3.131 \(\int \frac{\sec (e+f x) (a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^{9/2}} \, dx\)
Optimal. Leaf size=88 \[ -\frac{\tan (e+f x) (a \sec (e+f x)+a)^{5/2}}{48 c f (c-c \sec (e+f x))^{7/2}}-\frac{\tan (e+f x) (a \sec (e+f x)+a)^{5/2}}{8 f (c-c \sec (e+f x))^{9/2}} \]
[Out]
-((a + a*Sec[e + f*x])^(5/2)*Tan[e + f*x])/(8*f*(c - c*Sec[e + f*x])^(9/2)) - ((a + a*Sec[e + f*x])^(5/2)*Tan[
e + f*x])/(48*c*f*(c - c*Sec[e + f*x])^(7/2))
________________________________________________________________________________________
Rubi [A] time = 0.297633, antiderivative size = 88, 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 =
{3951, 3950} \[ -\frac{\tan (e+f x) (a \sec (e+f x)+a)^{5/2}}{48 c f (c-c \sec (e+f x))^{7/2}}-\frac{\tan (e+f x) (a \sec (e+f x)+a)^{5/2}}{8 f (c-c \sec (e+f x))^{9/2}} \]
Antiderivative was successfully verified.
[In]
Int[(Sec[e + f*x]*(a + a*Sec[e + f*x])^(5/2))/(c - c*Sec[e + f*x])^(9/2),x]
[Out]
-((a + a*Sec[e + f*x])^(5/2)*Tan[e + f*x])/(8*f*(c - c*Sec[e + f*x])^(9/2)) - ((a + a*Sec[e + f*x])^(5/2)*Tan[
e + f*x])/(48*c*f*(c - c*Sec[e + f*x])^(7/2))
Rule 3951
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))
^(n_.), x_Symbol] :> Simp[(b*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(c + d*Csc[e + f*x])^n)/(a*f*(2*m + 1)), x] +
Dist[(m + n + 1)/(a*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(c + d*Csc[e + f*x])^n, x], x]
/; FreeQ[{a, b, c, d, e, f, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && ILtQ[m + n + 1, 0] && NeQ[2
*m + 1, 0] && !LtQ[n, 0] && !(IGtQ[n + 1/2, 0] && LtQ[n + 1/2, -(m + n)])
Rule 3950
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))
^(n_.), x_Symbol] :> Simp[(b*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(c + d*Csc[e + f*x])^n)/(a*f*(2*m + 1)), x] /
; FreeQ[{a, b, c, d, e, f, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] && NeQ[2*m
+ 1, 0]
Rubi steps
\begin{align*} \int \frac{\sec (e+f x) (a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^{9/2}} \, dx &=-\frac{(a+a \sec (e+f x))^{5/2} \tan (e+f x)}{8 f (c-c \sec (e+f x))^{9/2}}+\frac{\int \frac{\sec (e+f x) (a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^{7/2}} \, dx}{8 c}\\ &=-\frac{(a+a \sec (e+f x))^{5/2} \tan (e+f x)}{8 f (c-c \sec (e+f x))^{9/2}}-\frac{(a+a \sec (e+f x))^{5/2} \tan (e+f x)}{48 c f (c-c \sec (e+f x))^{7/2}}\\ \end{align*}
Mathematica [A] time = 0.827386, size = 92, normalized size = 1.05 \[ -\frac{a^2 (17 \cos (e+f x)-3 \cos (2 (e+f x))+3 \cos (3 (e+f x))-5) \tan \left (\frac{1}{2} (e+f x)\right ) \sqrt{a (\sec (e+f x)+1)}}{12 c^4 f (\cos (e+f x)-1)^4 \sqrt{c-c \sec (e+f x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[(Sec[e + f*x]*(a + a*Sec[e + f*x])^(5/2))/(c - c*Sec[e + f*x])^(9/2),x]
[Out]
-(a^2*(-5 + 17*Cos[e + f*x] - 3*Cos[2*(e + f*x)] + 3*Cos[3*(e + f*x)])*Sqrt[a*(1 + Sec[e + f*x])]*Tan[(e + f*x
)/2])/(12*c^4*f*(-1 + Cos[e + f*x])^4*Sqrt[c - c*Sec[e + f*x]])
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Maple [A] time = 0.257, size = 85, normalized size = 1. \begin{align*} -{\frac{{a}^{2} \left ( 7\,\cos \left ( fx+e \right ) -1 \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{5}}{48\,f \left ( -1+\cos \left ( fx+e \right ) \right ) ^{2} \left ( \cos \left ( fx+e \right ) \right ) ^{4}}\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{9}{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))^(5/2)/(c-c*sec(f*x+e))^(9/2),x)
[Out]
-1/48/f*a^2*(7*cos(f*x+e)-1)*sin(f*x+e)^5*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/2)/(-1+cos(f*x+e))^2/cos(f*x+e)^4
/(c*(-1+cos(f*x+e))/cos(f*x+e))^(9/2)
