3.1273 \(\int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac {11}{2}}(c+d x) \, dx\)

Optimal. Leaf size=291 \[ \frac {2 a^2 (19 A+27 B+21 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{105 d}+\frac {4 a^2 (5 A+6 B+7 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{21 d}+\frac {4 a^2 (8 A+9 B+12 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{15 d}+\frac {4 a^2 (5 A+6 B+7 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}-\frac {4 a^2 (8 A+9 B+12 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 (4 A+9 B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{63 d}+\frac {2 A \sin (c+d x) \sec ^{\frac {9}{2}}(c+d x) (a \cos (c+d x)+a)^2}{9 d} \]

[Out]

4/21*a^2*(5*A+6*B+7*C)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/105*a^2*(19*A+27*B+21*C)*sec(d*x+c)^(5/2)*sin(d*x+c)/d+
2/63*(4*A+9*B)*(a^2+a^2*cos(d*x+c))*sec(d*x+c)^(7/2)*sin(d*x+c)/d+2/9*A*(a+a*cos(d*x+c))^2*sec(d*x+c)^(9/2)*si
n(d*x+c)/d+4/15*a^2*(8*A+9*B+12*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/d-4/15*a^2*(8*A+9*B+12*C)*(cos(1/2*d*x+1/2*c)^2
)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+4/21*a^2*
(5*A+6*B+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 = 0.65, antiderivative size = 291, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.209, Rules used = {4221, 3043, 2975, 2968, 3021, 2748, 2636, 2641, 2639} \[ \frac {2 a^2 (19 A+27 B+21 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{105 d}+\frac {4 a^2 (5 A+6 B+7 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{21 d}+\frac {4 a^2 (8 A+9 B+12 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{15 d}+\frac {4 a^2 (5 A+6 B+7 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}-\frac {4 a^2 (8 A+9 B+12 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 (4 A+9 B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{63 d}+\frac {2 A \sin (c+d x) \sec ^{\frac {9}{2}}(c+d x) (a \cos (c+d x)+a)^2}{9 d} \]

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)*Sec[c + d*x]^(11/2),x]

[Out]

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

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]

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

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Mathematica [A]  time = 2.30, size = 209, normalized size = 0.72 \[ \frac {a^2 \sec ^{\frac {9}{2}}(c+d x) \left (240 (5 A+6 B+7 C) \cos ^{\frac {9}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-336 (8 A+9 B+12 C) \cos ^{\frac {9}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+2 \sin (c+d x) (90 (9 A+8 B+7 C) \cos (c+d x)+14 (64 A+72 B+81 C) \cos (2 (c+d x))+150 A \cos (3 (c+d x))+168 A \cos (4 (c+d x))+868 A+180 B \cos (3 (c+d x))+189 B \cos (4 (c+d x))+819 B+210 C \cos (3 (c+d x))+252 C \cos (4 (c+d x))+882 C)\right )}{1260 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sec[c + d*x]^(9/2)*(-336*(8*A + 9*B + 12*C)*Cos[c + d*x]^(9/2)*EllipticE[(c + d*x)/2, 2] + 240*(5*A + 6*B
+ 7*C)*Cos[c + d*x]^(9/2)*EllipticF[(c + d*x)/2, 2] + 2*(868*A + 819*B + 882*C + 90*(9*A + 8*B + 7*C)*Cos[c +
d*x] + 14*(64*A + 72*B + 81*C)*Cos[2*(c + d*x)] + 150*A*Cos[3*(c + d*x)] + 180*B*Cos[3*(c + d*x)] + 210*C*Cos
[3*(c + d*x)] + 168*A*Cos[4*(c + d*x)] + 189*B*Cos[4*(c + d*x)] + 252*C*Cos[4*(c + d*x)])*Sin[c + d*x]))/(1260
*d)

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C a^{2} \cos \left (d x + c\right )^{4} + {\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + {\left (A + 2 \, B + C\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right ) + A a^{2}\right )} \sec \left (d x + c\right )^{\frac {11}{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

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)*sec(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*sec(d*x + c)^(11/2), x)

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maple [B]  time = 13.48, size = 1181, normalized size = 4.06 \[ \text {result too large to display} \]

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)*sec(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*C+1/4*B)*(-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)))+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+c
os(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+si
n(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))))-1/5*(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))*(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)+1/4*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)^(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)+(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))))/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 \[ \int {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {11}{2}}\,{d x} \]

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)*sec(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*sec(d*x + c)^(11/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((1/cos(c + d*x))^(11/2)*(a + a*cos(c + d*x))^2*(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+a*cos(d*x+c))**2*(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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