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

Optimal. Leaf size=85 \[ -\frac{8 c^2 \tan (e+f x) (a \sec (e+f x)+a)^2}{35 f \sqrt{c-c \sec (e+f x)}}-\frac{2 c \tan (e+f x) (a \sec (e+f x)+a)^2 \sqrt{c-c \sec (e+f x)}}{7 f} \]

[Out]

(-8*c^2*(a + a*Sec[e + f*x])^2*Tan[e + f*x])/(35*f*Sqrt[c - c*Sec[e + f*x]]) - (2*c*(a + a*Sec[e + f*x])^2*Sqr
t[c - c*Sec[e + f*x]]*Tan[e + f*x])/(7*f)

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Rubi [A]  time = 0.207793, 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} \[ -\frac{8 c^2 \tan (e+f x) (a \sec (e+f x)+a)^2}{35 f \sqrt{c-c \sec (e+f x)}}-\frac{2 c \tan (e+f x) (a \sec (e+f x)+a)^2 \sqrt{c-c \sec (e+f x)}}{7 f} \]

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]

(-8*c^2*(a + a*Sec[e + f*x])^2*Tan[e + f*x])/(35*f*Sqrt[c - c*Sec[e + f*x]]) - (2*c*(a + a*Sec[e + f*x])^2*Sqr
t[c - c*Sec[e + f*x]]*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))^2 (c-c \sec (e+f x))^{3/2} \, dx &=-\frac{2 c (a+a \sec (e+f x))^2 \sqrt{c-c \sec (e+f x)} \tan (e+f x)}{7 f}+\frac{1}{7} (4 c) \int \sec (e+f x) (a+a \sec (e+f x))^2 \sqrt{c-c \sec (e+f x)} \, dx\\ &=-\frac{8 c^2 (a+a \sec (e+f x))^2 \tan (e+f x)}{35 f \sqrt{c-c \sec (e+f x)}}-\frac{2 c (a+a \sec (e+f x))^2 \sqrt{c-c \sec (e+f x)} \tan (e+f x)}{7 f}\\ \end{align*}

Mathematica [A]  time = 0.73391, size = 66, normalized size = 0.78 \[ \frac{8 a^2 c \cos ^4\left (\frac{1}{2} (e+f x)\right ) (9 \cos (e+f x)-5) \cot \left (\frac{1}{2} (e+f x)\right ) \sec ^3(e+f x) \sqrt{c-c \sec (e+f x)}}{35 f} \]

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]

(8*a^2*c*Cos[(e + f*x)/2]^4*(-5 + 9*Cos[e + f*x])*Cot[(e + f*x)/2]*Sec[e + f*x]^3*Sqrt[c - c*Sec[e + f*x]])/(3
5*f)

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Maple [A]  time = 0.187, size = 65, normalized size = 0.8 \begin{align*} -{\frac{2\,{a}^{2} \left ( 9\,\cos \left ( fx+e \right ) -5 \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{5}}{35\,f \left ( -1+\cos \left ( fx+e \right ) \right ) ^{4} \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \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))^2*(c-c*sec(f*x+e))^(3/2),x)

[Out]

-2/35*a^2/f*(9*cos(f*x+e)-5)*sin(f*x+e)^5*(c*(-1+cos(f*x+e))/cos(f*x+e))^(3/2)/(-1+cos(f*x+e))^4/cos(f*x+e)^2

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

[Out]

Timed out

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Fricas [A]  time = 0.471815, size = 254, normalized size = 2.99 \begin{align*} \frac{2 \,{\left (9 \, a^{2} c \cos \left (f x + e\right )^{4} + 22 \, a^{2} c \cos \left (f x + e\right )^{3} + 12 \, a^{2} c \cos \left (f x + e\right )^{2} - 6 \, a^{2} c \cos \left (f x + e\right ) - 5 \, a^{2} c\right )} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{35 \, 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))^2*(c-c*sec(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

2/35*(9*a^2*c*cos(f*x + e)^4 + 22*a^2*c*cos(f*x + e)^3 + 12*a^2*c*cos(f*x + e)^2 - 6*a^2*c*cos(f*x + e) - 5*a^
2*c)*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))**2*(c-c*sec(f*x+e))**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 1.94869, size = 85, normalized size = 1. \begin{align*} -\frac{16 \, \sqrt{2}{\left (7 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )} c^{5} + 5 \, c^{6}\right )} a^{2}}{35 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{\frac{7}{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))^2*(c-c*sec(f*x+e))^(3/2),x, algorithm="giac")

[Out]

-16/35*sqrt(2)*(7*(c*tan(1/2*f*x + 1/2*e)^2 - c)*c^5 + 5*c^6)*a^2/((c*tan(1/2*f*x + 1/2*e)^2 - c)^(7/2)*c*f)