3.1284 \(\int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac{7}{2}}(c+d x) \, dx\)

Optimal. Leaf size=270 \[ -\frac{4 a^3 (21 A+20 B+5 C) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{2 (33 A+35 B+15 C) \sin (c+d x) \sqrt{\sec (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}{15 d}+\frac{4 a^3 (3 A+5 (B+C)) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{4 a^3 (9 A+5 B-5 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 (6 A+5 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{15 a d}+\frac{2 A \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+a)^3}{5 d} \]

[Out]

(-4*a^3*(9*A + 5*B - 5*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (4*a^3*(3*A
+ 5*(B + C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) - (4*a^3*(21*A + 20*B + 5
*C)*Sin[c + d*x])/(15*d*Sqrt[Sec[c + d*x]]) + (2*(33*A + 35*B + 15*C)*(a^3 + a^3*Cos[c + d*x])*Sqrt[Sec[c + d*
x]]*Sin[c + d*x])/(15*d) + (2*(6*A + 5*B)*(a^2 + a^2*Cos[c + d*x])^2*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(15*a*d)
+ (2*A*(a + a*Cos[c + d*x])^3*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(5*d)

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Rubi [A]  time = 0.790965, antiderivative size = 270, 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 = {4221, 3043, 2975, 2968, 3023, 2748, 2641, 2639} \[ -\frac{4 a^3 (21 A+20 B+5 C) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{2 (33 A+35 B+15 C) \sin (c+d x) \sqrt{\sec (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}{15 d}+\frac{4 a^3 (3 A+5 (B+C)) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{4 a^3 (9 A+5 B-5 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 (6 A+5 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{15 a d}+\frac{2 A \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+a)^3}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^3*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(7/2),x]

[Out]

(-4*a^3*(9*A + 5*B - 5*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (4*a^3*(3*A
+ 5*(B + C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) - (4*a^3*(21*A + 20*B + 5
*C)*Sin[c + d*x])/(15*d*Sqrt[Sec[c + d*x]]) + (2*(33*A + 35*B + 15*C)*(a^3 + a^3*Cos[c + d*x])*Sqrt[Sec[c + d*
x]]*Sin[c + d*x])/(15*d) + (2*(6*A + 5*B)*(a^2 + a^2*Cos[c + d*x])^2*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(15*a*d)
+ (2*A*(a + a*Cos[c + d*x])^3*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(5*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 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 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

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+a \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac{7}{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))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac{7}{2}}(c+d x)} \, dx\\ &=\frac{2 A (a+a \cos (c+d x))^3 \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{\left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+a \cos (c+d x))^3 \left (\frac{1}{2} a (6 A+5 B)-\frac{1}{2} a (3 A-5 C) \cos (c+d x)\right )}{\cos ^{\frac{5}{2}}(c+d x)} \, dx}{5 a}\\ &=\frac{2 (6 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 a d}+\frac{2 A (a+a \cos (c+d x))^3 \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{\left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+a \cos (c+d x))^2 \left (\frac{1}{4} a^2 (33 A+35 B+15 C)-\frac{3}{4} a^2 (9 A+5 B-5 C) \cos (c+d x)\right )}{\cos ^{\frac{3}{2}}(c+d x)} \, dx}{15 a}\\ &=\frac{2 (33 A+35 B+15 C) \left (a^3+a^3 \cos (c+d x)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 (6 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 a d}+\frac{2 A (a+a \cos (c+d x))^3 \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{\left (8 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+a \cos (c+d x)) \left (\frac{3}{4} a^3 (12 A+15 B+10 C)-\frac{3}{4} a^3 (21 A+20 B+5 C) \cos (c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx}{15 a}\\ &=\frac{2 (33 A+35 B+15 C) \left (a^3+a^3 \cos (c+d x)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 (6 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 a d}+\frac{2 A (a+a \cos (c+d x))^3 \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{\left (8 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{3}{4} a^4 (12 A+15 B+10 C)+\left (-\frac{3}{4} a^4 (21 A+20 B+5 C)+\frac{3}{4} a^4 (12 A+15 B+10 C)\right ) \cos (c+d x)-\frac{3}{4} a^4 (21 A+20 B+5 C) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)}} \, dx}{15 a}\\ &=-\frac{4 a^3 (21 A+20 B+5 C) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{2 (33 A+35 B+15 C) \left (a^3+a^3 \cos (c+d x)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 (6 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 a d}+\frac{2 A (a+a \cos (c+d x))^3 \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{\left (16 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{15}{8} a^4 (3 A+5 (B+C))-\frac{9}{8} a^4 (9 A+5 B-5 C) \cos (c+d x)}{\sqrt{\cos (c+d x)}} \, dx}{45 a}\\ &=-\frac{4 a^3 (21 A+20 B+5 C) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{2 (33 A+35 B+15 C) \left (a^3+a^3 \cos (c+d x)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 (6 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 a d}+\frac{2 A (a+a \cos (c+d x))^3 \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d}-\frac{1}{5} \left (2 a^3 (9 A+5 B-5 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx+\frac{1}{3} \left (2 a^3 (3 A+5 (B+C)) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{4 a^3 (9 A+5 B-5 C) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{5 d}+\frac{4 a^3 (3 A+5 (B+C)) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 d}-\frac{4 a^3 (21 A+20 B+5 C) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{2 (33 A+35 B+15 C) \left (a^3+a^3 \cos (c+d x)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 (6 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 a d}+\frac{2 A (a+a \cos (c+d x))^3 \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 2.10876, size = 157, normalized size = 0.58 \[ \frac{a^3 \sec ^{\frac{5}{2}}(c+d x) \left (80 (3 A+5 (B+C)) \cos ^{\frac{5}{2}}(c+d x) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )-48 (9 A+5 B-5 C) \cos ^{\frac{5}{2}}(c+d x) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+2 \sin (c+d x) (5 (12 A+4 B+3 C) \cos (c+d x)+6 (18 A+5 (3 B+C)) \cos (2 (c+d x))+5 (6 (4 A+3 B+C)+C \cos (3 (c+d x))))\right )}{60 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^3*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(7/2),x]

