3.1086 \(\int \frac{(a+b \cos (c+d x))^3 (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\cos ^{\frac{11}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=357 \[ \frac{2 F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (3 a^2 b (5 A+7 C)+5 a^3 B+21 a b^2 B+7 b^3 (A+3 C)\right )}{21 d}-\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^3 (7 A+9 C)+27 a^2 b B+9 a b^2 (3 A+5 C)+15 b^3 B\right )}{15 d}+\frac{2 \sin (c+d x) \left (9 a^2 b (5 A+7 C)+15 a^3 B+54 a b^2 B+8 A b^3\right )}{63 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 a \sin (c+d x) \left (7 a^2 (7 A+9 C)+99 a b B+24 A b^2\right )}{315 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 \sin (c+d x) \left (a^3 (7 A+9 C)+27 a^2 b B+9 a b^2 (3 A+5 C)+15 b^3 B\right )}{15 d \sqrt{\cos (c+d x)}}+\frac{2 (3 a B+2 A b) \sin (c+d x) (a+b \cos (c+d x))^2}{21 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a+b \cos (c+d x))^3}{9 d \cos ^{\frac{9}{2}}(c+d x)} \]

[Out]

(-2*(27*a^2*b*B + 15*b^3*B + 9*a*b^2*(3*A + 5*C) + a^3*(7*A + 9*C))*EllipticE[(c + d*x)/2, 2])/(15*d) + (2*(5*
a^3*B + 21*a*b^2*B + 7*b^3*(A + 3*C) + 3*a^2*b*(5*A + 7*C))*EllipticF[(c + d*x)/2, 2])/(21*d) + (2*a*(24*A*b^2
+ 99*a*b*B + 7*a^2*(7*A + 9*C))*Sin[c + d*x])/(315*d*Cos[c + d*x]^(5/2)) + (2*(8*A*b^3 + 15*a^3*B + 54*a*b^2*
B + 9*a^2*b*(5*A + 7*C))*Sin[c + d*x])/(63*d*Cos[c + d*x]^(3/2)) + (2*(27*a^2*b*B + 15*b^3*B + 9*a*b^2*(3*A +
5*C) + a^3*(7*A + 9*C))*Sin[c + d*x])/(15*d*Sqrt[Cos[c + d*x]]) + (2*(2*A*b + 3*a*B)*(a + b*Cos[c + d*x])^2*Si
n[c + d*x])/(21*d*Cos[c + d*x]^(7/2)) + (2*A*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(9*d*Cos[c + d*x]^(9/2))

________________________________________________________________________________________

Rubi [A]  time = 0.923861, antiderivative size = 357, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.163, Rules used = {3047, 3031, 3021, 2748, 2636, 2639, 2641} \[ \frac{2 F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (3 a^2 b (5 A+7 C)+5 a^3 B+21 a b^2 B+7 b^3 (A+3 C)\right )}{21 d}-\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^3 (7 A+9 C)+27 a^2 b B+9 a b^2 (3 A+5 C)+15 b^3 B\right )}{15 d}+\frac{2 \sin (c+d x) \left (9 a^2 b (5 A+7 C)+15 a^3 B+54 a b^2 B+8 A b^3\right )}{63 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 a \sin (c+d x) \left (7 a^2 (7 A+9 C)+99 a b B+24 A b^2\right )}{315 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 \sin (c+d x) \left (a^3 (7 A+9 C)+27 a^2 b B+9 a b^2 (3 A+5 C)+15 b^3 B\right )}{15 d \sqrt{\cos (c+d x)}}+\frac{2 (3 a B+2 A b) \sin (c+d x) (a+b \cos (c+d x))^2}{21 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a+b \cos (c+d x))^3}{9 d \cos ^{\frac{9}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(27*a^2*b*B + 15*b^3*B + 9*a*b^2*(3*A + 5*C) + a^3*(7*A + 9*C))*EllipticE[(c + d*x)/2, 2])/(15*d) + (2*(5*
a^3*B + 21*a*b^2*B + 7*b^3*(A + 3*C) + 3*a^2*b*(5*A + 7*C))*EllipticF[(c + d*x)/2, 2])/(21*d) + (2*a*(24*A*b^2
+ 99*a*b*B + 7*a^2*(7*A + 9*C))*Sin[c + d*x])/(315*d*Cos[c + d*x]^(5/2)) + (2*(8*A*b^3 + 15*a^3*B + 54*a*b^2*
B + 9*a^2*b*(5*A + 7*C))*Sin[c + d*x])/(63*d*Cos[c + d*x]^(3/2)) + (2*(27*a^2*b*B + 15*b^3*B + 9*a*b^2*(3*A +
5*C) + a^3*(7*A + 9*C))*Sin[c + d*x])/(15*d*Sqrt[Cos[c + d*x]]) + (2*(2*A*b + 3*a*B)*(a + b*Cos[c + d*x])^2*Si
n[c + d*x])/(21*d*Cos[c + d*x]^(7/2)) + (2*A*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(9*d*Cos[c + d*x]^(9/2))

