3.1145 \(\int \frac{\cos ^{\frac{3}{2}}(c+d x) (A+B \cos (c+d x)+C \cos ^2(c+d x))}{(a+b \cos (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=660 \[ -\frac{2 \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt{a+b \cos (c+d x)}}+\frac{\sin (c+d x) \sqrt{\cos (c+d x)} \left (5 a^2 C-4 a b B+4 A b^2-b^2 C\right ) \sqrt{a+b \cos (c+d x)}}{2 b^2 d \left (a^2-b^2\right )}+\frac{\sin (c+d x) \left (12 a^2 b B-15 a^3 C-a b^2 (8 A-7 C)-4 b^3 B\right ) \sqrt{a+b \cos (c+d x)}}{4 b^3 d \left (a^2-b^2\right ) \sqrt{\cos (c+d x)}}-\frac{\cot (c+d x) \left (15 a^2 C-a b (12 B-5 C)+8 A b^2-2 b^2 (2 B+C)\right ) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{4 b^3 d \sqrt{a+b}}-\frac{\cot (c+d x) \left (12 a^2 b B-15 a^3 C-a b^2 (8 A-7 C)-4 b^3 B\right ) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{4 a b^3 d \sqrt{a+b}}-\frac{\sqrt{a+b} \cot (c+d x) \left (15 a^2 C-12 a b B+8 A b^2+4 b^2 C\right ) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac{a+b}{b};\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{4 b^4 d} \]

[Out]

-((12*a^2*b*B - 4*b^3*B - a*b^2*(8*A - 7*C) - 15*a^3*C)*Cot[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)])/(4*a*b^3*Sqrt[a + b]*d) - ((8*A*b^2 - a*b*(12*B - 5*C) + 15*a^2*C - 2*b^2*(2*B + 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)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(4*b^3*Sqrt[a + b]*d) - (Sqrt[a + b]*(8
*A*b^2 - 12*a*b*B + 15*a^2*C + 4*b^2*C)*Cot[c + d*x]*EllipticPi[(a + b)/b, ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sq
rt[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)])/(4*b^4*d) - (2*(A*b^2 - a*(b*B - a*C))*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(b*(a^2 - b^2)*d*Sqrt
[a + b*Cos[c + d*x]]) + ((12*a^2*b*B - 4*b^3*B - a*b^2*(8*A - 7*C) - 15*a^3*C)*Sqrt[a + b*Cos[c + d*x]]*Sin[c
+ d*x])/(4*b^3*(a^2 - b^2)*d*Sqrt[Cos[c + d*x]]) + ((4*A*b^2 - 4*a*b*B + 5*a^2*C - b^2*C)*Sqrt[Cos[c + d*x]]*S
qrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(2*b^2*(a^2 - b^2)*d)

