3.1288 \(\int \frac{(a+a \cos (c+d x))^3 (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\sqrt{\sec (c+d x)}} \, dx\)

Optimal. Leaf size=307 \[ \frac{4 a^3 (264 A+253 B+210 C) \sin (c+d x)}{1155 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4 a^3 (143 A+121 B+105 C) \sin (c+d x)}{231 d \sqrt{\sec (c+d x)}}+\frac{2 (99 A+143 B+105 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{693 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4 a^3 (143 A+121 B+105 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{231 d}+\frac{4 a^3 (21 A+17 B+15 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{2 (11 B+6 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{99 a d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{11 d \sec ^{\frac{3}{2}}(c+d x)} \]

[Out]

(4*a^3*(21*A + 17*B + 15*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(15*d) + (4*a^3*(
143*A + 121*B + 105*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(231*d) + (4*a^3*(264*
A + 253*B + 210*C)*Sin[c + d*x])/(1155*d*Sec[c + d*x]^(3/2)) + (2*C*(a + a*Cos[c + d*x])^3*Sin[c + d*x])/(11*d
*Sec[c + d*x]^(3/2)) + (2*(11*B + 6*C)*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(99*a*d*Sec[c + d*x]^(3/2)) +
(2*(99*A + 143*B + 105*C)*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/(693*d*Sec[c + d*x]^(3/2)) + (4*a^3*(143*A +
121*B + 105*C)*Sin[c + d*x])/(231*d*Sqrt[Sec[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.826813, antiderivative size = 307, normalized size of antiderivative = 1., 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, 3045, 2976, 2968, 3023, 2748, 2639, 2635, 2641} \[ \frac{4 a^3 (264 A+253 B+210 C) \sin (c+d x)}{1155 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4 a^3 (143 A+121 B+105 C) \sin (c+d x)}{231 d \sqrt{\sec (c+d x)}}+\frac{2 (99 A+143 B+105 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{693 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4 a^3 (143 A+121 B+105 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{231 d}+\frac{4 a^3 (21 A+17 B+15 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{2 (11 B+6 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{99 a d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{11 d \sec ^{\frac{3}{2}}(c+d x)} \]

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))/Sqrt[Sec[c + d*x]],x]

[Out]

(4*a^3*(21*A + 17*B + 15*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(15*d) + (4*a^3*(
143*A + 121*B + 105*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(231*d) + (4*a^3*(264*
A + 253*B + 210*C)*Sin[c + d*x])/(1155*d*Sec[c + d*x]^(3/2)) + (2*C*(a + a*Cos[c + d*x])^3*Sin[c + d*x])/(11*d
*Sec[c + d*x]^(3/2)) + (2*(11*B + 6*C)*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(99*a*d*Sec[c + d*x]^(3/2)) +
(2*(99*A + 143*B + 105*C)*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/(693*d*Sec[c + d*x]^(3/2)) + (4*a^3*(143*A +
121*B + 105*C)*Sin[c + d*x])/(231*d*Sqrt[Sec[c + d*x]])

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 3045

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

Rule 2976

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*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])
^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x]
)^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*S
in[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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 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 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[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]

