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3.1346 \(\int \frac{(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac{9}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=333 \[ \frac{(19 A-15 B+11 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(11 A-7 B+7 C) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{14 a d \sqrt{a \cos (c+d x)+a}}-\frac{(A-B+C) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac{(67 A-63 B+35 C) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{70 a d \sqrt{a \cos (c+d x)+a}}+\frac{(397 A-273 B+245 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{210 a d \sqrt{a \cos (c+d x)+a}}-\frac{(1201 A-1029 B+665 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{210 a d \sqrt{a \cos (c+d x)+a}} \]

[Out]

((19*A - 15*B + 11*C)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])]*Sqr
t[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(2*Sqrt[2]*a^(3/2)*d) - ((1201*A - 1029*B + 665*C)*Sqrt[Sec[c + d*x]]*Sin[
c + d*x])/(210*a*d*Sqrt[a + a*Cos[c + d*x]]) + ((397*A - 273*B + 245*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(210*
a*d*Sqrt[a + a*Cos[c + d*x]]) - ((67*A - 63*B + 35*C)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(70*a*d*Sqrt[a + a*Cos[
c + d*x]]) - ((A - B + C)*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) + ((11*A - 7*B + 7
*C)*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(14*a*d*Sqrt[a + a*Cos[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 1.2265, antiderivative size = 333, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {4221, 3041, 2984, 12, 2782, 205} \[ \frac{(19 A-15 B+11 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(11 A-7 B+7 C) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{14 a d \sqrt{a \cos (c+d x)+a}}-\frac{(A-B+C) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac{(67 A-63 B+35 C) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{70 a d \sqrt{a \cos (c+d x)+a}}+\frac{(397 A-273 B+245 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{210 a d \sqrt{a \cos (c+d x)+a}}-\frac{(1201 A-1029 B+665 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{210 a d \sqrt{a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(9/2))/(a + a*Cos[c + d*x])^(3/2),x]

[Out]

((19*A - 15*B + 11*C)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])]*Sqr
t[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(2*Sqrt[2]*a^(3/2)*d) - ((1201*A - 1029*B + 665*C)*Sqrt[Sec[c + d*x]]*Sin[
c + d*x])/(210*a*d*Sqrt[a + a*Cos[c + d*x]]) + ((397*A - 273*B + 245*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(210*
a*d*Sqrt[a + a*Cos[c + d*x]]) - ((67*A - 63*B + 35*C)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(70*a*d*Sqrt[a + a*Cos[
c + d*x]]) - ((A - B + C)*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) + ((11*A - 7*B + 7
*C)*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(14*a*d*Sqrt[a + a*Cos[c + d*x]])

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 3041

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[((a*A - b*B + a*C)*Cos[e + f*x]*(
a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(f*(b*c - a*d)*(2*m + 1)), x] + Dist[1/(b*(b*c - a*d)*(2*m
 + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) + B*(
b*c*m + a*d*(n + 1)) - C*(a*c*m + b*d*(n + 1)) + (d*(a*A - b*B)*(m + n + 2) + C*(b*c*(2*m + 1) - a*d*(m - n -
1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^
2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)]

Rule 2984

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*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^(n + 1))/(f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin
[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e +
f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2
 - d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m + 1/2, 0])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2782

