3.529 \(\int \frac{(A+B \cos (c+d x)) \sec ^{\frac{9}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx\)
Optimal. Leaf size=317 \[ \frac{(19 A-15 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(11 A-7 B) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{14 a d \sqrt{a \cos (c+d x)+a}}-\frac{(A-B) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac{(67 A-63 B) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{70 a d \sqrt{a \cos (c+d x)+a}}+\frac{(397 A-273 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{210 a d \sqrt{a \cos (c+d x)+a}}-\frac{(1201 A-1029 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{210 a d \sqrt{a \cos (c+d x)+a}} \]
[Out]
((19*A - 15*B)*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 - 1029*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(210*
a*d*Sqrt[a + a*Cos[c + d*x]]) + ((397*A - 273*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(210*a*d*Sqrt[a + a*Cos[c +
d*x]]) - ((67*A - 63*B)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(70*a*d*Sqrt[a + a*Cos[c + d*x]]) - ((A - B)*Sec[c +
d*x]^(7/2)*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) + ((11*A - 7*B)*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(14
*a*d*Sqrt[a + a*Cos[c + d*x]])
________________________________________________________________________________________
Rubi [A] time = 1.10609, antiderivative size = 317, normalized size of antiderivative = 1.,
number of steps used = 9, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used =
{2961, 2978, 2984, 12, 2782, 205} \[ \frac{(19 A-15 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(11 A-7 B) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{14 a d \sqrt{a \cos (c+d x)+a}}-\frac{(A-B) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac{(67 A-63 B) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{70 a d \sqrt{a \cos (c+d x)+a}}+\frac{(397 A-273 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{210 a d \sqrt{a \cos (c+d x)+a}}-\frac{(1201 A-1029 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{210 a d \sqrt{a \cos (c+d x)+a}} \]
Antiderivative was successfully verified.
[In]
Int[((A + B*Cos[c + d*x])*Sec[c + d*x]^(9/2))/(a + a*Cos[c + d*x])^(3/2),x]
[Out]
((19*A - 15*B)*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 - 1029*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(210*
a*d*Sqrt[a + a*Cos[c + d*x]]) + ((397*A - 273*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(210*a*d*Sqrt[a + a*Cos[c +
d*x]]) - ((67*A - 63*B)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(70*a*d*Sqrt[a + a*Cos[c + d*x]]) - ((A - B)*Sec[c +
d*x]^(7/2)*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) + ((11*A - 7*B)*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(14
*a*d*Sqrt[a + a*Cos[c + d*x]])
Rule 2961
Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p, Int[((a + b*Sin[e + f*x])^m*(
c + d*Sin[e + f*x])^n)/(g*Sin[e + f*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d
, 0] && !IntegerQ[p] && !(IntegerQ[m] && IntegerQ[n])
Rule 2978
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*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*
x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[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] && LtQ[m, -2^(-1)] && !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
0])
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{(A+B \cos (c+d x)) \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)}{\cos ^{\frac{9}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx\\ &=-\frac{(A-B) \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)-4 a (A-B) \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) \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) \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)+\frac{3}{2} a^2 (11 A-7 B) \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) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{70 a d \sqrt{a+a \cos (c+d x)}}-\frac{(A-B) \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) \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)-\frac{1}{2} a^3 (67 A-63 B) \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) \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) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{70 a d \sqrt{a+a \cos (c+d x)}}-\frac{(A-B) \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) \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)+\frac{1}{8} a^4 (397 A-273 B) \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) \sqrt{\sec (c+d x)} \sin (c+d x)}{210 a d \sqrt{a+a \cos (c+d x)}}+\frac{(397 A-273 B) \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) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{70 a d \sqrt{a+a \cos (c+d x)}}-\frac{(A-B) \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) \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)}{32 \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{105 a^6}\\ &=-\frac{(1201 A-1029 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{210 a d \sqrt{a+a \cos (c+d x)}}+\frac{(397 A-273 B) \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) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{70 a d \sqrt{a+a \cos (c+d