3.125 \(\int \sec (e+f x) (a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=89 \[ -\frac{c^2 \tan (e+f x) (a \sec (e+f x)+a)^{5/2}}{6 f \sqrt{c-c \sec (e+f x)}}-\frac{c \tan (e+f x) (a \sec (e+f x)+a)^{5/2} \sqrt{c-c \sec (e+f x)}}{4 f} \]

[Out]

-(c^2*(a + a*Sec[e + f*x])^(5/2)*Tan[e + f*x])/(6*f*Sqrt[c - c*Sec[e + f*x]]) - (c*(a + a*Sec[e + f*x])^(5/2)*
Sqrt[c - c*Sec[e + f*x]]*Tan[e + f*x])/(4*f)

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Rubi [A]  time = 0.27494, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.056, Rules used = {3955, 3953} \[ -\frac{c^2 \tan (e+f x) (a \sec (e+f x)+a)^{5/2}}{6 f \sqrt{c-c \sec (e+f x)}}-\frac{c \tan (e+f x) (a \sec (e+f x)+a)^{5/2} \sqrt{c-c \sec (e+f x)}}{4 f} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]*(a + a*Sec[e + f*x])^(5/2)*(c - c*Sec[e + f*x])^(3/2),x]

[Out]

-(c^2*(a + a*Sec[e + f*x])^(5/2)*Tan[e + f*x])/(6*f*Sqrt[c - c*Sec[e + f*x]]) - (c*(a + a*Sec[e + f*x])^(5/2)*
Sqrt[c - c*Sec[e + f*x]]*Tan[e + f*x])/(4*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))^{5/2} (c-c \sec (e+f x))^{3/2} \, dx &=-\frac{c (a+a \sec (e+f x))^{5/2} \sqrt{c-c \sec (e+f x)} \tan (e+f x)}{4 f}+\frac{1}{2} c \int \sec (e+f x) (a+a \sec (e+f x))^{5/2} \sqrt{c-c \sec (e+f x)} \, dx\\ &=-\frac{c^2 (a+a \sec (e+f x))^{5/2} \tan (e+f x)}{6 f \sqrt{c-c \sec (e+f x)}}-\frac{c (a+a \sec (e+f x))^{5/2} \sqrt{c-c \sec (e+f x)} \tan (e+f x)}{4 f}\\ \end{align*}

Mathematica [A]  time = 0.599256, size = 96, normalized size = 1.08 \[ \frac{a^2 c (5 \cos (e+f x)+3 (\cos (2 (e+f x))+\cos (3 (e+f x)))) \csc \left (\frac{1}{2} (e+f x)\right ) \sec \left (\frac{1}{2} (e+f x)\right ) \sec ^3(e+f x) \sqrt{a (\sec (e+f x)+1)} \sqrt{c-c \sec (e+f x)}}{24 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]*(a + a*Sec[e + f*x])^(5/2)*(c - c*Sec[e + f*x])^(3/2),x]

[Out]

(a^2*c*(5*Cos[e + f*x] + 3*(Cos[2*(e + f*x)] + Cos[3*(e + f*x)]))*Csc[(e + f*x)/2]*Sec[(e + f*x)/2]*Sec[e + f*
x]^3*Sqrt[a*(1 + Sec[e + f*x])]*Sqrt[c - c*Sec[e + f*x]])/(24*f)

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Maple [A]  time = 0.261, size = 85, normalized size = 1. \begin{align*} -{\frac{{a}^{2} \left ( \sin \left ( fx+e \right ) \right ) ^{5} \left ( 5\,\cos \left ( fx+e \right ) -3 \right ) }{12\,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}}}\sqrt{{\frac{a \left ( 1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(a+a*sec(f*x+e))^(5/2)*(c-c*sec(f*x+e))^(3/2),x)

[Out]

-1/12/f*a^2*sin(f*x+e)^5*(5*cos(f*x+e)-3)*(c*(-1+cos(f*x+e))/cos(f*x+e))^(3/2)*(1/cos(f*x+e)*a*(1+cos(f*x+e)))
^(1/2)/(-1+cos(f*x+e))^4/cos(f*x+e)^2

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Maxima [B]  time = 1.885, size = 1493, normalized size = 16.78 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^(5/2)*(c-c*sec(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

