\(\int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \cos (c+d x)+C \cos ^2(c+d x))}{(a+b \cos (c+d x))^2} \, dx\) [1103]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [B] (verified)
   Fricas [F(-1)]
   Sympy [F(-1)]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 43, antiderivative size = 445 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=-\frac {\left (25 a^3 b B-20 a b^3 B-3 a^2 b^2 (5 A-8 C)-35 a^4 C+2 b^4 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^4 \left (a^2-b^2\right ) d}+\frac {\left (15 a^4 b B-16 a^2 b^3 B-2 b^5 B-a^3 b^2 (9 A-20 C)-21 a^5 C+4 a b^4 (3 A+C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b^5 \left (a^2-b^2\right ) d}-\frac {a^2 \left (5 A b^4+5 a^3 b B-7 a b^3 B-3 a^2 b^2 (A-3 C)-7 a^4 C\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{(a-b) b^5 (a+b)^2 d}+\frac {\left (5 a^2 b B-2 b^3 B-a b^2 (3 A-4 C)-7 a^3 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}+\frac {\left (5 A b^2-5 a b B+7 a^2 C-2 b^2 C\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac {\left (A b^2-a (b B-a C)\right ) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \]

[Out]

-1/5*(25*B*a^3*b-20*B*a*b^3-3*a^2*b^2*(5*A-8*C)-35*a^4*C+2*b^4*(5*A+3*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2
*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/b^4/(a^2-b^2)/d+1/3*(15*B*a^4*b-16*B*a^2*b^3-2*B*b^5-a^3*b^2
*(9*A-20*C)-21*C*a^5+4*a*b^4*(3*A+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^5/(a^2-b^2)/d-a^2*(5*A*b^4+5*B*a^3*b-7*B*a*b^3-3*a^2*b^2*(A-3*C)-7*a^4*C)*(cos(1/2*d*x+1/2*c)^
2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2*b/(a+b),2^(1/2))/(a-b)/b^5/(a+b)^2/d+1/5*(5*A*b^2-
5*B*a*b+7*C*a^2-2*C*b^2)*cos(d*x+c)^(3/2)*sin(d*x+c)/b^2/(a^2-b^2)/d-(A*b^2-a*(B*b-C*a))*cos(d*x+c)^(5/2)*sin(
d*x+c)/b/(a^2-b^2)/d/(a+b*cos(d*x+c))+1/3*(5*B*a^2*b-2*B*b^3-a*b^2*(3*A-4*C)-7*a^3*C)*sin(d*x+c)*cos(d*x+c)^(1
/2)/b^3/(a^2-b^2)/d

Rubi [A] (verified)

Time = 1.72 (sec) , antiderivative size = 445, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {3126, 3128, 3138, 2719, 3081, 2720, 2884} \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=-\frac {\sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (7 a^2 C-5 a b B+5 A b^2-2 b^2 C\right )}{5 b^2 d \left (a^2-b^2\right )}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \left (-7 a^3 C+5 a^2 b B-a b^2 (3 A-4 C)-2 b^3 B\right )}{3 b^3 d \left (a^2-b^2\right )}-\frac {E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-35 a^4 C+25 a^3 b B-3 a^2 b^2 (5 A-8 C)-20 a b^3 B+2 b^4 (5 A+3 C)\right )}{5 b^4 d \left (a^2-b^2\right )}-\frac {a^2 \left (-7 a^4 C+5 a^3 b B-3 a^2 b^2 (A-3 C)-7 a b^3 B+5 A b^4\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b^5 d (a-b) (a+b)^2}+\frac {\operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (-21 a^5 C+15 a^4 b B-a^3 b^2 (9 A-20 C)-16 a^2 b^3 B+4 a b^4 (3 A+C)-2 b^5 B\right )}{3 b^5 d \left (a^2-b^2\right )} \]

[In]

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

[Out]

-1/5*((25*a^3*b*B - 20*a*b^3*B - 3*a^2*b^2*(5*A - 8*C) - 35*a^4*C + 2*b^4*(5*A + 3*C))*EllipticE[(c + d*x)/2,
2])/(b^4*(a^2 - b^2)*d) + ((15*a^4*b*B - 16*a^2*b^3*B - 2*b^5*B - a^3*b^2*(9*A - 20*C) - 21*a^5*C + 4*a*b^4*(3
*A + C))*EllipticF[(c + d*x)/2, 2])/(3*b^5*(a^2 - b^2)*d) - (a^2*(5*A*b^4 + 5*a^3*b*B - 7*a*b^3*B - 3*a^2*b^2*
(A - 3*C) - 7*a^4*C)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/((a - b)*b^5*(a + b)^2*d) + ((5*a^2*b*B - 2*b^
3*B - a*b^2*(3*A - 4*C) - 7*a^3*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*b^3*(a^2 - b^2)*d) + ((5*A*b^2 - 5*a*b*
B + 7*a^2*C - 2*b^2*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*b^2*(a^2 - b^2)*d) - ((A*b^2 - a*(b*B - a*C))*Cos[c
 + d*x]^(5/2)*Sin[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Cos[c + d*x]))

