3.1179 \(\int (a+a \cos (c+d x))^3 (A+C \cos ^2(c+d x)) \sec ^{\frac {5}{2}}(c+d x) \, dx\)

Optimal. Leaf size=251 \[ -\frac {8 a^3 (10 A-3 C) \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}-\frac {2 (35 A-3 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{15 d \sqrt {\sec (c+d x)}}+\frac {4 a^3 (5 A+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}-\frac {4 a^3 (5 A-9 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 A \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \cos (c+d x)+a^2\right )^2}{a d}+\frac {2 A \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}{3 d} \]

[Out]

2/3*A*(a+a*cos(d*x+c))^3*sec(d*x+c)^(3/2)*sin(d*x+c)/d-8/15*a^3*(10*A-3*C)*sin(d*x+c)/d/sec(d*x+c)^(1/2)-2/15*
(35*A-3*C)*(a^3+a^3*cos(d*x+c))*sin(d*x+c)/d/sec(d*x+c)^(1/2)+4*A*(a^2+a^2*cos(d*x+c))^2*sin(d*x+c)*sec(d*x+c)
^(1/2)/a/d-4/5*a^3*(5*A-9*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+4/3*a^3*(5*A+3*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ell
ipticF(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 = 0.69, antiderivative size = 251, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {4221, 3044, 2975, 2976, 2968, 3023, 2748, 2641, 2639} \[ -\frac {8 a^3 (10 A-3 C) \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}-\frac {2 (35 A-3 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{15 d \sqrt {\sec (c+d x)}}+\frac {4 a^3 (5 A+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}-\frac {4 a^3 (5 A-9 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 A \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \cos (c+d x)+a^2\right )^2}{a d}+\frac {2 A \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^3*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(5/2),x]

[Out]

(-4*a^3*(5*A - 9*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (4*a^3*(5*A + 3*C
)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) - (8*a^3*(10*A - 3*C)*Sin[c + d*x])/(
15*d*Sqrt[Sec[c + d*x]]) - (2*(35*A - 3*C)*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/(15*d*Sqrt[Sec[c + d*x]]) +
(4*A*(a^2 + a^2*Cos[c + d*x])^2*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(a*d) + (2*A*(a + a*Cos[c + d*x])^3*Sec[c + d
*x]^(3/2)*Sin[c + d*x])/(3*d)

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 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 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 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 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))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{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 {5}{2}}(c+d x)} \, dx\\ &=\frac {2 A (a+a \cos (c+d x))^3 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 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 (5 A-3 C) \cos (c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{3 a}\\ &=\frac {4 A \left (a^2+a^2 \cos (c+d x)\right )^2 \sqrt {\sec (c+d x)} \sin (c+d x)}{a d}+\frac {2 A (a+a \cos (c+d x))^3 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 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 (25 A+3 C)-\frac {1}{4} a^2 (35 A-3 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx}{3 a}\\ &=-\frac {2 (35 A-3 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}+\frac {4 A \left (a^2+a^2 \cos (c+d x)\right )^2 \sqrt {\sec (c+d x)} \sin (c+d x)}{a d}+\frac {2 A (a+a \cos (c+d x))^3 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 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 {9}{4} a^3 (5 A+C)-\frac {3}{2} a^3 (10 A-3 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx}{15 a}\\ &=-\frac {2 (35 A-3 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}+\frac {4 A \left (a^2+a^2 \cos (c+d x)\right )^2 \sqrt {\sec (c+d x)} \sin (c+d x)}{a d}+\frac {2 A (a+a \cos (c+d x))^3 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {\left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {9}{4} a^4 (5 A+C)+\left (-\frac {3}{2} a^4 (10 A-3 C)+\frac {9}{4} a^4 (5 A+C)\right ) \cos (c+d x)-\frac {3}{2} a^4 (10 A-3 C) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{15 a}\\ &=-\frac {8 a^3 (10 A-3 C) \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}-\frac {2 (35 A-3 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}+\frac {4 A \left (a^2+a^2 \cos (c+d x)\right )^2 \sqrt {\sec (c+d x)} \sin (c+d x)}{a d}+\frac {2 A (a+a \cos (c+d x))^3 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {\left (16 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {15}{8} a^4 (5 A+3 C)-\frac {9}{8} a^4 (5 A-9 C) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{45 a}\\ &=-\frac {8 a^3 (10 A-3 C) \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}-\frac {2 (35 A-3 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}+\frac {4 A \left (a^2+a^2 \cos (c+d x)\right )^2 \sqrt {\sec (c+d x)} \sin (c+d x)}{a d}+\frac {2 A (a+a \cos (c+d x))^3 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}-\frac {1}{5} \left (2 a^3 (5 A-9 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 (5 A+3 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 (5 A-9 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 (5 A+3 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 {8 a^3 (10 A-3 C) \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}-\frac {2 (35 A-3 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}+\frac {4 A \left (a^2+a^2 \cos (c+d x)\right )^2 \sqrt {\sec (c+d x)} \sin (c+d x)}{a d}+\frac {2 A (a+a \cos (c+d x))^3 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}\\ \end {align*}

