3.1475 \(\int (a+b \cos (c+d x))^4 (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac {11}{2}}(c+d x) \, dx\)

Optimal. Leaf size=441 \[ \frac {2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (7 a^2 (7 A+9 C)+117 a b B+48 A b^2\right ) (a+b \cos (c+d x))^2}{315 d}+\frac {2 a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (75 a^3 B+a^2 (202 A b+294 b C)+261 a b^2 B+64 A b^3\right )}{315 d}+\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)} \left (21 a^4 (7 A+9 C)+756 a^3 b B+7 a^2 b^2 (155 A+261 C)+1098 a b^3 B+192 A b^4\right )}{315 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^4 B+4 a^3 b (5 A+7 C)+42 a^2 b^2 B+28 a b^3 (A+3 C)+21 b^4 B\right )}{21 d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^4 (7 A+9 C)+36 a^3 b B+18 a^2 b^2 (3 A+5 C)+60 a b^3 B+15 b^4 (A-C)\right )}{15 d}+\frac {2 (9 a B+8 A b) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))^3}{63 d}+\frac {2 A \sin (c+d x) \sec ^{\frac {9}{2}}(c+d x) (a+b \cos (c+d x))^4}{9 d} \]

[Out]

2/315*a*(64*A*b^3+75*a^3*B+261*a*b^2*B+a^2*(202*A*b+294*C*b))*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/315*(48*A*b^2+11
7*a*b*B+7*a^2*(7*A+9*C))*(a+b*cos(d*x+c))^2*sec(d*x+c)^(5/2)*sin(d*x+c)/d+2/63*(8*A*b+9*B*a)*(a+b*cos(d*x+c))^
3*sec(d*x+c)^(7/2)*sin(d*x+c)/d+2/9*A*(a+b*cos(d*x+c))^4*sec(d*x+c)^(9/2)*sin(d*x+c)/d+2/315*(192*A*b^4+756*a^
3*b*B+1098*a*b^3*B+21*a^4*(7*A+9*C)+7*a^2*b^2*(155*A+261*C))*sin(d*x+c)*sec(d*x+c)^(1/2)/d-2/15*(36*a^3*b*B+60
*a*b^3*B+15*b^4*(A-C)+18*a^2*b^2*(3*A+5*C)+a^4*(7*A+9*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Elli
pticE(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*(5*a^4*B+42*a^2*b^2*B+21*b^4*B+28*a
*b^3*(A+3*C)+4*a^3*b*(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

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Rubi [A]  time = 1.49, antiderivative size = 441, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {4221, 3047, 3031, 3021, 2748, 2641, 2639} \[ \frac {2 a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (a^2 (202 A b+294 b C)+75 a^3 B+261 a b^2 B+64 A b^3\right )}{315 d}+\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)} \left (7 a^2 b^2 (155 A+261 C)+21 a^4 (7 A+9 C)+756 a^3 b B+1098 a b^3 B+192 A b^4\right )}{315 d}+\frac {2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (7 a^2 (7 A+9 C)+117 a b B+48 A b^2\right ) (a+b \cos (c+d x))^2}{315 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (4 a^3 b (5 A+7 C)+42 a^2 b^2 B+5 a^4 B+28 a b^3 (A+3 C)+21 b^4 B\right )}{21 d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (18 a^2 b^2 (3 A+5 C)+a^4 (7 A+9 C)+36 a^3 b B+60 a b^3 B+15 b^4 (A-C)\right )}{15 d}+\frac {2 (9 a B+8 A b) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))^3}{63 d}+\frac {2 A \sin (c+d x) \sec ^{\frac {9}{2}}(c+d x) (a+b \cos (c+d x))^4}{9 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^4*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(11/2),x]

[Out]

