\(\int (a+b \cos (c+d x))^2 (A+C \cos ^2(c+d x)) \sec ^{\frac {9}{2}}(c+d x) \, dx\) [1368]

Optimal result

Integrand size = 35, antiderivative size = 243 \[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \, dx=-\frac {4 a b (3 A+5 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 \left (7 b^2 (A+3 C)+a^2 (5 A+7 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {4 a b (3 A+5 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 \left (4 A b^2+a^2 (5 A+7 C)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {8 a A b \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 A (a+b \cos (c+d x))^2 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d} \]

[Out]

2/21*(4*A*b^2+a^2*(5*A+7*C))*sec(d*x+c)^(3/2)*sin(d*x+c)/d+8/35*a*A*b*sec(d*x+c)^(5/2)*sin(d*x+c)/d+2/7*A*(a+b
*cos(d*x+c))^2*sec(d*x+c)^(7/2)*sin(d*x+c)/d+4/5*a*b*(3*A+5*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/d-4/5*a*b*(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*(7*b^2*(A+3*C)+a^2*(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 = 0.79 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {4306, 3127, 3110, 3100, 2827, 2716, 2719, 2720} \[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \, dx=\frac {2 \left (a^2 (5 A+7 C)+4 A b^2\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{21 d}+\frac {2 \left (a^2 (5 A+7 C)+7 b^2 (A+3 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {4 a b (3 A+5 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d}-\frac {4 a b (3 A+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {8 a A b \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{35 d}+\frac {2 A \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))^2}{7 d} \]

[In]

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

[Out]

(-4*a*b*(3*A + 5*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (2*(7*b^2*(A + 3*
C) + a^2*(5*A + 7*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (4*a*b*(3*A +
5*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(5*d) + (2*(4*A*b^2 + a^2*(5*A + 7*C))*Sec[c + d*x]^(3/2)*Sin[c + d*x])/
(21*d) + (8*a*A*b*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(35*d) + (2*A*(a + b*Cos[c + d*x])^2*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 3110

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

Rule 3127

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C + A*d^2))*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Si
n[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^
(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + c*C*(b*c*m + a*d*(n + 1)) - (A*d*(a*d*(n
+ 2) - b*c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*(A*d^2*(m + n + 2) + C*(c^2
*(m + 1) + d^2*(n + 1)))*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] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -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+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx \\ & = \frac {2 A (a+b \cos (c+d x))^2 \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 {(a+b \cos (c+d x)) \left (2 A b+\frac {1}{2} a (5 A+7 C) \cos (c+d x)+\frac {1}{2} b (A+7 C) \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx \\ & = \frac {8 a A b \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 A (a+b \cos (c+d x))^2 \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} \left (4 A b^2+a^2 (5 A+7 C)\right )-\frac {7}{2} a b (3 A+5 C) \cos (c+d x)-\frac {5}{4} b^2 (A+7 C) \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {2 \left (4 A b^2+a^2 (5 A+7 C)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {8 a A b \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 A (a+b \cos (c+d x))^2 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}-\frac {1}{105} \left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {21}{4} a b (3 A+5 C)-\frac {5}{8} \left (7 b^2 (A+3 C)+a^2 (5 A+7 C)\right ) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {2 \left (4 A b^2+a^2 (5 A+7 C)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {8 a A b \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 A (a+b \cos (c+d x))^2 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {1}{5} \left (2 a b (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}{21} \left (\left (7 b^2 (A+3 C)+a^2 (5 A+7 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx \\ & = \frac {2 \left (7 b^2 (A+3 C)+a^2 (5 A+7 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {4 a b (3 A+5 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 \left (4 A b^2+a^2 (5 A+7 C)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {8 a A b \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 A (a+b \cos (c+d x))^2 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}-\frac {1}{5} \left (2 a b (3 A+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx \\ & = -\frac {4 a b (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 \left (7 b^2 (A+3 C)+a^2 (5 A+7 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {4 a b (3 A+5 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 \left (4 A b^2+a^2 (5 A+7 C)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {8 a A b \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 A (a+b \cos (c+d x))^2 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.38 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.90 \[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \, dx=\frac {2 \sec ^{\frac {7}{2}}(c+d x) \left (-42 a b (3 A+5 C) \cos ^{\frac {7}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+5 \left (7 b^2 (A+3 C)+a^2 (5 A+7 C)\right ) \cos ^{\frac {7}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+15 a^2 A \sin (c+d x)+25 a^2 A \cos ^2(c+d x) \sin (c+d x)+35 A b^2 \cos ^2(c+d x) \sin (c+d x)+35 a^2 C \cos ^2(c+d x) \sin (c+d x)+126 a A b \cos ^3(c+d x) \sin (c+d x)+210 a b C \cos ^3(c+d x) \sin (c+d x)+21 a A b \sin (2 (c+d x))\right )}{105 d} \]

