3.114 \(\int \sec (e+f x) (a+a \sec (e+f x))^{3/2} (c-c \sec (e+f x))^{5/2} \, dx\)
Optimal. Leaf size=89 \[ \frac{a^2 \tan (e+f x) (c-c \sec (e+f x))^{5/2}}{6 f \sqrt{a \sec (e+f x)+a}}+\frac{a \tan (e+f x) \sqrt{a \sec (e+f x)+a} (c-c \sec (e+f x))^{5/2}}{4 f} \]
[Out]
(a^2*(c - c*Sec[e + f*x])^(5/2)*Tan[e + f*x])/(6*f*Sqrt[a + a*Sec[e + f*x]]) + (a*Sqrt[a + a*Sec[e + f*x]]*(c
- c*Sec[e + f*x])^(5/2)*Tan[e + f*x])/(4*f)
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Rubi [A] time = 0.275789, 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} \[ \frac{a^2 \tan (e+f x) (c-c \sec (e+f x))^{5/2}}{6 f \sqrt{a \sec (e+f x)+a}}+\frac{a \tan (e+f x) \sqrt{a \sec (e+f x)+a} (c-c \sec (e+f x))^{5/2}}{4 f} \]
Antiderivative was successfully verified.
[In]
Int[Sec[e + f*x]*(a + a*Sec[e + f*x])^(3/2)*(c - c*Sec[e + f*x])^(5/2),x]
[Out]
(a^2*(c - c*Sec[e + f*x])^(5/2)*Tan[e + f*x])/(6*f*Sqrt[a + a*Sec[e + f*x]]) + (a*Sqrt[a + a*Sec[e + f*x]]*(c
- c*Sec[e + f*x])^(5/2)*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))^{3/2} (c-c \sec (e+f x))^{5/2} \, dx &=\frac{a \sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2} \tan (e+f x)}{4 f}+\frac{1}{2} a \int \sec (e+f x) \sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2} \, dx\\ &=\frac{a^2 (c-c \sec (e+f x))^{5/2} \tan (e+f x)}{6 f \sqrt{a+a \sec (e+f x)}}+\frac{a \sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2} \tan (e+f x)}{4 f}\\ \end{align*}
Mathematica [A] time = 0.591755, size = 97, normalized size = 1.09 \[ \frac{a c^2 (5 \cos (e+f x)-3 \cos (2 (e+f x))+3 \cos (3 (e+f x))) \csc \left (\frac{1}{2} (e+f x)\right ) \sec \left (\frac{1}{2} (e+f x)\right ) \sec ^3(e+f x) \sqrt{a (\sec (e+f x)+1)} \sqrt{c-c \sec (e+f x)}}{24 f} \]
Antiderivative was successfully verified.
[In]
Integrate[Sec[e + f*x]*(a + a*Sec[e + f*x])^(3/2)*(c - c*Sec[e + f*x])^(5/2),x]
[Out]
(a*c^2*(5*Cos[e + f*x] - 3*Cos[2*(e + f*x)] + 3*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 = 0.263, size = 93, normalized size = 1. \begin{align*}{\frac{a \left ( 11\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}-10\,\cos \left ( fx+e \right ) +3 \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{3}}{12\,f \left ( -1+\cos \left ( fx+e \right ) \right ) ^{4}\cos \left ( fx+e \right ) }\sqrt{{\frac{a \left ( 1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }}} \left ({\frac{c \left ( -1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{{\frac{5}{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))^(3/2)*(c-c*sec(f*x+e))^(5/2),x)
[Out]
