3.124 \(\int \sec (e+f x) (a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^{5/2} \, dx\)
Optimal. Leaf size=134 \[ -\frac{2 c^3 \tan (e+f x) (a \sec (e+f x)+a)^{5/2}}{15 f \sqrt{c-c \sec (e+f x)}}-\frac{c^2 \tan (e+f x) (a \sec (e+f x)+a)^{5/2} \sqrt{c-c \sec (e+f x)}}{5 f}-\frac{c \tan (e+f x) (a \sec (e+f x)+a)^{5/2} (c-c \sec (e+f x))^{3/2}}{5 f} \]
[Out]
(-2*c^3*(a + a*Sec[e + f*x])^(5/2)*Tan[e + f*x])/(15*f*Sqrt[c - c*Sec[e + f*x]]) - (c^2*(a + a*Sec[e + f*x])^(
5/2)*Sqrt[c - c*Sec[e + f*x]]*Tan[e + f*x])/(5*f) - (c*(a + a*Sec[e + f*x])^(5/2)*(c - c*Sec[e + f*x])^(3/2)*T
an[e + f*x])/(5*f)
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Rubi [A] time = 0.424053, antiderivative size = 134, normalized size of antiderivative = 1.,
number of steps used = 3, number of rules used = 2, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.056, Rules used =
{3955, 3953} \[ -\frac{2 c^3 \tan (e+f x) (a \sec (e+f x)+a)^{5/2}}{15 f \sqrt{c-c \sec (e+f x)}}-\frac{c^2 \tan (e+f x) (a \sec (e+f x)+a)^{5/2} \sqrt{c-c \sec (e+f x)}}{5 f}-\frac{c \tan (e+f x) (a \sec (e+f x)+a)^{5/2} (c-c \sec (e+f x))^{3/2}}{5 f} \]
Antiderivative was successfully verified.
[In]
Int[Sec[e + f*x]*(a + a*Sec[e + f*x])^(5/2)*(c - c*Sec[e + f*x])^(5/2),x]
[Out]
(-2*c^3*(a + a*Sec[e + f*x])^(5/2)*Tan[e + f*x])/(15*f*Sqrt[c - c*Sec[e + f*x]]) - (c^2*(a + a*Sec[e + f*x])^(
5/2)*Sqrt[c - c*Sec[e + f*x]]*Tan[e + f*x])/(5*f) - (c*(a + a*Sec[e + f*x])^(5/2)*(c - c*Sec[e + f*x])^(3/2)*T
an[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))^{5/2} (c-c \sec (e+f x))^{5/2} \, dx &=-\frac{c (a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{5 f}+\frac{1}{5} (4 c) \int \sec (e+f x) (a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^{3/2} \, dx\\ &=-\frac{c^2 (a+a \sec (e+f x))^{5/2} \sqrt{c-c \sec (e+f x)} \tan (e+f x)}{5 f}-\frac{c (a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{5 f}+\frac{1}{5} \left (2 c^2\right ) \int \sec (e+f x) (a+a \sec (e+f x))^{5/2} \sqrt{c-c \sec (e+f x)} \, dx\\ &=-\frac{2 c^3 (a+a \sec (e+f x))^{5/2} \tan (e+f x)}{15 f \sqrt{c-c \sec (e+f x)}}-\frac{c^2 (a+a \sec (e+f x))^{5/2} \sqrt{c-c \sec (e+f x)} \tan (e+f x)}{5 f}-\frac{c (a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{5 f}\\ \end{align*}
Mathematica [A] time = 0.778556, size = 92, normalized size = 0.69 \[ \frac{a^2 c^2 (20 \cos (2 (e+f x))+15 \cos (4 (e+f x))+29) \csc \left (\frac{1}{2} (e+f x)\right ) \sec \left (\frac{1}{2} (e+f x)\right ) \sec ^4(e+f x) \sqrt{a (\sec (e+f x)+1)} \sqrt{c-c \sec (e+f x)}}{240 f} \]
Antiderivative was successfully verified.
