3.1130 \(\int \frac {(a+b \cos (c+d x))^{5/2} (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\sqrt {\cos (c+d x)}} \, dx\)

Optimal. Leaf size=700 \[ \frac {\sin (c+d x) \sqrt {\cos (c+d x)} \left (5 a^2 C+24 a b B+16 A b^2+12 b^2 C\right ) \sqrt {a+b \cos (c+d x)}}{32 d}+\frac {\sin (c+d x) \left (15 a^3 C+264 a^2 b B+4 a b^2 (108 A+71 C)+128 b^3 B\right ) \sqrt {a+b \cos (c+d x)}}{192 b d \sqrt {\cos (c+d x)}}+\frac {\sqrt {a+b} \cot (c+d x) \left (15 a^3 C+2 a^2 b (192 A+132 B+59 C)+4 a b^2 (108 A+52 B+71 C)+8 b^3 (12 A+16 B+9 C)\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{192 b d}-\frac {(a-b) \sqrt {a+b} \cot (c+d x) \left (15 a^3 C+264 a^2 b B+4 a b^2 (108 A+71 C)+128 b^3 B\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{192 a b d}-\frac {\sqrt {a+b} \cot (c+d x) \left (-5 a^4 C+40 a^3 b B+120 a^2 b^2 (2 A+C)+160 a b^3 B+16 b^4 (4 A+3 C)\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{b};\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{64 b^2 d}+\frac {(5 a C+8 b B) \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2}}{24 d}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{5/2}}{4 d} \]

[Out]

1/24*(8*B*b+5*C*a)*(a+b*cos(d*x+c))^(3/2)*sin(d*x+c)*cos(d*x+c)^(1/2)/d+1/4*C*(a+b*cos(d*x+c))^(5/2)*sin(d*x+c
)*cos(d*x+c)^(1/2)/d+1/192*(264*a^2*b*B+128*b^3*B+15*a^3*C+4*a*b^2*(108*A+71*C))*sin(d*x+c)*(a+b*cos(d*x+c))^(
1/2)/b/d/cos(d*x+c)^(1/2)+1/32*(16*A*b^2+24*B*a*b+5*C*a^2+12*C*b^2)*sin(d*x+c)*cos(d*x+c)^(1/2)*(a+b*cos(d*x+c
))^(1/2)/d-1/192*(a-b)*(264*a^2*b*B+128*b^3*B+15*a^3*C+4*a*b^2*(108*A+71*C))*cot(d*x+c)*EllipticE((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)*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1
+sec(d*x+c))/(a-b))^(1/2)/a/b/d+1/192*(15*a^3*C+8*b^3*(12*A+16*B+9*C)+2*a^2*b*(192*A+132*B+59*C)+4*a*b^2*(108*
A+52*B+71*C))*cot(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)*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/b/d-1/64*(40*a^3*b*B+160*a*b^3*B-5*a^
4*C+120*a^2*b^2*(2*A+C)+16*b^4*(4*A+3*C))*cot(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)*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/
b^2/d