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 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]

Rubi steps

\begin{align*} \int \frac{(a+b \cos (c+d x))^3 \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))^3 \sin (c+d x)}{9 d \cos ^{\frac{9}{2}}(c+d x)}+\frac{2}{9} \int \frac{(a+b \cos (c+d x))^2 \left (\frac{3}{2} (2 A b+3 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 (2 A b+3 a B) (a+b \cos (c+d x))^2 \sin (c+d x)}{21 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{9 d \cos ^{\frac{9}{2}}(c+d x)}+\frac{4}{63} \int \frac{(a+b \cos (c+d x)) \left (\frac{1}{4} \left (24 A b^2+99 a b B+7 a^2 (7 A+9 C)\right )+\frac{1}{4} \left (86 a A b+45 a^2 B+63 b^2 B+126 a b C\right ) \cos (c+d x)+\frac{1}{4} b (13 A b+9 a B+63 b C) \cos ^2(c+d x)\right )}{\cos ^{\frac{7}{2}}(c+d x)} \, dx\\ &=\frac{2 a \left (24 A b^2+99 a b B+7 a^2 (7 A+9 C)\right ) \sin (c+d x)}{315 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 (2 A b+3 a B) (a+b \cos (c+d x))^2 \sin (c+d x)}{21 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{9 d \cos ^{\frac{9}{2}}(c+d x)}-\frac{8}{315} \int \frac{-\frac{15}{8} \left (8 A b^3+15 a^3 B+54 a b^2 B+9 a^2 b (5 A+7 C)\right )-\frac{21}{8} \left (27 a^2 b B+15 b^3 B+9 a b^2 (3 A+5 C)+a^3 (7 A+9 C)\right ) \cos (c+d x)-\frac{5}{8} b^2 (13 A b+9 a B+63 b C) \cos ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 a \left (24 A b^2+99 a b B+7 a^2 (7 A+9 C)\right ) \sin (c+d x)}{315 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (8 A b^3+15 a^3 B+54 a b^2 B+9 a^2 b (5 A+7 C)\right ) \sin (c+d x)}{63 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 (2 A b+3 a B) (a+b \cos (c+d x))^2 \sin (c+d x)}{21 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{9 d \cos ^{\frac{9}{2}}(c+d x)}-\frac{16}{945} \int \frac{-\frac{63}{16} \left (27 a^2 b B+15 b^3 B+9 a b^2 (3 A+5 C)+a^3 (7 A+9 C)\right )-\frac{45}{16} \left (5 a^3 B+21 a b^2 B+7 b^3 (A+3 C)+3 a^2 b (5 A+7 C)\right ) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 a \left (24 A b^2+99 a b B+7 a^2 (7 A+9 C)\right ) \sin (c+d x)}{315 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (8 A b^3+15 a^3 B+54 a b^2 B+9 a^2 b (5 A+7 C)\right ) \sin (c+d x)}{63 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 (2 A b+3 a B) (a+b \cos (c+d x))^2 \sin (c+d x)}{21 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{9 d \cos ^{\frac{9}{2}}(c+d x)}-\frac{1}{21} \left (-5 a^3 B-21 a b^2 B-7 b^3 (A+3 C)-3 a^2 b (5 A+7 C)\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx-\frac{1}{15} \left (-27 a^2 b B-15 b^3 B-9 a b^2 (3 A+5 C)-a^3 (7 A+9 C)\right ) \int \frac{1}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 \left (5 a^3 B+21 a b^2 B+7 b^3 (A+3 C)+3 a^2 b (5 A+7 C)\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{2 a \left (24 A b^2+99 a b B+7 a^2 (7 A+9 C)\right ) \sin (c+d x)}{315 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (8 A b^3+15 a^3 B+54 a b^2 B+9 a^2 b (5 A+7 C)\right ) \sin (c+d x)}{63 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (27 a^2 b B+15 b^3 B+9 a b^2 (3 A+5 C)+a^3 (7 A+9 C)\right ) \sin (c+d x)}{15 d \sqrt{\cos (c+d x)}}+\frac{2 (2 A b+3 a B) (a+b \cos (c+d x))^2 \sin (c+d x)}{21 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{9 d \cos ^{\frac{9}{2}}(c+d x)}-\frac{1}{15} \left (27 a^2 b B+15 b^3 B+9 a b^2 (3 A+5 C)+a^3 (7 A+9 C)\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=-\frac{2 \left (27 a^2 b B+15 b^3 B+9 a b^2 (3 A+5 C)+a^3 (7 A+9 C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{2 \left (5 a^3 B+21 a b^2 B+7 b^3 (A+3 C)+3 a^2 b (5 A+7 C)\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{2 a \left (24 A b^2+99 a b B+7 a^2 (7 A+9 C)\right ) \sin (c+d x)}{315 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (8 A b^3+15 a^3 B+54 a b^2 B+9 a^2 b (5 A+7 C)\right ) \sin (c+d x)}{63 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (27 a^2 b B+15 b^3 B+9 a b^2 (3 A+5 C)+a^3 (7 A+9 C)\right ) \sin (c+d x)}{15 d \sqrt{\cos (c+d x)}}+\frac{2 (2 A b+3 a B) (a+b \cos (c+d x))^2 \sin (c+d x)}{21 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{9 d \cos ^{\frac{9}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 6.90144, size = 414, normalized