________________________________________________________________________________________
Maxima [B] time = 19.6824, size = 3671, normalized size = 41.72 \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))^(9/2),x, algorithm="maxima")
[Out]
2/3*(70*a^2*cos(6*f*x + 6*e)*sin(4*f*x + 4*e) - 70*a^2*cos(4*f*x + 4*e)*sin(2*f*x + 2*e) + 3*a^2*sin(2*f*x + 2
*e) + (3*a^2*sin(6*f*x + 6*e) + 10*a^2*sin(4*f*x + 4*e) + 3*a^2*sin(2*f*x + 2*e))*cos(8*f*x + 8*e) + (3*a^2*si
n(8*f*x + 8*e) + 60*a^2*sin(6*f*x + 6*e) + 130*a^2*sin(4*f*x + 4*e) + 60*a^2*sin(2*f*x + 2*e) - 32*a^2*sin(5/2
*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 32*a^2*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*c
os(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (17*a^2*sin(8*f*x + 8*e) + 308*a^2*sin(6*f*x + 6*e) + 63
0*a^2*sin(4*f*x + 4*e) + 308*a^2*sin(2*f*x + 2*e) + 32*a^2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))
))*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (17*a^2*sin(8*f*x + 8*e) + 308*a^2*sin(6*f*x + 6*e)
+ 630*a^2*sin(4*f*x + 4*e) + 308*a^2*sin(2*f*x + 2*e) + 32*a^2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2
*e))))*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (3*a^2*sin(8*f*x + 8*e) + 60*a^2*sin(6*f*x + 6*e
) + 130*a^2*sin(4*f*x + 4*e) + 60*a^2*sin(2*f*x + 2*e))*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) -
(3*a^2*cos(6*f*x + 6*e) + 10*a^2*cos(4*f*x + 4*e) + 3*a^2*cos(2*f*x + 2*e))*sin(8*f*x + 8*e) - (70*a^2*cos(4*
f*x + 4*e) - 3*a^2)*sin(6*f*x + 6*e) + 10*(7*a^2*cos(2*f*x + 2*e) + a^2)*sin(4*f*x + 4*e) - (3*a^2*cos(8*f*x +
8*e) + 60*a^2*cos(6*f*x + 6*e) + 130*a^2*cos(4*f*x + 4*e) + 60*a^2*cos(2*f*x + 2*e) - 32*a^2*cos(5/2*arctan2(
sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 32*a^2*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 3*a^2)*si
n(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (17*a^2*cos(8*f*x + 8*e) + 308*a^2*cos(6*f*x + 6*e) + 630
*a^2*cos(4*f*x + 4*e) + 308*a^2*cos(2*f*x + 2*e) + 32*a^2*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))
+ 17*a^2)*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (17*a^2*cos(8*f*x + 8*e) + 308*a^2*cos(6*f*x
+ 6*e) + 630*a^2*cos(4*f*x + 4*e) + 308*a^2*cos(2*f*x + 2*e) + 32*a^2*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2
*f*x + 2*e))) + 17*a^2)*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (3*a^2*cos(8*f*x + 8*e) + 60*a^
2*cos(6*f*x + 6*e) + 130*a^2*cos(4*f*x + 4*e) + 60*a^2*cos(2*f*x + 2*e) + 3*a^2)*sin(1/2*arctan2(sin(2*f*x + 2
*e), cos(2*f*x + 2*e))))*sqrt(a)*sqrt(c)/((c^5*cos(8*f*x + 8*e)^2 + 784*c^5*cos(6*f*x + 6*e)^2 + 4900*c^5*cos(
4*f*x + 4*e)^2 + 784*c^5*cos(2*f*x + 2*e)^2 + 64*c^5*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 +
3136*c^5*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 3136*c^5*cos(3/2*arctan2(sin(2*f*x + 2*e), c
os(2*f*x + 2*e)))^2 + 64*c^5*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + c^5*sin(8*f*x + 8*e)^2 +
784*c^5*sin(6*f*x + 6*e)^2 + 4900*c^5*sin(4*f*x + 4*e)^2 + 3920*c^5*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + 784*c
^5*sin(2*f*x + 2*e)^2 + 64*c^5*sin(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 3136*c^5*sin(5/2*arcta
n2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 3136*c^5*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 +
64*c^5*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 56*c^5*cos(2*f*x + 2*e) + c^5 + 2*(28*c^5*cos(
6*f*x + 6*e) + 70*c^5*cos(4*f*x + 4*e) + 28*c^5*cos(2*f*x + 2*e) + c^5)*cos(8*f*x + 8*e) + 56*(70*c^5*cos(4*f*
