3.1383 \(\int (a+b \cos (c+d x))^4 (A+C \cos ^2(c+d x)) \sec ^{\frac {13}{2}}(c+d x) \, dx\)

Optimal. Leaf size=417 \[ \frac {4 a b \left (a^2 (673 A+891 C)+96 A b^2\right ) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3465 d}+\frac {8 a b \left (a^2 (7 A+9 C)+3 b^2 (3 A+5 C)\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{15 d}+\frac {2 \left (3 a^2 (9 A+11 C)+16 A b^2\right ) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))^2}{231 d}-\frac {8 a b \left (a^2 (7 A+9 C)+3 b^2 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 \left (15 a^4 (9 A+11 C)+9 a^2 b^2 (101 A+143 C)+64 A b^4\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{693 d}+\frac {2 \left (5 a^4 (9 A+11 C)+66 a^2 b^2 (5 A+7 C)+77 b^4 (A+3 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{231 d}+\frac {2 A \sin (c+d x) \sec ^{\frac {11}{2}}(c+d x) (a+b \cos (c+d x))^4}{11 d}+\frac {16 A b \sin (c+d x) \sec ^{\frac {9}{2}}(c+d x) (a+b \cos (c+d x))^3}{99 d} \]

[Out]

2/693*(64*A*b^4+15*a^4*(9*A+11*C)+9*a^2*b^2*(101*A+143*C))*sec(d*x+c)^(3/2)*sin(d*x+c)/d+4/3465*a*b*(96*A*b^2+
a^2*(673*A+891*C))*sec(d*x+c)^(5/2)*sin(d*x+c)/d+2/231*(16*A*b^2+3*a^2*(9*A+11*C))*(a+b*cos(d*x+c))^2*sec(d*x+
c)^(7/2)*sin(d*x+c)/d+16/99*A*b*(a+b*cos(d*x+c))^3*sec(d*x+c)^(9/2)*sin(d*x+c)/d+2/11*A*(a+b*cos(d*x+c))^4*sec
(d*x+c)^(11/2)*sin(d*x+c)/d+8/15*a*b*(3*b^2*(3*A+5*C)+a^2*(7*A+9*C))*sin(d*x+c)*sec(d*x+c)^(1/2)/d-8/15*a*b*(3
*b^2*(3*A+5*C)+a^2*(7*A+9*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+2/231*(77*b^4*(A+3*C)+66*a^2*b^2*(5*A+7*C)+5*a^4*(9*A+11*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]  time = 1.38, antiderivative size = 417, 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, 3048, 3047, 3031, 3021, 2748, 2636, 2639, 2641} \[ \frac {4 a b \left (a^2 (673 A+891 C)+96 A b^2\right ) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3465 d}+\frac {2 \left (9 a^2 b^2 (101 A+143 C)+15 a^4 (9 A+11 C)+64 A b^4\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{693 d}+\frac {8 a b \left (a^2 (7 A+9 C)+3 b^2 (3 A+5 C)\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{15 d}+\frac {2 \left (3 a^2 (9 A+11 C)+16 A b^2\right ) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))^2}{231 d}+\frac {2 \left (66 a^2 b^2 (5 A+7 C)+5 a^4 (9 A+11 C)+77 b^4 (A+3 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{231 d}-\frac {8 a b \left (a^2 (7 A+9 C)+3 b^2 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 A \sin (c+d x) \sec ^{\frac {11}{2}}(c+d x) (a+b \cos (c+d x))^4}{11 d}+\frac {16 A b \sin (c+d x) \sec ^{\frac {9}{2}}(c+d x) (a+b \cos (c+d x))^3}{99 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^4*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(13/2),x]

[Out]

(-8*a*b*(3*b^2*(3*A + 5*C) + a^2*(7*A + 9*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])
/(15*d) + (2*(77*b^4*(A + 3*C) + 66*a^2*b^2*(5*A + 7*C) + 5*a^4*(9*A + 11*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c
+ d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(231*d) + (8*a*b*(3*b^2*(3*A + 5*C) + a^2*(7*A + 9*C))*Sqrt[Sec[c + d*x]]*Sin
[c + d*x])/(15*d) + (2*(64*A*b^4 + 15*a^4*(9*A + 11*C) + 9*a^2*b^2*(101*A + 143*C))*Sec[c + d*x]^(3/2)*Sin[c +
d*x])/(693*d) + (4*a*b*(96*A*b^2 + a^2*(673*A + 891*C))*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(3465*d) + (2*(16*A*
b^2 + 3*a^2*(9*A + 11*C))*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(231*d) + (16*A*b*(a + b*Cos
[c + d*x])^3*Sec[c + d*x]^(9/2)*Sin[c + d*x])/(99*d) + (2*A*(a + b*Cos[c + d*x])^4*Sec[c + d*x]^(11/2)*Sin[c +
d*x])/(11*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 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 3031

