\(\int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+C \sec ^2(c+d x)) \, dx\) [1146]

Optimal result

Integrand size = 37, antiderivative size = 216 \[ \int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {64 a^3 (13 A+21 C) \sin (c+d x)}{315 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {16 a^2 (13 A+21 C) \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{315 d}+\frac {2 a (13 A+21 C) \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{105 d}+\frac {10 A \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{63 d}+\frac {2 A \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{9 d} \]

[Out]

2/105*a*(13*A+21*C)*cos(d*x+c)^(3/2)*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d+10/63*A*cos(d*x+c)^(5/2)*(a+a*sec(d*x
+c))^(5/2)*sin(d*x+c)/d+2/9*A*cos(d*x+c)^(7/2)*(a+a*sec(d*x+c))^(5/2)*sin(d*x+c)/d+64/315*a^3*(13*A+21*C)*sin(
d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+16/315*a^2*(13*A+21*C)*sin(d*x+c)*cos(d*x+c)^(1/2)*(a+a*sec(d
*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {4350, 4172, 4098, 3894, 3889} \[ \int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {64 a^3 (13 A+21 C) \sin (c+d x)}{315 d \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {16 a^2 (13 A+21 C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}{315 d}+\frac {2 a (13 A+21 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{105 d}+\frac {2 A \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{9 d}+\frac {10 A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{63 d} \]

[In]

Int[Cos[c + d*x]^(9/2)*(a + a*Sec[c + d*x])^(5/2)*(A + C*Sec[c + d*x]^2),x]

[Out]

(64*a^3*(13*A + 21*C)*Sin[c + d*x])/(315*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*(13*A + 21*C
)*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(315*d) + (2*a*(13*A + 21*C)*Cos[c + d*x]^(3/2)*(a
+ a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(105*d) + (10*A*Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d
*x])/(63*d) + (2*A*Cos[c + d*x]^(7/2)*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(9*d)

Rule 3889

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)], x_Symbol] :> Simp[-2*a*(Co
t[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[d*Csc[e + f*x]])), x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^
2, 0]

Rule 3894

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-a)*C
ot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*m)), x] + Dist[b*((2*m - 1)/(d*m)), Int[(a + b
*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0
] && EqQ[m + n, 0] && GtQ[m, 1/2] && IntegerQ[2*m]

Rule 4098

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Dist[(
a*A*m - b*B*n)/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, A
, B, m, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] &&  !LeQ[m, -1]

Rule 4172

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Dis
t[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*(A*(m + n + 1) + C*n)*Csc[e +
f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2
^(-1)] || EqQ[m + n + 1, 0])

Rule 4350

Int[(cos[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cos[a + b*x])^m*(c*Sec[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Sec[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {9}{2}}(c+d x)} \, dx \\ & = \frac {2 A \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{9 d}+\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \sec (c+d x))^{5/2} \left (\frac {5 a A}{2}+\frac {1}{2} a (2 A+9 C) \sec (c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx}{9 a} \\ & = \frac {10 A \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{63 d}+\frac {2 A \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{9 d}+\frac {1}{21} \left ((13 A+21 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \sec (c+d x))^{5/2}}{\sec ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {2 a (13 A+21 C) \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{105 d}+\frac {10 A \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{63 d}+\frac {2 A \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{9 d}+\frac {1}{105} \left (8 a (13 A+21 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \sec (c+d x))^{3/2}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {16 a^2 (13 A+21 C) \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{315 d}+\frac {2 a (13 A+21 C) \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{105 d}+\frac {10 A \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{63 d}+\frac {2 A \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{9 d}+\frac {1}{315} \left (32 a^2 (13 A+21 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx \\ & = \frac {64 a^3 (13 A+21 C) \sin (c+d x)}{315 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {16 a^2 (13 A+21 C) \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{315 d}+\frac {2 a (13 A+21 C) \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{105 d}+\frac {10 A \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{63 d}+\frac {2 A \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{9 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.85 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.51 \[ \int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 \sqrt {\cos (c+d x)} (5653 A+7476 C+4 (779 A+588 C) \cos (c+d x)+4 (254 A+63 C) \cos (2 (c+d x))+260 A \cos (3 (c+d x))+35 A \cos (4 (c+d x))) \sqrt {a (1+\sec (c+d x))} \sin (c+d x)}{1260 d (1+\cos (c+d x))} \]

[In]

Integrate[Cos[c + d*x]^(9/2)*(a + a*Sec[c + d*x])^(5/2)*(A + C*Sec[c + d*x]^2),x]

[Out]

