3.1384 \(\int (a+b \cos (c+d x))^4 (A+C \cos ^2(c+d x)) \sec ^{\frac{11}{2}}(c+d x) \, dx\)

Optimal. Leaf size=365 \[ \frac{4 a b \left (a^2 (101 A+147 C)+32 A b^2\right ) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{315 d}+\frac{2 \left (7 a^2 b^2 (155 A+261 C)+21 a^4 (7 A+9 C)+192 A b^4\right ) \sin (c+d x) \sqrt{\sec (c+d x)}}{315 d}+\frac{2 \left (7 a^2 (7 A+9 C)+48 A b^2\right ) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x) (a+b \cos (c+d x))^2}{315 d}+\frac{8 a b \left (a^2 (5 A+7 C)+7 b^2 (A+3 C)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}-\frac{2 \left (18 a^2 b^2 (3 A+5 C)+a^4 (7 A+9 C)+15 b^4 (A-C)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{2 A \sin (c+d x) \sec ^{\frac{9}{2}}(c+d x) (a+b \cos (c+d x))^4}{9 d}+\frac{16 A b \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x) (a+b \cos (c+d x))^3}{63 d} \]

[Out]

(-2*(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]]*EllipticE[(c + d*x)/2, 2]*S
qrt[Sec[c + d*x]])/(15*d) + (8*a*b*(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) + (2*(192*A*b^4 + 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) + (4*a*b*(32*A*b^2 + a^2*(101*A + 147*C))*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(315*
d) + (2*(48*A*b^2 + 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) + (16*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)

________________________________________________________________________________________

Rubi [A]  time = 1.27745, antiderivative size = 365, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.229, Rules used = {4221, 3048, 3047, 3031, 3021, 2748, 2641, 2639} \[ \frac{4 a b \left (a^2 (101 A+147 C)+32 A b^2\right ) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{315 d}+\frac{2 \left (7 a^2 b^2 (155 A+261 C)+21 a^4 (7 A+9 C)+192 A b^4\right ) \sin (c+d x) \sqrt{\sec (c+d x)}}{315 d}+\frac{2 \left (7 a^2 (7 A+9 C)+48 A b^2\right ) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x) (a+b \cos (c+d x))^2}{315 d}+\frac{8 a b \left (a^2 (5 A+7 C)+7 b^2 (A+3 C)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}-\frac{2 \left (18 a^2 b^2 (3 A+5 C)+a^4 (7 A+9 C)+15 b^4 (A-C)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{2 A \sin (c+d x) \sec ^{\frac{9}{2}}(c+d x) (a+b \cos (c+d x))^4}{9 d}+\frac{16 A b \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x) (a+b \cos (c+d x))^3}{63 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(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]]*EllipticE[(c + d*x)/2, 2]*S
qrt[Sec[c + d*x]])/(15*d) + (8*a*b*(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) + (2*(192*A*b^4 + 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) + (4*a*b*(32*A*b^2 + a^2*(101*A + 147*C))*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(315*
d) + (2*(48*A*b^2 + 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) + (16*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 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]

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 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 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 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 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 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 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]

