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

Optimal. Leaf size=81 \[ -\frac{8 c^2 \tan (e+f x) (a \sec (e+f x)+a)}{15 f \sqrt{c-c \sec (e+f x)}}-\frac{2 c \tan (e+f x) (a \sec (e+f x)+a) \sqrt{c-c \sec (e+f x)}}{5 f} \]

[Out]

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

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Rubi [A]  time = 0.127802, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.062, Rules used = {3955, 3953} \[ -\frac{8 c^2 \tan (e+f x) (a \sec (e+f x)+a)}{15 f \sqrt{c-c \sec (e+f x)}}-\frac{2 c \tan (e+f x) (a \sec (e+f x)+a) \sqrt{c-c \sec (e+f x)}}{5 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*c^2*(a + a*Sec[e + f*x])*Tan[e + f*x])/(15*f*Sqrt[c - c*Sec[e + f*x]]) - (2*c*(a + a*Sec[e + f*x])*Sqrt[c
- c*Sec[e + f*x]]*Tan[e + f*x])/(5*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))^{3/2} \, dx &=-\frac{2 c (a+a \sec (e+f x)) \sqrt{c-c \sec (e+f x)} \tan (e+f x)}{5 f}+\frac{1}{5} (4 c) \int \sec (e+f x) (a+a \sec (e+f x)) \sqrt{c-c \sec (e+f x)} \, dx\\ &=-\frac{8 c^2 (a+a \sec (e+f x)) \tan (e+f x)}{15 f \sqrt{c-c \sec (e+f x)}}-\frac{2 c (a+a \sec (e+f x)) \sqrt{c-c \sec (e+f x)} \tan (e+f x)}{5 f}\\ \end{align*}

Mathematica [A]  time = 0.272285, size = 64, normalized size = 0.79 \[ \frac{4 a c \cos ^2\left (\frac{1}{2} (e+f x)\right ) (7 \cos (e+f x)-3) \cot \left (\frac{1}{2} (e+f x)\right ) \sec ^2(e+f x) \sqrt{c-c \sec (e+f x)}}{15 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a*c*Cos[(e + f*x)/2]^2*(-3 + 7*Cos[e + f*x])*Cot[(e + f*x)/2]*Sec[e + f*x]^2*Sqrt[c - c*Sec[e + f*x]])/(15*
f)

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Maple [A]  time = 0.171, size = 63, normalized size = 0.8 \begin{align*}{\frac{2\,a \left ( 7\,\cos \left ( fx+e \right ) -3 \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{3}}{15\,f \left ( -1+\cos \left ( fx+e \right ) \right ) ^{3}\cos \left ( fx+e \right ) } \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))*(c-c*sec(f*x+e))^(3/2),x)

[Out]

2/15*a/f*(7*cos(f*x+e)-3)*sin(f*x+e)^3*(c*(-1+cos(f*x+e))/cos(f*x+e))^(3/2)/(-1+cos(f*x+e))^3/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))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.466482, size = 205, normalized size = 2.53 \begin{align*} \frac{2 \,{\left (7 \, a c \cos \left (f x + e\right )^{3} + 11 \, a c \cos \left (f x + e\right )^{2} + a c \cos \left (f x + e\right ) - 3 \, a c\right )} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{15 \, f \cos \left (f x + e\right )^{2} \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))^(3/2),x, algorithm="fricas")

[Out]

2/15*(7*a*c*cos(f*x + e)^3 + 11*a*c*cos(f*x + e)^2 + a*c*cos(f*x + e) - 3*a*c)*sqrt((c*cos(f*x + e) - c)/cos(f
*x + e))/(f*cos(f*x + e)^2*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))**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 1.49286, size = 82, normalized size = 1.01 \begin{align*} \frac{8 \, \sqrt{2}{\left (5 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )} c^{4} + 3 \, c^{5}\right )} a}{15 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{\frac{5}{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))*(c-c*sec(f*x+e))^(3/2),x, algorithm="giac")

[Out]

8/15*sqrt(2)*(5*(c*tan(1/2*f*x + 1/2*e)^2 - c)*c^4 + 3*c^5)*a/((c*tan(1/2*f*x + 1/2*e)^2 - c)^(5/2)*c*f)