\(\int \frac {\cos ^2(c+d x) (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\sqrt {a+b \cos (c+d x)}} \, dx\) [1040]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [B] (verified)
   Fricas [C] (verification not implemented)
   Sympy [F(-1)]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 43, antiderivative size = 344 \[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx=\frac {2 \left (56 a^2 b B+63 b^3 B-48 a^3 C-2 a b^2 (35 A+22 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 b^4 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 \left (56 a^3 b B+49 a b^3 B-48 a^4 C-5 b^4 (7 A+5 C)-2 a^2 b^2 (35 A+16 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{105 b^4 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (35 A b^2-28 a b B+24 a^2 C+25 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{105 b^3 d}+\frac {2 (7 b B-6 a C) \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{35 b^2 d}+\frac {2 C \cos ^2(c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{7 b d} \]

[Out]

2/105*(35*A*b^2-28*B*a*b+24*C*a^2+25*C*b^2)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/b^3/d+2/35*(7*B*b-6*C*a)*cos(d*x
+c)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/b^2/d+2/7*C*cos(d*x+c)^2*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/b/d+2/105*(56
*B*a^2*b+63*B*b^3-48*a^3*C-2*a*b^2*(35*A+22*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/(a+b))^(1/2))*(a+b*cos(d*x+c))^(1/2)/b^4/d/((a+b*cos(d*x+c))/(a+b))^(1/2)-2/105*(56*
B*a^3*b+49*B*a*b^3-48*a^4*C-5*b^4*(7*A+5*C)-2*a^2*b^2*(35*A+16*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^4/d/(a+b*cos(d*x+c
))^(1/2)

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 344, 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 = {3128, 3102, 2831, 2742, 2740, 2734, 2732} \[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx=\frac {2 \sin (c+d x) \left (24 a^2 C-28 a b B+35 A b^2+25 b^2 C\right ) \sqrt {a+b \cos (c+d x)}}{105 b^3 d}+\frac {2 \left (-48 a^3 C+56 a^2 b B-2 a b^2 (35 A+22 C)+63 b^3 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 b^4 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 \left (-48 a^4 C+56 a^3 b B-2 a^2 b^2 (35 A+16 C)+49 a b^3 B-5 b^4 (7 A+5 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{105 b^4 d \sqrt {a+b \cos (c+d x)}}+\frac {2 (7 b B-6 a C) \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{35 b^2 d}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) \sqrt {a+b \cos (c+d x)}}{7 b d} \]

[In]

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

[Out]

(2*(56*a^2*b*B + 63*b^3*B - 48*a^3*C - 2*a*b^2*(35*A + 22*C))*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2,
(2*b)/(a + b)])/(105*b^4*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) - (2*(56*a^3*b*B + 49*a*b^3*B - 48*a^4*C - 5*b^
4*(7*A + 5*C) - 2*a^2*b^2*(35*A + 16*C))*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a +
b)])/(105*b^4*d*Sqrt[a + b*Cos[c + d*x]]) + (2*(35*A*b^2 - 28*a*b*B + 24*a^2*C + 25*b^2*C)*Sqrt[a + b*Cos[c +
d*x]]*Sin[c + d*x])/(105*b^3*d) + (2*(7*b*B - 6*a*C)*Cos[c + d*x]*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(35*b
^2*d) + (2*C*Cos[c + d*x]^2*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(7*b*d)

Rule 2732

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

Rule 2734

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/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
 b^2, 0] &&  !GtQ[a + b, 0]

Rule 2740

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

Rule 2742

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/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2831

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 3102

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 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])))

