3.1285 \(\int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac{5}{2}}(c+d x) \, dx\)

Optimal. Leaf size=267 \[ -\frac{4 a^3 (20 A+5 B-6 C) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}-\frac{2 (35 A+15 B-3 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{15 d \sqrt{\sec (c+d x)}}+\frac{4 a^3 (5 A+5 B+3 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{4 a^3 (5 A-5 B-9 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 (2 A+B) \sin (c+d x) \sqrt{\sec (c+d x)} \left (a^2 \cos (c+d x)+a^2\right )^2}{a d}+\frac{2 A \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^3}{3 d} \]

[Out]

(-4*a^3*(5*A - 5*B - 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
+ 5*B + 3*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) - (4*a^3*(20*A + 5*B - 6*
C)*Sin[c + d*x])/(15*d*Sqrt[Sec[c + d*x]]) - (2*(35*A + 15*B - 3*C)*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/(15
*d*Sqrt[Sec[c + d*x]]) + (2*(2*A + B)*(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)

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Rubi [A]  time = 0.762245, antiderivative size = 267, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.209, Rules used = {4221, 3043, 2975, 2976, 2968, 3023, 2748, 2641, 2639} \[ -\frac{4 a^3 (20 A+5 B-6 C) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}-\frac{2 (35 A+15 B-3 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{15 d \sqrt{\sec (c+d x)}}+\frac{4 a^3 (5 A+5 B+3 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{4 a^3 (5 A-5 B-9 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 (2 A+B) \sin (c+d x) \sqrt{\sec (c+d x)} \left (a^2 \cos (c+d x)+a^2\right )^2}{a d}+\frac{2 A \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^3}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*a^3*(5*A - 5*B - 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
+ 5*B + 3*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) - (4*a^3*(20*A + 5*B - 6*
C)*Sin[c + d*x])/(15*d*Sqrt[Sec[c + d*x]]) - (2*(35*A + 15*B - 3*C)*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/(15
*d*Sqrt[Sec[c + d*x]]) + (2*(2*A + B)*(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 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 3043

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/(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 -
B*d)*(a*c*m + b*d*(n + 1)) + b*(d*(B*c - A*d)*(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, B, 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 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 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+B \cos (c+d x)+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+B \cos (c+d x)+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 (\frac{3}{2} a (2 A+B)-\frac{1}{2} a (5 A-3 C) \cos (c+d x)\right )}{\cos ^{\frac{3}{2}}(c+d x)} \, dx}{3 a}\\ &=\frac{2 (2 A+B) \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+15 B+3 C)-\frac{1}{4} a^2 (35 A+15 B-3 C) \cos (c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx}{3 a}\\ &=-\frac{2 (35 A+15 B-3 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{2 (2 A+B) \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{3}{4} a^3 (15 A+10 B+3 C)-\frac{3}{4} a^3 (20 A+5 B-6 C) \cos (c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx}{15 a}\\ &=-\frac{2 (35 A+15 B-3 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{2 (2 A+B) \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{3}{4} a^4 (15 A+10 B+3 C)+\left (-\frac{3}{4} a^4 (20 A+5 B-6 C)+\frac{3}{4} a^4 (15 A+10 B+3 C)\right ) \cos (c+d x)-\frac{3}{4} a^4 (20 A+5 B-6 C) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)}} \, dx}{15 a}\\ &=-\frac{4 a^3 (20 A+5 B-6 C) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}-\frac{2 (35 A+15 B-3 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{2 (2 A+B) \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+5 B+3 C)-\frac{9}{8} a^4 (5 A-5 B-9 C) \cos (c+d x)}{\sqrt{\cos (c+d x)}} \, dx}{45 a}\\ &=-\frac{4 a^3 (20 A+5 B-6 C) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}-\frac{2 (35 A+15 B-3 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{2 (2 A+B) \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-5 B-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+5 B+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-5 B-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+5 B+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{4 a^3 (20 A+5 B-6 C) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}-\frac{2 (35 A+15 B-3 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{2 (2 A+B) \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*}

Mathematica [A]  time = 1.0416, size = 149, normalized size = 0.56 \[ \frac{a^3 \sec ^{\frac{3}{2}}(c+d x) \left (80 (5 A+5 B+3 C) \cos ^{\frac{3}{2}}(c+d x) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )-48 (5 A-5 B-9 C) \cos ^{\frac{3}{2}}(c+d x) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+2 \sin (c+d x) (3 (60 A+20 B+3 C) \cos (c+d x)+20 A+10 (B+3 C) \cos (2 (c+d x))+10 B+3 C \cos (3 (c+d x))+30 C)\right )}{60 d} \]

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)*Sec[c + d*x]^(5/2),x]

[Out]

(a^3*Sec[c + d*x]^(3/2)*(-48*(5*A - 5*B - 9*C)*Cos[c + d*x]^(3/2)*EllipticE[(c + d*x)/2, 2] + 80*(5*A + 5*B +
3*C)*Cos[c + d*x]^(3/2)*EllipticF[(c + d*x)/2, 2] + 2*(20*A + 10*B + 30*C + 3*(60*A + 20*B + 3*C)*Cos[c + d*x]
+ 10*(B + 3*C)*Cos[2*(c + d*x)] + 3*C*Cos[3*(c + d*x)])*Sin[c + d*x]))/(60*d)

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Maple [B]  time = 4.498, size = 950, normalized size = 3.6 \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)^(5/2),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/(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*(-24*C*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+20*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x
+1/2*c)^6+96*C*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+30*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+50*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2-90*A*cos(1/2*d*x+1/2*c)
*sin(1/2*d*x+1/2*c)^4-30*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*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+50*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellip
ticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2-50*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-54*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+30*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)^2-78*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-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))+50*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+15*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-25*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))+20*B*cos(1/2*d*x+1/
2*c)*sin(1/2*d*x+1/2*c)^2+27*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))-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)*EllipticF(cos(1/2*d*x+
1/2*c),2^(1/2))+18*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} + B \cos \left (d x + c\right ) + A\right )}{\left (a \cos \left (d x + c\right ) + a\right )}^{3} \sec \left (d x + c\right )^{\frac{5}{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+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)^3*sec(d*x + c)^(5/2), 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} +{\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}\right )} \sec \left (d x + c\right )^{\frac{5}{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+B*cos(d*x+c)+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 + (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)*sec(d*x + c)^(5/2), 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+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**(5/2),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 (a \cos \left (d x + c\right ) + a\right )}^{3} \sec \left (d x + c\right )^{\frac{5}{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+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2),x, algorithm="giac")

[Out]

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