2/3*(20*a^2*c*cos(3*f*x + 3*e)*sin(2*f*x + 2*e) - 12*a^2*c*cos(2*f*x + 2*e)*sin(f*x + e) - 3*a^2*c*sin(f*x + e
) - (3*a^2*c*sin(7*f*x + 7*e) + 3*a^2*c*sin(6*f*x + 6*e) + 5*a^2*c*sin(5*f*x + 5*e) + 5*a^2*c*sin(3*f*x + 3*e)
 + 3*a^2*c*sin(2*f*x + 2*e) + 3*a^2*c*sin(f*x + e))*cos(8*f*x + 8*e) + 6*(2*a^2*c*sin(6*f*x + 6*e) + 3*a^2*c*s
in(4*f*x + 4*e) + 2*a^2*c*sin(2*f*x + 2*e))*cos(7*f*x + 7*e) - 2*(10*a^2*c*sin(5*f*x + 5*e) - 9*a^2*c*sin(4*f*
x + 4*e) + 10*a^2*c*sin(3*f*x + 3*e) + 6*a^2*c*sin(f*x + e))*cos(6*f*x + 6*e) + 10*(3*a^2*c*sin(4*f*x + 4*e) +
 2*a^2*c*sin(2*f*x + 2*e))*cos(5*f*x + 5*e) - 6*(5*a^2*c*sin(3*f*x + 3*e) + 3*a^2*c*sin(2*f*x + 2*e) + 3*a^2*c
*sin(f*x + e))*cos(4*f*x + 4*e) + (3*a^2*c*cos(7*f*x + 7*e) + 3*a^2*c*cos(6*f*x + 6*e) + 5*a^2*c*cos(5*f*x + 5
*e) + 5*a^2*c*cos(3*f*x + 3*e) + 3*a^2*c*cos(2*f*x + 2*e) + 3*a^2*c*cos(f*x + e))*sin(8*f*x + 8*e) - 3*(4*a^2*
c*cos(6*f*x + 6*e) + 6*a^2*c*cos(4*f*x + 4*e) + 4*a^2*c*cos(2*f*x + 2*e) + a^2*c)*sin(7*f*x + 7*e) + (20*a^2*c
*cos(5*f*x + 5*e) - 18*a^2*c*cos(4*f*x + 4*e) + 20*a^2*c*cos(3*f*x + 3*e) + 12*a^2*c*cos(f*x + e) - 3*a^2*c)*s
in(6*f*x + 6*e) - 5*(6*a^2*c*cos(4*f*x + 4*e) + 4*a^2*c*cos(2*f*x + 2*e) + a^2*c)*sin(5*f*x + 5*e) + 6*(5*a^2*
c*cos(3*f*x + 3*e) + 3*a^2*c*cos(2*f*x + 2*e) + 3*a^2*c*cos(f*x + e))*sin(4*f*x + 4*e) - 5*(4*a^2*c*cos(2*f*x
+ 2*e) + a^2*c)*sin(3*f*x + 3*e) + 3*(4*a^2*c*cos(f*x + e) - a^2*c)*sin(2*f*x + 2*e))*sqrt(a)*sqrt(c)/((2*(4*c
os(6*f*x + 6*e) + 6*cos(4*f*x + 4*e) + 4*cos(2*f*x + 2*e) + 1)*cos(8*f*x + 8*e) + cos(8*f*x + 8*e)^2 + 8*(6*co
s(4*f*x + 4*e) + 4*cos(2*f*x + 2*e) + 1)*cos(6*f*x + 6*e) + 16*cos(6*f*x + 6*e)^2 + 12*(4*cos(2*f*x + 2*e) + 1
)*cos(4*f*x + 4*e) + 36*cos(4*f*x + 4*e)^2 + 16*cos(2*f*x + 2*e)^2 + 4*(2*sin(6*f*x + 6*e) + 3*sin(4*f*x + 4*e
) + 2*sin(2*f*x + 2*e))*sin(8*f*x + 8*e) + sin(8*f*x + 8*e)^2 + 16*(3*sin(4*f*x + 4*e) + 2*sin(2*f*x + 2*e))*s
in(6*f*x + 6*e) + 16*sin(6*f*x + 6*e)^2 + 36*sin(4*f*x + 4*e)^2 + 48*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + 16*si
n(2*f*x + 2*e)^2 + 8*cos(2*f*x + 2*e) + 1)*f)

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Fricas [A]  time = 0.479723, size = 273, normalized size = 3.07 \begin{align*} \frac{{\left (12 \, a^{2} c \cos \left (f x + e\right )^{3} + 6 \, a^{2} c \cos \left (f x + e\right )^{2} - 4 \, a^{2} c \cos \left (f x + e\right ) - 3 \, a^{2} c\right )} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{12 \, 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))^(5/2)*(c-c*sec(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

1/12*(12*a^2*c*cos(f*x + e)^3 + 6*a^2*c*cos(f*x + e)^2 - 4*a^2*c*cos(f*x + e) - 3*a^2*c)*sqrt((a*cos(f*x + e)
+ a)/cos(f*x + e))*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))**(5/2)*(c-c*sec(f*x+e))**(3/2),x)

[Out]

Timed out

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Giac [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))^(5/2)*(c-c*sec(f*x+e))^(3/2),x, algorithm="giac")

[Out]

Timed out