3.1176 \(\int (a+a \cos (c+d x))^3 (A+C \cos ^2(c+d x)) \sec ^{\frac {11}{2}}(c+d x) \, dx\)

Optimal. Leaf size=286 \[ \frac {8 a^3 (16 A+21 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{105 d}+\frac {4 a^3 (17 A+27 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{15 d}+\frac {2 (73 A+63 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{315 d}+\frac {4 a^3 (11 A+21 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}-\frac {4 a^3 (17 A+27 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {4 A \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{21 a d}+\frac {2 A \sin (c+d x) \sec ^{\frac {9}{2}}(c+d x) (a \cos (c+d x)+a)^3}{9 d} \]

[Out]

8/105*a^3*(16*A+21*C)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/315*(73*A+63*C)*(a^3+a^3*cos(d*x+c))*sec(d*x+c)^(5/2)*si
n(d*x+c)/d+4/21*A*(a^2+a^2*cos(d*x+c))^2*sec(d*x+c)^(7/2)*sin(d*x+c)/a/d+2/9*A*(a+a*cos(d*x+c))^3*sec(d*x+c)^(
9/2)*sin(d*x+c)/d+4/15*a^3*(17*A+27*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/d-4/15*a^3*(17*A+27*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+4/21*a^3
*(11*A+21*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 = 0.73, antiderivative size = 286, normalized size of antiderivative = 1.00, number of steps used = 10, 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} \[ \frac {8 a^3 (16 A+21 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{105 d}+\frac {4 a^3 (17 A+27 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{15 d}+\frac {2 (73 A+63 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{315 d}+\frac {4 a^3 (11 A+21 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}-\frac {4 a^3 (17 A+27 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {4 A \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{21 a d}+\frac {2 A \sin (c+d x) \sec ^{\frac {9}{2}}(c+d x) (a \cos (c+d x)+a)^3}{9 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^3*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(11/2),x]

[Out]

(-4*a^3*(17*A + 27*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(15*d) + (4*a^3*(11*A +
21*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (4*a^3*(17*A + 27*C)*Sqrt[Sec
[c + d*x]]*Sin[c + d*x])/(15*d) + (8*a^3*(16*A + 21*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(105*d) + (2*(73*A + 6
3*C)*(a^3 + a^3*Cos[c + d*x])*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(315*d) + (4*A*(a^2 + a^2*Cos[c + d*x])^2*Sec[c
+ d*x]^(7/2)*Sin[c + d*x])/(21*a*d) + (2*A*(a + a*Cos[c + d*x])^3*Sec[c + d*x]^(9/2)*Sin[c + d*x])/(9*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))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {11}{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))^3 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx\\ &=\frac {2 A (a+a \cos (c+d x))^3 \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \cos (c+d x))^3 \left (3 a A+\frac {1}{2} a (A+9 C) \cos (c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx}{9 a}\\ &=\frac {4 A \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{21 a d}+\frac {2 A (a+a \cos (c+d x))^3 \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac {\left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \cos (c+d x))^2 \left (\frac {1}{4} a^2 (73 A+63 C)+\frac {1}{4} a^2 (13 A+63 C) \cos (c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx}{63 a}\\ &=\frac {2 (73 A+63 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac {4 A \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{21 a d}+\frac {2 A (a+a \cos (c+d x))^3 \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac {\left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \cos (c+d x)) \left (\frac {9}{2} a^3 (16 A+21 C)+\frac {3}{4} a^3 (23 A+63 C) \cos (c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx}{315 a}\\ &=\frac {2 (73 A+63 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac {4 A \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{21 a d}+\frac {2 A (a+a \cos (c+d x))^3 \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac {\left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {9}{2} a^4 (16 A+21 C)+\left (\frac {9}{2} a^4 (16 A+21 C)+\frac {3}{4} a^4 (23 A+63 C)\right ) \cos (c+d x)+\frac {3}{4} a^4 (23 A+63 C) \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x)} \, dx}{315 a}\\ &=\frac {8 a^3 (16 A+21 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac {2 (73 A+63 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac {4 A \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{21 a d}+\frac {2 A (a+a \cos (c+d x))^3 \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac {\left (16 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {63}{8} a^4 (17 A+27 C)+\frac {45}{8} a^4 (11 A+21 C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{945 a}\\ &=\frac {8 a^3 (16 A+21 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac {2 (73 A+63 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac {4 A \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{21 a d}+\frac {2 A (a+a \cos (c+d x))^3 \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac {1}{21} \left (2 a^3 (11 A+21 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx+\frac {1}{15} \left (2 a^3 (17 A+27 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {4 a^3 (11 A+21 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^3 (17 A+27 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {8 a^3 (16 A+21 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac {2 (73 A+63 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac {4 A \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{21 a d}+\frac {2 A (a+a \cos (c+d x))^3 \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d}-\frac {1}{15} \left (2 a^3 (17 A+27 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=-\frac {4 a^3 (17 A+27 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{15 d}+\frac {4 a^3 (11 A+21 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^3 (17 A+27 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {8 a^3 (16 A+21 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac {2 (73 A+63 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac {4 A \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{21 a d}+\frac {2 A (a+a \cos (c+d x))^3 \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d}\\ \end {align*}

