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

Optimal. Leaf size=230 \[ \frac{4 a^3 (3 A+5 (B+C)) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{4 a^3 (9 A+5 B-5 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}-\frac{4 a^3 (21 A+20 B+5 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{15 d}+\frac{2 (33 A+35 B+15 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{15 d \sqrt{\cos (c+d x)}}+\frac{2 (6 A+5 B) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{15 a d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d \cos ^{\frac{5}{2}}(c+d x)} \]

[Out]

(-4*a^3*(9*A + 5*B - 5*C)*EllipticE[(c + d*x)/2, 2])/(5*d) + (4*a^3*(3*A + 5*(B + C))*EllipticF[(c + d*x)/2, 2
])/(3*d) - (4*a^3*(21*A + 20*B + 5*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(15*d) + (2*A*(a + a*Cos[c + d*x])^3*Si
n[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + (2*(6*A + 5*B)*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(15*a*d*Cos[c +
 d*x]^(3/2)) + (2*(33*A + 35*B + 15*C)*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/(15*d*Sqrt[Cos[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.677917, antiderivative size = 230, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.163, Rules used = {3043, 2975, 2968, 3023, 2748, 2641, 2639} \[ \frac{4 a^3 (3 A+5 (B+C)) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{4 a^3 (9 A+5 B-5 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}-\frac{4 a^3 (21 A+20 B+5 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{15 d}+\frac{2 (33 A+35 B+15 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{15 d \sqrt{\cos (c+d x)}}+\frac{2 (6 A+5 B) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{15 a d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d \cos ^{\frac{5}{2}}(c+d x)} \]

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))/Cos[c + d*x]^(7/2),x]

[Out]