(-2*(36*a^3*b*B + 60*a*b^3*B + 15*b^4*(A - C) + 18*a^2*b^2*(3*A + 5*C) + a^4*(7*A + 9*C))*Sqrt[Cos[c + d*x]]*E
llipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(15*d) + (2*(5*a^4*B + 42*a^2*b^2*B + 21*b^4*B + 28*a*b^3*(A + 3*
C) + 4*a^3*b*(5*A + 7*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (2*(192*A*
b^4 + 756*a^3*b*B + 1098*a*b^3*B + 21*a^4*(7*A + 9*C) + 7*a^2*b^2*(155*A + 261*C))*Sqrt[Sec[c + d*x]]*Sin[c +
d*x])/(315*d) + (2*a*(64*A*b^3 + 75*a^3*B + 261*a*b^2*B + a^2*(202*A*b + 294*b*C))*Sec[c + d*x]^(3/2)*Sin[c +
d*x])/(315*d) + (2*(48*A*b^2 + 117*a*b*B + 7*a^2*(7*A + 9*C))*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^(5/2)*Sin[c
+ d*x])/(315*d) + (2*(8*A*b + 9*a*B)*(a + b*Cos[c + d*x])^3*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(63*d) + (2*A*(a
+ b*Cos[c + d*x])^4*Sec[c + d*x]^(9/2)*Sin[c + d*x])/(9*d)

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 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+B \cos (c+d x)+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+b \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx\\ &=\frac {2 A (a+b \cos (c+d x))^4 \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac {1}{9} \left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+b \cos (c+d x))^3 \left (\frac {1}{2} (8 A b+9 a B)+\frac {1}{2} (7 a A+9 b B+9 a C) \cos (c+d x)-\frac {1}{2} b (A-9 C) \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\\ &=\frac {2 (8 A b+9 a B) (a+b \cos (c+d x))^3 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 A (a+b \cos (c+d x))^4 \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac {1}{63} \left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+b \cos (c+d x))^2 \left (\frac {1}{4} \left (48 A b^2+117 a b B+7 a^2 (7 A+9 C)\right )+\frac {1}{4} \left (82 a A b+45 a^2 B+63 b^2 B+126 a b C\right ) \cos (c+d x)-\frac {3}{4} b (5 A b+3 a B-21 b C) \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\\ &=\frac {2 \left (48 A b^2+117 a b B+7 a^2 (7 A+9 C)\right ) (a+b \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac {2 (8 A b+9 a B) (a+b \cos (c+d x))^3 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 A (a+b \cos (c+d x))^4 \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac {1}{315} \left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+b \cos (c+d x)) \left (\frac {3}{8} \left (64 A b^3+75 a^3 B+261 a b^2 B+a^2 (202 A b+294 b C)\right )+\frac {1}{8} \left (531 a^2 b B+315 b^3 B+21 a^3 (7 A+9 C)+a b^2 (479 A+945 C)\right ) \cos (c+d x)-\frac {1}{8} b \left (162 a b B+3 b^2 (41 A-105 C)+7 a^2 (7 A+9 C)\right ) \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 a \left (64 A b^3+75 a^3 B+261 a b^2 B+a^2 (202 A b+294 b C)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac {2 \left (48 A b^2+117 a b B+7 a^2 (7 A+9 C)\right ) (a+b \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac {2 (8 A b+9 a B) (a+b \cos (c+d x))^3 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 A (a+b \cos (c+d x))^4 \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d}-\frac {1}{945} \left (16 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {3}{16} \left (192 A b^4+756 a^3 b B+1098 a b^3 B+21 a^4 (7 A+9 C)+7 a^2 b^2 (155 A+261 C)\right )-\frac {45}{16} \left (5 a^4 B+42 a^2 b^2 B+21 b^4 B+28 a b^3 (A+3 C)+4 a^3 b (5 A+7 C)\right ) \cos (c+d x)+\frac {3}{16} b^2 \left (162 a b B+3 b^2 (41 A-105 C)+7 a^2 (7 A+9 C)\right ) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 \left (192 A b^4+756 a^3 b B+1098 a b^3 B+21 a^4 (7 A+9 C)+7 a^2 b^2 (155 A+261 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{315 d}+\frac {2 a \left (64 A b^3+75 a^3 B+261 a b^2 B+a^2 (202 A b+294 b C)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac {2 \left (48 A b^2+117 a b B+7 a^2 (7 A+9 C)\right ) (a+b \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac {2 (8 A b+9 a B) (a+b \cos (c+d x))^3 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 A (a+b \cos (c+d x))^4 \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d}-\frac {1}{945} \left (32 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {45}{32} \left (5 a^4 B+42 a^2 b^2 B+21 b^4 B+28 a b^3 (A+3 C)+4 a^3 b (5 A+7 C)\right )+\frac {63}{32} \left (36 a^3 b B+60 a b^3 B+15 b^4 (A-C)+18 a^2 b^2 (3 A+5 C)+a^4 (7 A+9 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 \left (192 A b^4+756 a^3 b B+1098 a b^3 B+21 a^4 (7 A+9 C)+7 a^2 b^2 (155 A+261 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{315 d}+\frac {2 a \left (64 A b^3+75 a^3 B+261 a b^2 B+a^2 (202 A b+294 b C)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac {2 \left (48 A b^2+117 a b B+7 a^2 (7 A+9 C)\right ) (a+b \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac {2 (8 A b+9 a B) (a+b \cos (c+d x))^3 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 A (a+b \cos (c+d x))^4 \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac {1}{21} \left (\left (5 a^4 B+42 a^2 b^2 B+21 b^4 B+28 a b^3 (A+3 C)+4 a^3 b (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 {1}{15} \left (\left (36 a^3 b B+60 a b^3 B+15 b^4 (A-C)+18 a^2 b^2 (3 A+5 C)+a^4 (7 A+9 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=-\frac {2 \left (36 a^3 b B+60 a b^3 B+15 b^4 (A-C)+18 a^2 b^2 (3 A+5 C)+a^4 (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 (5 a^4 B+42 a^2 b^2 B+21 b^4 B+28 a b^3 (A+3 C)+4 a^3 b (5 A+7 C)\right ) \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 {2 \left (192 A b^4+756 a^3 b B+1098 a b^3 B+21 a^4 (7 A+9 C)+7 a^2 b^2 (155 A+261 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{315 d}+\frac {2 a \left (64 A b^3+75 a^3 B+261 a b^2 B+a^2 (202 A b+294 b C)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac {2 \left (48 A b^2+117 a b B+7 a^2 (7 A+9 C)\right ) (a+b \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac {2 (8 A b+9 a B) (a+b \cos (c+d x))^3 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 A (a+b \cos (c+d x))^4 \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d}\\ \end {align*}