size = 1.16 \[ \frac{2 F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (75 a^2 A b+105 a^2 b C+25 a^3 B+105 a b^2 B+35 A b^3+105 b^3 C\right )+2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (-49 a^3 A-189 a^2 b B-63 a^3 C-189 a A b^2-315 a b^2 C-105 b^3 B\right )}{105 d}+\frac{\sqrt{\cos (c+d x)} \left (\frac{2}{45} \sec ^3(c+d x) \left (7 a^3 A \sin (c+d x)+27 a^2 b B \sin (c+d x)+9 a^3 C \sin (c+d x)+27 a A b^2 \sin (c+d x)\right )+\frac{2}{21} \sec ^2(c+d x) \left (15 a^2 A b \sin (c+d x)+21 a^2 b C \sin (c+d x)+5 a^3 B \sin (c+d x)+21 a b^2 B \sin (c+d x)+7 A b^3 \sin (c+d x)\right )+\frac{2}{15} \sec (c+d x) \left (7 a^3 A \sin (c+d x)+27 a^2 b B \sin (c+d x)+9 a^3 C \sin (c+d x)+27 a A b^2 \sin (c+d x)+45 a b^2 C \sin (c+d x)+15 b^3 B \sin (c+d x)\right )+\frac{2}{7} \sec ^4(c+d x) \left (3 a^2 A b \sin (c+d x)+a^3 B \sin (c+d x)\right )+\frac{2}{9} a^3 A \tan (c+d x) \sec ^4(c+d x)\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-49*a^3*A - 189*a*A*b^2 - 189*a^2*b*B - 105*b^3*B - 63*a^3*C - 315*a*b^2*C)*EllipticE[(c + d*x)/2, 2] + 2*
(75*a^2*A*b + 35*A*b^3 + 25*a^3*B + 105*a*b^2*B + 105*a^2*b*C + 105*b^3*C)*EllipticF[(c + d*x)/2, 2])/(105*d)
+ (Sqrt[Cos[c + d*x]]*((2*Sec[c + d*x]^4*(3*a^2*A*b*Sin[c + d*x] + a^3*B*Sin[c + d*x]))/7 + (2*Sec[c + d*x]^3*
(7*a^3*A*Sin[c + d*x] + 27*a*A*b^2*Sin[c + d*x] + 27*a^2*b*B*Sin[c + d*x] + 9*a^3*C*Sin[c + d*x]))/45 + (2*Sec
[c + d*x]^2*(15*a^2*A*b*Sin[c + d*x] + 7*A*b^3*Sin[c + d*x] + 5*a^3*B*Sin[c + d*x] + 21*a*b^2*B*Sin[c + d*x] +
21*a^2*b*C*Sin[c + d*x]))/21 + (2*Sec[c + d*x]*(7*a^3*A*Sin[c + d*x] + 27*a*A*b^2*Sin[c + d*x] + 27*a^2*b*B*S
in[c + d*x] + 15*b^3*B*Sin[c + d*x] + 9*a^3*C*Sin[c + d*x] + 45*a*b^2*C*Sin[c + d*x]))/15 + (2*a^3*A*Sec[c + d
*x]^4*Tan[c + d*x])/9))/d

________________________________________________________________________________________

Maple [B]  time = 4.806, size = 1292, normalized size = 3.6 \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))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(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^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*a^2*(3*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+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+s
in(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+2*b*(A*b^2+3*B*a*b+3*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)*Elliptic
F(cos(1/2*d*x+1/2*c),2^(1/2)))-2/5*a*(3*A*b^2+3*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))*(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*Ell
ipticE(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*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^3*(-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*(B*b+3*C*a)*(-(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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\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 )}^{3}}{\cos \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))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(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)^3/cos(d*x + c)^(11/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{C b^{3} \cos \left (d x + c\right )^{5} +{\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )^{4} + A a^{3} +{\left (3 \, C a^{2} b + 3 \, B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right )^{3} +{\left (C a^{3} + 3 \, B a^{2} b + 3 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2} +{\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )}{\cos \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))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(11/2),x, algorithm="fricas")

[Out]

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

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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))**3*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(11/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\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 )}^{3}}{\cos \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))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(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)^3/cos(d*x + c)^(11/2), x)