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 2884

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

Rule 3081

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

Rule 3126

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 3128

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 3138

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) +
(f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[C/(b*d), Int[Sqrt[a + b*Sin[e + f*x]]
, x], x] - Dist[1/(b*d), Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[e + f*x], x]/(Sqrt[a + b*Sin[e +
 f*x]]*(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*} \text {integral}& = -\frac {\left (A b^2-a (b B-a C)\right ) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac {\int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (\frac {5}{2} \left (A b^2-a (b B-a C)\right )+b (b B-a (A+C)) \cos (c+d x)-\frac {1}{2} \left (5 A b^2-5 a b B+7 a^2 C-2 b^2 C\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{b \left (a^2-b^2\right )} \\ & = \frac {\left (5 A b^2-5 a b B+7 a^2 C-2 b^2 C\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac {\left (A b^2-a (b B-a C)\right ) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac {2 \int \frac {\sqrt {\cos (c+d x)} \left (-\frac {3}{4} a \left (5 A b^2-5 a b B+7 a^2 C-2 b^2 C\right )+\frac {1}{2} b \left (5 A b^2-5 a b B+2 a^2 C+3 b^2 C\right ) \cos (c+d x)-\frac {5}{4} \left (5 a^2 b B-2 b^3 B-a b^2 (3 A-4 C)-7 a^3 C\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{5 b^2 \left (a^2-b^2\right )} \\ & = \frac {\left (5 a^2 b B-2 b^3 B-a b^2 (3 A-4 C)-7 a^3 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}+\frac {\left (5 A b^2-5 a b B+7 a^2 C-2 b^2 C\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac {\left (A b^2-a (b B-a C)\right ) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac {4 \int \frac {-\frac {5}{8} a \left (5 a^2 b B-2 b^3 B-a b^2 (3 A-4 C)-7 a^3 C\right )+\frac {1}{4} b \left (10 a^2 b B+5 b^3 B-14 a^3 C-a b^2 (15 A+C)\right ) \cos (c+d x)+\frac {3}{8} \left (25 a^3 b B-20 a b^3 B-3 a^2 b^2 (5 A-8 C)-35 a^4 C+2 b^4 (5 A+3 C)\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{15 b^3 \left (a^2-b^2\right )} \\ & = \frac {\left (5 a^2 b B-2 b^3 B-a b^2 (3 A-4 C)-7 a^3 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}+\frac {\left (5 A b^2-5 a b B+7 a^2 C-2 b^2 C\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac {\left (A b^2-a (b B-a C)\right ) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {4 \int \frac {\frac {5}{8} a b \left (5 a^2 b B-2 b^3 B-a b^2 (3 A-4 C)-7 a^3 C\right )+\frac {5}{8} \left (15 a^4 b B-16 a^2 b^3 B-2 b^5 B-a^3 b^2 (9 A-20 C)-21 a^5 C+4 a b^4 (3 A+C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{15 b^4 \left (a^2-b^2\right )}-\frac {\left (25 a^3 b B-20 a b^3 B-3 a^2 b^2 (5 A-8 C)-35 a^4 C+2 b^4 (5 A+3 C)\right ) \int \sqrt {\cos (c+d x)} \, dx}{10 b^4 \left (a^2-b^2\right )} \\ & = -\frac {\left (25 a^3 b B-20 a b^3 B-3 a^2 b^2 (5 A-8 C)-35 a^4 C+2 b^4 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^4 \left (a^2-b^2\right ) d}+\frac {\left (5 a^2 b B-2 b^3 B-a b^2 (3 A-4 C)-7 a^3 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}+\frac {\left (5 A b^2-5 a b B+7 a^2 C-2 b^2 C\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac {\left (A b^2-a (b B-a C)\right ) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac {\left (a^2 \left (5 A b^4+5 a^3 b B-7 a b^3 B-3 a^2 b^2 (A-3 C)-7 a^4 C\right )\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{2 b^5 \left (a^2-b^2\right )}+\frac {\left (15 a^4 b B-16 a^2 b^3 B-2 b^5 B-a^3 b^2 (9 A-20 C)-21 a^5 C+4 a b^4 (3 A+C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 b^5 \left (a^2-b^2\right )} \\ & = -\frac {\left (25 a^3 b B-20 a b^3 B-3 a^2 b^2 (5 A-8 C)-35 a^4 C+2 b^4 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^4 \left (a^2-b^2\right ) d}+\frac {\left (15 a^4 b B-16 a^2 b^3 B-2 b^5 B-a^3 b^2 (9 A-20 C)-21 a^5 C+4 a b^4 (3 A+C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b^5 \left (a^2-b^2\right ) d}-\frac {a^2 \left (5 A b^4+5 a^3 b B-7 a b^3 B-3 a^2 b^2 (A-3 C)-7 a^4 C\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{(a-b) b^5 (a+b)^2 d}+\frac {\left (5 a^2 b B-2 b^3 B-a b^2 (3 A-4 C)-7 a^3 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}+\frac {\left (5 A b^2-5 a b B+7 a^2 C-2 b^2 C\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac {\left (A b^2-a (b B-a C)\right ) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.42 (sec) , antiderivative size = 404, normalized size of antiderivative = 0.91 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\frac {\frac {\frac {2 \left (-25 a^3 b B+40 a b^3 B+a^2 b^2 (15 A-32 C)+35 a^4 C-6 b^4 (5 A+3 C)\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {8 \left (-10 a^2 b B-5 b^3 B+14 a^3 C+a b^2 (15 A+C)\right ) \left ((a+b) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )\right )}{a+b}+\frac {6 \left (-25 a^3 b B+20 a b^3 B+3 a^2 b^2 (5 A-8 C)+35 a^4 C-2 b^4 (5 A+3 C)\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b^2 \sqrt {\sin ^2(c+d x)}}}{(a-b) (a+b)}+4 \sqrt {\cos (c+d x)} \left (10 (b B-2 a C) \sin (c+d x)-\frac {15 a^2 \left (A b^2+a (-b B+a C)\right ) \sin (c+d x)}{\left (a^2-b^2\right ) (a+b \cos (c+d x))}+3 b C \sin (2 (c+d x))\right )}{60 b^3 d} \]

