3.1053 \(\int \frac {\cos ^3(c+d x) (A+B \cos (c+d x)+C \cos ^2(c+d x))}{(a+b \cos (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=622 \[ -\frac {2 \sin (c+d x) \cos ^3(c+d x) \left (A b^2-a (b B-a C)\right )}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}+\frac {2 \sin (c+d x) \cos ^2(c+d x) \left (-8 a^4 C+5 a^3 b B-2 a^2 b^2 (A-6 C)-9 a b^3 B+6 A b^4\right )}{3 b^2 d \left (a^2-b^2\right )^2 \sqrt {a+b \cos (c+d x)}}-\frac {2 \sin (c+d x) \cos (c+d x) \left (-48 a^4 C+30 a^3 b B-a^2 b^2 (15 A-71 C)-50 a b^3 B+b^4 (35 A-3 C)\right ) \sqrt {a+b \cos (c+d x)}}{15 b^3 d \left (a^2-b^2\right )^2}+\frac {2 \sin (c+d x) \left (-64 a^5 C+40 a^4 b B-2 a^3 b^2 (10 A-49 C)-65 a^2 b^3 B+2 a b^4 (20 A-7 C)+5 b^5 B\right ) \sqrt {a+b \cos (c+d x)}}{15 b^4 d \left (a^2-b^2\right )^2}+\frac {2 \left (-128 a^5 C+80 a^4 b B-4 a^3 b^2 (10 A-29 C)-80 a^2 b^3 B+a b^4 (45 A+17 C)-5 b^5 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^5 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (-128 a^6 C+80 a^5 b B-4 a^4 b^2 (10 A-53 C)-140 a^3 b^3 B+5 a^2 b^4 (15 A-11 C)+40 a b^5 B-3 b^6 (5 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^5 d \left (a^2-b^2\right )^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}} \]

[Out]

