3.1470 \(\int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac{3}{2}}(c+d x) \, dx\)

Optimal. Leaf size=319 \[ \frac{2 b \sin (c+d x) \left (-6 a^2 (7 A-3 C)+21 a b B+b^2 (7 A+5 C)\right )}{21 d \sqrt{\sec (c+d x)}}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (21 a^2 b (3 A+C)+21 a^3 B+21 a b^2 B+b^3 (7 A+5 C)\right )}{21 d}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (-5 a^3 (A-C)+15 a^2 b B+3 a b^2 (5 A+3 C)+3 b^3 B\right )}{5 d}-\frac{2 b^2 \sin (c+d x) (35 a A-11 a C-7 b B)}{35 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{2 b (7 A-C) \sin (c+d x) (a+b \cos (c+d x))^2}{7 d \sqrt{\sec (c+d x)}}+\frac{2 A \sin (c+d x) \sqrt{\sec (c+d x)} (a+b \cos (c+d x))^3}{d} \]

[Out]

(2*(15*a^2*b*B + 3*b^3*B - 5*a^3*(A - C) + 3*a*b^2*(5*A + 3*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*S
qrt[Sec[c + d*x]])/(5*d) + (2*(21*a^3*B + 21*a*b^2*B + 21*a^2*b*(3*A + C) + b^3*(7*A + 5*C))*Sqrt[Cos[c + d*x]
]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) - (2*b^2*(35*a*A - 7*b*B - 11*a*C)*Sin[c + d*x])/(35*d*
Sec[c + d*x]^(3/2)) + (2*b*(21*a*b*B - 6*a^2*(7*A - 3*C) + b^2*(7*A + 5*C))*Sin[c + d*x])/(21*d*Sqrt[Sec[c + d
*x]]) - (2*b*(7*A - C)*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(7*d*Sqrt[Sec[c + d*x]]) + (2*A*(a + b*Cos[c + d*x
])^3*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d

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Rubi [A]  time = 0.994247, antiderivative size = 319, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.186, Rules used = {4221, 3047, 3049, 3033, 3023, 2748, 2641, 2639} \[ \frac{2 b \sin (c+d x) \left (-6 a^2 (7 A-3 C)+21 a b B+b^2 (7 A+5 C)\right )}{21 d \sqrt{\sec (c+d x)}}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (21 a^2 b (3 A+C)+21 a^3 B+21 a b^2 B+b^3 (7 A+5 C)\right )}{21 d}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (-5 a^3 (A-C)+15 a^2 b B+3 a b^2 (5 A+3 C)+3 b^3 B\right )}{5 d}-\frac{2 b^2 \sin (c+d x) (35 a A-11 a C-7 b B)}{35 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{2 b (7 A-C) \sin (c+d x) (a+b \cos (c+d x))^2}{7 d \sqrt{\sec (c+d x)}}+\frac{2 A \sin (c+d x) \sqrt{\sec (c+d x)} (a+b \cos (c+d x))^3}{d} \]

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)*Sec[c + d*x]^(3/2),x]

[Out]

(2*(15*a^2*b*B + 3*b^3*B - 5*a^3*(A - C) + 3*a*b^2*(5*A + 3*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*S
qrt[Sec[c + d*x]])/(5*d) + (2*(21*a^3*B + 21*a*b^2*B + 21*a^2*b*(3*A + C) + b^3*(7*A + 5*C))*Sqrt[Cos[c + d*x]
]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) - (2*b^2*(35*a*A - 7*b*B - 11*a*C)*Sin[c + d*x])/(35*d*
Sec[c + d*x]^(3/2)) + (2*b*(21*a*b*B - 6*a^2*(7*A - 3*C) + b^2*(7*A + 5*C))*Sin[c + d*x])/(21*d*Sqrt[Sec[c + d
*x]]) - (2*b*(7*A - C)*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(7*d*Sqrt[Sec[c + d*x]]) + (2*A*(a + b*Cos[c + d*x
])^3*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/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 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 3049

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

Rule 3033

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[(C*d*Cos[e + f*x]*Sin[e + f*x]*(a + b
*Sin[e + f*x])^(m + 1))/(b*f*(m + 3)), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*c*
(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e +
f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&
!LtQ[m, -1]

