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

Optimal. Leaf size=251 \[ \frac{4 a^2 (5 A+6 B+7 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}-\frac{4 a^2 (8 A+9 B+12 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{4 a^2 (5 A+6 B+7 C) \sin (c+d x)}{21 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 a^2 (19 A+27 B+21 C) \sin (c+d x)}{105 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{4 a^2 (8 A+9 B+12 C) \sin (c+d x)}{15 d \sqrt{\cos (c+d x)}}+\frac{2 (4 A+9 B) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{63 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{9 d \cos ^{\frac{9}{2}}(c+d x)} \]

[Out]

(-4*a^2*(8*A + 9*B + 12*C)*EllipticE[(c + d*x)/2, 2])/(15*d) + (4*a^2*(5*A + 6*B + 7*C)*EllipticF[(c + d*x)/2,
2])/(21*d) + (2*a^2*(19*A + 27*B + 21*C)*Sin[c + d*x])/(105*d*Cos[c + d*x]^(5/2)) + (4*a^2*(5*A + 6*B + 7*C)*
Sin[c + d*x])/(21*d*Cos[c + d*x]^(3/2)) + (4*a^2*(8*A + 9*B + 12*C)*Sin[c + d*x])/(15*d*Sqrt[Cos[c + d*x]]) +
(2*A*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/(9*d*Cos[c + d*x]^(9/2)) + (2*(4*A + 9*B)*(a^2 + a^2*Cos[c + d*x])*S
in[c + d*x])/(63*d*Cos[c + d*x]^(7/2))

________________________________________________________________________________________

Rubi [A]  time = 0.551878, antiderivative size = 251, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.186, Rules used = {3043, 2975, 2968, 3021, 2748, 2636, 2641, 2639} \[ \frac{4 a^2 (5 A+6 B+7 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}-\frac{4 a^2 (8 A+9 B+12 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{4 a^2 (5 A+6 B+7 C) \sin (c+d x)}{21 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 a^2 (19 A+27 B+21 C) \sin (c+d x)}{105 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{4 a^2 (8 A+9 B+12 C) \sin (c+d x)}{15 d \sqrt{\cos (c+d x)}}+\frac{2 (4 A+9 B) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{63 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{9 d \cos ^{\frac{9}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*a^2*(8*A + 9*B + 12*C)*EllipticE[(c + d*x)/2, 2])/(15*d) + (4*a^2*(5*A + 6*B + 7*C)*EllipticF[(c + d*x)/2,
2])/(21*d) + (2*a^2*(19*A + 27*B + 21*C)*Sin[c + d*x])/(105*d*Cos[c + d*x]^(5/2)) + (4*a^2*(5*A + 6*B + 7*C)*
Sin[c + d*x])/(21*d*Cos[c + d*x]^(3/2)) + (4*a^2*(8*A + 9*B + 12*C)*Sin[c + d*x])/(15*d*Sqrt[Cos[c + d*x]]) +
(2*A*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/(9*d*Cos[c + d*x]^(9/2)) + (2*(4*A + 9*B)*(a^2 + a^2*Cos[c + d*x])*S
in[c + d*x])/(63*d*Cos[c + d*x]^(7/2))

Rule 3043

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/(b*d*(n + 1)
*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + (c*C -
B*d)*(a*c*m + b*d*(n + 1)) + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2,
0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0])

Rule 2975

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b^2*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*S
in[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c + a*d)), x] - Dist[b/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])
^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n +
1) - B*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d
, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n]
|| EqQ[c, 0])

Rule 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

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 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 \frac{(a+a \cos (c+d x))^2 \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+a \cos (c+d x))^2 \sin (c+d x)}{9 d \cos ^{\frac{9}{2}}(c+d x)}+\frac{2 \int \frac{(a+a \cos (c+d x))^2 \left (\frac{1}{2} a (4 A+9 B)+\frac{3}{2} a (A+3 C) \cos (c+d x)\right )}{\cos ^{\frac{9}{2}}(c+d x)} \, dx}{9 a}\\ &=\frac{2 A (a+a \cos (c+d x))^2 \sin (c+d x)}{9 d \cos ^{\frac{9}{2}}(c+d x)}+\frac{2 (4 A+9 B) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{63 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{4 \int \frac{(a+a \cos (c+d x)) \left (\frac{3}{4} a^2 (19 A+27 B+21 C)+\frac{3}{4} a^2 (11 A+9 B+21 C) \cos (c+d x)\right )}{\cos ^{\frac{7}{2}}(c+d x)} \, dx}{63 a}\\ &=\frac{2 A (a+a \cos (c+d x))^2 \sin (c+d x)}{9 d \cos ^{\frac{9}{2}}(c+d x)}+\frac{2 (4 A+9 B) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{63 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{4 \int \frac{\frac{3}{4} a^3 (19 A+27 B+21 C)+\left (\frac{3}{4} a^3 (11 A+9 B+21 C)+\frac{3}{4} a^3 (19 A+27 B+21 C)\right ) \cos (c+d x)+\frac{3}{4} a^3 (11 A+9 B+21 C) \cos ^2(c+d x)}{\cos ^{\frac{7}{2}}(c+d x)} \, dx}{63 a}\\ &=\frac{2 a^2 (19 A+27 B+21 C) \sin (c+d x)}{105 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 A (a+a \cos (c+d x))^2 \sin (c+d x)}{9 d \cos ^{\frac{9}{2}}(c+d x)}+\frac{2 (4 A+9 B) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{63 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{8 \int \frac{\frac{45}{4} a^3 (5 A+6 B+7 C)+\frac{21}{4} a^3 (8 A+9 B+12 C) \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x)} \, dx}{315 a}\\ &=\frac{2 a^2 (19 A+27 B+21 C) \sin (c+d x)}{105 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 A (a+a \cos (c+d x))^2 \sin (c+d x)}{9 d \cos ^{\frac{9}{2}}(c+d x)}+\frac{2 (4 A+9 B) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{63 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{1}{7} \left (2 a^2 (5 A+6 B+7 C)\right ) \int \frac{1}{\cos ^{\frac{5}{2}}(c+d x)} \, dx+\frac{1}{15} \left (2 a^2 (8 A+9 B+12 C)\right ) \int \frac{1}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 a^2 (19 A+27 B+21 C) \sin (c+d x)}{105 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{4 a^2 (5 A+6 B+7 C) \sin (c+d x)}{21 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 (8 A+9 B+12 C) \sin (c+d x)}{15 d \sqrt{\cos (c+d x)}}+\frac{2 A (a+a \cos (c+d x))^2 \sin (c+d x)}{9 d \cos ^{\frac{9}{2}}(c+d x)}+\frac{2 (4 A+9 B) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{63 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{1}{21} \left (2 a^2 (5 A+6 B+7 C)\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx-\frac{1}{15} \left (2 a^2 (8 A+9 B+12 C)\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=-\frac{4 a^2 (8 A+9 B+12 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{4 a^2 (5 A+6 B+7 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{2 a^2 (19 A+27 B+21 C) \sin (c+d x)}{105 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{4 a^2 (5 A+6 B+7 C) \sin (c+d x)}{21 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 (8 A+9 B+12 C) \sin (c+d x)}{15 d \sqrt{\cos (c+d x)}}+\frac{2 A (a+a \cos (c+d x))^2 \sin (c+d x)}{9 d \cos ^{\frac{9}{2}}(c+d x)}+\frac{2 (4 A+9 B) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{63 d \cos ^{\frac{7}{2}}(c+d x)}\\ \end{align*}