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Mathematica [C]  time = 6.91, size = 655, normalized size = 2.29 \[ \sqrt {\sec (c+d x)} \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^3 \left (\frac {(17 A+27 C) \csc (c) \cos (d x)}{30 d}+\frac {\sec (c) \sec ^2(c+d x) (135 A \sin (c)+238 A \sin (d x)+63 C \sin (d x))}{1260 d}+\frac {\sec (c) \sec (c+d x) (238 A \sin (c)+330 A \sin (d x)+63 C \sin (c)+315 C \sin (d x))}{1260 d}+\frac {(22 A+21 C) \tan (c)}{84 d}+\frac {A \sec (c) \sin (d x) \sec ^4(c+d x)}{36 d}+\frac {\sec (c) \sec ^3(c+d x) (7 A \sin (c)+27 A \sin (d x))}{252 d}\right )+\frac {17 A \csc (c) e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \left (\left (-1+e^{2 i c}\right ) e^{2 i d x} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )-3 \sqrt {1+e^{2 i (c+d x)}}\right ) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^3}{90 \sqrt {2} d}+\frac {11 A \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) (a \cos (c+d x)+a)^3}{42 d}+\frac {3 C \csc (c) e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \left (\left (-1+e^{2 i c}\right ) e^{2 i d x} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )-3 \sqrt {1+e^{2 i (c+d x)}}\right ) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^3}{10 \sqrt {2} d}+\frac {C \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) (a \cos (c+d x)+a)^3}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^3*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(11/2),x]

[Out]

(17*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*(a + a*Cos[c + d*x])^3*Csc
[c]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^(
(2*I)*(c + d*x))])*Sec[c/2 + (d*x)/2]^6)/(90*Sqrt[2]*d*E^(I*d*x)) + (3*C*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c
+ d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*(a + a*Cos[c + d*x])^3*Csc[c]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^(
(2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*Sec[c/2 + (d*x)/2]^6)/(1
0*Sqrt[2]*d*E^(I*d*x)) + (11*A*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^3*EllipticF[(c + d*x)/2, 2]*Sec[c/2 + (
d*x)/2]^6*Sqrt[Sec[c + d*x]])/(42*d) + (C*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^3*EllipticF[(c + d*x)/2, 2]*
Sec[c/2 + (d*x)/2]^6*Sqrt[Sec[c + d*x]])/(2*d) + (a + a*Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*Sqrt[Sec[c + d*x]
]*(((17*A + 27*C)*Cos[d*x]*Csc[c])/(30*d) + (A*Sec[c]*Sec[c + d*x]^4*Sin[d*x])/(36*d) + (Sec[c]*Sec[c + d*x]^3
*(7*A*Sin[c] + 27*A*Sin[d*x]))/(252*d) + (Sec[c]*Sec[c + d*x]^2*(135*A*Sin[c] + 238*A*Sin[d*x] + 63*C*Sin[d*x]
))/(1260*d) + (Sec[c]*Sec[c + d*x]*(238*A*Sin[c] + 63*C*Sin[c] + 330*A*Sin[d*x] + 315*C*Sin[d*x]))/(1260*d) +
((22*A + 21*C)*Tan[c])/(84*d))

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fricas [F]  time = 1.17, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C a^{3} \cos \left (d x + c\right )^{5} + 3 \, C a^{3} \cos \left (d x + c\right )^{4} + {\left (A + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + {\left (3 \, A + C\right )} a^{3} \cos \left (d x + c\right )^{2} + 3 \, A a^{3} \cos \left (d x + c\right ) + A a^{3}\right )} \sec \left (d x + c\right )^{\frac {11}{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(11/2),x, algorithm="fricas")

[Out]

integral((C*a^3*cos(d*x + c)^5 + 3*C*a^3*cos(d*x + c)^4 + (A + 3*C)*a^3*cos(d*x + c)^3 + (3*A + C)*a^3*cos(d*x
+ c)^2 + 3*A*a^3*cos(d*x + c) + A*a^3)*sec(d*x + c)^(11/2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{3} \sec \left (d x + c\right )^{\frac {11}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(11/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(a*cos(d*x + c) + a)^3*sec(d*x + c)^(11/2), x)

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maple [B]  time = 11.12, size = 1246, normalized size = 4.36 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(11/2),x)

[Out]

-16*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^3*(1/8*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))+3/8*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+cos(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)))+3/8*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)+(1/8*A+3/8*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*(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)))-1/5*(3/8*A+1/8*
C)/(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*Elliptic
E(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*sin(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)*EllipticE(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)+1/8*A*(-1/144*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)^5-7/180*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)^3-14/15*sin(1/2*d*x+1/2*c)^
2*cos(1/2*d*x+1/2*c)/(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)+7/15*(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))-7/15*(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))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))))/s
in(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 \[ \int {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{3} \sec \left (d x + c\right )^{\frac {11}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(11/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(a*cos(d*x + c) + a)^3*sec(d*x + c)^(11/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{11/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(11/2)*(a + a*cos(c + d*x))^3,x)

[Out]

int((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(11/2)*(a + a*cos(c + d*x))^3, 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))**3*(A+C*cos(d*x+c)**2)*sec(d*x+c)**(11/2),x)

[Out]

Timed out

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