(-4*a^3*(9*A + 5*B - 5*C)*EllipticE[(c + d*x)/2, 2])/(5*d) + (4*a^3*(3*A + 5*(B + C))*EllipticF[(c + d*x)/2, 2
])/(3*d) - (4*a^3*(21*A + 20*B + 5*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(15*d) + (2*A*(a + a*Cos[c + d*x])^3*Si
n[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + (2*(6*A + 5*B)*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(15*a*d*Cos[c +
 d*x]^(3/2)) + (2*(33*A + 35*B + 15*C)*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/(15*d*Sqrt[Cos[c + d*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 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 \frac{(a+a \cos (c+d x))^3 \left (A+B \cos (c+d x)+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 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 \int \frac{(a+a \cos (c+d x))^3 \left (\frac{1}{2} a (6 A+5 B)-\frac{1}{2} a (3 A-5 C) \cos (c+d x)\right )}{\cos ^{\frac{5}{2}}(c+d x)} \, dx}{5 a}\\ &=\frac{2 A (a+a \cos (c+d x))^3 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 (6 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{15 a d \cos ^{\frac{3}{2}}(c+d x)}+\frac{4 \int \frac{(a+a \cos (c+d x))^2 \left (\frac{1}{4} a^2 (33 A+35 B+15 C)-\frac{3}{4} a^2 (9 A+5 B-5 C) \cos (c+d x)\right )}{\cos ^{\frac{3}{2}}(c+d x)} \, dx}{15 a}\\ &=\frac{2 A (a+a \cos (c+d x))^3 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 (6 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{15 a d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 (33 A+35 B+15 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d \sqrt{\cos (c+d x)}}+\frac{8 \int \frac{(a+a \cos (c+d x)) \left (\frac{3}{4} a^3 (12 A+15 B+10 C)-\frac{3}{4} a^3 (21 A+20 B+5 C) \cos (c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx}{15 a}\\ &=\frac{2 A (a+a \cos (c+d x))^3 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 (6 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{15 a d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 (33 A+35 B+15 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d \sqrt{\cos (c+d x)}}+\frac{8 \int \frac{\frac{3}{4} a^4 (12 A+15 B+10 C)+\left (-\frac{3}{4} a^4 (21 A+20 B+5 C)+\frac{3}{4} a^4 (12 A+15 B+10 C)\right ) \cos (c+d x)-\frac{3}{4} a^4 (21 A+20 B+5 C) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)}} \, dx}{15 a}\\ &=-\frac{4 a^3 (21 A+20 B+5 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{15 d}+\frac{2 A (a+a \cos (c+d x))^3 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 (6 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{15 a d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 (33 A+35 B+15 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d \sqrt{\cos (c+d x)}}+\frac{16 \int \frac{\frac{15}{8} a^4 (3 A+5 (B+C))-\frac{9}{8} a^4 (9 A+5 B-5 C) \cos (c+d x)}{\sqrt{\cos (c+d x)}} \, dx}{45 a}\\ &=-\frac{4 a^3 (21 A+20 B+5 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{15 d}+\frac{2 A (a+a \cos (c+d x))^3 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 (6 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{15 a d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 (33 A+35 B+15 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d \sqrt{\cos (c+d x)}}-\frac{1}{5} \left (2 a^3 (9 A+5 B-5 C)\right ) \int \sqrt{\cos (c+d x)} \, dx+\frac{1}{3} \left (2 a^3 (3 A+5 (B+C))\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{4 a^3 (9 A+5 B-5 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{4 a^3 (3 A+5 (B+C)) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{4 a^3 (21 A+20 B+5 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{15 d}+\frac{2 A (a+a \cos (c+d x))^3 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 (6 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{15 a d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 (33 A+35 B+15 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d \sqrt{\cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 6.80837, size = 1298, normalized size = 5.64 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*(-((-36*A - 25*B + 5*C + 5*B*Cos[2*c] + 15*C*Co
s[2*c])*Csc[c]*Sec[c])/(40*d) + (C*Cos[d*x]*Sin[c])/(12*d) + (C*Cos[c]*Sin[d*x])/(12*d) + (A*Sec[c]*Sec[c + d*
x]^3*Sin[d*x])/(20*d) + (Sec[c]*Sec[c + d*x]^2*(3*A*Sin[c] + 15*A*Sin[d*x] + 5*B*Sin[d*x]))/(60*d) + (Sec[c]*S
ec[c + d*x]*(15*A*Sin[c] + 5*B*Sin[c] + 54*A*Sin[d*x] + 45*B*Sin[d*x] + 15*C*Sin[d*x]))/(60*d)) - (A*(a + a*Co
s[c + d*x])^3*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^6*Se
c[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[
Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(2*d*Sqrt[1 + Cot[c]^2]) - (5*B*(a + a*Cos[c + d*x])^3*Csc[c]*
HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^6*Sec[d*x - ArcTan[Cot[c]
]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + S
in[d*x - ArcTan[Cot[c]]]])/(6*d*Sqrt[1 + Cot[c]^2]) - (5*C*(a + a*Cos[c + d*x])^3*Csc[c]*HypergeometricPFQ[{1/
4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^6*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x -
 ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c
]]]])/(6*d*Sqrt[1 + Cot[c]^2]) + (9*A*(a + a*Cos[c + d*x])^3*Csc[c]*Sec[c/2 + (d*x)/2]^6*((HypergeometricPFQ[{
-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[
Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1
 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]
*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(20*d)
 + (B*(a + a*Cos[c + d*x])^3*Csc[c]*Sec[c/2 + (d*x)/2]^6*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + Ar
cTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + Arc
Tan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcT
an[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 +
 Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(4*d) - (C*(a + a*Cos[c + d*x])^3*Csc[
c]*Sec[c/2 + (d*x)/2]^6*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan
[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x
 + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[
c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x +
 ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(4*d)

________________________________________________________________________________________

Maple [B]  time = 0.893, size = 1328, normalized size = 5.8 \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)/cos(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*(-216*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+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))+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))+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))-50*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*
c)^2-120*C*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+90*C*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-20*C*sin(1/2*d
*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+60*B*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)*(si
n(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4+100*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4-60*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2-60*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+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))*si
n(1/2*d*x+1/2*c)^4+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)^4-100*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)*Ellipt
icF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2-60*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)*EllipticF(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)*(s
in(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)^2-100*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+190*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+246*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-72*A*cos(
1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+40*C*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)-180*B*cos(1/2*d*x+1/2*c)*sin(
1/2*d*x+1/2*c)^6)*(-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 \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}}{\cos \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+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

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

[Out]

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