Rubi steps

\begin{align*} \int \frac{(a+a \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\\ &=\frac{2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{11 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{\left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^3 \left (\frac{1}{2} a (11 A+3 C)+\frac{1}{2} a (11 B+6 C) \cos (c+d x)\right ) \, dx}{11 a}\\ &=\frac{2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{11 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (11 B+6 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{99 a d \sec ^{\frac{3}{2}}(c+d x)}+\frac{\left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^2 \left (\frac{3}{4} a^2 (33 A+11 B+15 C)+\frac{1}{4} a^2 (99 A+143 B+105 C) \cos (c+d x)\right ) \, dx}{99 a}\\ &=\frac{2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{11 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (11 B+6 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{99 a d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (99 A+143 B+105 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{693 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{\left (8 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} (a+a \cos (c+d x)) \left (\frac{15}{4} a^3 (33 A+22 B+21 C)+\frac{3}{4} a^3 (264 A+253 B+210 C) \cos (c+d x)\right ) \, dx}{693 a}\\ &=\frac{2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{11 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (11 B+6 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{99 a d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (99 A+143 B+105 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{693 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{\left (8 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \left (\frac{15}{4} a^4 (33 A+22 B+21 C)+\left (\frac{15}{4} a^4 (33 A+22 B+21 C)+\frac{3}{4} a^4 (264 A+253 B+210 C)\right ) \cos (c+d x)+\frac{3}{4} a^4 (264 A+253 B+210 C) \cos ^2(c+d x)\right ) \, dx}{693 a}\\ &=\frac{4 a^3 (264 A+253 B+210 C) \sin (c+d x)}{1155 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{11 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (11 B+6 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{99 a d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (99 A+143 B+105 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{693 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{\left (16 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \left (\frac{231}{8} a^4 (21 A+17 B+15 C)+\frac{45}{8} a^4 (143 A+121 B+105 C) \cos (c+d x)\right ) \, dx}{3465 a}\\ &=\frac{4 a^3 (264 A+253 B+210 C) \sin (c+d x)}{1155 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{11 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (11 B+6 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{99 a d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (99 A+143 B+105 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{693 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{1}{15} \left (2 a^3 (21 A+17 B+15 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx+\frac{1}{77} \left (2 a^3 (143 A+121 B+105 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \cos ^{\frac{3}{2}}(c+d x) \, dx\\ &=\frac{4 a^3 (21 A+17 B+15 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^3 (264 A+253 B+210 C) \sin (c+d x)}{1155 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{11 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (11 B+6 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{99 a d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (99 A+143 B+105 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{693 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4 a^3 (143 A+121 B+105 C) \sin (c+d x)}{231 d \sqrt{\sec (c+d x)}}+\frac{1}{231} \left (2 a^3 (143 A+121 B+105 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 (21 A+17 B+15 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^3 (143 A+121 B+105 C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{231 d}+\frac{4 a^3 (264 A+253 B+210 C) \sin (c+d x)}{1155 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{11 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (11 B+6 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{99 a d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (99 A+143 B+105 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{693 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4 a^3 (143 A+121 B+105 C) \sin (c+d x)}{231 d \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 1.56686, size = 174, normalized size = 0.57 \[ \frac{a^3 \sqrt{\sec (c+d x)} \left (2 \sin (2 (c+d x)) (154 (108 A+151 B+165 C) \cos (c+d x)+5 (36 (11 A+33 B+49 C) \cos (2 (c+d x))+7260 A+154 (B+3 C) \cos (3 (c+d x))+6996 B+63 C \cos (4 (c+d x))+6741 C))+960 (143 A+121 B+105 C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+14784 (21 A+17 B+15 C) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{55440 d} \]

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))/Sqrt[Sec[c + d*x]],x]

[Out]

(a^3*Sqrt[Sec[c + d*x]]*(14784*(21*A + 17*B + 15*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 960*(143*A
+ 121*B + 105*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + 2*(154*(108*A + 151*B + 165*C)*Cos[c + d*x] +
5*(7260*A + 6996*B + 6741*C + 36*(11*A + 33*B + 49*C)*Cos[2*(c + d*x)] + 154*(B + 3*C)*Cos[3*(c + d*x)] + 63*C
*Cos[4*(c + d*x)]))*Sin[2*(c + d*x)]))/(55440*d)

________________________________________________________________________________________

Maple [A]  time = 1.523, size = 545, normalized size = 1.8 \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)^(1/2),x)

[Out]

-4/3465*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^3*(10080*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/
2*c)^12+(-6160*B-43680*C)*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)+(3960*A+24200*B+77280*C)*sin(1/2*d*x+1/2*c)
^8*cos(1/2*d*x+1/2*c)+(-14256*A-37532*B-72240*C)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(19866*A+29722*B+3927
0*C)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-6864*A-8118*B-8820*C)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+2
145*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))-4851
*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))+1815*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))-3927*B*(si
n(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))+1575*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))-3465*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)))/(-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 )}^{3}}{\sqrt{\sec \left (d x + c\right )}}\,{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)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{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}}{\sqrt{\sec \left (d x + c\right )}}, 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)^(1/2),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)/sqrt(sec(d*x + c)), 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)**(1/2),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 )}^{3}}{\sqrt{\sec \left (d x + c\right )}}\,{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)^(1/2),x, algorithm="giac")

[Out]

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