(a^2*Sqrt[Cos[c + d*x]]*(5653*A + 7476*C + 4*(779*A + 588*C)*Cos[c + d*x] + 4*(254*A + 63*C)*Cos[2*(c + d*x)]
+ 260*A*Cos[3*(c + d*x)] + 35*A*Cos[4*(c + d*x)])*Sqrt[a*(1 + Sec[c + d*x])]*Sin[c + d*x])/(1260*d*(1 + Cos[c
+ d*x]))

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.53

\[\frac {2 a^{2} \sin \left (d x +c \right ) \left (35 A \cos \left (d x +c \right )^{4}+130 A \cos \left (d x +c \right )^{3}+219 A \cos \left (d x +c \right )^{2}+63 C \cos \left (d x +c \right )^{2}+292 A \cos \left (d x +c \right )+294 C \cos \left (d x +c \right )+584 A +903 C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {\cos \left (d x +c \right )}}{315 d \left (1+\cos \left (d x +c \right )\right )}\]

[In]

int(cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x)

[Out]

2/315*a^2/d*sin(d*x+c)*(35*A*cos(d*x+c)^4+130*A*cos(d*x+c)^3+219*A*cos(d*x+c)^2+63*C*cos(d*x+c)^2+292*A*cos(d*
x+c)+294*C*cos(d*x+c)+584*A+903*C)*(a*(1+sec(d*x+c)))^(1/2)*cos(d*x+c)^(1/2)/(1+cos(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.59 \[ \int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (35 \, A a^{2} \cos \left (d x + c\right )^{4} + 130 \, A a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (73 \, A + 21 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (146 \, A + 147 \, C\right )} a^{2} \cos \left (d x + c\right ) + {\left (584 \, A + 903 \, C\right )} a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

[In]

integrate(cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

2/315*(35*A*a^2*cos(d*x + c)^4 + 130*A*a^2*cos(d*x + c)^3 + 3*(73*A + 21*C)*a^2*cos(d*x + c)^2 + 2*(146*A + 14
7*C)*a^2*cos(d*x + c) + (584*A + 903*C)*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*
x + c)/(d*cos(d*x + c) + d)

Sympy [F(-1)]

Timed out. \[ \int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**(9/2)*(a+a*sec(d*x+c))**(5/2)*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 580 vs. \(2 (186) = 372\).

Time = 0.48 (sec) , antiderivative size = 580, normalized size of antiderivative = 2.69 \[ \int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \]

[In]

integrate(cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/5040*(sqrt(2)*(8190*a^2*cos(8/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c)))*sin(9/2*d*x + 9/2*c) +
2100*a^2*cos(2/3*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c)))*sin(9/2*d*x + 9/2*c) + 756*a^2*cos(4/9*a
rctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c)))*sin(9/2*d*x + 9/2*c) + 225*a^2*cos(2/9*arctan2(sin(9/2*d*x
+ 9/2*c), cos(9/2*d*x + 9/2*c)))*sin(9/2*d*x + 9/2*c) - 8190*a^2*cos(9/2*d*x + 9/2*c)*sin(8/9*arctan2(sin(9/2
*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) - 2100*a^2*cos(9/2*d*x + 9/2*c)*sin(2/3*arctan2(sin(9/2*d*x + 9/2*c), co
s(9/2*d*x + 9/2*c))) - 756*a^2*cos(9/2*d*x + 9/2*c)*sin(4/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c)
)) - 225*a^2*cos(9/2*d*x + 9/2*c)*sin(2/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 70*a^2*sin(9/
2*d*x + 9/2*c) + 225*a^2*sin(7/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 756*a^2*sin(5/9*arctan
2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 2100*a^2*sin(1/3*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x +
9/2*c))) + 8190*a^2*sin(1/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))))*A*sqrt(a) - 168*(75*sqrt(2)*
a^2*cos(5/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(2*d*x + 2*c) - 25*sqrt(2)*a^2*sin(3/4*arctan2(sin
(2*d*x + 2*c), cos(2*d*x + 2*c))) - 75*sqrt(2)*a^2*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 3*(2
5*sqrt(2)*a^2*cos(2*d*x + 2*c) + sqrt(2)*a^2)*sin(5/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*C*sqrt(a))
/d

Giac [F]

\[ \int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{\frac {9}{2}} \,d x } \]

[In]

integrate(cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int {\cos \left (c+d\,x\right )}^{9/2}\,\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \]

[In]

int(cos(c + d*x)^(9/2)*(A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^(5/2),x)

[Out]

int(cos(c + d*x)^(9/2)*(A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^(5/2), x)