________________________________________________________________________________________

Rubi [A]  time = 2.53, antiderivative size = 700, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {3049, 3061, 3053, 2809, 2998, 2816, 2994} \[ \frac {\sin (c+d x) \sqrt {\cos (c+d x)} \left (5 a^2 C+24 a b B+16 A b^2+12 b^2 C\right ) \sqrt {a+b \cos (c+d x)}}{32 d}+\frac {\sin (c+d x) \left (264 a^2 b B+15 a^3 C+4 a b^2 (108 A+71 C)+128 b^3 B\right ) \sqrt {a+b \cos (c+d x)}}{192 b d \sqrt {\cos (c+d x)}}+\frac {\sqrt {a+b} \cot (c+d x) \left (2 a^2 b (192 A+132 B+59 C)+15 a^3 C+4 a b^2 (108 A+52 B+71 C)+8 b^3 (12 A+16 B+9 C)\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{192 b d}-\frac {(a-b) \sqrt {a+b} \cot (c+d x) \left (264 a^2 b B+15 a^3 C+4 a b^2 (108 A+71 C)+128 b^3 B\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{192 a b d}-\frac {\sqrt {a+b} \cot (c+d x) \left (120 a^2 b^2 (2 A+C)+40 a^3 b B-5 a^4 C+160 a b^3 B+16 b^4 (4 A+3 C)\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{b};\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{64 b^2 d}+\frac {(5 a C+8 b B) \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2}}{24 d}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{5/2}}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a - b)*Sqrt[a + b]*(264*a^2*b*B + 128*b^3*B + 15*a^3*C + 4*a*b^2*(108*A + 71*C))*Cot[c + d*x]*EllipticE[Arc
Sin[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)])/(192*a*b*d) + (Sqrt[a + b]*(15*a^3*C + 8*b^3*(12*A + 16*B + 9
*C) + 2*a^2*b*(192*A + 132*B + 59*C) + 4*a*b^2*(108*A + 52*B + 71*C))*Cot[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)]*Sqr
t[(a*(1 + Sec[c + d*x]))/(a - b)])/(192*b*d) - (Sqrt[a + b]*(40*a^3*b*B + 160*a*b^3*B - 5*a^4*C + 120*a^2*b^2*
(2*A + C) + 16*b^4*(4*A + 3*C))*Cot[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)])/(64*b^2*d) + ((264*a^2*b*B + 128*b^3*B + 15*a^3*C + 4*a*b^2*(108*A + 71*C))*Sqrt[a + b*Cos[c + d*x]]*
Sin[c + d*x])/(192*b*d*Sqrt[Cos[c + d*x]]) + ((16*A*b^2 + 24*a*b*B + 5*a^2*C + 12*b^2*C)*Sqrt[Cos[c + d*x]]*Sq
rt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(32*d) + ((8*b*B + 5*a*C)*Sqrt[Cos[c + d*x]]*(a + b*Cos[c + d*x])^(3/2)*S
in[c + d*x])/(24*d) + (C*Sqrt[Cos[c + d*x]]*(a + b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(4*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 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} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx &=\frac {C \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{4 d}+\frac {1}{4} \int \frac {(a+b \cos (c+d x))^{3/2} \left (\frac {1}{2} a (8 A+C)+(4 A b+4 a B+3 b C) \cos (c+d x)+\frac {1}{2} (8 b B+5 a C) \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {(8 b B+5 a C) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{24 d}+\frac {C \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{4 d}+\frac {1}{12} \int \frac {\sqrt {a+b \cos (c+d x)} \left (\frac {1}{4} a (48 a A+8 b B+11 a C)+\frac {1}{2} \left (24 a^2 B+16 b^2 B+a b (48 A+31 C)\right ) \cos (c+d x)+\frac {3}{4} \left (16 A b^2+24 a b B+5 a^2 C+12 b^2 C\right ) \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {\left (16 A b^2+24 a b B+5 a^2 C+12 b^2 C\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{32 d}+\frac {(8 b B+5 a C) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{24 d}+\frac {C \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{4 d}+\frac {1}{24} \int \frac {\frac {1}{8} a \left (192 a^2 A+48 A b^2+104 a b B+59 a^2 C+36 b^2 C\right )+\frac {1}{4} \left (96 a^3 B+152 a b^2 B+12 b^3 (4 A+3 C)+a^2 b (288 A+161 C)\right ) \cos (c+d x)+\frac {1}{8} \left (264 a^2 b B+128 b^3 B+15 a^3 C+4 a b^2 (108 A+71 C)\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}} \, dx\\ &=\frac {\left (264 a^2 b B+128 b^3 B+15 a^3 C+4 a b^2 (108 A+71 C)\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{192 b d \sqrt {\cos (c+d x)}}+\frac {\left (16 A b^2+24 a b B+5 a^2 C+12 b^2 C\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{32 d}+\frac {(8 b B+5 a C) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{24 d}+\frac {C \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{4 d}+\frac {\int \frac {-\frac {1}{8} a \left (264 a^2 b B+128 b^3 B+15 a^3 C+4 a b^2 (108 A+71 C)\right )+\frac {1}{4} a b \left (104 a b B+12 b^2 (4 A+3 C)+a^2 (192 A+59 C)\right ) \cos (c+d x)+\frac {3}{8} \left (40 a^3 b B+160 a b^3 B-5 a^4 C+120 a^2 b^2 (2 A+C)+16 b^4 (4 A+3 C)\right ) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{48 b}\\ &=\frac {\left (264 a^2 b B+128 b^3 B+15 a^3 C+4 a b^2 (108 A+71 C)\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{192 b d \sqrt {\cos (c+d x)}}+\frac {\left (16 A b^2+24 a b B+5 a^2 C+12 b^2 C\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{32 d}+\frac {(8 b B+5 a C) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{24 d}+\frac {C \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{4 d}+\frac {\int \frac {-\frac {1}{8} a \left (264 a^2 b B+128 b^3 B+15 a^3 C+4 a b^2 (108 A+71 C)\right )+\frac {1}{4} a b \left (104 a b B+12 b^2 (4 A+3 C)+a^2 (192 A+59 C)\right ) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{48 b}+\frac {\left (40 a^3 b B+160 a b^3 B-5 a^4 C+120 a^2 b^2 (2 A+C)+16 b^4 (4 A+3 C)\right ) \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {a+b \cos (c+d x)}} \, dx}{128 b}\\ &=-\frac {\sqrt {a+b} \left (40 a^3 b B+160 a b^3 B-5 a^4 C+120 a^2 b^2 (2 A+C)+16 b^4 (4 A+3 C)\right ) \cot (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}}}{64 b^2 d}+\frac {\left (264 a^2 b B+128 b^3 B+15 a^3 C+4 a b^2 (108 A+71 C)\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{192 b d \sqrt {\cos (c+d x)}}+\frac {\left (16 A b^2+24 a b B+5 a^2 C+12 b^2 C\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{32 d}+\frac {(8 b B+5 a C) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{24 d}+\frac {C \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{4 d}-\frac {\left (a \left (264 a^2 b B+128 b^3 B+15 a^3 C+4 a b^2 (108 A+71 C)\right )\right ) \int \frac {1+\cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{384 b}+\frac {\left (a \left (15 a^3 C+8 b^3 (12 A+16 B+9 C)+2 a^2 b (192 A+132 B+59 C)+4 a b^2 (108 A+52 B+71 C)\right )\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}} \, dx}{384 b}\\ &=-\frac {(a-b) \sqrt {a+b} \left (264 a^2 b B+128 b^3 B+15 a^3 C+4 a b^2 (108 A+71 C)\right ) \cot (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}}}{192 a b d}+\frac {\sqrt {a+b} \left (15 a^3 C+8 b^3 (12 A+16 B+9 C)+2 a^2 b (192 A+132 B+59 C)+4 a b^2 (108 A+52 B+71 C)\right ) \cot (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}}}{192 b d}-\frac {\sqrt {a+b} \left (40 a^3 b B+160 a b^3 B-5 a^4 C+120 a^2 b^2 (2 A+C)+16 b^4 (4 A+3 C)\right ) \cot (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}}}{64 b^2 d}+\frac {\left (264 a^2 b B+128 b^3 B+15 a^3 C+4 a b^2 (108 A+71 C)\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{192 b d \sqrt {\cos (c+d x)}}+\frac {\left (16 A b^2+24 a b B+5 a^2 C+12 b^2 C\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{32 d}+\frac {(8 b B+5 a C) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{24 d}+\frac {C \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{4 d}\\ \end {align*}