Rubi steps \begin{align*} \text {integral}& = \frac {2 C \cos ^2(c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{7 b d}+\frac {2 \int \frac {\cos (c+d x) \left (2 a C+\frac {1}{2} b (7 A+5 C) \cos (c+d x)+\frac {1}{2} (7 b B-6 a C) \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx}{7 b} \\ & = \frac {2 (7 b B-6 a C) \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{35 b^2 d}+\frac {2 C \cos ^2(c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{7 b d}+\frac {4 \int \frac {\frac {1}{2} a (7 b B-6 a C)+\frac {1}{4} b (21 b B+2 a C) \cos (c+d x)+\frac {1}{4} \left (35 A b^2-28 a b B+24 a^2 C+25 b^2 C\right ) \cos ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{35 b^2} \\ & = \frac {2 \left (35 A b^2-28 a b B+24 a^2 C+25 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{105 b^3 d}+\frac {2 (7 b B-6 a C) \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{35 b^2 d}+\frac {2 C \cos ^2(c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{7 b d}+\frac {8 \int \frac {\frac {1}{8} b \left (35 A b^2+14 a b B-12 a^2 C+25 b^2 C\right )+\frac {1}{8} \left (56 a^2 b B+63 b^3 B-48 a^3 C-2 a b^2 (35 A+22 C)\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{105 b^3} \\ & = \frac {2 \left (35 A b^2-28 a b B+24 a^2 C+25 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{105 b^3 d}+\frac {2 (7 b B-6 a C) \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{35 b^2 d}+\frac {2 C \cos ^2(c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{7 b d}-\frac {\left (56 a^3 b B+49 a b^3 B-48 a^4 C-5 b^4 (7 A+5 C)-2 a^2 b^2 (35 A+16 C)\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx}{105 b^4}+\frac {\left (56 a^2 b B+63 b^3 B-48 a^3 C-2 a b^2 (35 A+22 C)\right ) \int \sqrt {a+b \cos (c+d x)} \, dx}{105 b^4} \\ & = \frac {2 \left (35 A b^2-28 a b B+24 a^2 C+25 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{105 b^3 d}+\frac {2 (7 b B-6 a C) \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{35 b^2 d}+\frac {2 C \cos ^2(c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{7 b d}+\frac {\left (\left (56 a^2 b B+63 b^3 B-48 a^3 C-2 a b^2 (35 A+22 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}{105 b^4 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (\left (56 a^3 b B+49 a b^3 B-48 a^4 C-5 b^4 (7 A+5 C)-2 a^2 b^2 (35 A+16 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}{105 b^4 \sqrt {a+b \cos (c+d x)}} \\ & = \frac {2 \left (56 a^2 b B+63 b^3 B-48 a^3 C-2 a b^2 (35 A+22 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 b^4 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 \left (56 a^3 b B+49 a b^3 B-48 a^4 C-5 b^4 (7 A+5 C)-2 a^2 b^2 (35 A+16 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{105 b^4 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (35 A b^2-28 a b B+24 a^2 C+25 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{105 b^3 d}+\frac {2 (7 b B-6 a C) \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{35 b^2 d}+\frac {2 C \cos ^2(c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{7 b d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.54 (sec) , antiderivative size = 252, normalized size of antiderivative = 0.73 \[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx=\frac {4 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \left (b^2 \left (35 A b^2+14 a b B-12 a^2 C+25 b^2 C\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )-\left (-56 a^2 b B-63 b^3 B+48 a^3 C+2 a b^2 (35 A+22 C)\right ) \left ((a+b) E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-a \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )\right )\right )+b (a+b \cos (c+d x)) \left (\left (140 A b^2-112 a b B+96 a^2 C+115 b^2 C\right ) \sin (c+d x)+3 b (2 (7 b B-6 a C) \sin (2 (c+d x))+5 b C \sin (3 (c+d x)))\right )}{210 b^4 d \sqrt {a+b \cos (c+d x)}} \]

[In]

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

[Out]

(4*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*(b^2*(35*A*b^2 + 14*a*b*B - 12*a^2*C + 25*b^2*C)*EllipticF[(c + d*x)/2,
(2*b)/(a + b)] - (-56*a^2*b*B - 63*b^3*B + 48*a^3*C + 2*a*b^2*(35*A + 22*C))*((a + b)*EllipticE[(c + d*x)/2, (
2*b)/(a + b)] - a*EllipticF[(c + d*x)/2, (2*b)/(a + b)])) + b*(a + b*Cos[c + d*x])*((140*A*b^2 - 112*a*b*B + 9
6*a^2*C + 115*b^2*C)*Sin[c + d*x] + 3*b*(2*(7*b*B - 6*a*C)*Sin[2*(c + d*x)] + 5*b*C*Sin[3*(c + d*x)])))/(210*b
^4*d*Sqrt[a + b*Cos[c + d*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1634\) vs. \(2(378)=756\).

Time = 13.17 (sec) , antiderivative size = 1635, normalized size of antiderivative = 4.75

method result size
default \(\text {Expression too large to display}\) \(1635\)
parts \(\text {Expression too large to display}\) \(1947\)

[In]

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

[Out]