-2/3*(A*b^2-a*(B*b-C*a))*cos(d*x+c)^3*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*cos(d*x+c))^(3/2)+2/3*(6*A*b^4+5*a^3*b*B-9
*a*b^3*B-2*a^2*b^2*(A-6*C)-8*a^4*C)*cos(d*x+c)^2*sin(d*x+c)/b^2/(a^2-b^2)^2/d/(a+b*cos(d*x+c))^(1/2)+2/15*(40*
a^4*b*B-65*a^2*b^3*B+5*b^5*B-2*a^3*b^2*(10*A-49*C)+2*a*b^4*(20*A-7*C)-64*a^5*C)*sin(d*x+c)*(a+b*cos(d*x+c))^(1
/2)/b^4/(a^2-b^2)^2/d-2/15*(30*a^3*b*B-50*a*b^3*B-a^2*b^2*(15*A-71*C)+b^4*(35*A-3*C)-48*a^4*C)*cos(d*x+c)*sin(
d*x+c)*(a+b*cos(d*x+c))^(1/2)/b^3/(a^2-b^2)^2/d-2/15*(80*a^5*b*B-140*a^3*b^3*B+40*a*b^5*B-4*a^4*b^2*(10*A-53*C
)+5*a^2*b^4*(15*A-11*C)-128*a^6*C-3*b^6*(5*A+3*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(s
in(1/2*d*x+1/2*c),2^(1/2)*(b/(a+b))^(1/2))*(a+b*cos(d*x+c))^(1/2)/b^5/(a^2-b^2)^2/d/((a+b*cos(d*x+c))/(a+b))^(
1/2)+2/15*(80*a^4*b*B-80*a^2*b^3*B-5*b^5*B-4*a^3*b^2*(10*A-29*C)-128*a^5*C+a*b^4*(45*A+17*C))*(cos(1/2*d*x+1/2
*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2)*(b/(a+b))^(1/2))*((a+b*cos(d*x+c))/(a+b))
^(1/2)/b^5/(a^2-b^2)/d/(a+b*cos(d*x+c))^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 1.67, antiderivative size = 622, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.186, Rules used = {3047, 3049, 3023, 2752, 2663, 2661, 2655, 2653} \[ -\frac {2 \sin (c+d x) \cos ^3(c+d x) \left (A b^2-a (b B-a C)\right )}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}+\frac {2 \sin (c+d x) \cos ^2(c+d x) \left (-2 a^2 b^2 (A-6 C)+5 a^3 b B-8 a^4 C-9 a b^3 B+6 A b^4\right )}{3 b^2 d \left (a^2-b^2\right )^2 \sqrt {a+b \cos (c+d x)}}-\frac {2 \sin (c+d x) \cos (c+d x) \left (-a^2 b^2 (15 A-71 C)+30 a^3 b B-48 a^4 C-50 a b^3 B+b^4 (35 A-3 C)\right ) \sqrt {a+b \cos (c+d x)}}{15 b^3 d \left (a^2-b^2\right )^2}+\frac {2 \sin (c+d x) \left (-2 a^3 b^2 (10 A-49 C)-65 a^2 b^3 B+40 a^4 b B-64 a^5 C+2 a b^4 (20 A-7 C)+5 b^5 B\right ) \sqrt {a+b \cos (c+d x)}}{15 b^4 d \left (a^2-b^2\right )^2}+\frac {2 \left (-4 a^3 b^2 (10 A-29 C)-80 a^2 b^3 B+80 a^4 b B-128 a^5 C+a b^4 (45 A+17 C)-5 b^5 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^5 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (-4 a^4 b^2 (10 A-53 C)+5 a^2 b^4 (15 A-11 C)-140 a^3 b^3 B+80 a^5 b B-128 a^6 C+40 a b^5 B-3 b^6 (5 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^5 d \left (a^2-b^2\right )^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(80*a^5*b*B - 140*a^3*b^3*B + 40*a*b^5*B - 4*a^4*b^2*(10*A - 53*C) + 5*a^2*b^4*(15*A - 11*C) - 128*a^6*C -
3*b^6*(5*A + 3*C))*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(15*b^5*(a^2 - b^2)^2*d*Sq
rt[(a + b*Cos[c + d*x])/(a + b)]) + (2*(80*a^4*b*B - 80*a^2*b^3*B - 5*b^5*B - 4*a^3*b^2*(10*A - 29*C) - 128*a^
5*C + a*b^4*(45*A + 17*C))*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(15*b^5*(
a^2 - b^2)*d*Sqrt[a + b*Cos[c + d*x]]) - (2*(A*b^2 - a*(b*B - a*C))*Cos[c + d*x]^3*Sin[c + d*x])/(3*b*(a^2 - b
^2)*d*(a + b*Cos[c + d*x])^(3/2)) + (2*(6*A*b^4 + 5*a^3*b*B - 9*a*b^3*B - 2*a^2*b^2*(A - 6*C) - 8*a^4*C)*Cos[c
+ d*x]^2*Sin[c + d*x])/(3*b^2*(a^2 - b^2)^2*d*Sqrt[a + b*Cos[c + d*x]]) + (2*(40*a^4*b*B - 65*a^2*b^3*B + 5*b
^5*B - 2*a^3*b^2*(10*A - 49*C) + 2*a*b^4*(20*A - 7*C) - 64*a^5*C)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(15*b
^4*(a^2 - b^2)^2*d) - (2*(30*a^3*b*B - 50*a*b^3*B - a^2*b^2*(15*A - 71*C) + b^4*(35*A - 3*C) - 48*a^4*C)*Cos[c
+ d*x]*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(15*b^3*(a^2 - b^2)^2*d)

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2752

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

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((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 + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

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])))

