65, normalized size = 0.8 \begin{align*}{\frac{2\,{a}^{3} \left ( 11\,\cos \left ( fx+e \right ) -7 \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{7}}{63\,f \left ( -1+\cos \left ( fx+e \right ) \right ) ^{5} \left ( \cos \left ( fx+e \right ) \right ) ^{3}} \left ({\frac{c \left ( -1+\cos \left ( fx+e \right ) \right ) }{\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))^3*(c-c*sec(f*x+e))^(3/2),x)

[Out]

2/63*a^3/f*(11*cos(f*x+e)-7)*sin(f*x+e)^7*(c*(-1+cos(f*x+e))/cos(f*x+e))^(3/2)/(-1+cos(f*x+e))^5/cos(f*x+e)^3

________________________________________________________________________________________

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))^3*(c-c*sec(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

Fricas [A]  time = 0.479186, size = 290, normalized size = 3.41 \begin{align*} \frac{2 \,{\left (11 \, a^{3} c \cos \left (f x + e\right )^{5} + 37 \, a^{3} c \cos \left (f x + e\right )^{4} + 38 \, a^{3} c \cos \left (f x + e\right )^{3} + 2 \, a^{3} c \cos \left (f x + e\right )^{2} - 17 \, a^{3} c \cos \left (f x + e\right ) - 7 \, a^{3} c\right )} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{63 \, f \cos \left (f x + e\right )^{4} \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))^3*(c-c*sec(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

2/63*(11*a^3*c*cos(f*x + e)^5 + 37*a^3*c*cos(f*x + e)^4 + 38*a^3*c*cos(f*x + e)^3 + 2*a^3*c*cos(f*x + e)^2 - 1
7*a^3*c*cos(f*x + e) - 7*a^3*c)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))/(f*cos(f*x + e)^4*sin(f*x + e))

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 2.44048, size = 85, normalized size = 1. \begin{align*} \frac{32 \, \sqrt{2}{\left (9 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )} c^{6} + 7 \, c^{7}\right )} a^{3}}{63 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{\frac{9}{2}} c 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))^3*(c-c*sec(f*x+e))^(3/2),x, algorithm="giac")

[Out]

32/63*sqrt(2)*(9*(c*tan(1/2*f*x + 1/2*e)^2 - c)*c^6 + 7*c^7)*a^3/((c*tan(1/2*f*x + 1/2*e)^2 - c)^(9/2)*c*f)