-2/105*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(240*C*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1
/2*c)^8+(-168*B*b^4+24*C*a*b^3-360*C*b^4)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(140*A*b^4-28*B*a*b^3+168*B*
b^4+24*C*a^2*b^2-24*C*a*b^3+280*C*b^4)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-70*A*a*b^3-70*A*b^4+56*B*a^2*
b^2+14*B*a*b^3-42*B*b^4-48*C*a^3*b-12*C*a^2*b^2-44*C*a*b^3-80*C*b^4)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+7
0*A*(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)*EllipticF(cos(1/2*d*x+1/2
*c),(-2*b/(a-b))^(1/2))*a^2*b^2+35*A*b^4*(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)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-70*A*(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))*a^2*b^2+70*A*(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))*a*b^3-56*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)*EllipticF
(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3*b-49*B*a*(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)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^3+56*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))*a^3*b-56
*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))*a^2*b^2+63*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))*a*b^3-63*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*si
n(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))*b^4+48*C*(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)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(
1/2))*a^4+32*C*(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)*EllipticF(cos(
1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b^2+25*C*b^4*(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)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-48*C*(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))*a^4+48*C*(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))*a^3*b-44*C*(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)*El
lipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b^2+44*C*(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))*a*b^3)/b^4/(-2*b*sin(1/2*d*x+1
/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(-2*b*sin(1/2*d*x+1/2*c)^2+a+b)^(1/2)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.16 (sec) , antiderivative size = 604, normalized size of antiderivative = 1.76 \[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx=\frac {\sqrt {2} {\left (-96 i \, C a^{4} + 112 i \, B a^{3} b - 4 i \, {\left (35 \, A + 13 \, C\right )} a^{2} b^{2} + 84 i \, B a b^{3} - 15 i \, {\left (7 \, A + 5 \, C\right )} b^{4}\right )} \sqrt {b} {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) + 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right ) + \sqrt {2} {\left (96 i \, C a^{4} - 112 i \, B a^{3} b + 4 i \, {\left (35 \, A + 13 \, C\right )} a^{2} b^{2} - 84 i \, B a b^{3} + 15 i \, {\left (7 \, A + 5 \, C\right )} b^{4}\right )} \sqrt {b} {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) - 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right ) - 3 \, \sqrt {2} {\left (48 i \, C a^{3} b - 56 i \, B a^{2} b^{2} + 2 i \, {\left (35 \, A + 22 \, C\right )} a b^{3} - 63 i \, B b^{4}\right )} \sqrt {b} {\rm weierstrassZeta}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) + 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right )\right ) - 3 \, \sqrt {2} {\left (-48 i \, C a^{3} b + 56 i \, B a^{2} b^{2} - 2 i \, {\left (35 \, A + 22 \, C\right )} a b^{3} + 63 i \, B b^{4}\right )} \sqrt {b} {\rm weierstrassZeta}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) - 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right )\right ) + 6 \, {\left (15 \, C b^{4} \cos \left (d x + c\right )^{2} + 24 \, C a^{2} b^{2} - 28 \, B a b^{3} + 5 \, {\left (7 \, A + 5 \, C\right )} b^{4} - 3 \, {\left (6 \, C a b^{3} - 7 \, B b^{4}\right )} \cos \left (d x + c\right )\right )} \sqrt {b \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{315 \, b^{5} d} \]

[In]

integrate(cos(d*x+c)^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/315*(sqrt(2)*(-96*I*C*a^4 + 112*I*B*a^3*b - 4*I*(35*A + 13*C)*a^2*b^2 + 84*I*B*a*b^3 - 15*I*(7*A + 5*C)*b^4)
*sqrt(b)*weierstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) + 3*I
*b*sin(d*x + c) + 2*a)/b) + sqrt(2)*(96*I*C*a^4 - 112*I*B*a^3*b + 4*I*(35*A + 13*C)*a^2*b^2 - 84*I*B*a*b^3 + 1
5*I*(7*A + 5*C)*b^4)*sqrt(b)*weierstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*
b*cos(d*x + c) - 3*I*b*sin(d*x + c) + 2*a)/b) - 3*sqrt(2)*(48*I*C*a^3*b - 56*I*B*a^2*b^2 + 2*I*(35*A + 22*C)*a
*b^3 - 63*I*B*b^4)*sqrt(b)*weierstrassZeta(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, weierstrassPI
nverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) + 3*I*b*sin(d*x + c) + 2*a)
/b)) - 3*sqrt(2)*(-48*I*C*a^3*b + 56*I*B*a^2*b^2 - 2*I*(35*A + 22*C)*a*b^3 + 63*I*B*b^4)*sqrt(b)*weierstrassZe
ta(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, weierstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8
*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) - 3*I*b*sin(d*x + c) + 2*a)/b)) + 6*(15*C*b^4*cos(d*x + c)^2 + 24*C
*a^2*b^2 - 28*B*a*b^3 + 5*(7*A + 5*C)*b^4 - 3*(6*C*a*b^3 - 7*B*b^4)*cos(d*x + c))*sqrt(b*cos(d*x + c) + a)*sin
(d*x + c))/(b^5*d)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

integrate(cos(d*x+c)^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate(cos(d*x+c)^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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