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> D
ist[(-2*a)/f, Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, (b*Cos[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c
+ d*Sin[e + f*x]])], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 -
 d^2, 0]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac{9}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac{9}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx\\ &=-\frac{(A-B+C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{2} a (11 A-7 B+7 C)-2 a (2 A-2 B+C) \cos (c+d x)}{\cos ^{\frac{9}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{2 a^2}\\ &=-\frac{(A-B+C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(11 A-7 B+7 C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{14 a d \sqrt{a+a \cos (c+d x)}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{1}{4} a^2 (67 A-63 B+35 C)+\frac{3}{2} a^2 (11 A-7 B+7 C) \cos (c+d x)}{\cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{7 a^3}\\ &=-\frac{(67 A-63 B+35 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{70 a d \sqrt{a+a \cos (c+d x)}}-\frac{(A-B+C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(11 A-7 B+7 C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{14 a d \sqrt{a+a \cos (c+d x)}}+\frac{\left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{8} a^3 (397 A-273 B+245 C)-\frac{1}{2} a^3 (67 A-63 B+35 C) \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{35 a^4}\\ &=\frac{(397 A-273 B+245 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{210 a d \sqrt{a+a \cos (c+d x)}}-\frac{(67 A-63 B+35 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{70 a d \sqrt{a+a \cos (c+d x)}}-\frac{(A-B+C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(11 A-7 B+7 C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{14 a d \sqrt{a+a \cos (c+d x)}}+\frac{\left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{1}{16} a^4 (1201 A-1029 B+665 C)+\frac{1}{8} a^4 (397 A-273 B+245 C) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{105 a^5}\\ &=-\frac{(1201 A-1029 B+665 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{210 a d \sqrt{a+a \cos (c+d x)}}+\frac{(397 A-273 B+245 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{210 a d \sqrt{a+a \cos (c+d x)}}-\frac{(67 A-63 B+35 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{70 a d \sqrt{a+a \cos (c+d x)}}-\frac{(A-B+C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(11 A-7 B+7 C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{14 a d \sqrt{a+a \cos (c+d x)}}+\frac{\left (8 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{105 a^5 (19 A-15 B+11 C)}{32 \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{105 a^6}\\ &=-\frac{(1201 A-1029 B+665 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{210 a d \sqrt{a+a \cos (c+d x)}}+\frac{(397 A-273 B+245 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{210 a d \sqrt{a+a \cos (c+d x)}}-\frac{(67 A-63 B+35 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{70 a d \sqrt{a+a \cos (c+d x)}}-\frac{(A-B+C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(11 A-7 B+7 C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{14 a d \sqrt{a+a \cos (c+d x)}}+\frac{\left ((19 A-15 B+11 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{4 a}\\ &=-\frac{(1201 A-1029 B+665 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{210 a d \sqrt{a+a \cos (c+d x)}}+\frac{(397 A-273 B+245 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{210 a d \sqrt{a+a \cos (c+d x)}}-\frac{(67 A-63 B+35 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{70 a d \sqrt{a+a \cos (c+d x)}}-\frac{(A-B+C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(11 A-7 B+7 C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{14 a d \sqrt{a+a \cos (c+d x)}}-\frac{\left ((19 A-15 B+11 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{2 a^2+a x^2} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\right )}{2 d}\\ &=\frac{(19 A-15 B+11 C) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}}{2 \sqrt{2} a^{3/2} d}-\frac{(1201 A-1029 B+665 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{210 a d \sqrt{a+a \cos (c+d x)}}+\frac{(397 A-273 B+245 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{210 a d \sqrt{a+a \cos (c+d x)}}-\frac{(67 A-63 B+35 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{70 a d \sqrt{a+a \cos (c+d x)}}-\frac{(A-B+C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(11 A-7 B+7 C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{14 a d \sqrt{a+a \cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 10.1321, size = 3136, normalized size = 9.42 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(9/2))/(a + a*Cos[c + d*x])^(3/2),x]

[Out]