[In]
Integrate[Sec[e + f*x]*(a + a*Sec[e + f*x])^(5/2)*(c - c*Sec[e + f*x])^(5/2),x]
[Out]
(a^2*c^2*(29 + 20*Cos[2*(e + f*x)] + 15*Cos[4*(e + f*x)])*Csc[(e + f*x)/2]*Sec[(e + f*x)/2]*Sec[e + f*x]^4*Sqr
t[a*(1 + Sec[e + f*x])]*Sqrt[c - c*Sec[e + f*x]])/(240*f)
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Maple [A] time = 0.26, size = 95, normalized size = 0.7 \begin{align*} -{\frac{{a}^{2} \left ( \sin \left ( fx+e \right ) \right ) ^{5} \left ( 8\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}-9\,\cos \left ( fx+e \right ) +3 \right ) }{15\,f \left ( -1+\cos \left ( fx+e \right ) \right ) ^{5} \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{5}{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))^(5/2),x)
[Out]
-1/15/f*a^2*sin(f*x+e)^5*(8*cos(f*x+e)^2-9*cos(f*x+e)+3)*(c*(-1+cos(f*x+e))/cos(f*x+e))^(5/2)*(1/cos(f*x+e)*a*
(1+cos(f*x+e)))^(1/2)/(-1+cos(f*x+e))^5/cos(f*x+e)^2
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Maxima [B] time = 1.90967, size = 2060, normalized size = 15.37 \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))^(5/2),x, algorithm="maxima")
[Out]
2/15*(100*a^2*c^2*cos(3*f*x + 3*e)*sin(2*f*x + 2*e) + 75*a^2*c^2*cos(f*x + e)*sin(2*f*x + 2*e) - 75*a^2*c^2*co
s(2*f*x + 2*e)*sin(f*x + e) - 15*a^2*c^2*sin(f*x + e) - (15*a^2*c^2*sin(9*f*x + 9*e) + 20*a^2*c^2*sin(7*f*x +
7*e) + 58*a^2*c^2*sin(5*f*x + 5*e) + 20*a^2*c^2*sin(3*f*x + 3*e) + 15*a^2*c^2*sin(f*x + e))*cos(10*f*x + 10*e)
+ 75*(a^2*c^2*sin(8*f*x + 8*e) + 2*a^2*c^2*sin(6*f*x + 6*e) + 2*a^2*c^2*sin(4*f*x + 4*e) + a^2*c^2*sin(2*f*x
+ 2*e))*cos(9*f*x + 9*e) - 5*(20*a^2*c^2*sin(7*f*x + 7*e) + 58*a^2*c^2*sin(5*f*x + 5*e) + 20*a^2*c^2*sin(3*f*x
+ 3*e) + 15*a^2*c^2*sin(f*x + e))*cos(8*f*x + 8*e) + 100*(2*a^2*c^2*sin(6*f*x + 6*e) + 2*a^2*c^2*sin(4*f*x +
4*e) + a^2*c^2*sin(2*f*x + 2*e))*cos(7*f*x + 7*e) - 10*(58*a^2*c^2*sin(5*f*x + 5*e) + 20*a^2*c^2*sin(3*f*x + 3
*e) + 15*a^2*c^2*sin(f*x + e))*cos(6*f*x + 6*e) + 290*(2*a^2*c^2*sin(4*f*x + 4*e) + a^2*c^2*sin(2*f*x + 2*e))*
cos(5*f*x + 5*e) - 50*(4*a^2*c^2*sin(3*f*x + 3*e) + 3*a^2*c^2*sin(f*x + e))*cos(4*f*x + 4*e) + (15*a^2*c^2*cos
(9*f*x + 9*e) + 20*a^2*c^2*cos(7*f*x + 7*e) + 58*a^2*c^2*cos(5*f*x + 5*e) + 20*a^2*c^2*cos(3*f*x + 3*e) + 15*a
^2*c^2*cos(f*x + e))*sin(10*f*x + 10*e) - 15*(5*a^2*c^2*cos(8*f*x + 8*e) + 10*a^2*c^2*cos(6*f*x + 6*e) + 10*a^
