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

Optimal. Leaf size=839 \[ \frac {B \sin (c+d x) (a+b \cos (c+d x))^{7/2}}{5 b d \sqrt {\sec (c+d x)}}+\frac {(10 A b-3 a B) \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{40 b d \sqrt {\sec (c+d x)}}+\frac {\left (-15 B a^2+50 A b a+64 b^2 B\right ) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{240 b d \sqrt {\sec (c+d x)}}+\frac {\left (-45 B a^4+150 A b a^3+1692 b^2 B a^2+2840 A b^3 a+1024 b^4 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{1920 b^2 d}+\frac {\left (-15 B a^3+50 A b a^2+172 b^2 B a+120 A b^3\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{320 b d \sqrt {\sec (c+d x)}}-\frac {(a-b) \sqrt {a+b} \left (-45 B a^4+150 A b a^3+1692 b^2 B a^2+2840 A b^3 a+1024 b^4 B\right ) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}}}{1920 a b^2 d \sqrt {\sec (c+d x)}}-\frac {\sqrt {a+b} \left (45 B a^4-30 b (5 A+B) a^3-4 b^2 (295 A+423 B) a^2-8 b^3 (355 A+193 B) a-16 b^4 (45 A+64 B)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}}}{1920 b^2 d \sqrt {\sec (c+d x)}}+\frac {\sqrt {a+b} \left (-3 B a^5+10 A b a^4-40 b^2 B a^3-240 A b^3 a^2-240 b^4 B a-96 A b^5\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \Pi \left (\frac {a+b}{b};\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}}}{128 b^3 d \sqrt {\sec (c+d x)}} \]

[Out]

