3.1170 \(\int (a+a \cos (c+d x))^2 (A+C \cos ^2(c+d x)) \sec ^{\frac {5}{2}}(c+d x) \, dx\)

Optimal. Leaf size=196 \[ -\frac {2 a^2 (5 A-C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {8 a^2 (A+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}-\frac {4 a^2 (A-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {8 A \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \cos (c+d x)+a^2\right )}{3 d}+\frac {2 A \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}{3 d} \]

[Out]

2/3*A*(a+a*cos(d*x+c))^2*sec(d*x+c)^(3/2)*sin(d*x+c)/d-2/3*a^2*(5*A-C)*sin(d*x+c)/d/sec(d*x+c)^(1/2)+8/3*A*(a^
2+a^2*cos(d*x+c))*sin(d*x+c)*sec(d*x+c)^(1/2)/d-4*a^2*(A-C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*El
lipticE(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+8/3*a^2*(A+C)*(cos(1/2*d*x+1/2*c)^2)^(
1/2)/cos(1/2*d*x+1/2*c)*EllipticF(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 = 0.54, antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ -\frac {2 a^2 (5 A-C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {8 a^2 (A+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}-\frac {4 a^2 (A-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {8 A \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \cos (c+d x)+a^2\right )}{3 d}+\frac {2 A \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*a^2*(A - C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (8*a^2*(A + C)*Sqrt[Cos[c
+ d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) - (2*a^2*(5*A - C)*Sin[c + d*x])/(3*d*Sqrt[Sec[c
+ d*x]]) + (8*A*(a^2 + a^2*Cos[c + d*x])*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3*d) + (2*A*(a + a*Cos[c + d*x])^2*
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 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))^2 \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))^2 \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))^2 \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))^2 \left (2 a A-\frac {3}{2} a (A-C) \cos (c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{3 a}\\ &=\frac {8 A \left (a^2+a^2 \cos (c+d x)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+a \cos (c+d x))^2 \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)) \left (\frac {3}{4} a^2 (3 A+C)-\frac {3}{4} a^2 (5 A-C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx}{3 a}\\ &=\frac {8 A \left (a^2+a^2 \cos (c+d x)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+a \cos (c+d x))^2 \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 {\frac {3}{4} a^3 (3 A+C)+\left (-\frac {3}{4} a^3 (5 A-C)+\frac {3}{4} a^3 (3 A+C)\right ) \cos (c+d x)-\frac {3}{4} a^3 (5 A-C) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{3 a}\\ &=-\frac {2 a^2 (5 A-C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {8 A \left (a^2+a^2 \cos (c+d x)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+a \cos (c+d x))^2 \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 {3}{2} a^3 (A+C)-\frac {9}{4} a^3 (A-C) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{9 a}\\ &=-\frac {2 a^2 (5 A-C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {8 A \left (a^2+a^2 \cos (c+d x)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+a \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}-\left (2 a^2 (A-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{3} \left (4 a^2 (A+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {4 a^2 (A-C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {8 a^2 (A+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 {2 a^2 (5 A-C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {8 A \left (a^2+a^2 \cos (c+d x)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+a \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}\\ \end {align*}

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Mathematica [C]  time = 1.70, size = 191, normalized size = 0.97 \[ \frac {a^2 e^{-i d x} \sec ^{\frac {3}{2}}(c+d x) (\cos (d x)+i \sin (d x)) \left (4 i (A-C) \left (1+e^{2 i (c+d x)}\right )^{3/2} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )+16 (A+C) \cos ^{\frac {3}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+4 A \sin (c+d x)+12 A \sin (2 (c+d x))-12 i A \cos (2 (c+d x))-12 i A+C \sin (c+d x)+C \sin (3 (c+d x))+12 i C \cos (2 (c+d x))+12 i C\right )}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sec[c + d*x]^(3/2)*(Cos[d*x] + I*Sin[d*x])*((-12*I)*A + (12*I)*C - (12*I)*A*Cos[2*(c + d*x)] + (12*I)*C*C
os[2*(c + d*x)] + 16*(A + C)*Cos[c + d*x]^(3/2)*EllipticF[(c + d*x)/2, 2] + (4*I)*(A - C)*(1 + E^((2*I)*(c + d
*x)))^(3/2)*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))] + 4*A*Sin[c + d*x] + C*Sin[c + d*x] + 12*A*
Sin[2*(c + d*x)] + C*Sin[3*(c + d*x)]))/(6*d*E^(I*d*x))

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C a^{2} \cos \left (d x + c\right )^{4} + 2 \, C a^{2} \cos \left (d x + c\right )^{3} + {\left (A + C\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \, A a^{2} \cos \left (d x + c\right ) + A a^{2}\right )} \sec \left (d x + c\right )^{\frac {5}{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [B]  time = 6.68, size = 651, normalized size = 3.32 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/3*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2/(4*sin(1/2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c)
^2+1)/sin(1/2*d*x+1/2*c)^3*(4*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+4*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*E
llipticF(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)^2+6*A*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)^2-12*A*cos
(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+4*C*(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)^2-6*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)^2-4*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4
-2*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))-3*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))+7*A*cos(1/2
*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-2*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(c
os(1/2*d*x+1/2*c),2^(1/2))+3*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))+C*cos(1/2*d*x+1/2*c)*sin(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)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^2 \,d x \]

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))^2,x)

[Out]

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

[Out]

Timed out

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