\(\int \frac {\cos (c+d x) (B \cos (c+d x)+C \cos ^2(c+d x))}{\sqrt {a+b \cos (c+d x)}} \, dx\) [835]

Optimal result

Integrand size = 40, antiderivative size = 246 \[ \int \frac {\cos (c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx=-\frac {2 \left (10 a b B-8 a^2 C-9 b^2 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^3 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 \left (10 a^2 b B+5 b^3 B-8 a^3 C-7 a b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{15 b^3 d \sqrt {a+b \cos (c+d x)}}+\frac {2 (5 b B-4 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 b^2 d}+\frac {2 C \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 b d} \]

[Out]

2/15*(5*B*b-4*C*a)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/b^2/d+2/5*C*cos(d*x+c)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/
b/d-2/15*(10*B*a*b-8*C*a^2-9*C*b^2)*(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^3/d/((a+b*cos(d*x+c))/(a+b))^(1/2)+2/15*(10*B*a^2*b+5*B*b
^3-8*C*a^3-7*C*a*b^2)*(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^3/d/(a+b*cos(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3108, 3069, 3102, 2831, 2742, 2740, 2734, 2732} \[ \int \frac {\cos (c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx=-\frac {2 \left (-8 a^2 C+10 a b B-9 b^2 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^3 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 \left (-8 a^3 C+10 a^2 b B-7 a b^2 C+5 b^3 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{15 b^3 d \sqrt {a+b \cos (c+d x)}}+\frac {2 (5 b B-4 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{15 b^2 d}+\frac {2 C \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d} \]

[In]

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

[Out]

(-2*(10*a*b*B - 8*a^2*C - 9*b^2*C)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(15*b^3*d*S
qrt[(a + b*Cos[c + d*x])/(a + b)]) + (2*(10*a^2*b*B + 5*b^3*B - 8*a^3*C - 7*a*b^2*C)*Sqrt[(a + b*Cos[c + d*x])
/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(15*b^3*d*Sqrt[a + b*Cos[c + d*x]]) + (2*(5*b*B - 4*a*C)*Sqrt
[a + b*Cos[c + d*x]]*Sin[c + d*x])/(15*b^2*d) + (2*C*Cos[c + d*x]*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(5*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 3069

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*
x])^(n + 1)/(d*f*(m + n + 1))), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f
*x])^n*Simp[a^2*A*d*(m + n + 1) + b*B*(b*c*(m - 1) + a*d*(n + 1)) + (a*d*(2*A*b + a*B)*(m + n + 1) - b*B*(a*c
- b*d*(m + n)))*Sin[e + f*x] + b*(A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m
, 1] &&  !(IGtQ[n, 1] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 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 3108

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] :> Dist[1/b^2, Int[(a + b*Sin[e + f*x])
^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\cos ^2(c+d x) (B+C \cos (c+d x))}{\sqrt {a+b \cos (c+d x)}} \, dx \\ & = \frac {2 C \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 b d}+\frac {2 \int \frac {a C+\frac {3}{2} b C \cos (c+d x)+\frac {1}{2} (5 b B-4 a C) \cos ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{5 b} \\ & = \frac {2 (5 b B-4 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 b^2 d}+\frac {2 C \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 b d}+\frac {4 \int \frac {\frac {1}{4} b (5 b B+2 a C)-\frac {1}{4} \left (10 a b B-8 a^2 C-9 b^2 C\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{15 b^2} \\ & = \frac {2 (5 b B-4 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 b^2 d}+\frac {2 C \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 b d}-\frac {\left (10 a b B-8 a^2 C-9 b^2 C\right ) \int \sqrt {a+b \cos (c+d x)} \, dx}{15 b^3}+\frac {\left (10 a^2 b B+5 b^3 B-8 a^3 C-7 a b^2 C\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx}{15 b^3} \\ & = \frac {2 (5 b B-4 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 b^2 d}+\frac {2 C \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 b d}-\frac {\left (\left (10 a b B-8 a^2 C-9 b^2 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^3 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\left (\left (10 a^2 b B+5 b^3 B-8 a^3 C-7 a b^2 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^3 \sqrt {a+b \cos (c+d x)}} \\ & = -\frac {2 \left (10 a b B-8 a^2 C-9 b^2 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^3 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 \left (10 a^2 b B+5 b^3 B-8 a^3 C-7 a b^2 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 )}{15 b^3 d \sqrt {a+b \cos (c+d x)}}+\frac {2 (5 b B-4 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 b^2 d}+\frac {2 C \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 b d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.69 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.73 \[ \int \frac {\cos (c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx=\frac {2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \left (b^2 (5 b B+2 a C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )+\left (-10 a b B+8 a^2 C+9 b^2 C\right ) \left ((a+b) E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-a \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )\right )\right )+2 b (a+b \cos (c+d x)) (5 b B-4 a C+3 b C \cos (c+d x)) \sin (c+d x)}{15 b^3 d \sqrt {a+b \cos (c+d x)}} \]

