3.1168 \(\int (a+a \cos (c+d x))^2 (A+C \cos ^2(c+d x)) \sec ^{\frac {9}{2}}(c+d x) \, dx\)

Optimal. Leaf size=237 \[ \frac {2 a^2 (33 A+35 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{105 d}+\frac {4 a^2 (3 A+5 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d}+\frac {8 a^2 (3 A+7 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}-\frac {4 a^2 (3 A+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {8 A \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{35 d}+\frac {2 A \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+a)^2}{7 d} \]

[Out]

2/105*a^2*(33*A+35*C)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+8/35*A*(a^2+a^2*cos(d*x+c))*sec(d*x+c)^(5/2)*sin(d*x+c)/d+
2/7*A*(a+a*cos(d*x+c))^2*sec(d*x+c)^(7/2)*sin(d*x+c)/d+4/5*a^2*(3*A+5*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/d-4/5*a^2
*(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+8/21*a^2*(3*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

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Rubi [A]  time = 0.56, antiderivative size = 237, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {4221, 3044, 2975, 2968, 3021, 2748, 2636, 2639, 2641} \[ \frac {2 a^2 (33 A+35 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{105 d}+\frac {4 a^2 (3 A+5 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d}+\frac {8 a^2 (3 A+7 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}-\frac {4 a^2 (3 A+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {8 A \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{35 d}+\frac {2 A \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+a)^2}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*a^2*(3*A + 5*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (8*a^2*(3*A + 7*C
)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (4*a^2*(3*A + 5*C)*Sqrt[Sec[c + d*
x]]*Sin[c + d*x])/(5*d) + (2*a^2*(33*A + 35*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(105*d) + (8*A*(a^2 + a^2*Cos[
c + d*x])*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(35*d) + (2*A*(a + a*Cos[c + d*x])^2*Sec[c + d*x]^(7/2)*Sin[c + d*x
])/(7*d)

Rule 2636

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 2639

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

Rule 2641

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

Rule 2748

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 2968

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

Rule 2975

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b^2*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*S
in[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c + a*d)), x] - Dist[b/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])
^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n +
1) - B*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d
, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n]
|| EqQ[c, 0])

Rule 3021

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 3044

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*Sin[
e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^
m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + c*C*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(m + n +
2) + C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m}, x] && NeQ[b
*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0
])

Rule 4221

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*} \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \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+a \cos (c+d x))^2 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \cos (c+d x))^2 \left (2 a A+\frac {1}{2} a (A+7 C) \cos (c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx}{7 a}\\ &=\frac {8 A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 A (a+a \cos (c+d x))^2 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {\left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \cos (c+d x)) \left (\frac {1}{4} a^2 (33 A+35 C)+\frac {1}{4} a^2 (9 A+35 C) \cos (c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx}{35 a}\\ &=\frac {8 A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 A (a+a \cos (c+d x))^2 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {\left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{4} a^3 (33 A+35 C)+\left (\frac {1}{4} a^3 (9 A+35 C)+\frac {1}{4} a^3 (33 A+35 C)\right ) \cos (c+d x)+\frac {1}{4} a^3 (9 A+35 C) \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x)} \, dx}{35 a}\\ &=\frac {2 a^2 (33 A+35 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac {8 A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 A (a+a \cos (c+d x))^2 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {\left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {21}{4} a^3 (3 A+5 C)+\frac {5}{2} a^3 (3 A+7 C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{105 a}\\ &=\frac {2 a^2 (33 A+35 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac {8 A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 A (a+a \cos (c+d x))^2 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {1}{5} \left (2 a^2 (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 (4 a^2 (3 A+7 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {8 a^2 (3 A+7 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {4 a^2 (3 A+5 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a^2 (33 A+35 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac {8 A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 A (a+a \cos (c+d x))^2 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}-\frac {1}{5} \left (2 a^2 (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^2 (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 {8 a^2 (3 A+7 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {4 a^2 (3 A+5 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a^2 (33 A+35 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac {8 A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 A (a+a \cos (c+d x))^2 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}\\ \end {align*}

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Mathematica [C]  time = 4.38, size = 399, normalized size = 1.68 \[ \frac {a^2 \csc (c) e^{-i d x} (\cos (c+d x)+1)^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (7 \sqrt {2} \left (-1+e^{2 i c}\right ) (3 A+5 C) e^{2 i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )-\frac {\left (-1+e^{2 i c}\right ) e^{-i (c-d x)} \sqrt {\sec (c+d x)} \left (6 A \left (7 e^{i (c+d x)}-20 e^{2 i (c+d x)}+63 e^{3 i (c+d x)}+20 e^{4 i (c+d x)}+77 e^{5 i (c+d x)}+10 e^{6 i (c+d x)}+21 e^{7 i (c+d x)}-10\right )+35 C \left (6 e^{i (c+d x)}+e^{2 i (c+d x)}+6 e^{3 i (c+d x)}-1\right ) \left (1+e^{2 i (c+d x)}\right )^2\right )}{2 \left (1+e^{2 i (c+d x)}\right )^3}+20 (3 A+7 C) \sin (c) e^{i d x} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{210 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(1 + Cos[c + d*x])^2*Csc[c]*Sec[(c + d*x)/2]^4*(7*Sqrt[2]*(3*A + 5*C)*E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Sq
rt[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 + 6*E^(I*(c + d*x)) + E^((2*I
)*(c + d*x)) + 6*E^((3*I)*(c + d*x))) + 6*A*(-10 + 7*E^(I*(c + d*x)) - 20*E^((2*I)*(c + d*x)) + 63*E^((3*I)*(c
+ d*x)) + 20*E^((4*I)*(c + d*x)) + 77*E^((5*I)*(c + d*x)) + 10*E^((6*I)*(c + d*x)) + 21*E^((7*I)*(c + d*x))))
*Sqrt[Sec[c + d*x]])/(2*E^(I*(c - d*x))*(1 + E^((2*I)*(c + d*x)))^3) + 20*(3*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))

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C a^{2} \cos \left (d x + c\right )^{4} + 2 \, C a^{2} \cos \left (d x + c\right )^{3} + {\left (A + C\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \, A a^{2} \cos \left (d x + c\right ) + A a^{2}\right )} \sec \left (d x + c\right )^{\frac {9}{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*a^2*cos(d*x + c)^4 + 2*C*a^2*cos(d*x + c)^3 + (A + C)*a^2*cos(d*x + c)^2 + 2*A*a^2*cos(d*x + c) +
A*a^2)*sec(d*x + c)^(9/2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {9}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [B]  time = 8.71, size = 918, normalized size = 3.87 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*(1/4*C*(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))+1/2*C*(-(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(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*(-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)*sin(1/2*d*x+1/2*c)^2)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)+(1/4*A+1/4*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)/(-1/2+cos(1/2*d*x+1/2*c)^2)^2+1/3*(si
n(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)))-1/10*A/(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*(12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(
sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4-24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-12*EllipticE(cos(1
/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+24*s
in(1/2*d*x+1/2*c)^4*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)*Ellipti
cE(cos(1/2*d*x+1/2*c),2^(1/2))-8*sin(1/2*d*x+1/2*c)^2*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)+1/4*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)/(-1/2+co
s(1/2*d*x+1/2*c)^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)/(-1/2+cos(1
/2*d*x+1/2*c)^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))))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)
^2-1)^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {9}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{9/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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