x)}}-\frac{(A-B) \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) \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) \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) \sqrt{\sec (c+d x)} \sin (c+d x)}{210 a d \sqrt{a+a \cos (c+d x)}}+\frac{(397 A-273 B) \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) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{70 a d \sqrt{a+a \cos (c+d x)}}-\frac{(A-B) \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) \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) \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) \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) \sqrt{\sec (c+d x)} \sin (c+d x)}{210 a d \sqrt{a+a \cos (c+d x)}}+\frac{(397 A-273 B) \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) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{70 a d \sqrt{a+a \cos (c+d x)}}-\frac{(A-B) \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) \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 = 10.3343, size = 2966, normalized size = 9.36 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[((A + B*Cos[c + d*x])*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]*(-((A - B)*(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)*(1 + 2*S
in[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)*(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)*(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)*Csc[c/2 + (d*x)/2]^9*(363825*Si
n[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 - 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 + 213120
160*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*Hypergeom
etric2F1[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*Hyper
geometricPFQ[{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]^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*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*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*Ar
cTanh[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*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*Hypergeomet
ricPFQ[{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]^1
2*(-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))))/(d*(a*(1 + Cos[c + d
*x]))^(3/2))
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Maple [B] time = 0.72, size = 731, 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+B*cos(d*x+c))*sec(d*x+c)^(9/2)/(a+cos(d*x+c)*a)^(3/2),x)
[Out]
1/420/d*2^(1/2)/a^2*(1995*A*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*cos(d*x+c)^4*sin(d*x+c)*arcsin((-1+cos(d*x+c))/s
in(d*x+c))-1575*B*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*cos(d*x+c)^4*sin(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))
+7980*A*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*cos(d*x+c)^3*sin(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))-6300*B*(c
os(d*x+c)/(1+cos(d*x+c)))^(7/2)*cos(d*x+c)^3*sin(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+11970*A*(cos(d*x+c)
/(1+cos(d*x+c)))^(7/2)*cos(d*x+c)^2*sin(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))-9450*B*(cos(d*x+c)/(1+cos(d*
x+c)))^(7/2)*cos(d*x+c)^2*sin(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+7980*A*(cos(d*x+c)/(1+cos(d*x+c)))^(7/
2)*cos(d*x+c)*sin(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))-6300*B*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*cos(d*x+c
)*sin(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+1995*A*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*sin(d*x+c)*arcsin((-1
+cos(d*x+c))/sin(d*x+c))-1575*B*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*sin(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c)
)-1201*A*cos(d*x+c)^5*2^(1/2)+1029*B*cos(d*x+c)^5*2^(1/2)+397*A*cos(d*x+c)^4*2^(1/2)-273*B*cos(d*x+c)^4*2^(1/2
)+1000*A*cos(d*x+c)^3*2^(1/2)-840*B*cos(d*x+c)^3*2^(1/2)-232*A*cos(d*x+c)^2*2^(1/2)+168*B*cos(d*x+c)^2*2^(1/2)
+96*A*cos(d*x+c)*2^(1/2)-84*B*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+B*cos(d*x+c))*sec(d*x+c)^(9/2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="maxima")
[Out]
Timed out
________________________________________________________________________________________
Fricas [A] time = 2.90661, size = 645, normalized size = 2.03 \begin{align*} -\frac{105 \, \sqrt{2}{\left ({\left (19 \, A - 15 \, B\right )} \cos \left (d x + c\right )^{5} + 2 \,{\left (19 \, A - 15 \, B\right )} \cos \left (d x + c\right )^{4} +{\left (19 \, A - 15 \, B\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\right )} \cos \left (d x + c\right )^{4} + 12 \,{\left (67 \, A - 63 \, B\right )} \cos \left (d x + c\right )^{3} - 28 \,{\left (7 \, A - 3 \, B\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))*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)*cos(d*x + c)^5 + 2*(19*A - 15*B)*cos(d*x + c)^4 + (19*A - 15*B)*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)*cos(d*x + c)^4 + 12*(67*A - 63*B)*cos(d*x + c)^3 - 28*(7*A - 3*B)*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*co
s(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))*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 (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))*sec(d*x+c)^(9/2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="giac")
[Out]
integrate((B*cos(d*x + c) + A)*sec(d*x + c)^(9/2)/(a*cos(d*x + c) + a)^(3/2), x)