[In]

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

[Out]

(((2*(-25*a^3*b*B + 40*a*b^3*B + a^2*b^2*(15*A - 32*C) + 35*a^4*C - 6*b^4*(5*A + 3*C))*EllipticPi[(2*b)/(a + b
), (c + d*x)/2, 2])/(a + b) + (8*(-10*a^2*b*B - 5*b^3*B + 14*a^3*C + a*b^2*(15*A + C))*((a + b)*EllipticF[(c +
 d*x)/2, 2] - a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2]))/(a + b) + (6*(-25*a^3*b*B + 20*a*b^3*B + 3*a^2*b^2
*(5*A - 8*C) + 35*a^4*C - 2*b^4*(5*A + 3*C))*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*a*(a + b)*E
llipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + (-2*a^2 + b^2)*EllipticPi[-(b/a), ArcSin[Sqrt[Cos[c + d*x]]], -1])*
Sin[c + d*x])/(a*b^2*Sqrt[Sin[c + d*x]^2]))/((a - b)*(a + b)) + 4*Sqrt[Cos[c + d*x]]*(10*(b*B - 2*a*C)*Sin[c +
 d*x] - (15*a^2*(A*b^2 + a*(-(b*B) + a*C))*Sin[c + d*x])/((a^2 - b^2)*(a + b*Cos[c + d*x])) + 3*b*C*Sin[2*(c +
 d*x)]))/(60*b^3*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1381\) vs. \(2(509)=1018\).

Time = 13.50 (sec) , antiderivative size = 1382, normalized size of antiderivative = 3.11

method result size
default \(\text {Expression too large to display}\) \(1382\)

[In]

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

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-4/5*C/b^2/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/
2*c)^2)^(1/2)*(4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-14*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+6*sin(1/2*
d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(
1/2*d*x+1/2*c),2^(1/2))+9*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+
1/2*c),2^(1/2)))+4/3/b^3*(B*b-2*C*a-3*C*b)*(2*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-sin(1/2*d*x+1/2*c)^2*cos
(1/2*d*x+1/2*c)+2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2
^(1/2))-3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))
/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)-4*a^2/b^4*(3*A*b^2-4*B*a*b+5*C*a^2)/(-2*a*b+2*b^2)*(sin(
1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)
*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))-2*a^3*(A*b^2-B*a*b+C*a^2)/b^5*(-b^2/a/(a^2-b^2)*cos(1/2*d*x
+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2*b+a-b)-1/2/(a+b)/a*(sin(1
/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*
EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-1/2*b/a/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+
1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1/2*b/a/(a
^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/
2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-3*a/(a^2-b^2)/(-2*a*b+2*b^2)*b*(sin(1/2*d*x+1/2*c)^2)^(1/2
)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*
x+1/2*c),-2*b/(a-b),2^(1/2))+1/a/(a^2-b^2)/(-2*a*b+2*b^2)*b^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2
*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2
^(1/2)))+2/b^4*(A*b^2-2*B*a*b-2*B*b^2+3*C*a^2+4*C*a*b+3*C*b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/
2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-El
lipticE(cos(1/2*d*x+1/2*c),2^(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+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x
+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))/sin(1/2*d*x+1/2*c)/(-1+2*cos(1/2*d*x+1/2*c)^2)^(1/2)/d

Fricas [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{5/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]

[In]

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

[Out]

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