Rubi steps

\begin {align*} \int \frac {\cos ^3(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{5/2}} \, dx &=-\frac {2 \left (A b^2-a (b B-a C)\right ) \cos ^3(c+d x) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac {2 \int \frac {\cos ^2(c+d x) \left (3 \left (A b^2-a (b B-a C)\right )+\frac {3}{2} b (b B-a (A+C)) \cos (c+d x)-\frac {1}{2} \left (5 A b^2-5 a b B+8 a^2 C-3 b^2 C\right ) \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{3/2}} \, dx}{3 b \left (a^2-b^2\right )}\\ &=-\frac {2 \left (A b^2-a (b B-a C)\right ) \cos ^3(c+d x) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac {2 \left (6 A b^4+5 a^3 b B-9 a b^3 B-2 a^2 b^2 (A-6 C)-8 a^4 C\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}+\frac {4 \int \frac {\cos (c+d x) \left (6 A b^4+5 a^3 b B-9 a b^3 B-2 a^2 b^2 (A-6 C)-8 a^4 C+\frac {1}{4} b \left (a^2 b B+3 b^3 B+2 a^3 C-2 a b^2 (2 A+3 C)\right ) \cos (c+d x)-\frac {1}{4} \left (30 a^3 b B-50 a b^3 B-a^2 b^2 (15 A-71 C)+b^4 (35 A-3 C)-48 a^4 C\right ) \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx}{3 b^2 \left (a^2-b^2\right )^2}\\ &=-\frac {2 \left (A b^2-a (b B-a C)\right ) \cos ^3(c+d x) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac {2 \left (6 A b^4+5 a^3 b B-9 a b^3 B-2 a^2 b^2 (A-6 C)-8 a^4 C\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (30 a^3 b B-50 a b^3 B-a^2 b^2 (15 A-71 C)+b^4 (35 A-3 C)-48 a^4 C\right ) \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 b^3 \left (a^2-b^2\right )^2 d}+\frac {8 \int \frac {-\frac {1}{4} a \left (30 a^3 b B-50 a b^3 B-a^2 b^2 (15 A-71 C)+b^4 (35 A-3 C)-48 a^4 C\right )+\frac {1}{8} b \left (10 a^3 b B-30 a b^3 B-16 a^4 C+3 b^4 (5 A+3 C)+a^2 b^2 (5 A+27 C)\right ) \cos (c+d x)+\frac {3}{8} \left (40 a^4 b B-65 a^2 b^3 B+5 b^5 B-2 a^3 b^2 (10 A-49 C)+2 a b^4 (20 A-7 C)-64 a^5 C\right ) \cos ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{15 b^3 \left (a^2-b^2\right )^2}\\ &=-\frac {2 \left (A b^2-a (b B-a C)\right ) \cos ^3(c+d x) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac {2 \left (6 A b^4+5 a^3 b B-9 a b^3 B-2 a^2 b^2 (A-6 C)-8 a^4 C\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (40 a^4 b B-65 a^2 b^3 B+5 b^5 B-2 a^3 b^2 (10 A-49 C)+2 a b^4 (20 A-7 C)-64 a^5 C\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 b^4 \left (a^2-b^2\right )^2 d}-\frac {2 \left (30 a^3 b B-50 a b^3 B-a^2 b^2 (15 A-71 C)+b^4 (35 A-3 C)-48 a^4 C\right ) \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 b^3 \left (a^2-b^2\right )^2 d}+\frac {16 \int \frac {-\frac {3}{16} b \left (20 a^4 b B-35 a^2 b^3 B-5 b^5 B-2 a^3 b^2 (5 A-22 C)-32 a^5 C+2 a b^4 (15 A+4 C)\right )-\frac {3}{16} \left (80 a^5 b B-140 a^3 b^3 B+40 a b^5 B-4 a^4 b^2 (10 A-53 C)+5 a^2 b^4 (15 A-11 C)-128 a^6 C-3 b^6 (5 A+3 C)\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{45 b^4 \left (a^2-b^2\right )^2}\\ &=-\frac {2 \left (A b^2-a (b B-a C)\right ) \cos ^3(c+d x) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac {2 \left (6 A b^4+5 a^3 b B-9 a b^3 B-2 a^2 b^2 (A-6 C)-8 a^4 C\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (40 a^4 b B-65 a^2 b^3 B+5 b^5 B-2 a^3 b^2 (10 A-49 C)+2 a b^4 (20 A-7 C)-64 a^5 C\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 b^4 \left (a^2-b^2\right )^2 d}-\frac {2 \left (30 a^3 b B-50 a b^3 B-a^2 b^2 (15 A-71 C)+b^4 (35 A-3 C)-48 a^4 C\right ) \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 b^3 \left (a^2-b^2\right )^2 d}-\frac {\left (80 a^5 b B-140 a^3 b^3 B+40 a b^5 B-4 a^4 b^2 (10 A-53 C)+5 a^2 b^4 (15 A-11 C)-128 a^6 C-3 b^6 (5 A+3 C)\right ) \int \sqrt {a+b \cos (c+d x)} \, dx}{15 b^5 \left (a^2-b^2\right )^2}+\frac {\left (80 a^4 b B-80 a^2 b^3 B-5 b^5 B-4 a^3 b^2 (10 A-29 C)-128 a^5 C+a b^4 (45 A+17 C)\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx}{15 b^5 \left (a^2-b^2\right )}\\ &=-\frac {2 \left (A b^2-a (b B-a C)\right ) \cos ^3(c+d x) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac {2 \left (6 A b^4+5 a^3 b B-9 a b^3 B-2 a^2 b^2 (A-6 C)-8 a^4 C\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (40 a^4 b B-65 a^2 b^3 B+5 b^5 B-2 a^3 b^2 (10 A-49 C)+2 a b^4 (20 A-7 C)-64 a^5 C\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 b^4 \left (a^2-b^2\right )^2 d}-\frac {2 \left (30 a^3 b B-50 a b^3 B-a^2 b^2 (15 A-71 C)+b^4 (35 A-3 C)-48 a^4 C\right ) \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 b^3 \left (a^2-b^2\right )^2 d}-\frac {\left (\left (80 a^5 b B-140 a^3 b^3 B+40 a b^5 B-4 a^4 b^2 (10 A-53 C)+5 a^2 b^4 (15 A-11 C)-128 a^6 C-3 b^6 (5 A+3 C)\right ) \sqrt {a+b \cos (c+d x)}\right ) \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}} \, dx}{15 b^5 \left (a^2-b^2\right )^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\left (\left (80 a^4 b B-80 a^2 b^3 B-5 b^5 B-4 a^3 b^2 (10 A-29 C)-128 a^5 C+a b^4 (45 A+17 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}}\right ) \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}} \, dx}{15 b^5 \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\\ &=-\frac {2 \left (80 a^5 b B-140 a^3 b^3 B+40 a b^5 B-4 a^4 b^2 (10 A-53 C)+5 a^2 b^4 (15 A-11 C)-128 a^6 C-3 b^6 (5 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^5 \left (a^2-b^2\right )^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 \left (80 a^4 b B-80 a^2 b^3 B-5 b^5 B-4 a^3 b^2 (10 A-29 C)-128 a^5 C+a b^4 (45 A+17 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^5 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (A b^2-a (b B-a C)\right ) \cos ^3(c+d x) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac {2 \left (6 A b^4+5 a^3 b B-9 a b^3 B-2 a^2 b^2 (A-6 C)-8 a^4 C\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (40 a^4 b B-65 a^2 b^3 B+5 b^5 B-2 a^3 b^2 (10 A-49 C)+2 a b^4 (20 A-7 C)-64 a^5 C\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 b^4 \left (a^2-b^2\right )^2 d}-\frac {2 \left (30 a^3 b B-50 a b^3 B-a^2 b^2 (15 A-71 C)+b^4 (35 A-3 C)-48 a^4 C\right ) \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 b^3 \left (a^2-b^2\right )^2 d}\\ \end {align*}