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] && Ne
Q[a^2 - b^2, 0] && LtQ[m, -1]

Rule 3047

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C - B*c*d + 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/(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*d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*c*
d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] + b*(d*(B*c - A*d)*(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, B, 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 3048

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/(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 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+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {13}{2}}(c+d x) \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {13}{2}}(c+d x)} \, dx\\ &=\frac {2 A (a+b \cos (c+d x))^4 \sec ^{\frac {11}{2}}(c+d x) \sin (c+d x)}{11 d}+\frac {1}{11} \left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+b \cos (c+d x))^3 \left (4 A b+\frac {1}{2} a (9 A+11 C) \cos (c+d x)+\frac {1}{2} b (A+11 C) \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx\\ &=\frac {16 A b (a+b \cos (c+d x))^3 \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{99 d}+\frac {2 A (a+b \cos (c+d x))^4 \sec ^{\frac {11}{2}}(c+d x) \sin (c+d x)}{11 d}+\frac {1}{99} \left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+b \cos (c+d x))^2 \left (\frac {3}{4} \left (16 A b^2+3 a^2 (9 A+11 C)\right )+\frac {1}{2} a b (73 A+99 C) \cos (c+d x)+\frac {1}{4} b^2 (17 A+99 C) \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\\ &=\frac {2 \left (16 A b^2+3 a^2 (9 A+11 C)\right ) (a+b \cos (c+d x))^2 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{231 d}+\frac {16 A b (a+b \cos (c+d x))^3 \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{99 d}+\frac {2 A (a+b \cos (c+d x))^4 \sec ^{\frac {11}{2}}(c+d x) \sin (c+d x)}{11 d}+\frac {1}{693} \left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+b \cos (c+d x)) \left (\frac {1}{4} b \left (96 A b^2+a^2 (673 A+891 C)\right )+\frac {1}{8} a \left (45 a^2 (9 A+11 C)+b^2 (1381 A+2079 C)\right ) \cos (c+d x)+\frac {1}{8} b \left (9 a^2 (9 A+11 C)+b^2 (167 A+693 C)\right ) \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\\ &=\frac {4 a b \left (96 A b^2+a^2 (673 A+891 C)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3465 d}+\frac {2 \left (16 A b^2+3 a^2 (9 A+11 C)\right ) (a+b \cos (c+d x))^2 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{231 d}+\frac {16 A b (a+b \cos (c+d x))^3 \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{99 d}+\frac {2 A (a+b \cos (c+d x))^4 \sec ^{\frac {11}{2}}(c+d x) \sin (c+d x)}{11 d}-\frac {\left (16 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {15}{16} \left (64 A b^4+15 a^4 (9 A+11 C)+9 a^2 b^2 (101 A+143 C)\right )-\frac {231}{4} a b \left (3 b^2 (3 A+5 C)+a^2 (7 A+9 C)\right ) \cos (c+d x)-\frac {5}{16} b^2 \left (9 a^2 (9 A+11 C)+b^2 (167 A+693 C)\right ) \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x)} \, dx}{3465}\\ &=\frac {2 \left (64 A b^4+15 a^4 (9 A+11 C)+9 a^2 b^2 (101 A+143 C)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{693 d}+\frac {4 a b \left (96 A b^2+a^2 (673 A+891 C)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3465 d}+\frac {2 \left (16 A b^2+3 a^2 (9 A+11 C)\right ) (a+b \cos (c+d x))^2 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{231 d}+\frac {16 A b (a+b \cos (c+d x))^3 \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{99 d}+\frac {2 A (a+b \cos (c+d x))^4 \sec ^{\frac {11}{2}}(c+d x) \sin (c+d x)}{11 d}-\frac {\left (32 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {693}{8} a b \left (3 b^2 (3 A+5 C)+a^2 (7 A+9 C)\right )-\frac {45}{32} \left (77 b^4 (A+3 C)+66 a^2 b^2 (5 A+7 C)+5 a^4 (9 A+11 C)\right ) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{10395}\\ &=\frac {2 \left (64 A b^4+15 a^4 (9 A+11 C)+9 a^2 b^2 (101 A+143 C)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{693 d}+\frac {4 a b \left (96 A b^2+a^2 (673 A+891 C)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3465 d}+\frac {2 \left (16 A b^2+3 a^2 (9 A+11 C)\right ) (a+b \cos (c+d x))^2 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{231 d}+\frac {16 A b (a+b \cos (c+d x))^3 \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{99 d}+\frac {2 A (a+b \cos (c+d x))^4 \sec ^{\frac {11}{2}}(c+d x) \sin (c+d x)}{11 d}+\frac {1}{15} \left (4 a b \left (3 b^2 (3 A+5 C)+a^2 (7 A+9 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)} \, dx+\frac {1}{231} \left (\left (77 b^4 (A+3 C)+66 a^2 b^2 (5 A+7 C)+5 a^4 (9 A+11 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 (77 b^4 (A+3 C)+66 a^2 b^2 (5 A+7 C)+5 a^4 (9 A+11 C)\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{231 d}+\frac {8 a b \left (3 b^2 (3 A+5 C)+a^2 (7 A+9 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 \left (64 A b^4+15 a^4 (9 A+11 C)+9 a^2 b^2 (101 A+143 C)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{693 d}+\frac {4 a b \left (96 A b^2+a^2 (673 A+891 C)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3465 d}+\frac {2 \left (16 A b^2+3 a^2 (9 A+11 C)\right ) (a+b \cos (c+d x))^2 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{231 d}+\frac {16 A b (a+b \cos (c+d x))^3 \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{99 d}+\frac {2 A (a+b \cos (c+d x))^4 \sec ^{\frac {11}{2}}(c+d x) \sin (c+d x)}{11 d}-\frac {1}{15} \left (4 a b \left (3 b^2 (3 A+5 C)+a^2 (7 A+9 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=-\frac {8 a b \left (3 b^2 (3 A+5 C)+a^2 (7 A+9 C)\right ) \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 {2 \left (77 b^4 (A+3 C)+66 a^2 b^2 (5 A+7 C)+5 a^4 (9 A+11 C)\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{231 d}+\frac {8 a b \left (3 b^2 (3 A+5 C)+a^2 (7 A+9 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 \left (64 A b^4+15 a^4 (9 A+11 C)+9 a^2 b^2 (101 A+143 C)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{693 d}+\frac {4 a b \left (96 A b^2+a^2 (673 A+891 C)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3465 d}+\frac {2 \left (16 A b^2+3 a^2 (9 A+11 C)\right ) (a+b \cos (c+d x))^2 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{231 d}+\frac {16 A b (a+b \cos (c+d x))^3 \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{99 d}+\frac {2 A (a+b \cos (c+d x))^4 \sec ^{\frac {11}{2}}(c+d x) \sin (c+d x)}{11 d}\\ \end {align*}