[Out]

(a^3*Sec[c + d*x]^(5/2)*(-48*(9*A + 5*B - 5*C)*Cos[c + d*x]^(5/2)*EllipticE[(c + d*x)/2, 2] + 80*(3*A + 5*(B +
C))*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2] + 2*(5*(12*A + 4*B + 3*C)*Cos[c + d*x] + 6*(18*A + 5*(3*B +
C))*Cos[2*(c + d*x)] + 5*(6*(4*A + 3*B + C) + C*Cos[3*(c + d*x)]))*Sin[c + d*x]))/(60*d)

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Maple [B]  time = 4.743, size = 1328, normalized size = 4.9 \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))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x)

[Out]

4/15*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^3/(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)^3*(-216*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+15*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+27*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+25*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-15*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+15*B*(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))+25*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(
2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-50*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*
c)^2-120*C*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+90*C*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-20*C*sin(1/2*d
*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-60*C*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)*(si
n(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4+60*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4+108*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4+100*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4-100*B*(2*sin(
1/2*d*x+1/2*c)^2-1)^(1/2)*(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
)^2-60*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*s
in(1/2*d*x+1/2*c)^2-108*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+
1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2-100*C*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellip
ticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2+60*C*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c
)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2+60*B*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)^4-60*B*(sin(1/2*d*x+1/2*c)
^2)^(1/2)*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+190*B*co
s(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+246*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-72*A*cos(1/2*d*x+1/2*c)*si
n(1/2*d*x+1/2*c)^2+40*C*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)-180*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+
100*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(
1/2*d*x+1/2*c)^4)*(-2*sin(1/2*d*x+1/2*c)^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)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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 )}^{3} \sec \left (d x + c\right )^{\frac{7}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)^3*sec(d*x + c)^(7/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C a^{3} \cos \left (d x + c\right )^{5} +{\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} +{\left (A + 3 \, B + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} +{\left (3 \, A + 3 \, B + C\right )} a^{3} \cos \left (d x + c\right )^{2} +{\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right ) + A a^{3}\right )} \sec \left (d x + c\right )^{\frac{7}{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

integral((C*a^3*cos(d*x + c)^5 + (B + 3*C)*a^3*cos(d*x + c)^4 + (A + 3*B + 3*C)*a^3*cos(d*x + c)^3 + (3*A + 3*
B + C)*a^3*cos(d*x + c)^2 + (3*A + B)*a^3*cos(d*x + c) + A*a^3)*sec(d*x + c)^(7/2), 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))**3*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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 )}^{3} \sec \left (d x + c\right )^{\frac{7}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)^3*sec(d*x + c)^(7/2), x)