________________________________________________________________________________________

Rubi [A]  time = 2.04702, antiderivative size = 660, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.178, Rules used = {3047, 3049, 3061, 3053, 2809, 2998, 2816, 2994} \[ -\frac{2 \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt{a+b \cos (c+d x)}}+\frac{\sin (c+d x) \sqrt{\cos (c+d x)} \left (5 a^2 C-4 a b B+4 A b^2-b^2 C\right ) \sqrt{a+b \cos (c+d x)}}{2 b^2 d \left (a^2-b^2\right )}+\frac{\sin (c+d x) \left (12 a^2 b B-15 a^3 C-a b^2 (8 A-7 C)-4 b^3 B\right ) \sqrt{a+b \cos (c+d x)}}{4 b^3 d \left (a^2-b^2\right ) \sqrt{\cos (c+d x)}}-\frac{\cot (c+d x) \left (15 a^2 C-a b (12 B-5 C)+8 A b^2-2 b^2 (2 B+C)\right ) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{4 b^3 d \sqrt{a+b}}-\frac{\cot (c+d x) \left (12 a^2 b B-15 a^3 C-a b^2 (8 A-7 C)-4 b^3 B\right ) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{4 a b^3 d \sqrt{a+b}}-\frac{\sqrt{a+b} \cot (c+d x) \left (15 a^2 C-12 a b B+8 A b^2+4 b^2 C\right ) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac{a+b}{b};\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{4 b^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((12*a^2*b*B - 4*b^3*B - a*b^2*(8*A - 7*C) - 15*a^3*C)*Cot[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)])/(4*a*b^3*Sqrt[a + b]*d) - ((8*A*b^2 - a*b*(12*B - 5*C) + 15*a^2*C - 2*b^2*(2*B + 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)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(4*b^3*Sqrt[a + b]*d) - (Sqrt[a + b]*(8
*A*b^2 - 12*a*b*B + 15*a^2*C + 4*b^2*C)*Cot[c + d*x]*EllipticPi[(a + b)/b, ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sq
rt[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)])/(4*b^4*d) - (2*(A*b^2 - a*(b*B - a*C))*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(b*(a^2 - b^2)*d*Sqrt
[a + b*Cos[c + d*x]]) + ((12*a^2*b*B - 4*b^3*B - a*b^2*(8*A - 7*C) - 15*a^3*C)*Sqrt[a + b*Cos[c + d*x]]*Sin[c
+ d*x])/(4*b^3*(a^2 - b^2)*d*Sqrt[Cos[c + d*x]]) + ((4*A*b^2 - 4*a*b*B + 5*a^2*C - b^2*C)*Sqrt[Cos[c + d*x]]*S
qrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(2*b^2*(a^2 - b^2)*d)

Rule 3047

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

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 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]

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 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 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 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]