1/240*(50*A*a*b-15*B*a^2+64*B*b^2)*(a+b*cos(d*x+c))^(3/2)*sin(d*x+c)/b/d/sec(d*x+c)^(1/2)+1/40*(10*A*b-3*B*a)*
(a+b*cos(d*x+c))^(5/2)*sin(d*x+c)/b/d/sec(d*x+c)^(1/2)+1/5*B*(a+b*cos(d*x+c))^(7/2)*sin(d*x+c)/b/d/sec(d*x+c)^
(1/2)+1/320*(50*A*a^2*b+120*A*b^3-15*B*a^3+172*B*a*b^2)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/b/d/sec(d*x+c)^(1/2)
+1/1920*(150*A*a^3*b+2840*A*a*b^3-45*B*a^4+1692*B*a^2*b^2+1024*B*b^4)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)*sec(d*
x+c)^(1/2)/b^2/d-1/1920*(a-b)*(150*A*a^3*b+2840*A*a*b^3-45*B*a^4+1692*B*a^2*b^2+1024*B*b^4)*csc(d*x+c)*Ellipti
cE((a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),((-a-b)/(a-b))^(1/2))*(a+b)^(1/2)*cos(d*x+c)^(1/2)*(a*(
1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/a/b^2/d/sec(d*x+c)^(1/2)-1/1920*(45*a^4*B-30*a^3*b*(
5*A+B)-16*b^4*(45*A+64*B)-8*a*b^3*(355*A+193*B)-4*a^2*b^2*(295*A+423*B))*csc(d*x+c)*EllipticF((a+b*cos(d*x+c))
^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),((-a-b)/(a-b))^(1/2))*(a+b)^(1/2)*cos(d*x+c)^(1/2)*(a*(1-sec(d*x+c))/(a+b)
)^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/b^2/d/sec(d*x+c)^(1/2)+1/128*(10*A*a^4*b-240*A*a^2*b^3-96*A*b^5-3*B*a^5
-40*B*a^3*b^2-240*B*a*b^4)*csc(d*x+c)*EllipticPi((a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),(a+b)/b,(
(-a-b)/(a-b))^(1/2))*(a+b)^(1/2)*cos(d*x+c)^(1/2)*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2
)/b^3/d/sec(d*x+c)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 3.60, antiderivative size = 839, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {2961, 2990, 3049, 3061, 3053, 2809, 2998, 2816, 2994} \[ \frac {B \sin (c+d x) (a+b \cos (c+d x))^{7/2}}{5 b d \sqrt {\sec (c+d x)}}+\frac {(10 A b-3 a B) \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{40 b d \sqrt {\sec (c+d x)}}+\frac {\left (-15 B a^2+50 A b a+64 b^2 B\right ) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{240 b d \sqrt {\sec (c+d x)}}+\frac {\left (-45 B a^4+150 A b a^3+1692 b^2 B a^2+2840 A b^3 a+1024 b^4 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{1920 b^2 d}+\frac {\left (-15 B a^3+50 A b a^2+172 b^2 B a+120 A b^3\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{320 b d \sqrt {\sec (c+d x)}}-\frac {(a-b) \sqrt {a+b} \left (-45 B a^4+150 A b a^3+1692 b^2 B a^2+2840 A b^3 a+1024 b^4 B\right ) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}}}{1920 a b^2 d \sqrt {\sec (c+d x)}}-\frac {\sqrt {a+b} \left (45 B a^4-30 b (5 A+B) a^3-4 b^2 (295 A+423 B) a^2-8 b^3 (355 A+193 B) a-16 b^4 (45 A+64 B)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}}}{1920 b^2 d \sqrt {\sec (c+d x)}}+\frac {\sqrt {a+b} \left (-3 B a^5+10 A b a^4-40 b^2 B a^3-240 A b^3 a^2-240 b^4 B a-96 A b^5\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \Pi \left (\frac {a+b}{b};\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}}}{128 b^3 d \sqrt {\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a - b)*Sqrt[a + b]*(150*a^3*A*b + 2840*a*A*b^3 - 45*a^4*B + 1692*a^2*b^2*B + 1024*b^4*B)*Sqrt[Cos[c + d*x]]
*Csc[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]
*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(1920*a*b^2*d*Sqrt[Sec[c + d*x]])
- (Sqrt[a + b]*(45*a^4*B - 30*a^3*b*(5*A + B) - 16*b^4*(45*A + 64*B) - 8*a*b^3*(355*A + 193*B) - 4*a^2*b^2*(29
5*A + 423*B))*Sqrt[Cos[c + d*x]]*Csc[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[
c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(1
920*b^2*d*Sqrt[Sec[c + d*x]]) + (Sqrt[a + b]*(10*a^4*A*b - 240*a^2*A*b^3 - 96*A*b^5 - 3*a^5*B - 40*a^3*b^2*B -
240*a*b^4*B)*Sqrt[Cos[c + d*x]]*Csc[c + d*x]*EllipticPi[(a + b)/b, ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a +
b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/
(a - b)])/(128*b^3*d*Sqrt[Sec[c + d*x]]) + ((50*a^2*A*b + 120*A*b^3 - 15*a^3*B + 172*a*b^2*B)*Sqrt[a + b*Cos[c
+ d*x]]*Sin[c + d*x])/(320*b*d*Sqrt[Sec[c + d*x]]) + ((50*a*A*b - 15*a^2*B + 64*b^2*B)*(a + b*Cos[c + d*x])^(
3/2)*Sin[c + d*x])/(240*b*d*Sqrt[Sec[c + d*x]]) + ((10*A*b - 3*a*B)*(a + b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(
40*b*d*Sqrt[Sec[c + d*x]]) + (B*(a + b*Cos[c + d*x])^(7/2)*Sin[c + d*x])/(5*b*d*Sqrt[Sec[c + d*x]]) + ((150*a^
3*A*b + 2840*a*A*b^3 - 45*a^4*B + 1692*a^2*b^2*B + 1024*b^4*B)*Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*Sin
[c + d*x])/(1920*b^2*d)

Rule 2809

Int[Sqrt[(b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[(2*b*Tan
[e + f*x]*Rt[(c + d)/b, 2]*Sqrt[(c*(1 + Csc[e + f*x]))/(c - d)]*Sqrt[(c*(1 - Csc[e + f*x]))/(c + d)]*EllipticP
i[(c + d)/d, ArcSin[Sqrt[c + d*Sin[e + f*x]]/(Sqrt[b*Sin[e + f*x]]*Rt[(c + d)/b, 2])], -((c + d)/(c - d))])/(d
*f), x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2, 0] && PosQ[(c + d)/b]

