3.1178 \(\int (a+a \cos (c+d x))^3 (A+C \cos ^2(c+d x)) \sec ^{\frac{7}{2}}(c+d x) \, dx\)

Optimal. Leaf size=253 \[ -\frac{4 a^3 (21 A+5 C) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{2 (11 A+5 C) \sin (c+d x) \sqrt{\sec (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}{5 d}+\frac{4 a^3 (3 A+5 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{4 a^3 (9 A-5 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{4 A \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{5 a d}+\frac{2 A \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+a)^3}{5 d} \]

[Out]

(-4*a^3*(9*A - 5*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*C
)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) - (4*a^3*(21*A + 5*C)*Sin[c + d*x])/(
15*d*Sqrt[Sec[c + d*x]]) + (2*(11*A + 5*C)*(a^3 + a^3*Cos[c + d*x])*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(5*d) + (
4*A*(a^2 + a^2*Cos[c + d*x])^2*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(5*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 = 0.682042, antiderivative size = 253, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.229, Rules used = {4221, 3044, 2975, 2968, 3023, 2748, 2641, 2639} \[ -\frac{4 a^3 (21 A+5 C) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{2 (11 A+5 C) \sin (c+d x) \sqrt{\sec (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}{5 d}+\frac{4 a^3 (3 A+5 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{4 a^3 (9 A-5 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{4 A \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{5 a d}+\frac{2 A \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+a)^3}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*a^3*(9*A - 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*C
)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) - (4*a^3*(21*A + 5*C)*Sin[c + d*x])/(
15*d*Sqrt[Sec[c + d*x]]) + (2*(11*A + 5*C)*(a^3 + a^3*Cos[c + d*x])*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(5*d) + (
4*A*(a^2 + a^2*Cos[c + d*x])^2*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(5*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 3044

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C + 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*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(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, 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+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+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 (3 a A-\frac{1}{2} a (3 A-5 C) \cos (c+d x)\right )}{\cos ^{\frac{5}{2}}(c+d x)} \, dx}{5 a}\\ &=\frac{4 A \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 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{3}{4} a^2 (11 A+5 C)-\frac{3}{4} a^2 (9 A-5 C) \cos (c+d x)\right )}{\cos ^{\frac{3}{2}}(c+d x)} \, dx}{15 a}\\ &=\frac{2 (11 A+5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d}+\frac{4 A \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 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}{2} a^3 (6 A+5 C)-\frac{3}{4} a^3 (21 A+5 C) \cos (c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx}{15 a}\\ &=\frac{2 (11 A+5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d}+\frac{4 A \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 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}{2} a^4 (6 A+5 C)+\left (\frac{3}{2} a^4 (6 A+5 C)-\frac{3}{4} a^4 (21 A+5 C)\right ) \cos (c+d x)-\frac{3}{4} a^4 (21 A+5 C) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)}} \, dx}{15 a}\\ &=-\frac{4 a^3 (21 A+5 C) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{2 (11 A+5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d}+\frac{4 A \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 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 C)-\frac{9}{8} a^4 (9 A-5 C) \cos (c+d x)}{\sqrt{\cos (c+d x)}} \, dx}{45 a}\\ &=-\frac{4 a^3 (21 A+5 C) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{2 (11 A+5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d}+\frac{4 A \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 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 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 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 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 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+5 C) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{2 (11 A+5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d}+\frac{4 A \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 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 [C]  time = 3.03102, size = 279, normalized size = 1.1 \[ \frac{a^3 \csc (c) \sec (c) e^{-i d x} \sqrt{\sec (c+d x)} (\cos (d x)+i \sin (d x)) \left (4 \left (-1+e^{4 i c}\right ) (9 A-5 C) e^{-i (c-d x)} \sqrt{1+e^{2 i (c+d x)}} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-e^{2 i (c+d x)}\right )+\frac{1}{2} \sin (2 c) \sec ^2(c+d x) \left (-36 i (9 A-5 C) \cos (c+d x)+80 (3 A+5 C) \cos ^{\frac{5}{2}}(c+d x) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+132 A \sin (c+d x)+60 A \sin (2 (c+d x))+108 A \sin (3 (c+d x))-108 i A \cos (3 (c+d x))+30 C \sin (c+d x)+10 C \sin (2 (c+d x))+30 C \sin (3 (c+d x))+5 C \sin (4 (c+d x))+60 i C \cos (3 (c+d x))\right )\right )}{60 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*Csc[c]*Sec[c]*Sqrt[Sec[c + d*x]]*(Cos[d*x] + I*Sin[d*x])*((4*(9*A - 5*C)*(-1 + E^((4*I)*c))*Sqrt[1 + E^((
2*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])/E^(I*(c - d*x)) + (Sec[c + d*x]^2*Sin
[2*c]*((-36*I)*(9*A - 5*C)*Cos[c + d*x] - (108*I)*A*Cos[3*(c + d*x)] + (60*I)*C*Cos[3*(c + d*x)] + 80*(3*A + 5
*C)*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2] + 132*A*Sin[c + d*x] + 30*C*Sin[c + d*x] + 60*A*Sin[2*(c + d*
x)] + 10*C*Sin[2*(c + d*x)] + 108*A*Sin[3*(c + d*x)] + 30*C*Sin[3*(c + d*x)] + 5*C*Sin[4*(c + d*x)]))/2))/(60*
d*E^(I*d*x))

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Maple [B]  time = 3.352, size = 939, normalized size = 3.7 \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))^3*(A+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*(40*C*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+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)^4-216*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+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)^4-60*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)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4-120*C*sin(
1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-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)*Ellipti
cF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(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+246*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*
c)^4-100*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)*EllipticF(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+90*C*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+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))-72*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1
/2*c)^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))-20
*C*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)/(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} + 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+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + 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} + 3 \, C a^{3} \cos \left (d x + c\right )^{4} +{\left (A + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} +{\left (3 \, A + C\right )} a^{3} \cos \left (d x + c\right )^{2} + 3 \, A 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+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 + 3*C*a^3*cos(d*x + c)^4 + (A + 3*C)*a^3*cos(d*x + c)^3 + (3*A + C)*a^3*cos(d*x
+ c)^2 + 3*A*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+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} + 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+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x, algorithm="giac")

[Out]

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