Rubi steps

\begin{align*} \int (a+b \cos (c+d x))^4 \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+b \cos (c+d x))^4 \left (A+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 (4 A b+\frac{1}{2} a (7 A+9 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{16 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+7 a^2 (7 A+9 C)\right )+\frac{1}{2} a b (41 A+63 C) \cos (c+d x)-\frac{3}{4} b^2 (5 A-21 C) \cos ^2(c+d x)\right )}{\cos ^{\frac{7}{2}}(c+d x)} \, dx\\ &=\frac{2 \left (48 A b^2+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{16 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}{4} b \left (32 A b^2+a^2 (101 A+147 C)\right )+\frac{1}{8} a \left (21 a^2 (7 A+9 C)+b^2 (479 A+945 C)\right ) \cos (c+d x)-\frac{1}{8} b \left (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{4 a b \left (32 A b^2+a^2 (101 A+147 C)\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac{2 \left (48 A b^2+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{16 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+21 a^4 (7 A+9 C)+7 a^2 b^2 (155 A+261 C)\right )-\frac{45}{4} a b \left (7 b^2 (A+3 C)+a^2 (5 A+7 C)\right ) \cos (c+d x)+\frac{3}{16} b^2 \left (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+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{4 a b \left (32 A b^2+a^2 (101 A+147 C)\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac{2 \left (48 A b^2+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{16 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}{8} a b \left (7 b^2 (A+3 C)+a^2 (5 A+7 C)\right )+\frac{63}{32} \left (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+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{4 a b \left (32 A b^2+a^2 (101 A+147 C)\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac{2 \left (48 A b^2+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{16 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 (4 a b \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{1}{15} \left (\left (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 (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{8 a b \left (7 b^2 (A+3 C)+a^2 (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+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{4 a b \left (32 A b^2+a^2 (101 A+147 C)\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac{2 \left (48 A b^2+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{16 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*}

Mathematica [A]  time = 6.74148, size = 356, normalized size = 0.98 \[ \frac{\sqrt{\sec (c+d x)} \left (\frac{2}{15} \left (54 a^2 A b^2+7 a^4 A+90 a^2 b^2 C+9 a^4 C+15 A b^4\right ) \sin (c+d x)+\frac{2}{45} \sec ^2(c+d x) \left (54 a^2 A b^2 \sin (c+d x)+7 a^4 A \sin (c+d x)+9 a^4 C \sin (c+d x)\right )+\frac{8}{21} \sec (c+d x) \left (5 a^3 A b \sin (c+d x)+7 a^3 b C \sin (c+d x)+7 a A b^3 \sin (c+d x)\right )+\frac{8}{7} a^3 A b \tan (c+d x) \sec ^2(c+d x)+\frac{2}{9} a^4 A \tan (c+d x) \sec ^3(c+d x)\right )}{d}+\frac{2 \left (100 a^3 A b+140 a^3 b C+140 a A b^3+420 a b^3 C\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+\frac{2 \left (-378 a^2 A b^2-49 a^4 A-630 a^2 b^2 C-63 a^4 C-105 A b^4+105 b^4 C\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}}}{105 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^4*(A + 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 - 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 + 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 + 9*a^4*C + 90*a^2*b^2*C)*Sin[c + d*x])/15 + (2*Sec[c + d*x]^2*(7*a^4*A*Sin[c + d*x] + 54*a^2*A*
b^2*Sin[c + d*x] + 9*a^4*C*Sin[c + d*x]))/45 + (8*Sec[c + d*x]*(5*a^3*A*b*Sin[c + d*x] + 7*a*A*b^3*Sin[c + d*x
] + 7*a^3*b*C*Sin[c + d*x]))/21 + (8*a^3*A*b*Sec[c + d*x]^2*Tan[c + d*x])/7 + (2*a^4*A*Sec[c + d*x]^3*Tan[c +
d*x])/9))/d

________________________________________________________________________________________

Maple [B]  time = 6.142, size = 1451, normalized size = 4. \begin{align*} \text{result too large to display} \end{align*}

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)^(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)))+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))+8*a*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+si
n(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*(-1/56*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)))+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+s
in(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*c
os(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))))-2/5*a^2*(6*A*b^2+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))*(sin(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)^4-24*sin
(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12*EllipticE(cos(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)*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*EllipticE(cos(
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)-8*sin(1/2*d*x+1/2*c)^2*c
os(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*b^2*(A*b^2+6*C*a^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)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(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))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^
(1/2)/d

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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{11}{2}}\,{d x} \end{align*}

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)^(11/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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{11}{2}}, x\right ) \end{align*}

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)^(11/2),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
)^(11/2), x)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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)**(11/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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{11}{2}}\,{d x} \end{align*}

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)^(11/2),x, algorithm="giac")

[Out]

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