2*c^2*cos(4*f*x + 4*e) + 5*a^2*c^2*cos(2*f*x + 2*e) + a^2*c^2)*sin(9*f*x + 9*e) + 5*(20*a^2*c^2*cos(7*f*x + 7*
e) + 58*a^2*c^2*cos(5*f*x + 5*e) + 20*a^2*c^2*cos(3*f*x + 3*e) + 15*a^2*c^2*cos(f*x + e))*sin(8*f*x + 8*e) - 2
0*(10*a^2*c^2*cos(6*f*x + 6*e) + 10*a^2*c^2*cos(4*f*x + 4*e) + 5*a^2*c^2*cos(2*f*x + 2*e) + a^2*c^2)*sin(7*f*x
+ 7*e) + 10*(58*a^2*c^2*cos(5*f*x + 5*e) + 20*a^2*c^2*cos(3*f*x + 3*e) + 15*a^2*c^2*cos(f*x + e))*sin(6*f*x +
6*e) - 58*(10*a^2*c^2*cos(4*f*x + 4*e) + 5*a^2*c^2*cos(2*f*x + 2*e) + a^2*c^2)*sin(5*f*x + 5*e) + 50*(4*a^2*c
^2*cos(3*f*x + 3*e) + 3*a^2*c^2*cos(f*x + e))*sin(4*f*x + 4*e) - 20*(5*a^2*c^2*cos(2*f*x + 2*e) + a^2*c^2)*sin
(3*f*x + 3*e))*sqrt(a)*sqrt(c)/((2*(5*cos(8*f*x + 8*e) + 10*cos(6*f*x + 6*e) + 10*cos(4*f*x + 4*e) + 5*cos(2*f
*x + 2*e) + 1)*cos(10*f*x + 10*e) + cos(10*f*x + 10*e)^2 + 10*(10*cos(6*f*x + 6*e) + 10*cos(4*f*x + 4*e) + 5*c
os(2*f*x + 2*e) + 1)*cos(8*f*x + 8*e) + 25*cos(8*f*x + 8*e)^2 + 20*(10*cos(4*f*x + 4*e) + 5*cos(2*f*x + 2*e) +
1)*cos(6*f*x + 6*e) + 100*cos(6*f*x + 6*e)^2 + 20*(5*cos(2*f*x + 2*e) + 1)*cos(4*f*x + 4*e) + 100*cos(4*f*x +
4*e)^2 + 25*cos(2*f*x + 2*e)^2 + 10*(sin(8*f*x + 8*e) + 2*sin(6*f*x + 6*e) + 2*sin(4*f*x + 4*e) + sin(2*f*x +
2*e))*sin(10*f*x + 10*e) + sin(10*f*x + 10*e)^2 + 50*(2*sin(6*f*x + 6*e) + 2*sin(4*f*x + 4*e) + sin(2*f*x + 2
*e))*sin(8*f*x + 8*e) + 25*sin(8*f*x + 8*e)^2 + 100*(2*sin(4*f*x + 4*e) + sin(2*f*x + 2*e))*sin(6*f*x + 6*e) +
100*sin(6*f*x + 6*e)^2 + 100*sin(4*f*x + 4*e)^2 + 100*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + 25*sin(2*f*x + 2*e)
^2 + 10*cos(2*f*x + 2*e) + 1)*f)
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Fricas [A] time = 0.483366, size = 251, normalized size = 1.87 \begin{align*} \frac{{\left (15 \, a^{2} c^{2} \cos \left (f x + e\right )^{4} - 10 \, a^{2} c^{2} \cos \left (f x + e\right )^{2} + 3 \, a^{2} c^{2}\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 )}}}{15 \, 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))^(5/2)*(c-c*sec(f*x+e))^(5/2),x, algorithm="fricas")
[Out]
1/15*(15*a^2*c^2*cos(f*x + e)^4 - 10*a^2*c^2*cos(f*x + e)^2 + 3*a^2*c^2)*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)^4*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))**(5/2),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))^(5/2),x, algorithm="giac")
[Out]
Timed out