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Mathematica [A]  time = 5.78, size = 422, normalized size = 0.68 \[ \frac {b \left (\frac {10 a^3 \sin (c+d x) \left (a (a C-b B)+A b^2\right )}{a^2-b^2}-\frac {10 a^2 \sin (c+d x) \left (11 a^4 C-8 a^3 b B+5 a^2 b^2 (A-3 C)+12 a b^3 B-9 A b^4\right ) (a+b \cos (c+d x))}{\left (a^2-b^2\right )^2}+2 (5 b B-14 a C) \sin (c+d x) (a+b \cos (c+d x))^2+3 b C \sin (2 (c+d x)) (a+b \cos (c+d x))^2\right )+\frac {2 \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{3/2} \left (b^2 \left (32 a^5 C-20 a^4 b B+2 a^3 b^2 (5 A-22 C)+35 a^2 b^3 B-2 a b^4 (15 A+4 C)+5 b^5 B\right ) F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )+\left (128 a^6 C-80 a^5 b B+4 a^4 b^2 (10 A-53 C)+140 a^3 b^3 B+5 a^2 b^4 (11 C-15 A)-40 a b^5 B+3 b^6 (5 A+3 C)\right ) \left ((a+b) E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-a F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )\right )\right )}{(a-b)^2 (a+b)}}{15 b^5 d (a+b \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*((a + b*Cos[c + d*x])/(a + b))^(3/2)*(b^2*(-20*a^4*b*B + 35*a^2*b^3*B + 5*b^5*B + 2*a^3*b^2*(5*A - 22*C) +
32*a^5*C - 2*a*b^4*(15*A + 4*C))*EllipticF[(c + d*x)/2, (2*b)/(a + b)] + (-80*a^5*b*B + 140*a^3*b^3*B - 40*a*
b^5*B + 4*a^4*b^2*(10*A - 53*C) + 128*a^6*C + 3*b^6*(5*A + 3*C) + 5*a^2*b^4*(-15*A + 11*C))*((a + b)*EllipticE
[(c + d*x)/2, (2*b)/(a + b)] - a*EllipticF[(c + d*x)/2, (2*b)/(a + b)])))/((a - b)^2*(a + b)) + b*((10*a^3*(A*
b^2 + a*(-(b*B) + a*C))*Sin[c + d*x])/(a^2 - b^2) - (10*a^2*(-9*A*b^4 - 8*a^3*b*B + 12*a*b^3*B + 5*a^2*b^2*(A
- 3*C) + 11*a^4*C)*(a + b*Cos[c + d*x])*Sin[c + d*x])/(a^2 - b^2)^2 + 2*(5*b*B - 14*a*C)*(a + b*Cos[c + d*x])^
2*Sin[c + d*x] + 3*b*C*(a + b*Cos[c + d*x])^2*Sin[2*(c + d*x)]))/(15*b^5*d*(a + b*Cos[c + d*x])^(3/2))

