3.1078 \(\int \frac{(a+b \cos (c+d x))^2 (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\cos ^{\frac{9}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=248 \[ \frac{2 F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^2 (5 A+7 C)+14 a b B+7 b^2 (A+3 C)\right )}{21 d}-\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (3 a^2 B+6 a A b+10 a b C+5 b^2 B\right )}{5 d}+\frac{2 \sin (c+d x) \left (a^2 (5 A+7 C)+14 a b B+4 A b^2\right )}{21 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 \sin (c+d x) \left (3 a^2 B+6 a A b+10 a b C+5 b^2 B\right )}{5 d \sqrt{\cos (c+d x)}}+\frac{2 a (7 a B+4 A b) \sin (c+d x)}{35 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a+b \cos (c+d x))^2}{7 d \cos ^{\frac{7}{2}}(c+d x)} \]

[Out]

(-2*(6*a*A*b + 3*a^2*B + 5*b^2*B + 10*a*b*C)*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*(14*a*b*B + 7*b^2*(A + 3*C)
+ a^2*(5*A + 7*C))*EllipticF[(c + d*x)/2, 2])/(21*d) + (2*a*(4*A*b + 7*a*B)*Sin[c + d*x])/(35*d*Cos[c + d*x]^
(5/2)) + (2*(4*A*b^2 + 14*a*b*B + a^2*(5*A + 7*C))*Sin[c + d*x])/(21*d*Cos[c + d*x]^(3/2)) + (2*(6*a*A*b + 3*a
^2*B + 5*b^2*B + 10*a*b*C)*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]]) + (2*A*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/
(7*d*Cos[c + d*x]^(7/2))

________________________________________________________________________________________

Rubi [A]  time = 0.560113, antiderivative size = 248, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ \frac{2 F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^2 (5 A+7 C)+14 a b B+7 b^2 (A+3 C)\right )}{21 d}-\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (3 a^2 B+6 a A b+10 a b C+5 b^2 B\right )}{5 d}+\frac{2 \sin (c+d x) \left (a^2 (5 A+7 C)+14 a b B+4 A b^2\right )}{21 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 \sin (c+d x) \left (3 a^2 B+6 a A b+10 a b C+5 b^2 B\right )}{5 d \sqrt{\cos (c+d x)}}+\frac{2 a (7 a B+4 A b) \sin (c+d x)}{35 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a+b \cos (c+d x))^2}{7 d \cos ^{\frac{7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*Cos[c + d*x])^2*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos[c + d*x]^(9/2),x]

[Out]

(-2*(6*a*A*b + 3*a^2*B + 5*b^2*B + 10*a*b*C)*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*(14*a*b*B + 7*b^2*(A + 3*C)
+ a^2*(5*A + 7*C))*EllipticF[(c + d*x)/2, 2])/(21*d) + (2*a*(4*A*b + 7*a*B)*Sin[c + d*x])/(35*d*Cos[c + d*x]^
(5/2)) + (2*(4*A*b^2 + 14*a*b*B + a^2*(5*A + 7*C))*Sin[c + d*x])/(21*d*Cos[c + d*x]^(3/2)) + (2*(6*a*A*b + 3*a
^2*B + 5*b^2*B + 10*a*b*C)*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]]) + (2*A*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/
(7*d*Cos[c + d*x]^(7/2))