Rule 2816

Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(-2*
Tan[e + f*x]*Rt[(a + b)/d, 2]*Sqrt[(a*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[(a*(1 + Csc[e + f*x]))/(a - b)]*Ellipt
icF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/(Sqrt[d*Sin[e + f*x]]*Rt[(a + b)/d, 2])], -((a + b)/(a - b))])/(a*f), x] /
; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && PosQ[(a + b)/d]

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 2990

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 - 2)*(c + d*Sin[e + f*x
])^n*Simp[a^2*A*d*(m + n + 1) + b*B*(b*c*(m - 1) + a*d*(n + 1)) + (a*d*(2*A*b + a*B)*(m + n + 1) - b*B*(a*c -
b*d*(m + n)))*Sin[e + f*x] + b*(A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + f*x]^2, x], x], x] /; F
reeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m,
1] &&  !(IGtQ[n, 1] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

Rule 2994

Int[((A_) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]]), x_Symbol] :> Simp[(-2*A*(c - d)*Tan[e + f*x]*Rt[(c + d)/b, 2]*Sqrt[(c*(1 + Csc[e + f*x]))/(c
- d)]*Sqrt[(c*(1 - Csc[e + f*x]))/(c + d)]*EllipticE[ArcSin[Sqrt[c + d*Sin[e + f*x]]/(Sqrt[b*Sin[e + f*x]]*Rt[
(c + d)/b, 2])], -((c + d)/(c - d))])/(f*b*c^2), x] /; FreeQ[{b, c, d, e, f, A, B}, x] && NeQ[c^2 - d^2, 0] &&
EqQ[A, B] && PosQ[(c + d)/b]

Rule 2998

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*s
in[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[(A - B)/(a - b), Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e
+ f*x]]), x], x] - Dist[(A*b - a*B)/(a - b), Int[(1 + Sin[e + f*x])/((a + b*Sin[e + f*x])^(3/2)*Sqrt[c + d*Sin
[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2
- d^2, 0] && NeQ[A, B]

Rule 3049

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((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*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x]
)^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2)
- C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2
, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

Rule 3053

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(((a_.) + (b_.)*sin[(e_.) + (f_.
)*(x_)])^(3/2)*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[C/b^2, Int[Sqrt[a + b*Sin[e + f
*x]]/Sqrt[c + d*Sin[e + f*x]], x], x] + Dist[1/b^2, Int[(A*b^2 - a^2*C + b*(b*B - 2*a*C)*Sin[e + f*x])/((a + b
*Sin[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a
*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3061

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) +
(f_.)*(x_)]]*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> -Simp[(C*Cos[e + f*x]*Sqrt[c + d*Sin[e
+ f*x]])/(d*f*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[1/(2*d), Int[(1*Simp[2*a*A*d - C*(b*c - a*d) - 2*(a*c*C - d
*(A*b + a*B))*Sin[e + f*x] + (2*b*B*d - C*(b*c + a*d))*Sin[e + f*x]^2, x])/((a + b*Sin[e + f*x])^(3/2)*Sqrt[c
+ d*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
&& NeQ[c^2 - d^2, 0]