x + 4*e) + 28*c^5*cos(2*f*x + 2*e) + c^5)*cos(6*f*x + 6*e) + 140*(28*c^5*cos(2*f*x + 2*e) + c^5)*cos(4*f*x + 4
*e) - 16*(c^5*cos(8*f*x + 8*e) + 28*c^5*cos(6*f*x + 6*e) + 70*c^5*cos(4*f*x + 4*e) + 28*c^5*cos(2*f*x + 2*e) -
56*c^5*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 56*c^5*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*
f*x + 2*e))) - 8*c^5*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + c^5)*cos(7/2*arctan2(sin(2*f*x + 2
*e), cos(2*f*x + 2*e))) - 112*(c^5*cos(8*f*x + 8*e) + 28*c^5*cos(6*f*x + 6*e) + 70*c^5*cos(4*f*x + 4*e) + 28*c
^5*cos(2*f*x + 2*e) - 56*c^5*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 8*c^5*cos(1/2*arctan2(sin(
2*f*x + 2*e), cos(2*f*x + 2*e))) + c^5)*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 112*(c^5*cos(8*
f*x + 8*e) + 28*c^5*cos(6*f*x + 6*e) + 70*c^5*cos(4*f*x + 4*e) + 28*c^5*cos(2*f*x + 2*e) - 8*c^5*cos(1/2*arcta
n2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + c^5)*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 16*(c^5*
cos(8*f*x + 8*e) + 28*c^5*cos(6*f*x + 6*e) + 70*c^5*cos(4*f*x + 4*e) + 28*c^5*cos(2*f*x + 2*e) + c^5)*cos(1/2*
arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 28*(2*c^5*sin(6*f*x + 6*e) + 5*c^5*sin(4*f*x + 4*e) + 2*c^5*sin
(2*f*x + 2*e))*sin(8*f*x + 8*e) + 784*(5*c^5*sin(4*f*x + 4*e) + 2*c^5*sin(2*f*x + 2*e))*sin(6*f*x + 6*e) - 16*
(c^5*sin(8*f*x + 8*e) + 28*c^5*sin(6*f*x + 6*e) + 70*c^5*sin(4*f*x + 4*e) + 28*c^5*sin(2*f*x + 2*e) - 56*c^5*s
in(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 56*c^5*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e
))) - 8*c^5*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sin(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x
+ 2*e))) - 112*(c^5*sin(8*f*x + 8*e) + 28*c^5*sin(6*f*x + 6*e) + 70*c^5*sin(4*f*x + 4*e) + 28*c^5*sin(2*f*x +
2*e) - 56*c^5*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 8*c^5*sin(1/2*arctan2(sin(2*f*x + 2*e), c
os(2*f*x + 2*e))))*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 112*(c^5*sin(8*f*x + 8*e) + 28*c^5*s
in(6*f*x + 6*e) + 70*c^5*sin(4*f*x + 4*e) + 28*c^5*sin(2*f*x + 2*e) - 8*c^5*sin(1/2*arctan2(sin(2*f*x + 2*e),
cos(2*f*x + 2*e))))*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 16*(c^5*sin(8*f*x + 8*e) + 28*c^5*s
in(6*f*x + 6*e) + 70*c^5*sin(4*f*x + 4*e) + 28*c^5*sin(2*f*x + 2*e))*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f
*x + 2*e))))*f)
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Fricas [B] time = 0.492618, size = 394, normalized size = 4.48 \begin{align*} \frac{{\left (6 \, a^{2} \cos \left (f x + e\right )^{4} - 3 \, a^{2} \cos \left (f x + e\right )^{3} + 4 \, a^{2} \cos \left (f x + e\right )^{2} - a^{2} \cos \left (f x + e\right )\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 )}}}{6 \,{\left (c^{5} f \cos \left (f x + e\right )^{4} - 4 \, c^{5} f \cos \left (f x + e\right )^{3} + 6 \, c^{5} f \cos \left (f x + e\right )^{2} - 4 \, c^{5} f \cos \left (f x + e\right ) + c^{5} f\right )} \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))^(9/2),x, algorithm="fricas")
[Out]
1/6*(6*a^2*cos(f*x + e)^4 - 3*a^2*cos(f*x + e)^3 + 4*a^2*cos(f*x + e)^2 - a^2*cos(f*x + e))*sqrt((a*cos(f*x +
e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))/((c^5*f*cos(f*x + e)^4 - 4*c^5*f*cos(f*x + e)^3
+ 6*c^5*f*cos(f*x + e)^2 - 4*c^5*f*cos(f*x + e) + c^5*f)*sin(f*x + e))
________________________________________________________________________________________
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))**(9/2),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))^(9/2),x, algorithm="giac")
[Out]
Timed out