\(\int (a+a \cos (c+d x)) (A+C \cos ^2(c+d x)) \sec ^{\frac {9}{2}}(c+d x) \, dx\) [1160]

Optimal result

Integrand size = 33, antiderivative size = 205 \[ \int (a+a \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \, dx=-\frac {2 a (3 A+5 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 a (5 A+7 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {2 a (3 A+5 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a (5 A+7 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {2 a A \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a A \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d} \]

[Out]

2/21*a*(5*A+7*C)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/5*a*A*sec(d*x+c)^(5/2)*sin(d*x+c)/d+2/7*a*A*sec(d*x+c)^(7/2)*
sin(d*x+c)/d+2/5*a*(3*A+5*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/d-2/5*a*(3*A+5*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/
2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+2/21*a*(5*A+7*C)*(cos(1
/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1
/2)/d

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {4306, 3111, 3100, 2827, 2716, 2720, 2719} \[ \int (a+a \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \, dx=\frac {2 a (5 A+7 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{21 d}+\frac {2 a (3 A+5 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 a (5 A+7 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}-\frac {2 a (3 A+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a A \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{7 d}+\frac {2 a A \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d} \]

[In]

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

[Out]

(-2*a*(3*A + 5*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (2*a*(5*A + 7*C)*Sq
rt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (2*a*(3*A + 5*C)*Sqrt[Sec[c + d*x]]*Si
n[c + d*x])/(5*d) + (2*a*(5*A + 7*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(21*d) + (2*a*A*Sec[c + d*x]^(5/2)*Sin[c
+ d*x])/(5*d) + (2*a*A*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(7*d)

Rule 2716

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
))), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3100

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m
+ 1)*(a^2 - b^2))), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B +
a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b,
e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rule 3111

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (C_.)*sin[(e
_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(b*c - a*d))*(A*b^2 + a^2*C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m +
1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] + Dist[1/(b^2*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp
[b*(m + 1)*(a*C*(b*c - a*d) + A*b*(a*c - b*d)) - ((b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e +
f*x] + b*C*d*(m + 1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]

Rule 4306

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

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx \\ & = \frac {2 a A \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {1}{7} \left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {7 a A}{2}+\frac {1}{2} a (5 A+7 C) \cos (c+d x)+\frac {7}{2} a C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x)} \, dx \\ & = \frac {2 a A \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a A \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {1}{35} \left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {5}{4} a (5 A+7 C)+\frac {7}{4} a (3 A+5 C) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {2 a A \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a A \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {1}{5} \left (a (3 A+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)} \, dx+\frac {1}{7} \left (a (5 A+7 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {2 a (3 A+5 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a (5 A+7 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {2 a A \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a A \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}-\frac {1}{5} \left (a (3 A+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{21} \left (a (5 A+7 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx \\ & = -\frac {2 a (3 A+5 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 a (5 A+7 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {2 a (3 A+5 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a (5 A+7 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {2 a A \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a A \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 6.03 (sec) , antiderivative size = 392, normalized size of antiderivative = 1.91 \[ \int (a+a \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \, dx=\frac {a e^{-i d x} (1+\cos (c+d x)) \csc (c) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (7 \sqrt {2} (3 A+5 C) e^{2 i d x} \left (-1+e^{2 i c}\right ) \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )-\frac {e^{-i (c-d x)} \left (-1+e^{2 i c}\right ) \left (35 C \left (1+e^{2 i (c+d x)}\right )^2 \left (-1+3 e^{i (c+d x)}+e^{2 i (c+d x)}+3 e^{3 i (c+d x)}\right )+A \left (-25+21 e^{i (c+d x)}-85 e^{2 i (c+d x)}+189 e^{3 i (c+d x)}+85 e^{4 i (c+d x)}+231 e^{5 i (c+d x)}+25 e^{6 i (c+d x)}+63 e^{7 i (c+d x)}\right )\right ) \sqrt {\sec (c+d x)}}{\left (1+e^{2 i (c+d x)}\right )^3}+10 (5 A+7 C) e^{i d x} \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)} \sin (c)\right )}{210 d} \]