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))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac{9}{2}}(c+d x)} \, dx &=\frac{2 A (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2}{7} \int \frac{(a+b \cos (c+d x)) \left (\frac{1}{2} (4 A b+7 a B)+\frac{1}{2} (5 a A+7 b B+7 a C) \cos (c+d x)+\frac{1}{2} b (A+7 C) \cos ^2(c+d x)\right )}{\cos ^{\frac{7}{2}}(c+d x)} \, dx\\ &=\frac{2 a (4 A b+7 a B) \sin (c+d x)}{35 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 A (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x)}-\frac{4}{35} \int \frac{-\frac{5}{4} \left (4 A b^2+14 a b B+a^2 (5 A+7 C)\right )-\frac{7}{4} \left (6 a A b+3 a^2 B+5 b^2 B+10 a b C\right ) \cos (c+d x)-\frac{5}{4} b^2 (A+7 C) \cos ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 a (4 A b+7 a B) \sin (c+d x)}{35 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (4 A b^2+14 a b B+a^2 (5 A+7 C)\right ) \sin (c+d x)}{21 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 A (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x)}-\frac{8}{105} \int \frac{-\frac{21}{8} \left (6 a A b+3 a^2 B+5 b^2 B+10 a b C\right )-\frac{5}{8} \left (14 a b B+7 b^2 (A+3 C)+a^2 (5 A+7 C)\right ) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 a (4 A b+7 a B) \sin (c+d x)}{35 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (4 A b^2+14 a b B+a^2 (5 A+7 C)\right ) \sin (c+d x)}{21 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 A (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x)}-\frac{1}{5} \left (-6 a A b-3 a^2 B-5 b^2 B-10 a b C\right ) \int \frac{1}{\cos ^{\frac{3}{2}}(c+d x)} \, dx-\frac{1}{21} \left (-14 a b B-7 b^2 (A+3 C)-a^2 (5 A+7 C)\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 \left (14 a b B+7 b^2 (A+3 C)+a^2 (5 A+7 C)\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{2 a (4 A b+7 a B) \sin (c+d x)}{35 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (4 A b^2+14 a b B+a^2 (5 A+7 C)\right ) \sin (c+d x)}{21 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (6 a A b+3 a^2 B+5 b^2 B+10 a b C\right ) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{2 A (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x)}-\frac{1}{5} \left (6 a A b+3 a^2 B+5 b^2 B+10 a b C\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=-\frac{2 \left (6 a A b+3 a^2 B+5 b^2 B+10 a b C\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 \left (14 a b B+7 b^2 (A+3 C)+a^2 (5 A+7 C)\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{2 a (4 A b+7 a B) \sin (c+d x)}{35 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (4 A b^2+14 a b B+a^2 (5 A+7 C)\right ) \sin (c+d x)}{21 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (6 a A b+3 a^2 B+5 b^2 B+10 a b C\right ) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{2 A (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 4.56026, size = 217, normalized size = 0.88 \[ \frac{2 \left (5 F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^2 (5 A+7 C)+14 a b B+7 b^2 (A+3 C)\right )-21 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (3 a^2 B+2 a b (3 A+5 C)+5 b^2 B\right )+\frac{5 \sin (c+d x) \left (a^2 (5 A+7 C)+14 a b B+7 A b^2\right )}{\cos ^{\frac{3}{2}}(c+d x)}+\frac{21 \sin (c+d x) \left (3 a^2 B+2 a b (3 A+5 C)+5 b^2 B\right )}{\sqrt{\cos (c+d x)}}+\frac{15 a^2 A \sin (c+d x)}{\cos ^{\frac{7}{2}}(c+d x)}+\frac{21 a (a B+2 A b) \sin (c+d x)}{\cos ^{\frac{5}{2}}(c+d x)}\right )}{105 d} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*Cos[c + d*x])^2*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos[c + d*x]^(9/2),x]

[Out]

(2*(-21*(3*a^2*B + 5*b^2*B + 2*a*b*(3*A + 5*C))*EllipticE[(c + d*x)/2, 2] + 5*(14*a*b*B + 7*b^2*(A + 3*C) + a^
2*(5*A + 7*C))*EllipticF[(c + d*x)/2, 2] + (15*a^2*A*Sin[c + d*x])/Cos[c + d*x]^(7/2) + (21*a*(2*A*b + a*B)*Si
n[c + d*x])/Cos[c + d*x]^(5/2) + (5*(7*A*b^2 + 14*a*b*B + a^2*(5*A + 7*C))*Sin[c + d*x])/Cos[c + d*x]^(3/2) +
(21*(3*a^2*B + 5*b^2*B + 2*a*b*(3*A + 5*C))*Sin[c + d*x])/Sqrt[Cos[c + d*x]]))/(105*d)

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Maple [B]  time = 3.714, size = 947, normalized size = 3.8 \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))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2),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*b^2*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)
)+2*A*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+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*b^2+2*B*a*b+C*a^2)*(-1/6*cos(1/2*d*x+1/2*c)*(-2*s
in(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)))-2/5*a*(2*A*b+B*a)/(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)/s
in(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*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*b*(B*b+2*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+si
n(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 )}^{2}}{\cos \left (d x + c\right )^{\frac{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^2/cos(d*x + c)^(9/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{C b^{2} \cos \left (d x + c\right )^{4} +{\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )^{3} + A a^{2} +{\left (C a^{2} + 2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{2} +{\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{\frac{9}{2}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2),x, algorithm="fricas")

[Out]

integral((C*b^2*cos(d*x + c)^4 + (2*C*a*b + B*b^2)*cos(d*x + c)^3 + A*a^2 + (C*a^2 + 2*B*a*b + A*b^2)*cos(d*x
+ c)^2 + (B*a^2 + 2*A*a*b)*cos(d*x + c))/cos(d*x + c)^(9/2), 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))**2*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(9/2),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 )}^{2}}{\cos \left (d x + c\right )^{\frac{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^2/cos(d*x + c)^(9/2), x)