[In]

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

[Out]

(2*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*(b^2*(5*b*B + 2*a*C)*EllipticF[(c + d*x)/2, (2*b)/(a + b)] + (-10*a*b*B
+ 8*a^2*C + 9*b^2*C)*((a + b)*EllipticE[(c + d*x)/2, (2*b)/(a + b)] - a*EllipticF[(c + d*x)/2, (2*b)/(a + b)])
) + 2*b*(a + b*Cos[c + d*x])*(5*b*B - 4*a*C + 3*b*C*Cos[c + d*x])*Sin[c + d*x])/(15*b^3*d*Sqrt[a + b*Cos[c + d
*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(992\) vs. \(2(284)=568\).

Time = 13.87 (sec) , antiderivative size = 993, normalized size of antiderivative = 4.04

method	result	size
default	\(\text {Expression too large to display}\)	\(993\)
parts	\(\text {Expression too large to display}\)	\(1120\)

[In]

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

[Out]

-2/15*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-24*C*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*
c)*b^3+(20*B*b^3-4*C*a*b^2+24*C*b^3)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-10*B*a*b^2-10*B*b^3+8*C*a^2*b+2
*C*a*b^2-6*C*b^3)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+10*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^2*b+5*B*b^3*(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))-10*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+10*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^2-8*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^3-7*a*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))*b^2+8*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)*Elliptic
E(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3-8*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^2*b+9*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^2-9*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))*b^3)/b^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)/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 = 0.13 (sec) , antiderivative size = 493, normalized size of antiderivative = 2.00 \[ \int \frac {\cos (c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx=\frac {\sqrt {2} {\left (16 i \, C a^{3} - 20 i \, B a^{2} b + 12 i \, C a b^{2} - 15 i \, B b^{3}\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 (-16 i \, C a^{3} + 20 i \, B a^{2} b - 12 i \, C a b^{2} + 15 i \, B b^{3}\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 (-8 i \, C a^{2} b + 10 i \, B a b^{2} - 9 i \, C b^{3}\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 (8 i \, C a^{2} b - 10 i \, B a b^{2} + 9 i \, C b^{3}\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 (3 \, C b^{3} \cos \left (d x + c\right ) - 4 \, C a b^{2} + 5 \, B b^{3}\right )} \sqrt {b \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{45 \, b^{4} d} \]

[In]

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

[Out]

1/45*(sqrt(2)*(16*I*C*a^3 - 20*I*B*a^2*b + 12*I*C*a*b^2 - 15*I*B*b^3)*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)*(-16*
I*C*a^3 + 20*I*B*a^2*b - 12*I*C*a*b^2 + 15*I*B*b^3)*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)*(-8*I*C*a^2*b + 10*I*
B*a*b^2 - 9*I*C*b^3)*sqrt(b)*weierstrassZeta(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, weierstrass
PInverse(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)*(8*I*C*a^2*b - 10*I*B*a*b^2 + 9*I*C*b^3)*sqrt(b)*weierstrassZeta(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*(3*C*b^3*cos(d*x + c) - 4*C*a*b^2 + 5*B*b^3)*sqrt(b*cos(d*x
+ c) + a)*sin(d*x + c))/(b^4*d)

Sympy [F(-1)]

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

[In]

integrate(cos(d*x+c)*(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]

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

[In]

integrate(cos(d*x+c)*(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))*cos(d*x + c)/sqrt(b*cos(d*x + c) + a), x)

Giac [F]

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

[In]

integrate(cos(d*x+c)*(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))*cos(d*x + c)/sqrt(b*cos(d*x + c) + a), x)

Mupad [F(-1)]

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

[In]

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