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fricas [F]  time = 1.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \cos \left (d x + c\right )^{5} + B \cos \left (d x + c\right )^{4} + A \cos \left (d x + c\right )^{3}\right )} \sqrt {b \cos \left (d x + c\right ) + a}}{b^{3} \cos \left (d x + c\right )^{3} + 3 \, a b^{2} \cos \left (d x + c\right )^{2} + 3 \, a^{2} b \cos \left (d x + c\right ) + a^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*cos(d*x + c)^5 + B*cos(d*x + c)^4 + A*cos(d*x + c)^3)*sqrt(b*cos(d*x + c) + a)/(b^3*cos(d*x + c)^3
+ 3*a*b^2*cos(d*x + c)^2 + 3*a^2*b*cos(d*x + c) + a^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 17.58, size = 1780, normalized size = 2.86 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2*b-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(16/b^2*C*(-1/10/b*cos(1/2*d*x+1/2*c)^3*(-2*sin
(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)-1/60/b^2*(-4*a+12*b)*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+
1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)+1/60/b^2*(-4*a+12*b)*(a-b)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1
/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos
(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-1/60*(4*a^2-15*a*b+27*b^2)/b^3*(a-b)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(
1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(c
os(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))))+8/b^3*(B*b-2*C*a-3*C*
b)*(-1/6/b*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)+1/6/b*(a-b)*(sin(1/
2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*
x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-1/12/b^2*(-2*a+6*b)*(a-b)*(sin(1/2*d*x+1/2*
c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2
)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))))-2
/b^5*(A*b^2-2*B*a*b-2*B*b^2+3*C*a^2+4*C*a*b+3*C*b^2)*(a-b)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)
^2*b+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2
*c),(-2*b/(a-b))^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2)))-2*(2*A*a*b^2+A*b^3-3*B*a^2*b-2*B*a*b
^2-B*b^3+4*C*a^3+3*C*a^2*b+2*C*a*b^2+C*b^3)/b^5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(
a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a
-b))^(1/2))+2*a^2/b^5*(3*A*b^2-4*B*a*b+5*C*a^2)/sin(1/2*d*x+1/2*c)^2/(-2*sin(1/2*d*x+1/2*c)^2*b+a+b)/(a^2-b^2)
*(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*((-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(
1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a-(-2*b/(a-b)*sin(1/2*d*x+1
/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*b+2*b
*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2)-2*a^3*(A*b^2-B*a*b+C*a^2)/b^5*(1/6/b/(a-b)/(a+b)*cos(1/2*d*x+1/2*c)*
(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2+1/2/b*(a-b))^2+8/3*b*sin(1/
2*d*x+1/2*c)^2/(a-b)^2/(a+b)^2*cos(1/2*d*x+1/2*c)*a/(-(-2*cos(1/2*d*x+1/2*c)^2*b-a+b)*sin(1/2*d*x+1/2*c)^2)^(1
/2)+(3*a-b)/(3*a^3+3*a^2*b-3*a*b^2-3*b^3)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^
(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(
1/2))-4/3*a/(a-b)/(a+b)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*sin(1/
2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-EllipticE
(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2)))))/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*b+a+b)^(1/2)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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