(2*Cos[c/2 + (d*x)/2]^3*Sqrt[(1 - 2*Sin[c/2 + (d*x)/2]^2)^(-1)]*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]*((4*C*Sin[c/2
 + (d*x)/2])/(7*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(7/2)) - ((A - B + C)*(1 - 2*Sin[c/2 + (d*x)/2]))/(28*(1 + Sin[c/
2 + (d*x)/2])*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(7/2)) + ((A - B + C)*(1 + 2*Sin[c/2 + (d*x)/2]))/(28*(1 - Sin[c/2
+ (d*x)/2])*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(7/2)) - ((A - B + C)*(315*ArcTan[(1 - 2*Sin[c/2 + (d*x)/2])/Sqrt[1 -
 2*Sin[c/2 + (d*x)/2]^2]] + (5 + 3*Sin[c/2 + (d*x)/2])/((1 - Sin[c/2 + (d*x)/2])*(1 - 2*Sin[c/2 + (d*x)/2]^2)^
(5/2)) - (11 + 17*Sin[c/2 + (d*x)/2])/((1 - Sin[c/2 + (d*x)/2])*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(3/2)) + (61 + 71
*Sin[c/2 + (d*x)/2])/((1 - Sin[c/2 + (d*x)/2])*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]) + (193*Sqrt[1 - 2*Sin[c/2 + (
d*x)/2]^2])/(1 - Sin[c/2 + (d*x)/2])))/70 + ((A - B + C)*(315*ArcTan[(1 + 2*Sin[c/2 + (d*x)/2])/Sqrt[1 - 2*Sin
[c/2 + (d*x)/2]^2]] + (5 - 3*Sin[c/2 + (d*x)/2])/((1 + Sin[c/2 + (d*x)/2])*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(5/2))
 - (11 - 17*Sin[c/2 + (d*x)/2])/((1 + Sin[c/2 + (d*x)/2])*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(3/2)) + (61 - 71*Sin[c
/2 + (d*x)/2])/((1 + Sin[c/2 + (d*x)/2])*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]) + (193*Sqrt[1 - 2*Sin[c/2 + (d*x)/2
]^2])/(1 + Sin[c/2 + (d*x)/2])))/70 - ((-A - 3*B + 7*C)*Csc[c/2 + (d*x)/2]^9*(363825*Sin[c/2 + (d*x)/2]^2 - 47
29725*Sin[c/2 + (d*x)/2]^4 + 26785605*Sin[c/2 + (d*x)/2]^6 - 86790165*Sin[c/2 + (d*x)/2]^8 + 177677808*Sin[c/2
 + (d*x)/2]^10 - 239283044*Sin[c/2 + (d*x)/2]^12 + 52080*Hypergeometric2F1[2, 11/2, 13/2, Sin[c/2 + (d*x)/2]^2
/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^12 + 560*HypergeometricPFQ[{2, 2, 2, 2, 11/2}, {1, 1, 1, 13
/2}, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^12 + 213120160*Sin[c/2 + (d*x)/2]^
14 - 168280*Hypergeometric2F1[2, 11/2, 13/2, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*
x)/2]^14 - 2240*HypergeometricPFQ[{2, 2, 2, 2, 11/2}, {1, 1, 1, 13/2}, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 +
(d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^14 - 121497024*Sin[c/2 + (d*x)/2]^16 + 212520*Hypergeometric2F1[2, 11/2, 13/2,
 Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^16 + 3360*HypergeometricPFQ[{2, 2, 2,
2, 11/2}, {1, 1, 1, 13/2}, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^16 + 4012518
4*Sin[c/2 + (d*x)/2]^18 - 124320*Hypergeometric2F1[2, 11/2, 13/2, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)
/2]^2)]*Sin[c/2 + (d*x)/2]^18 - 2240*HypergeometricPFQ[{2, 2, 2, 2, 11/2}, {1, 1, 1, 13/2}, Sin[c/2 + (d*x)/2]
^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^18 - 5840384*Sin[c/2 + (d*x)/2]^20 + 28000*Hypergeometric
2F1[2, 11/2, 13/2, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^20 + 560*Hypergeomet
ricPFQ[{2, 2, 2, 2, 11/2}, {1, 1, 1, 13/2}, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x
)/2]^20 + 363825*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sqrt[Sin[c/2 + (d*x)/2]^2/(
-1 + 2*Sin[c/2 + (d*x)/2]^2)] - 5336100*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[
c/2 + (d*x)/2]^2*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] + 34636140*ArcTanh[Sqrt[Sin[c/2 + (d
*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^4*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x
)/2]^2)] - 131060160*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^6*Sq
rt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] + 320535600*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*S
in[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^8*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] - 5306716
80*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^10*Sqrt[Sin[c/2 + (d*x
)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] + 604296000*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]
^2)]]*Sin[c/2 + (d*x)/2]^12*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] - 468948480*ArcTanh[Sqrt[
Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^14*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*S
in[c/2 + (d*x)/2]^2)] + 237726720*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 +
(d*x)/2]^16*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] - 70963200*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2
]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^18*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]
^2)] + 9461760*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^20*Sqrt[Si
n[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] - 1120*Cos[(c + d*x)/2]^6*HypergeometricPFQ[{2, 2, 2, 11/2},
 {1, 1, 13/2}, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^12*(-6 + 5*Sin[c/2 + (d*
x)/2]^2) + 280*Cos[(c + d*x)/2]^4*HypergeometricPFQ[{2, 2, 11/2}, {1, 13/2}, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[
c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^12*(103 - 164*Sin[c/2 + (d*x)/2]^2 + 70*Sin[c/2 + (d*x)/2]^4)))/(80850*(
1 - 2*Sin[c/2 + (d*x)/2]^2)^(9/2)*(-1 + 2*Sin[c/2 + (d*x)/2]^2)) + (8*C*((3*Sin[c/2 + (d*x)/2])/(1 - 2*Sin[c/2
 + (d*x)/2]^2)^(5/2) + 4*(Sin[c/2 + (d*x)/2]/(1 - 2*Sin[c/2 + (d*x)/2]^2)^(3/2) + (2*Sin[c/2 + (d*x)/2])/Sqrt[
1 - 2*Sin[c/2 + (d*x)/2]^2])))/35))/(d*(a*(1 + Cos[c + d*x]))^(3/2))

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Maple [B]  time = 0.2, size = 1047, normalized size = 3.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(9/2)/(a+a*cos(d*x+c))^(3/2),x)

[Out]