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Mathematica [A]  time = 7.26, size = 459, normalized size = 1.04 \[ \frac {\sqrt {\sec (c+d x)} \left (\frac {2}{9} a^4 A \tan (c+d x) \sec ^3(c+d x)+\frac {2}{7} \sec ^3(c+d x) \left (a^4 B \sin (c+d x)+4 a^3 A b \sin (c+d x)\right )+\frac {2}{45} \sec ^2(c+d x) \left (7 a^4 A \sin (c+d x)+9 a^4 C \sin (c+d x)+36 a^3 b B \sin (c+d x)+54 a^2 A b^2 \sin (c+d x)\right )+\frac {2}{21} \sec (c+d x) \left (5 a^4 B \sin (c+d x)+20 a^3 A b \sin (c+d x)+28 a^3 b C \sin (c+d x)+42 a^2 b^2 B \sin (c+d x)+28 a A b^3 \sin (c+d x)\right )+\frac {2}{15} \sin (c+d x) \left (7 a^4 A+9 a^4 C+36 a^3 b B+54 a^2 A b^2+90 a^2 b^2 C+60 a b^3 B+15 A b^4\right )\right )}{d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (25 a^4 B+100 a^3 A b+140 a^3 b C+210 a^2 b^2 B+140 a A b^3+420 a b^3 C+105 b^4 B\right )+\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-49 a^4 A-63 a^4 C-252 a^3 b B-378 a^2 A b^2-630 a^2 b^2 C-420 a b^3 B-105 A b^4+105 b^4 C\right )}{\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}}{105 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^4*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(11/2),x]

[Out]