1/12/f*a*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/2)*(c*(-1+cos(f*x+e))/cos(f*x+e))^(5/2)*(11*cos(f*x+e)^2-10*cos(f*
x+e)+3)*sin(f*x+e)^3/(-1+cos(f*x+e))^4/cos(f*x+e)
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Maxima [B] time = 1.85881, size = 1492, normalized size = 16.76 \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))^(3/2)*(c-c*sec(f*x+e))^(5/2),x, algorithm="maxima")
[Out]
2/3*(20*a*c^2*cos(3*f*x + 3*e)*sin(2*f*x + 2*e) - 12*a*c^2*cos(2*f*x + 2*e)*sin(f*x + e) - 3*a*c^2*sin(f*x + e
) - (3*a*c^2*sin(7*f*x + 7*e) - 3*a*c^2*sin(6*f*x + 6*e) + 5*a*c^2*sin(5*f*x + 5*e) + 5*a*c^2*sin(3*f*x + 3*e)
- 3*a*c^2*sin(2*f*x + 2*e) + 3*a*c^2*sin(f*x + e))*cos(8*f*x + 8*e) + 6*(2*a*c^2*sin(6*f*x + 6*e) + 3*a*c^2*s
in(4*f*x + 4*e) + 2*a*c^2*sin(2*f*x + 2*e))*cos(7*f*x + 7*e) - 2*(10*a*c^2*sin(5*f*x + 5*e) + 9*a*c^2*sin(4*f*
x + 4*e) + 10*a*c^2*sin(3*f*x + 3*e) + 6*a*c^2*sin(f*x + e))*cos(6*f*x + 6*e) + 10*(3*a*c^2*sin(4*f*x + 4*e) +
2*a*c^2*sin(2*f*x + 2*e))*cos(5*f*x + 5*e) - 6*(5*a*c^2*sin(3*f*x + 3*e) - 3*a*c^2*sin(2*f*x + 2*e) + 3*a*c^2
*sin(f*x + e))*cos(4*f*x + 4*e) + (3*a*c^2*cos(7*f*x + 7*e) - 3*a*c^2*cos(6*f*x + 6*e) + 5*a*c^2*cos(5*f*x + 5
*e) + 5*a*c^2*cos(3*f*x + 3*e) - 3*a*c^2*cos(2*f*x + 2*e) + 3*a*c^2*cos(f*x + e))*sin(8*f*x + 8*e) - 3*(4*a*c^
2*cos(6*f*x + 6*e) + 6*a*c^2*cos(4*f*x + 4*e) + 4*a*c^2*cos(2*f*x + 2*e) + a*c^2)*sin(7*f*x + 7*e) + (20*a*c^2
*cos(5*f*x + 5*e) + 18*a*c^2*cos(4*f*x + 4*e) + 20*a*c^2*cos(3*f*x + 3*e) + 12*a*c^2*cos(f*x + e) + 3*a*c^2)*s
in(6*f*x + 6*e) - 5*(6*a*c^2*cos(4*f*x + 4*e) + 4*a*c^2*cos(2*f*x + 2*e) + a*c^2)*sin(5*f*x + 5*e) + 6*(5*a*c^
2*cos(3*f*x + 3*e) - 3*a*c^2*cos(2*f*x + 2*e) + 3*a*c^2*cos(f*x + e))*sin(4*f*x + 4*e) - 5*(4*a*c^2*cos(2*f*x
+ 2*e) + a*c^2)*sin(3*f*x + 3*e) + 3*(4*a*c^2*cos(f*x + e) + a*c^2)*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 = 0.481349, size = 273, normalized size = 3.07 \begin{align*} \frac{{\left (12 \, a c^{2} \cos \left (f x + e\right )^{3} - 6 \, a c^{2} \cos \left (f x + e\right )^{2} - 4 \, a c^{2} \cos \left (f x + e\right ) + 3 \, a 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 )}}}{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))^(3/2)*(c-c*sec(f*x+e))^(5/2),x, algorithm="fricas")
[Out]
1/12*(12*a*c^2*cos(f*x + e)^3 - 6*a*c^2*cos(f*x + e)^2 - 4*a*c^2*cos(f*x + e) + 3*a*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)^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))**(3/2)*(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))^(3/2)*(c-c*sec(f*x+e))^(5/2),x, algorithm="giac")
[Out]
Timed out