1/420/d*2^(1/2)/a^2*(-1201*A*cos(d*x+c)^5*2^(1/2)-1575*B*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arcsin((-1+cos(d*x+
c))/sin(d*x+c))*sin(d*x+c)+1155*C*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x
+c)+1995*A*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)-60*A*2^(1/2)-665*C*
cos(d*x+c)^5*2^(1/2)+4620*C*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^3*
sin(d*x+c)+11970*A*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^2*sin(d*x+c
)-9450*B*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^2*sin(d*x+c)+6930*C*(
cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^2*sin(d*x+c)+7980*A*(cos(d*x+c)
/(1+cos(d*x+c)))^(7/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)*sin(d*x+c)-6300*B*(cos(d*x+c)/(1+cos(d*x+
c)))^(7/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)*sin(d*x+c)+4620*C*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*a
rcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)*sin(d*x+c)-840*B*2^(1/2)*cos(d*x+c)^3+168*B*2^(1/2)*cos(d*x+c)^2-
84*B*2^(1/2)*cos(d*x+c)+1029*B*cos(d*x+c)^5*2^(1/2)-273*B*cos(d*x+c)^4*2^(1/2)+397*A*2^(1/2)*cos(d*x+c)^4+245*
C*2^(1/2)*cos(d*x+c)^4+1000*A*cos(d*x+c)^3*2^(1/2)+560*C*cos(d*x+c)^3*2^(1/2)-232*A*cos(d*x+c)^2*2^(1/2)-140*C
*cos(d*x+c)^2*2^(1/2)+96*A*cos(d*x+c)*2^(1/2)+1995*A*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arcsin((-1+cos(d*x+c))/
sin(d*x+c))*cos(d*x+c)^4*sin(d*x+c)-1575*B*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c)
)*cos(d*x+c)^4*sin(d*x+c)+1155*C*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+
c)^4*sin(d*x+c)+7980*A*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^3*sin(d
*x+c)-6300*B*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^3*sin(d*x+c))*cos
(d*x+c)*(1/cos(d*x+c))^(9/2)*(a*(1+cos(d*x+c)))^(1/2)*sin(d*x+c)^5/(-1+cos(d*x+c))^3/(1+cos(d*x+c))^4

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Maxima [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+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(9/2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 2.74924, size = 702, normalized size = 2.11 \begin{align*} -\frac{105 \, \sqrt{2}{\left ({\left (19 \, A - 15 \, B + 11 \, C\right )} \cos \left (d x + c\right )^{5} + 2 \,{\left (19 \, A - 15 \, B + 11 \, C\right )} \cos \left (d x + c\right )^{4} +{\left (19 \, A - 15 \, B + 11 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \sqrt{a} \arctan \left (\frac{\sqrt{2} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )}}{\sqrt{a} \sin \left (d x + c\right )}\right ) + \frac{2 \,{\left ({\left (1201 \, A - 1029 \, B + 665 \, C\right )} \cos \left (d x + c\right )^{4} + 12 \,{\left (67 \, A - 63 \, B + 35 \, C\right )} \cos \left (d x + c\right )^{3} - 28 \,{\left (7 \, A - 3 \, B + 5 \, C\right )} \cos \left (d x + c\right )^{2} + 12 \,{\left (3 \, A - 7 \, B\right )} \cos \left (d x + c\right ) - 60 \, A\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt{\cos \left (d x + c\right )}}}{420 \,{\left (a^{2} d \cos \left (d x + c\right )^{5} + 2 \, a^{2} d \cos \left (d x + c\right )^{4} + a^{2} d \cos \left (d x + c\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(9/2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

-1/420*(105*sqrt(2)*((19*A - 15*B + 11*C)*cos(d*x + c)^5 + 2*(19*A - 15*B + 11*C)*cos(d*x + c)^4 + (19*A - 15*
B + 11*C)*cos(d*x + c)^3)*sqrt(a)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x
+ c))) + 2*((1201*A - 1029*B + 665*C)*cos(d*x + c)^4 + 12*(67*A - 63*B + 35*C)*cos(d*x + c)^3 - 28*(7*A - 3*B
+ 5*C)*cos(d*x + c)^2 + 12*(3*A - 7*B)*cos(d*x + c) - 60*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x
 + c)))/(a^2*d*cos(d*x + c)^5 + 2*a^2*d*cos(d*x + c)^4 + a^2*d*cos(d*x + c)^3)

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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+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**(9/2)/(a+a*cos(d*x+c))**(3/2),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 )} \sec \left (d x + c\right )^{\frac{9}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(9/2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sec(d*x + c)^(9/2)/(a*cos(d*x + c) + a)^(3/2), x)

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