3.1239 \(\int \frac{(A+C \cos ^2(c+d x)) \sec ^{\frac{9}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx\)
Optimal. Leaf size=315 \[ \frac{(19 A+11 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(11 A+7 C) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{14 a d \sqrt{a \cos (c+d x)+a}}-\frac{(A+C) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac{(67 A+35 C) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{70 a d \sqrt{a \cos (c+d x)+a}}+\frac{(397 A+245 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{210 a d \sqrt{a \cos (c+d x)+a}}-\frac{(1201 A+665 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{210 a d \sqrt{a \cos (c+d x)+a}} \]
[Out]
((19*A + 11*C)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])]*Sqrt[Cos[c
+ d*x]]*Sqrt[Sec[c + d*x]])/(2*Sqrt[2]*a^(3/2)*d) - ((1201*A + 665*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(210*a
*d*Sqrt[a + a*Cos[c + d*x]]) + ((397*A + 245*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(210*a*d*Sqrt[a + a*Cos[c + d
*x]]) - ((67*A + 35*C)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(70*a*d*Sqrt[a + a*Cos[c + d*x]]) - ((A + C)*Sec[c + d
*x]^(7/2)*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) + ((11*A + 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 = 1.08455, antiderivative size = 315, normalized size of antiderivative = 1.,
number of steps used = 9, number of rules used = 6, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.162, Rules used =
{4221, 3042, 2984, 12, 2782, 205} \[ \frac{(19 A+11 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(11 A+7 C) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{14 a d \sqrt{a \cos (c+d x)+a}}-\frac{(A+C) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac{(67 A+35 C) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{70 a d \sqrt{a \cos (c+d x)+a}}+\frac{(397 A+245 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{210 a d \sqrt{a \cos (c+d x)+a}}-\frac{(1201 A+665 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{210 a d \sqrt{a \cos (c+d x)+a}} \]
Antiderivative was successfully verified.
[In]
Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(9/2))/(a + a*Cos[c + d*x])^(3/2),x]
[Out]
((19*A + 11*C)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])]*Sqrt[Cos[c
+ d*x]]*Sqrt[Sec[c + d*x]])/(2*Sqrt[2]*a^(3/2)*d) - ((1201*A + 665*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(210*a
*d*Sqrt[a + a*Cos[c + d*x]]) + ((397*A + 245*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(210*a*d*Sqrt[a + a*Cos[c + d
*x]]) - ((67*A + 35*C)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(70*a*d*Sqrt[a + a*Cos[c + d*x]]) - ((A + C)*Sec[c + d
*x]^(7/2)*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) + ((11*A + 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 3042
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[(a*(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)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(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, 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+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+C \cos ^2(c+d x)}{\cos ^{\frac{9}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx\\ &=-\frac{(A+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 C)-2 a (2 A+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+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 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+35 C)+\frac{3}{2} a^2 (11 A+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+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+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 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+245 C)-\frac{1}{2} a^3 (67 A+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+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+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+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 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+665 C)+\frac{1}{8} a^4 (397 A+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+665 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{210 a d \sqrt{a+a \cos (c+d x)}}+\frac{(397 A+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+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+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 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+11 C)}{32 \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{105 a^6}\\ &=-\frac{(1201 A+665 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{210 a d \sqrt{a+a \cos (c+d x)}}+\frac{(397 A+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+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+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 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+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+665 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{210 a d \sqrt{a+a \cos (c+d x)}}+\frac{(397 A+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+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+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 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+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+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+665 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{210 a d \sqrt{a+a \cos (c+d x)}}+\frac{(397 A+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+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+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 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 = 10.5468, size = 3121, normalized size = 9.91 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[((A + 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 + 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 + 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 + 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 + 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 + 7*C)*Csc[c/2 + (d*x)/2]^9*(363825*Sin[c/2 + (d*x)/2]^2 - 4729725*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 - 23928
3044*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*Hypergeome
tric2F1[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*Hyperg
eometricPFQ[{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 + 40125184*Sin[c/2 + (d*x)/2]^1
8 - 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*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]^20 + 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]^20 + 363825*ArcTa
nh[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*A
rcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^6*Sqrt[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*Sin[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)] - 530671680*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*Sin[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[Sin[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 = 0.244, size = 719, normalized size = 2.3 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+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*(-1995*A*cos(d*x+c)^4*sin(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*(cos(d*x+c)/(1+cos(d*
x+c)))^(7/2)-1155*C*cos(d*x+c)^4*sin(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(7/
2)-7980*A*cos(d*x+c)^3*sin(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)-4620*C*
cos(d*x+c)^3*sin(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)-11970*A*cos(d*x+c
)^2*sin(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)-6930*C*cos(d*x+c)^2*sin(d*
x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)-7980*A*cos(d*x+c)*sin(d*x+c)*arcsin(
(-1+cos(d*x+c))/sin(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)-4620*C*cos(d*x+c)*sin(d*x+c)*arcsin((-1+cos(d*x+
c))/sin(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)-1995*A*sin(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*(cos(d*
x+c)/(1+cos(d*x+c)))^(7/2)-1155*C*sin(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(7
/2)+1201*A*cos(d*x+c)^5*2^(1/2)+665*C*cos(d*x+c)^5*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)+60*A*2^(1/2))*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+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 = 2.23885, size = 630, normalized size = 2. \begin{align*} -\frac{105 \, \sqrt{2}{\left ({\left (19 \, A + 11 \, C\right )} \cos \left (d x + c\right )^{5} + 2 \,{\left (19 \, A + 11 \, C\right )} \cos \left (d x + c\right )^{4} +{\left (19 \, A + 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 + 665 \, C\right )} \cos \left (d x + c\right )^{4} + 12 \,{\left (67 \, A + 35 \, C\right )} \cos \left (d x + c\right )^{3} - 28 \,{\left (7 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{2} + 36 \, A \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+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 + 11*C)*cos(d*x + c)^5 + 2*(19*A + 11*C)*cos(d*x + c)^4 + (19*A + 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 +
665*C)*cos(d*x + c)^4 + 12*(67*A + 35*C)*cos(d*x + c)^3 - 28*(7*A + 5*C)*cos(d*x + c)^2 + 36*A*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+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} + 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+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 + A)*sec(d*x + c)^(9/2)/(a*cos(d*x + c) + a)^(3/2), x)