[In]

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

[Out]

(2*Sec[c + d*x]^(7/2)*(-42*a*b*(3*A + 5*C)*Cos[c + d*x]^(7/2)*EllipticE[(c + d*x)/2, 2] + 5*(7*b^2*(A + 3*C) +
a^2*(5*A + 7*C))*Cos[c + d*x]^(7/2)*EllipticF[(c + d*x)/2, 2] + 15*a^2*A*Sin[c + d*x] + 25*a^2*A*Cos[c + d*x]
^2*Sin[c + d*x] + 35*A*b^2*Cos[c + d*x]^2*Sin[c + d*x] + 35*a^2*C*Cos[c + d*x]^2*Sin[c + d*x] + 126*a*A*b*Cos[
c + d*x]^3*Sin[c + d*x] + 210*a*b*C*Cos[c + d*x]^3*Sin[c + d*x] + 21*a*A*b*Sin[2*(c + d*x)]))/(105*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(902\) vs. \(2(267)=534\).

Time = 1406.65 (sec) , antiderivative size = 903, normalized size of antiderivative = 3.72

method	result	size
default	\(\text {Expression too large to display}\)	\(903\)
parts	\(\text {Expression too large to display}\)	\(1151\)

[In]

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

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*C*b^2*(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)
)+2*A*a^2*(-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*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)))+4*a*b*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)))+4/5*a*A*b/(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)/sin(1/2*d*x+1/2*c)^2*(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)+2*(A*b^2+C*a^2)*(-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 = 289, normalized size of antiderivative = 1.19 \[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \, dx=\frac {-42 i \, \sqrt {2} {\left (3 \, A + 5 \, C\right )} a b \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 42 i \, \sqrt {2} {\left (3 \, A + 5 \, C\right )} a b \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 5 \, \sqrt {2} {\left (i \, {\left (5 \, A + 7 \, C\right )} a^{2} + 7 i \, {\left (A + 3 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 \, \sqrt {2} {\left (-i \, {\left (5 \, A + 7 \, C\right )} a^{2} - 7 i \, {\left (A + 3 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + \frac {2 \, {\left (42 \, {\left (3 \, A + 5 \, C\right )} a b \cos \left (d x + c\right )^{3} + 42 \, A a b \cos \left (d x + c\right ) + 15 \, A a^{2} + 5 \, {\left ({\left (5 \, A + 7 \, C\right )} a^{2} + 7 \, A b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{105 \, d \cos \left (d x + c\right )^{3}} \]

[In]

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

[Out]

1/105*(-42*I*sqrt(2)*(3*A + 5*C)*a*b*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x
+ c) + I*sin(d*x + c))) + 42*I*sqrt(2)*(3*A + 5*C)*a*b*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInver
se(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 5*sqrt(2)*(I*(5*A + 7*C)*a^2 + 7*I*(A + 3*C)*b^2)*cos(d*x + c)^3*w
eierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 5*sqrt(2)*(-I*(5*A + 7*C)*a^2 - 7*I*(A + 3*C)*b^2)*
cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*(42*(3*A + 5*C)*a*b*cos(d*x + c)^
3 + 42*A*a*b*cos(d*x + c) + 15*A*a^2 + 5*((5*A + 7*C)*a^2 + 7*A*b^2)*cos(d*x + c)^2)*sin(d*x + c)/sqrt(cos(d*x
+ c)))/(d*cos(d*x + c)^3)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

Giac [F]

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

[In]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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