Rubi steps

\begin {align*} \int \frac {(a+b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx\\ &=\frac {B (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{5 b d \sqrt {\sec (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+b \cos (c+d x))^{5/2} \left (\frac {a B}{2}+4 b B \cos (c+d x)+\frac {1}{2} (10 A b-3 a B) \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx}{5 b}\\ &=\frac {(10 A b-3 a B) (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{40 b d \sqrt {\sec (c+d x)}}+\frac {B (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{5 b d \sqrt {\sec (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+b \cos (c+d x))^{3/2} \left (\frac {5}{4} a (2 A b+a B)+\frac {3}{2} b (10 A b+9 a B) \cos (c+d x)+\frac {1}{4} \left (50 a A b-15 a^2 B+64 b^2 B\right ) \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx}{20 b}\\ &=\frac {\left (50 a A b-15 a^2 B+64 b^2 B\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{240 b d \sqrt {\sec (c+d x)}}+\frac {(10 A b-3 a B) (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{40 b d \sqrt {\sec (c+d x)}}+\frac {B (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{5 b d \sqrt {\sec (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+b \cos (c+d x)} \left (\frac {1}{8} a \left (110 a A b+15 a^2 B+64 b^2 B\right )+\frac {1}{4} b \left (310 a A b+147 a^2 B+128 b^2 B\right ) \cos (c+d x)+\frac {3}{8} \left (50 a^2 A b+120 A b^3-15 a^3 B+172 a b^2 B\right ) \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx}{60 b}\\ &=\frac {\left (50 a^2 A b+120 A b^3-15 a^3 B+172 a b^2 B\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{320 b d \sqrt {\sec (c+d x)}}+\frac {\left (50 a A b-15 a^2 B+64 b^2 B\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{240 b d \sqrt {\sec (c+d x)}}+\frac {(10 A b-3 a B) (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{40 b d \sqrt {\sec (c+d x)}}+\frac {B (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{5 b d \sqrt {\sec (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{16} a \left (590 a^2 A b+360 A b^3+15 a^3 B+772 a b^2 B\right )+\frac {1}{8} b \left (1610 a^2 A b+360 A b^3+573 a^3 B+1156 a b^2 B\right ) \cos (c+d x)+\frac {1}{16} \left (150 a^3 A b+2840 a A b^3-45 a^4 B+1692 a^2 b^2 B+1024 b^4 B\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}} \, dx}{120 b}\\ &=\frac {\left (50 a^2 A b+120 A b^3-15 a^3 B+172 a b^2 B\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{320 b d \sqrt {\sec (c+d x)}}+\frac {\left (50 a A b-15 a^2 B+64 b^2 B\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{240 b d \sqrt {\sec (c+d x)}}+\frac {(10 A b-3 a B) (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{40 b d \sqrt {\sec (c+d x)}}+\frac {B (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{5 b d \sqrt {\sec (c+d x)}}+\frac {\left (150 a^3 A b+2840 a A b^3-45 a^4 B+1692 a^2 b^2 B+1024 b^4 B\right ) \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{1920 b^2 d}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {1}{16} a \left (150 a^3 A b+2840 a A b^3-45 a^4 B+1692 a^2 b^2 B+1024 b^4 B\right )+\frac {1}{8} a b \left (590 a^2 A b+360 A b^3+15 a^3 B+772 a b^2 B\right ) \cos (c+d x)-\frac {15}{16} \left (10 a^4 A b-240 a^2 A b^3-96 A b^5-3 a^5 B-40 a^3 b^2 B-240 a b^4 B\right ) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{240 b^2}\\ &=\frac {\left (50 a^2 A b+120 A b^3-15 a^3 B+172 a b^2 B\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{320 b d \sqrt {\sec (c+d