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Mathematica [A]  time = 6.88, size = 425, normalized size = 1.02 \[ \frac {\sqrt {\sec (c+d x)} \left (\frac {2}{11} a^4 A \tan (c+d x) \sec ^4(c+d x)+\frac {8}{45} \sec ^2(c+d x) \left (7 a^3 A b \sin (c+d x)+9 a^3 b C \sin (c+d x)+9 a A b^3 \sin (c+d x)\right )+\frac {8}{9} a^3 A b \tan (c+d x) \sec ^3(c+d x)+\frac {8}{15} a b \left (7 a^2 A+9 a^2 C+9 A b^2+15 b^2 C\right ) \sin (c+d x)+\frac {2}{77} \sec ^3(c+d x) \left (9 a^4 A \sin (c+d x)+11 a^4 C \sin (c+d x)+66 a^2 A b^2 \sin (c+d x)\right )+\frac {2}{231} \sec (c+d x) \left (45 a^4 A \sin (c+d x)+55 a^4 C \sin (c+d x)+330 a^2 A b^2 \sin (c+d x)+462 a^2 b^2 C \sin (c+d x)+77 A b^4 \sin (c+d x)\right )\right )}{d}+\frac {\frac {2 \left (-2156 a^3 A b-2772 a^3 b C-2772 a A b^3-4620 a b^3 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}+2 \left (225 a^4 A+275 a^4 C+1650 a^2 A b^2+2310 a^2 b^2 C+385 A b^4+1155 b^4 C\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{1155 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^4*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(13/2),x]