Rubi steps

\begin{align*} \int \frac{\cos ^{\frac{3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{3/2}} \, dx &=-\frac{2 \left (A b^2-a (b B-a C)\right ) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}-\frac{2 \int \frac{\sqrt{\cos (c+d x)} \left (\frac{3}{2} \left (A b^2-a (b B-a C)\right )+\frac{1}{2} b (b B-a (A+C)) \cos (c+d x)-\frac{1}{2} \left (4 A b^2-4 a b B+5 a^2 C-b^2 C\right ) \cos ^2(c+d x)\right )}{\sqrt{a+b \cos (c+d x)}} \, dx}{b \left (a^2-b^2\right )}\\ &=-\frac{2 \left (A b^2-a (b B-a C)\right ) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}+\frac{\left (4 A b^2-4 a b B+5 a^2 C-b^2 C\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d}-\frac{\int \frac{-\frac{1}{4} a \left (4 A b^2-4 a b B+5 a^2 C-b^2 C\right )+\frac{1}{2} b \left (2 A b^2-2 a b B+a^2 C+b^2 C\right ) \cos (c+d x)-\frac{1}{4} \left (12 a^2 b B-4 b^3 B-a b^2 (8 A-7 C)-15 a^3 C\right ) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a+b \cos (c+d x)}} \, dx}{b^2 \left (a^2-b^2\right )}\\ &=-\frac{2 \left (A b^2-a (b B-a C)\right ) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}+\frac{\left (12 a^2 b B-4 b^3 B-a b^2 (8 A-7 C)-15 a^3 C\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{4 b^3 \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)}}+\frac{\left (4 A b^2-4 a b B+5 a^2 C-b^2 C\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d}-\frac{\int \frac{\frac{1}{4} a \left (12 a^2 b B-4 b^3 B-a b^2 (8 A-7 C)-15 a^3 C\right )-\frac{1}{2} a b \left (4 A b^2-4 a b B+5 a^2 C-b^2 C\right ) \cos (c+d x)-\frac{1}{4} \left (a^2-b^2\right ) \left (8 A b^2-12 a b B+15 a^2 C+4 b^2 C\right ) \cos ^2(c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \cos (c+d x)}} \, dx}{2 b^3 \left (a^2-b^2\right )}\\ &=-\frac{2 \left (A b^2-a (b B-a C)\right ) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}+\frac{\left (12 a^2 b B-4 b^3 B-a b^2 (8 A-7 C)-15 a^3 C\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{4 b^3 \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)}}+\frac{\left (4 A b^2-4 a b B+5 a^2 C-b^2 C\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d}-\frac{\int \frac{\frac{1}{4} a \left (12 a^2 b B-4 b^3 B-a b^2 (8 A-7 C)-15 a^3 C\right )-\frac{1}{2} a b \left (4 A b^2-4 a b B+5 a^2 C-b^2 C\right ) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \cos (c+d x)}} \, dx}{2 b^3 \left (a^2-b^2\right )}+\frac{\left (8 A b^2-12 a b B+15 a^2 C+4 b^2 C\right ) \int \frac{\sqrt{\cos (c+d x)}}{\sqrt{a+b \cos (c+d x)}} \, dx}{8 b^3}\\ &=-\frac{\sqrt{a+b} \left (8 A b^2-12 a b B+15 a^2 C+4 b^2 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}}}{4 b^4 d}-\frac{2 \left (A b^2-a (b B-a C)\right ) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}+\frac{\left (12 a^2 b B-4 b^3 B-a b^2 (8 A-7 C)-15 a^3 C\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{4 b^3 \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)}}+\frac{\left (4 A b^2-4 a b B+5 a^2 C-b^2 C\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d}-\frac{\left (a \left (12 a^2 b B-4 b^3 B-a b^2 (8 A-7 C)-15 a^3 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}{8 b^3 \left (a^2-b^2\right )}-\frac{\left (a \left (8 A b^2-a b (12 B-5 C)+15 a^2 C-2 b^2 (2 B+C)\right )\right ) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+b \cos (c+d x)}} \, dx}{8 b^3 (a+b)}\\ &=-\frac{\left (12 a^2 b B-4 b^3 B-a b^2 (8 A-7 C)-15 a^3 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}}}{4 a b^3 \sqrt{a+b} d}-\frac{\left (8 A b^2-a b (12 B-5 C)+15 a^2 C-2 b^2 (2 B+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}}}{4 b^3 \sqrt{a+b} d}-\frac{\sqrt{a+b} \left (8 A b^2-12 a b B+15 a^2 C+4 b^2 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}}}{4 b^4 d}-\frac{2 \left (A b^2-a (b B-a C)\right ) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}+\frac{\left (12 a^2 b B-4 b^3 B-a b^2 (8 A-7 C)-15 a^3 C\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{4 b^3 \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)}}+\frac{\left (4 A b^2-4 a b B+5 a^2 C-b^2 C\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d}\\ \end{align*}

Mathematica [C]  time = 6.72388, size = 1322, normalized size = 2. \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]*((C*Sin[c + d*x])/(2*b^2) - (2*(a*A*b^2*Sin[c + d*x] - a^2*b*B*Si
n[c + d*x] + a^3*C*Sin[c + d*x]))/(b^2*(-a^2 + b^2)*(a + b*Cos[c + d*x]))))/d - ((-4*a*(-4*a^2*b*B + 4*b^3*B +
5*a^3*C - 5*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*(8*A*b^3 - 8*a*b^2*B + 4*a^2*b*C + 4*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]*C
sc[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*(8*a*A*b^2 - 12*a^2*b*B + 4*b^3
*B + 15*a^3*C - 7*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[ArcSi
n[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)*Sq
rt[Cos[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)*Co
s[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]])))/b + (Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(b*Sqrt[Cos[c +
d*x]])))/(8*(a - b)*b^2*(a + b)*d)

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Maple [B]  time = 0.259, size = 5209, normalized size = 7.9 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [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 )} \cos \left (d x + c\right )^{\frac{3}{2}}}{{\left (b \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(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [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(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

Timed out

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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(cos(d*x+c)**(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/(a+b*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 )} \cos \left (d x + c\right )^{\frac{3}{2}}}{{\left (b \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(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(3/2),x, algorithm="giac")

[Out]

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