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Mathematica [C]  time = 6.91, size = 1326, normalized size = 1.89 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-4*a*(384*a^3*A + 528*a*A*b^2 + 472*a^2*b*B + 128*b^3*B + 133*a^3*C + 356*a*b^2*C)*Sqrt[((a + b)*Cot[(c + d*
x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*
x)/2]^2)/a]*Csc[c + d*x]*EllipticF[ArcSin[Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]/Sqrt[2]], (-2*a)/(
-a + b)]*Sin[(c + d*x)/2]^4)/((a + b)*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]) - 4*a*(1152*a^2*A*b + 192*A
*b^3 + 384*a^3*B + 608*a*b^2*B + 644*a^2*b*C + 144*b^3*C)*((Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-
(((a + b)*Cos[c + d*x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]*Csc[c + d*x]*
EllipticF[ArcSin[Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]/Sqrt[2]], (-2*a)/(-a + b)]*Sin[(c + d*x)/2]
^4)/((a + b)*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]) - (Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[
-(((a + b)*Cos[c + d*x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]*Csc[c + d*x]
*EllipticPi[-(a/b), ArcSin[Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]/Sqrt[2]], (-2*a)/(-a + b)]*Sin[(c
+ d*x)/2]^4)/(b*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]])) + 2*(432*a*A*b^2 + 264*a^2*b*B + 128*b^3*B + 15
*a^3*C + 284*a*b^2*C)*((I*Cos[(c + d*x)/2]*Sqrt[a + b*Cos[c + d*x]]*EllipticE[I*ArcSinh[Sin[(c + d*x)/2]/Sqrt[
Cos[c + d*x]]], (-2*a)/(-a - b)]*Sec[c + d*x])/(b*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*Sqrt[((a + b*Cos[c + d
*x])*Sec[c + d*x])/(a + b)]) + (2*a*((a*Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*
x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]*Csc[c + d*x]*EllipticF[ArcSin[Sqr
t[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]/Sqrt[2]], (-2*a)/(-a + b)]*Sin[(c + d*x)/2]^4)/((a + b)*Sqrt[Co
s[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]) - (a*Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c +
d*x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]*Csc[c + d*x]*EllipticPi[-(a/b)
, ArcSin[Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]/Sqrt[2]], (-2*a)/(-a + b)]*Sin[(c + d*x)/2]^4)/(b*S
qrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]])))/b + (Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(b*Sqrt[Cos[c + d*x]
])))/(384*d) + (Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]*(((48*A*b^2 + 104*a*b*B + 59*a^2*C + 42*b^2*C)*Sin
[c + d*x])/96 + (b*(8*b*B + 17*a*C)*Sin[2*(c + d*x)])/48 + (b^2*C*Sin[3*(c + d*x)])/16))/d

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fricas [F]  time = 97.02, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C b^{2} \cos \left (d x + c\right )^{4} + {\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )^{3} + A a^{2} + {\left (C a^{2} + 2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )\right )} \sqrt {b \cos \left (d x + c\right ) + a}}{\sqrt {\cos \left (d x + c\right )}}, x\right ) \]

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)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

integral((C*b^2*cos(d*x + c)^4 + (2*C*a*b + B*b^2)*cos(d*x + c)^3 + A*a^2 + (C*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))*sqrt(b*cos(d*x + c) + a)/sqrt(cos(d*x + c)), x)

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giac [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)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 1.30, size = 5873, normalized size = 8.39 \[ \text {output too large to display} \]

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)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2),x)

[Out]

result too large to display

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

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)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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