x)}}+\frac {\left (50 a A b-15 a^2 B+64 b^2 B\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{240 b d \sqrt {\sec (c+d x)}}+\frac {(10 A b-3 a B) (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{40 b d \sqrt {\sec (c+d x)}}+\frac {B (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{5 b d \sqrt {\sec (c+d x)}}+\frac {\left (150 a^3 A b+2840 a A b^3-45 a^4 B+1692 a^2 b^2 B+1024 b^4 B\right ) \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{1920 b^2 d}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {1}{16} a \left (150 a^3 A b+2840 a A b^3-45 a^4 B+1692 a^2 b^2 B+1024 b^4 B\right )+\frac {1}{8} a b \left (590 a^2 A b+360 A b^3+15 a^3 B+772 a b^2 B\right ) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{240 b^2}-\frac {\left (\left (10 a^4 A b-240 a^2 A b^3-96 A b^5-3 a^5 B-40 a^3 b^2 B-240 a b^4 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {a+b \cos (c+d x)}} \, dx}{256 b^2}\\ &=\frac {\sqrt {a+b} \left (10 a^4 A b-240 a^2 A b^3-96 A b^5-3 a^5 B-40 a^3 b^2 B-240 a b^4 B\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \Pi \left (\frac {a+b}{b};\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{128 b^3 d \sqrt {\sec (c+d x)}}+\frac {\left (50 a^2 A b+120 A b^3-15 a^3 B+172 a b^2 B\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{320 b d \sqrt {\sec (c+d x)}}+\frac {\left (50 a A b-15 a^2 B+64 b^2 B\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{240 b d \sqrt {\sec (c+d x)}}+\frac {(10 A b-3 a B) (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{40 b d \sqrt {\sec (c+d x)}}+\frac {B (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{5 b d \sqrt {\sec (c+d x)}}+\frac {\left (150 a^3 A b+2840 a A b^3-45 a^4 B+1692 a^2 b^2 B+1024 b^4 B\right ) \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{1920 b^2 d}-\frac {\left (a \left (150 a^3 A b+2840 a A b^3-45 a^4 B+1692 a^2 b^2 B+1024 b^4 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1+\cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{3840 b^2}-\frac {\left (a \left (45 a^4 B-30 a^3 b (5 A+B)-16 b^4 (45 A+64 B)-8 a b^3 (355 A+193 B)-4 a^2 b^2 (295 A+423 B)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}} \, dx}{3840 b^2}\\ &=-\frac {(a-b) \sqrt {a+b} \left (150 a^3 A b+2840 a A b^3-45 a^4 B+1692 a^2 b^2 B+1024 b^4 B\right ) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{1920 a b^2 d \sqrt {\sec (c+d x)}}-\frac {\sqrt {a+b} \left (45 a^4 B-30 a^3 b (5 A+B)-16 b^4 (45 A+64 B)-8 a b^3 (355 A+193 B)-4 a^2 b^2 (295 A+423 B)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{1920 b^2 d \sqrt {\sec (c+d x)}}+\frac {\sqrt {a+b} \left (10 a^4 A b-240 a^2 A b^3-96 A b^5-3 a^5 B-40 a^3 b^2 B-240 a b^4 B\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \Pi \left (\frac {a+b}{b};\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{128 b^3 d \sqrt {\sec (c+d x)}}+\frac {\left (50 a^2 A b+120 A b^3-15 a^3 B+172 a b^2 B\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{320 b d \sqrt {\sec (c+d x)}}+\frac {\left (50 a A b-15 a^2 B+64 b^2 B\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{240 b d \sqrt {\sec (c+d x)}}+\frac {(10 A b-3 a B) (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{40 b d \sqrt {\sec (c+d x)}}+\frac {B (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{5 b d \sqrt {\sec (c+d x)}}+\frac {\left (150 a^3 A b+2840 a A b^3-45 a^4 B+1692 a^2 b^2 B+1024 b^4 B\right ) \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{1920 b^2 d}\\ \end {align*}