[In]

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

[Out]

(a*(1 + Cos[c + d*x])*Csc[c]*Sec[(c + d*x)/2]^2*(7*Sqrt[2]*(3*A + 5*C)*E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Sqrt[E
^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((
2*I)*(c + d*x))] - ((-1 + E^((2*I)*c))*(35*C*(1 + E^((2*I)*(c + d*x)))^2*(-1 + 3*E^(I*(c + d*x)) + E^((2*I)*(c
+ d*x)) + 3*E^((3*I)*(c + d*x))) + A*(-25 + 21*E^(I*(c + d*x)) - 85*E^((2*I)*(c + d*x)) + 189*E^((3*I)*(c + d
*x)) + 85*E^((4*I)*(c + d*x)) + 231*E^((5*I)*(c + d*x)) + 25*E^((6*I)*(c + d*x)) + 63*E^((7*I)*(c + d*x))))*Sq
rt[Sec[c + d*x]])/(E^(I*(c - d*x))*(1 + E^((2*I)*(c + d*x)))^3) + 10*(5*A + 7*C)*E^(I*d*x)*Sqrt[Cos[c + d*x]]*
EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]]*Sin[c]))/(210*d*E^(I*d*x))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(810\) vs. \(2(229)=458\).

Time = 94.91 (sec) , antiderivative size = 811, normalized size of antiderivative = 3.96

method	result	size
default	\(\text {Expression too large to display}\)	\(811\)
parts	\(\text {Expression too large to display}\)	\(1002\)

[In]

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

[Out]

-4*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a*(1/2*C/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*
c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-(sin(1
/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+1/10*A/sin(1/2*
d*x+1/2*c)^2/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)*(24*cos(1/2*d*x+1/2*c)*
sin(1/2*d*x+1/2*c)^6-12*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/
2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+12*(2*sin(1/2*d*x+1/2*c)^2-1)^(1
/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2+8*sin(1/2*d*x+1/2*
c)^2*cos(1/2*d*x+1/2*c)-3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+
1/2*c),2^(1/2)))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+1/2*A*(-1/56*cos(1/2*d*x+1/2*c)*(-2*sin(
1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^4-5/42*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*
d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+5/21*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*co
s(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),
2^(1/2)))+1/2*C*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2
*c)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1
/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))))/sin(1/2*d*x+1/2*c)/(-1+2*cos(1/2*d*x+1/2*c)^2)^
(1/2)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.11 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.19 \[ \int (a+a \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \, dx=\frac {-5 i \, \sqrt {2} {\left (5 \, A + 7 \, C\right )} a \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} {\left (5 \, A + 7 \, C\right )} a \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 21 i \, \sqrt {2} {\left (3 \, A + 5 \, C\right )} a \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 21 i \, \sqrt {2} {\left (3 \, A + 5 \, C\right )} a \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \frac {2 \, {\left (21 \, {\left (3 \, A + 5 \, C\right )} a \cos \left (d x + c\right )^{3} + 5 \, {\left (5 \, A + 7 \, C\right )} a \cos \left (d x + c\right )^{2} + 21 \, A a \cos \left (d x + c\right ) + 15 \, A a\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{105 \, d \cos \left (d x + c\right )^{3}} \]

[In]

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

[Out]

1/105*(-5*I*sqrt(2)*(5*A + 7*C)*a*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5
*I*sqrt(2)*(5*A + 7*C)*a*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 21*I*sqrt(
2)*(3*A + 5*C)*a*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c
))) + 21*I*sqrt(2)*(3*A + 5*C)*a*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c)
- I*sin(d*x + c))) + 2*(21*(3*A + 5*C)*a*cos(d*x + c)^3 + 5*(5*A + 7*C)*a*cos(d*x + c)^2 + 21*A*a*cos(d*x + c
) + 15*A*a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^3)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \, dx=\int \left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{9/2}\,\left (a+a\,\cos \left (c+d\,x\right )\right ) \,d x \]

[In]

int((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(9/2)*(a + a*cos(c + d*x)),x)

[Out]

int((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(9/2)*(a + a*cos(c + d*x)), x)