Mathematica [C]  time = 6.93691, size = 1364, normalized size = 5.43 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^2*Sec[c/2 + (d*x)/2]^4*(((8*A + 9*B + 12*C)*Csc[c]*Sec[c])/(15*d) + (A
*Sec[c]*Sec[c + d*x]^5*Sin[d*x])/(18*d) + (Sec[c]*Sec[c + d*x]^4*(7*A*Sin[c] + 18*A*Sin[d*x] + 9*B*Sin[d*x]))/
(126*d) + (Sec[c]*Sec[c + d*x]^3*(90*A*Sin[c] + 45*B*Sin[c] + 112*A*Sin[d*x] + 126*B*Sin[d*x] + 63*C*Sin[d*x])
)/(630*d) + (Sec[c]*Sec[c + d*x]*(25*A*Sin[c] + 30*B*Sin[c] + 35*C*Sin[c] + 56*A*Sin[d*x] + 63*B*Sin[d*x] + 84
*C*Sin[d*x]))/(105*d) + (Sec[c]*Sec[c + d*x]^2*(112*A*Sin[c] + 126*B*Sin[c] + 63*C*Sin[c] + 150*A*Sin[d*x] + 1
80*B*Sin[d*x] + 210*C*Sin[d*x]))/(630*d)) - (5*A*(a + a*Cos[c + d*x])^2*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {
5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^4*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Co
t[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(21*
d*Sqrt[1 + Cot[c]^2]) - (2*B*(a + a*Cos[c + d*x])^2*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcT
an[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^4*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[
1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(7*d*Sqrt[1 + Cot[c]^2])
- (C*(a + a*Cos[c + d*x])^2*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2
+ (d*x)/2]^4*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Si
n[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*Sqrt[1 + Cot[c]^2]) + (4*A*(a + a*Cos[c +
d*x])^2*Csc[c]*Sec[c/2 + (d*x)/2]^4*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[
d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Co
s[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/S
qrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[
c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(15*d) + (3*B*(a + a*Cos[c + d*x])^2*Csc[c]*Sec[c/2 + (d*x)
/2]^4*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/
(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]
*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^
2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*S
qrt[1 + Tan[c]^2]]))/(10*d) + (2*C*(a + a*Cos[c + d*x])^2*Csc[c]*Sec[c/2 + (d*x)/2]^4*((HypergeometricPFQ[{-1/
2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan
[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 +
Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sq
rt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(5*d)

________________________________________________________________________________________

Maple [B]  time = 1.128, size = 1181, normalized size = 4.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*((1/2*A+1/4*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)))+(1/4*B+1/2*C)*(-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)))+1/4*A*(-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*co
s(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))))+1/4*C*
(-(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)-1/5*(1/4*A+1/2*B+1/4*C)/(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*s
in(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*co
s(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))/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 (a \cos \left (d x + c\right ) + a\right )}^{2}}{\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+a*cos(d*x+c))^2*(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)*(a*cos(d*x + c) + a)^2/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 a^{2} \cos \left (d x + c\right )^{4} +{\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} +{\left (A + 2 \, B + C\right )} a^{2} \cos \left (d x + c\right )^{2} +{\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right ) + A a^{2}}{\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+a*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(11/2),x, algorithm="fricas")

[Out]

integral((C*a^2*cos(d*x + c)^4 + (B + 2*C)*a^2*cos(d*x + c)^3 + (A + 2*B + C)*a^2*cos(d*x + c)^2 + (2*A + B)*a
^2*cos(d*x + c) + A*a^2)/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+a*cos(d*x+c))**2*(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 (a \cos \left (d x + c\right ) + a\right )}^{2}}{\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+a*cos(d*x+c))^2*(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)*(a*cos(d*x + c) + a)^2/cos(d*x + c)^(11/2), x)