\(\int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+C \sec ^2(c+d x)) \, dx\) [1136]

Optimal result

Integrand size = 37, antiderivative size = 219 \[ \int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {4 a^2 (136 A+189 C) \sin (c+d x)}{315 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (136 A+189 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{315 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (52 A+63 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{315 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a A \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{21 d}+\frac {2 A \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{9 d} \]

[Out]

2/9*A*cos(d*x+c)^(7/2)*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d+2/315*a^2*(52*A+63*C)*cos(d*x+c)^(3/2)*sin(d*x+c)/d
/(a+a*sec(d*x+c))^(1/2)+4/315*a^2*(136*A+189*C)*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+2/315*a^2
*(136*A+189*C)*sin(d*x+c)*cos(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)+2/21*a*A*cos(d*x+c)^(5/2)*sin(d*x+c)*(a+a*
sec(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {4350, 4172, 4102, 4100, 3890, 3889} \[ \int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 a^2 (52 A+63 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{315 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 (136 A+189 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{315 d \sqrt {a \sec (c+d x)+a}}+\frac {4 a^2 (136 A+189 C) \sin (c+d x)}{315 d \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {2 A \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d}+\frac {2 a A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{21 d} \]

[In]

Int[Cos[c + d*x]^(9/2)*(a + a*Sec[c + d*x])^(3/2)*(A + C*Sec[c + d*x]^2),x]

[Out]

(4*a^2*(136*A + 189*C)*Sin[c + d*x])/(315*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(136*A + 189
*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(315*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(52*A + 63*C)*Cos[c + d*x]^(3/2
)*Sin[c + d*x])/(315*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*A*Cos[c + d*x]^(5/2)*Sqrt[a + a*Sec[c + d*x]]*Sin[c +
d*x])/(21*d) + (2*A*Cos[c + d*x]^(7/2)*(a + a*Sec[c + d*x])^(3/2)*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 3890

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

Rule 4100

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

Rule 4102

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[a*A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*n)), x]
- Dist[b/(a*d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*(m - n - 1) - b*B*n - (a*
B*n + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2
- b^2, 0] && GtQ[m, 1/2] && LtQ[n, -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))^{3/2} \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))^{3/2} \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))^{3/2} \left (\frac {3 a A}{2}+\frac {1}{2} a (4 A+9 C) \sec (c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx}{9 a} \\ & = \frac {2 a A \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{21 d}+\frac {2 A \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{9 d}+\frac {\left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \sec (c+d x)} \left (\frac {1}{4} a^2 (52 A+63 C)+\frac {1}{4} a^2 (40 A+63 C) \sec (c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx}{63 a} \\ & = \frac {2 a^2 (52 A+63 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{315 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a A \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{21 d}+\frac {2 A \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{9 d}+\frac {1}{105} \left (a (136 A+189 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {2 a^2 (136 A+189 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{315 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (52 A+63 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{315 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a A \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{21 d}+\frac {2 A \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{9 d}+\frac {1}{315} \left (2 a (136 A+189 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 {4 a^2 (136 A+189 C) \sin (c+d x)}{315 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (136 A+189 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{315 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (52 A+63 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{315 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a A \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{21 d}+\frac {2 A \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{9 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.35 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.50 \[ \int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a \sqrt {\cos (c+d x)} (2689 A+3276 C+2 (799 A+756 C) \cos (c+d x)+4 (137 A+63 C) \cos (2 (c+d x))+170 A \cos (3 (c+d x))+35 A \cos (4 (c+d x))) \sqrt {a (1+\sec (c+d x))} \sin (c+d x)}{1260 d (1+\cos (c+d x))} \]

[In]

Integrate[Cos[c + d*x]^(9/2)*(a + a*Sec[c + d*x])^(3/2)*(A + C*Sec[c + d*x]^2),x]

[Out]

(a*Sqrt[Cos[c + d*x]]*(2689*A + 3276*C + 2*(799*A + 756*C)*Cos[c + d*x] + 4*(137*A + 63*C)*Cos[2*(c + d*x)] +
170*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 = 0.59 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.49

method	result	size
default	\(-\frac {2 a \left (\left (35 \cos \left (d x +c \right )^{4}+85 \cos \left (d x +c \right )^{3}+102 \cos \left (d x +c \right )^{2}+136 \cos \left (d x +c \right )+272\right ) A +\left (63 \cos \left (d x +c \right )^{2}+189 \cos \left (d x +c \right )+378\right ) C \right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )}{315 d}\)	\(107\)

[In]

int(cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^(3/2)*(A+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

-2/315*a/d*((35*cos(d*x+c)^4+85*cos(d*x+c)^3+102*cos(d*x+c)^2+136*cos(d*x+c)+272)*A+(63*cos(d*x+c)^2+189*cos(d
*x+c)+378)*C)*cos(d*x+c)^(1/2)*(a*(1+sec(d*x+c)))^(1/2)*(cot(d*x+c)-csc(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.53 \[ \int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (35 \, A a \cos \left (d x + c\right )^{4} + 85 \, A a \cos \left (d x + c\right )^{3} + 3 \, {\left (34 \, A + 21 \, C\right )} a \cos \left (d x + c\right )^{2} + {\left (136 \, A + 189 \, C\right )} a \cos \left (d x + c\right ) + 2 \, {\left (136 \, A + 189 \, C\right )} a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

[In]

integrate(cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^(3/2)*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

2/315*(35*A*a*cos(d*x + c)^4 + 85*A*a*cos(d*x + c)^3 + 3*(34*A + 21*C)*a*cos(d*x + c)^2 + (136*A + 189*C)*a*co
s(d*x + c) + 2*(136*A + 189*C)*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/(d*c
os(d*x + c) + d)

Sympy [F(-1)]

Timed out. \[ \int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**(9/2)*(a+a*sec(d*x+c))**(3/2)*(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. 544 vs. \(2 (189) = 378\).

Time = 0.47 (sec) , antiderivative size = 544, normalized size of antiderivative = 2.48 \[ \int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \]

[In]

integrate(cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^(3/2)*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/5040*(sqrt(2)*(3780*a*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) + 10
50*a*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) + 378*a*cos(4/9*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) + 135*a*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) - 3780*a*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))) - 1050*a*cos(9/2*d*x + 9/2*c)*sin(2/3*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x +
9/2*c))) - 378*a*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))) - 135*a*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*sin(9/2*d*x + 9/2*c) + 1
35*a*sin(7/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 378*a*sin(5/9*arctan2(sin(9/2*d*x + 9/2*c)
, cos(9/2*d*x + 9/2*c))) + 1050*a*sin(1/3*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 3780*a*sin(1/
9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))))*A*sqrt(a) - 504*(10*sqrt(2)*a*cos(5/4*arctan2(sin(2*d*
x + 2*c), cos(2*d*x + 2*c)))*sin(2*d*x + 2*c) - 5*sqrt(2)*a*sin(3/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)
)) - 10*sqrt(2)*a*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - (10*sqrt(2)*a*cos(2*d*x + 2*c) + sqrt
(2)*a)*sin(5/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*C*sqrt(a))/d

Giac [F]

\[ \int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{\frac {9}{2}} \,d x } \]

[In]

integrate(cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^(3/2)*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int {\cos \left (c+d\,x\right )}^{9/2}\,\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \]

[In]

int(cos(c + d*x)^(9/2)*(A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^(3/2),x)

[Out]

int(cos(c + d*x)^(9/2)*(A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^(3/2), x)