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Mathematica [C]  time = 2.34, size = 221, normalized size = 0.88 \[ \frac {a^3 e^{-i d x} \sec ^{\frac {3}{2}}(c+d x) (\cos (d x)+i \sin (d x)) \left (8 i (5 A-9 C) \left (1+e^{2 i (c+d x)}\right )^{3/2} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )+80 (5 A+3 C) \cos ^{\frac {3}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+40 A \sin (c+d x)+180 A \sin (2 (c+d x))-120 i A \cos (2 (c+d x))-120 i A+30 C \sin (c+d x)+6 C \sin (2 (c+d x))+30 C \sin (3 (c+d x))+3 C \sin (4 (c+d x))+216 i C \cos (2 (c+d x))+216 i C\right )}{60 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^3*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(5/2),x]

[Out]

(a^3*Sec[c + d*x]^(3/2)*(Cos[d*x] + I*Sin[d*x])*((-120*I)*A + (216*I)*C - (120*I)*A*Cos[2*(c + d*x)] + (216*I)
*C*Cos[2*(c + d*x)] + 80*(5*A + 3*C)*Cos[c + d*x]^(3/2)*EllipticF[(c + d*x)/2, 2] + (8*I)*(5*A - 9*C)*(1 + E^(
(2*I)*(c + d*x)))^(3/2)*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))] + 40*A*Sin[c + d*x] + 30*C*Sin[
c + d*x] + 180*A*Sin[2*(c + d*x)] + 6*C*Sin[2*(c + d*x)] + 30*C*Sin[3*(c + d*x)] + 3*C*Sin[4*(c + d*x)]))/(60*
d*E^(I*d*x))

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C a^{3} \cos \left (d x + c\right )^{5} + 3 \, C a^{3} \cos \left (d x + c\right )^{4} + {\left (A + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + {\left (3 \, A + C\right )} a^{3} \cos \left (d x + c\right )^{2} + 3 \, A a^{3} \cos \left (d x + c\right ) + A a^{3}\right )} \sec \left (d x + c\right )^{\frac {5}{2}}, x\right ) \]

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)^(5/2),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)^(5/2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{3} \sec \left (d x + c\right )^{\frac {5}{2}}\,{d x} \]

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)^(5/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(a*cos(d*x + c) + a)^3*sec(d*x + c)^(5/2), x)

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maple [B]  time = 3.50, size = 704, normalized size = 2.80 \[ \text {result too large to display} \]

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)^(5/2),x)

[Out]

-4/15*(24*C*(-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)^8-96*C*
(-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)^6+6*(-2*sin(1/2*d*x
+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(15*A+13*C)*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-2*(-2*sin(1/2*d*x+1/
2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(25*A+9*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)^2)
^(1/2)*(2*sin(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)*(25*A*EllipticF(c
os(1/2*d*x+1/2*c),2^(1/2))+15*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+15*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2
))-27*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*sin(1/2*d*x+1/2*c)^2+25*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))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^
(1/2)+15*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))+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)*(-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))-27*C*(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)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2
)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*a^3/(-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)^(3/2)/sin(1/2*d*x+1/2*c)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{3} \sec \left (d x + c\right )^{\frac {5}{2}}\,{d x} \]

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)^(5/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(a*cos(d*x + c) + a)^3*sec(d*x + c)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(5/2)*(a + a*cos(c + d*x))^3,x)

[Out]

int((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(5/2)*(a + a*cos(c + d*x))^3, 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))**3*(A+C*cos(d*x+c)**2)*sec(d*x+c)**(5/2),x)

[Out]

Timed out

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