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Mathematica [A]  time = 16.07, size = 703, normalized size = 0.84 \[ \frac {\sqrt {\sec (c+d x)} \sqrt {a+b \cos (c+d x)} \left (\frac {1}{960} \left (93 a^2 B+170 a A b+88 b^2 B\right ) \sin (c+d x)+\frac {1}{960} \left (93 a^2 B+170 a A b+100 b^2 B\right ) \sin (3 (c+d x))+\frac {\left (15 a^3 B+590 a^2 A b+1024 a b^2 B+480 A b^3\right ) \sin (2 (c+d x))}{1920 b}+\frac {1}{320} b (21 a B+10 A b) \sin (4 (c+d x))+\frac {1}{80} b^2 B \sin (5 (c+d x))\right )}{d}-\frac {-b \left (-45 a^4 B+150 a^3 A b+1692 a^2 b^2 B+2840 a A b^3+1024 b^4 B\right ) \tan \left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )\right )^{3/2} (a+b \cos (c+d x))+a (a+b) \left (45 a^4 B-30 a^3 b (5 A+3 B)+60 a^2 b^2 (5 A+11 B)+8 a b^3 (265 A+129 B)+16 b^4 (45 A+64 B)\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) (a+b \cos (c+d x))}{a+b}} F\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )-b (a+b) \left (-45 a^4 B+150 a^3 A b+1692 a^2 b^2 B+2840 a A b^3+1024 b^4 B\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) (a+b \cos (c+d x))}{a+b}} E\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )+15 \left (-3 a^5 B+10 a^4 A b-40 a^3 b^2 B-240 a^2 A b^3-240 a b^4 B-96 A b^5\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) (a+b \cos (c+d x))}{a+b}} \left ((a-b) F\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )+2 b \Pi \left (-1;\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )\right )}{1920 b^3 d \sec ^{\frac {3}{2}}(c+d x) \left (\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )\right )^{3/2} \sqrt {a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(((170*a*A*b + 93*a^2*B + 88*b^2*B)*Sin[c + d*x])/960 + ((590*a^2
*A*b + 480*A*b^3 + 15*a^3*B + 1024*a*b^2*B)*Sin[2*(c + d*x)])/(1920*b) + ((170*a*A*b + 93*a^2*B + 100*b^2*B)*S
in[3*(c + d*x)])/960 + (b*(10*A*b + 21*a*B)*Sin[4*(c + d*x)])/320 + (b^2*B*Sin[5*(c + d*x)])/80))/d - (-(b*(a
+ b)*(150*a^3*A*b + 2840*a*A*b^3 - 45*a^4*B + 1692*a^2*b^2*B + 1024*b^4*B)*EllipticE[ArcSin[Tan[(c + d*x)/2]],
(-a + b)/(a + b)]*Sec[(c + d*x)/2]^2*Sqrt[((a + b*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)]) + a*(a + b)*(45
*a^4*B - 30*a^3*b*(5*A + 3*B) + 60*a^2*b^2*(5*A + 11*B) + 16*b^4*(45*A + 64*B) + 8*a*b^3*(265*A + 129*B))*Elli
pticF[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x)/2]^2*Sqrt[((a + b*Cos[c + d*x])*Sec[(c + d*x)/
2]^2)/(a + b)] + 15*(10*a^4*A*b - 240*a^2*A*b^3 - 96*A*b^5 - 3*a^5*B - 40*a^3*b^2*B - 240*a*b^4*B)*((a - b)*El
lipticF[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)] + 2*b*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a
+ b)])*Sec[(c + d*x)/2]^2*Sqrt[((a + b*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)] - b*(150*a^3*A*b + 2840*a*A
*b^3 - 45*a^4*B + 1692*a^2*b^2*B + 1024*b^4*B)*(a + b*Cos[c + d*x])*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(3/2)*Se
c[c + d*x]*Tan[(c + d*x)/2])/(1920*b^3*d*Sqrt[a + b*Cos[c + d*x]]*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(3/2)*Sec[
c + d*x]^(3/2))

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fricas [F]  time = 5.79, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B b^{2} \cos \left (d x + c\right )^{3} + A a^{2} + {\left (2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )\right )} \sqrt {b \cos \left (d x + c\right ) + a}}{\sec \left (d x + c\right )^{\frac {3}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))/sec(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

integral((B*b^2*cos(d*x + c)^3 + A*a^2 + (2*B*a*b + A*b^2)*cos(d*x + c)^2 + (B*a^2 + 2*A*a*b)*cos(d*x + c))*sq
rt(b*cos(d*x + c) + a)/sec(d*x + c)^(3/2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\sec \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))/sec(d*x+c)^(3/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.84, size = 5172, normalized size = 6.16 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\sec \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))/sec(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*cos(c + d*x))*(a + b*cos(c + d*x))^(5/2))/(1/cos(c + d*x))^(3/2),x)

[Out]

int(((A + B*cos(c + d*x))*(a + b*cos(c + d*x))^(5/2))/(1/cos(c + d*x))^(3/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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