[Out]

((2*(-2156*a^3*A*b - 2772*a*A*b^3 - 2772*a^3*b*C - 4620*a*b^3*C)*EllipticE[(c + d*x)/2, 2])/(Sqrt[Cos[c + d*x]
]*Sqrt[Sec[c + d*x]]) + 2*(225*a^4*A + 1650*a^2*A*b^2 + 385*A*b^4 + 275*a^4*C + 2310*a^2*b^2*C + 1155*b^4*C)*S
qrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(1155*d) + (Sqrt[Sec[c + d*x]]*((8*a*b*(7*a^2*
A + 9*A*b^2 + 9*a^2*C + 15*b^2*C)*Sin[c + d*x])/15 + (2*Sec[c + d*x]^3*(9*a^4*A*Sin[c + d*x] + 66*a^2*A*b^2*Si
n[c + d*x] + 11*a^4*C*Sin[c + d*x]))/77 + (8*Sec[c + d*x]^2*(7*a^3*A*b*Sin[c + d*x] + 9*a*A*b^3*Sin[c + d*x] +
9*a^3*b*C*Sin[c + d*x]))/45 + (2*Sec[c + d*x]*(45*a^4*A*Sin[c + d*x] + 330*a^2*A*b^2*Sin[c + d*x] + 77*A*b^4*
Sin[c + d*x] + 55*a^4*C*Sin[c + d*x] + 462*a^2*b^2*C*Sin[c + d*x]))/231 + (8*a^3*A*b*Sec[c + d*x]^3*Tan[c + d*
x])/9 + (2*a^4*A*Sec[c + d*x]^4*Tan[c + d*x])/11))/d

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fricas [F]  time = 0.94, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C b^{4} \cos \left (d x + c\right )^{6} + 4 \, C a b^{3} \cos \left (d x + c\right )^{5} + 4 \, A a^{3} b \cos \left (d x + c\right ) + A a^{4} + {\left (6 \, C a^{2} b^{2} + A b^{4}\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (C a^{3} b + A a b^{3}\right )} \cos \left (d x + c\right )^{3} + {\left (C a^{4} + 6 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sec \left (d x + c\right )^{\frac {13}{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(13/2),x, algorithm="fricas")

[Out]

integral((C*b^4*cos(d*x + c)^6 + 4*C*a*b^3*cos(d*x + c)^5 + 4*A*a^3*b*cos(d*x + c) + A*a^4 + (6*C*a^2*b^2 + A*
b^4)*cos(d*x + c)^4 + 4*(C*a^3*b + A*a*b^3)*cos(d*x + c)^3 + (C*a^4 + 6*A*a^2*b^2)*cos(d*x + c)^2)*sec(d*x + c
)^(13/2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{4} \sec \left (d x + c\right )^{\frac {13}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(13/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)^4*sec(d*x + c)^(13/2), x)

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maple [B]  time = 16.50, size = 1521, normalized size = 3.65 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(13/2),x)

[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^4*(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^2*(6*A*b^2+C*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)/(-1/2+c
os(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)))+2*b^2*(A*b^2+6*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)/(-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)))+8*A*a^3*b*(-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))))-8/5*a*b*(A*b^2+C*a^2)/(8*sin(1/2*d*x+1/2*c)^6-12*s
in(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*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)+8*C*a*b^3*(-(-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)+2*A*a^4*(-1/352*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)^6-9/616*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)^4-15/154*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+15/77*(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*co
s(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 \[ \int {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{4} \sec \left (d x + c\right )^{\frac {13}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(13/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)^4*sec(d*x + c)^(13/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{13/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^4 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(13/2)*(a + b*cos(c + d*x))^4,x)

[Out]

int((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(13/2)*(a + b*cos(c + d*x))^4, 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+b*cos(d*x+c))**4*(A+C*cos(d*x+c)**2)*sec(d*x+c)**(13/2),x)

[Out]

Timed out

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