((2*(-49*a^4*A - 378*a^2*A*b^2 - 105*A*b^4 - 252*a^3*b*B - 420*a*b^3*B - 63*a^4*C - 630*a^2*b^2*C + 105*b^4*C)
*EllipticE[(c + d*x)/2, 2])/(Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + 2*(100*a^3*A*b + 140*a*A*b^3 + 25*a^4*B
+ 210*a^2*b^2*B + 105*b^4*B + 140*a^3*b*C + 420*a*b^3*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec
[c + d*x]])/(105*d) + (Sqrt[Sec[c + d*x]]*((2*(7*a^4*A + 54*a^2*A*b^2 + 15*A*b^4 + 36*a^3*b*B + 60*a*b^3*B + 9
*a^4*C + 90*a^2*b^2*C)*Sin[c + d*x])/15 + (2*Sec[c + d*x]^3*(4*a^3*A*b*Sin[c + d*x] + a^4*B*Sin[c + d*x]))/7 +
(2*Sec[c + d*x]^2*(7*a^4*A*Sin[c + d*x] + 54*a^2*A*b^2*Sin[c + d*x] + 36*a^3*b*B*Sin[c + d*x] + 9*a^4*C*Sin[c
+ d*x]))/45 + (2*Sec[c + d*x]*(20*a^3*A*b*Sin[c + d*x] + 28*a*A*b^3*Sin[c + d*x] + 5*a^4*B*Sin[c + d*x] + 42*
a^2*b^2*B*Sin[c + d*x] + 28*a^3*b*C*Sin[c + d*x]))/21 + (2*a^4*A*Sec[c + d*x]^3*Tan[c + d*x])/9))/d

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fricas [F]  time = 1.13, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C b^{4} \cos \left (d x + c\right )^{6} + {\left (4 \, C a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )^{5} + A a^{4} + {\left (6 \, C a^{2} b^{2} + 4 \, B a b^{3} + A b^{4}\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (2 \, C a^{3} b + 3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} \cos \left (d x + c\right )^{3} + {\left (C a^{4} + 4 \, B a^{3} b + 6 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right )\right )} \sec \left (d x + c\right )^{\frac {11}{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(11/2),x, algorithm="fricas")

[Out]

integral((C*b^4*cos(d*x + c)^6 + (4*C*a*b^3 + B*b^4)*cos(d*x + c)^5 + A*a^4 + (6*C*a^2*b^2 + 4*B*a*b^3 + A*b^4
)*cos(d*x + c)^4 + 2*(2*C*a^3*b + 3*B*a^2*b^2 + 2*A*a*b^3)*cos(d*x + c)^3 + (C*a^4 + 4*B*a^3*b + 6*A*a^2*b^2)*
cos(d*x + c)^2 + (B*a^4 + 4*A*a^3*b)*cos(d*x + c))*sec(d*x + c)^(11/2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{4} \sec \left (d x + c\right )^{\frac {11}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(11/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^4*sec(d*x + c)^(11/2), x)

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maple [B]  time = 16.95, size = 1550, normalized size = 3.51 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(11/2),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
))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+2*B*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))+8*C*a*b^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))-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))-2/5*a^2*(6*A*b^2+4*B
*a*b+C*a^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)/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)+2*A*a^4*(-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))))+2*b^2*(A*b^2+4*B*a*b+6*C*a^2)*(-(-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^3*(4*A*b+B*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+co
s(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)))+4*a*b*(2*A*b^2+3*B*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)/(-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))))/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 \[ \int {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{4} \sec \left (d x + c\right )^{\frac {11}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(11/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^4*sec(d*x + c)^(11/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{11/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^4\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/cos(c + d*x))^(11/2)*(a + b*cos(c + d*x))^4*(A + B*cos(c + d*x) + C*cos(c + d*x)^2),x)

[Out]

int((1/cos(c + d*x))^(11/2)*(a + b*cos(c + d*x))^4*(A + B*cos(c + d*x) + C*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+b*cos(d*x+c))**4*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**(11/2),x)

[Out]

Timed out

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