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+b \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac{3}{2}}(c+d x) \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \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{3}{2}}(c+d x)} \, dx\\ &=\frac{2 A (a+b \cos (c+d x))^3 \sqrt{\sec (c+d x)} \sin (c+d x)}{d}+\left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+b \cos (c+d x))^2 \left (\frac{1}{2} (6 A b+a B)+\frac{1}{2} (b B-a (A-C)) \cos (c+d x)-\frac{1}{2} b (7 A-C) \cos ^2(c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{2 b (7 A-C) (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+b \cos (c+d x))^3 \sqrt{\sec (c+d x)} \sin (c+d x)}{d}+\frac{1}{7} \left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+b \cos (c+d x)) \left (\frac{1}{4} a (35 A b+7 a B+b C)+\frac{1}{4} \left (14 a b B-7 a^2 (A-C)+b^2 (7 A+5 C)\right ) \cos (c+d x)-\frac{1}{4} b (35 a A-7 b B-11 a C) \cos ^2(c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{2 b^2 (35 a A-7 b B-11 a C) \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{2 b (7 A-C) (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+b \cos (c+d x))^3 \sqrt{\sec (c+d x)} \sin (c+d x)}{d}+\frac{1}{35} \left (8 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{5}{8} a^2 (35 A b+7 a B+b C)+\frac{7}{8} \left (15 a^2 b B+3 b^3 B-5 a^3 (A-C)+3 a b^2 (5 A+3 C)\right ) \cos (c+d x)+\frac{5}{8} b \left (21 a b B-6 a^2 (7 A-3 C)+b^2 (7 A+5 C)\right ) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{2 b^2 (35 a A-7 b B-11 a C) \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 b \left (21 a b B-6 a^2 (7 A-3 C)+b^2 (7 A+5 C)\right ) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}-\frac{2 b (7 A-C) (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+b \cos (c+d x))^3 \sqrt{\sec (c+d x)} \sin (c+d x)}{d}+\frac{1}{105} \left (16 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{5}{16} \left (21 a^3 B+21 a b^2 B+21 a^2 b (3 A+C)+b^3 (7 A+5 C)\right )+\frac{21}{16} \left (15 a^2 b B+3 b^3 B-5 a^3 (A-C)+3 a b^2 (5 A+3 C)\right ) \cos (c+d x)}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{2 b^2 (35 a A-7 b B-11 a C) \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 b \left (21 a b B-6 a^2 (7 A-3 C)+b^2 (7 A+5 C)\right ) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}-\frac{2 b (7 A-C) (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+b \cos (c+d x))^3 \sqrt{\sec (c+d x)} \sin (c+d x)}{d}+\frac{1}{5} \left (\left (15 a^2 b B+3 b^3 B-5 a^3 (A-C)+3 a b^2 (5 A+3 C)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx+\frac{1}{21} \left (\left (21 a^3 B+21 a b^2 B+21 a^2 b (3 A+C)+b^3 (7 A+5 C)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 \left (15 a^2 b B+3 b^3 B-5 a^3 (A-C)+3 a b^2 (5 A+3 C)\right ) \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{2 \left (21 a^3 B+21 a b^2 B+21 a^2 b (3 A+C)+b^3 (7 A+5 C)\right ) \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{2 b^2 (35 a A-7 b B-11 a C) \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 b \left (21 a b B-6 a^2 (7 A-3 C)+b^2 (7 A+5 C)\right ) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}-\frac{2 b (7 A-C) (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+b \cos (c+d x))^3 \sqrt{\sec (c+d x)} \sin (c+d x)}{d}\\ \end{align*}

Mathematica [A]  time = 1.92404, size = 233, normalized size = 0.73 \[ \frac{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \left (40 F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (21 a^2 b (3 A+C)+21 a^3 B+21 a b^2 B+b^3 (7 A+5 C)\right )-168 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (5 a^3 (A-C)-15 a^2 b B-3 a b^2 (5 A+3 C)-3 b^3 B\right )+\frac{2 \sin (c+d x) \left (5 b \cos (c+d x) \left (84 a^2 C+84 a b B+28 A b^2+29 b^2 C\right )+420 a^3 A+42 b^2 (3 a C+b B) \cos (2 (c+d x))+126 a b^2 C+42 b^3 B+15 b^3 C \cos (3 (c+d x))\right )}{\sqrt{\cos (c+d x)}}\right )}{420 d} \]

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)*Sec[c + d*x]^(3/2),x]

[Out]

(Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-168*(-15*a^2*b*B - 3*b^3*B + 5*a^3*(A - C) - 3*a*b^2*(5*A + 3*C))*Ell
ipticE[(c + d*x)/2, 2] + 40*(21*a^3*B + 21*a*b^2*B + 21*a^2*b*(3*A + C) + b^3*(7*A + 5*C))*EllipticF[(c + d*x)
/2, 2] + (2*(420*a^3*A + 42*b^3*B + 126*a*b^2*C + 5*b*(28*A*b^2 + 84*a*b*B + 84*a^2*C + 29*b^2*C)*Cos[c + d*x]
+ 42*b^2*(b*B + 3*a*C)*Cos[2*(c + d*x)] + 15*b^3*C*Cos[3*(c + d*x)])*Sin[c + d*x])/Sqrt[Cos[c + d*x]]))/(420*
d)

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Maple [B]  time = 1.794, size = 1278, normalized size = 4. \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)*sec(d*x+c)^(3/2),x)

[Out]

-2/105*(240*C*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8
-24*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*b^2*(7*B*b+21*C*a+15*C*b)*sin(1/2*d*x+1/2*c)^6*cos(1/
2*d*x+1/2*c)+28*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*b*(5*A*b^2+15*B*a*b+6*B*b^2+15*C*a^2+18*C
*a*b+10*C*b^2)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*
(105*A*a^3+35*A*b^3+105*B*a*b^2+21*B*b^3+105*C*a^2*b+63*C*a*b^2+40*C*b^3)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2
*c)+315*A*a^2*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))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+35*A*b^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)
+105*A*(-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))*a^3-315*A*(-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))*a*b^2+1
05*a^3*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))*(
-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+105*a*b^2*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))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)-315
*B*(-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))*a^2*b-63*B*(-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))*b^3+105*a^
2*b*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))*(-2*
sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+25*C*b^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)-105*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))*a^3-189*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))*a*b^2)/(-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{\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} \sec \left (d x + c\right )^{\frac{3}{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)*sec(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

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

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

[Out]

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