\(\int \sqrt {a+b \cos (c+d x)} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\) [1016]

Optimal result

Integrand size = 35, antiderivative size = 237 \[ \int \sqrt {a+b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {2 \left (3 b^2 (5 A+3 C)+a (5 b B-2 a C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 \left (a^2-b^2\right ) (5 b B-2 a C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{15 b^2 d \sqrt {a+b \cos (c+d x)}}+\frac {2 (5 b B-2 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 b d}+\frac {2 C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 b d} \]

[Out]

2/5*C*(a+b*cos(d*x+c))^(3/2)*sin(d*x+c)/b/d+2/15*(5*B*b-2*C*a)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/b/d+2/15*(3*b
^2*(5*A+3*C)+a*(5*B*b-2*C*a))*(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^2/d/((a+b*cos(d*x+c))/(a+b))^(1/2)-2/15*(a^2-b^2)*(5*B*b-2*C*a)
*(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*c
os(d*x+c))/(a+b))^(1/2)/b^2/d/(a+b*cos(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3102, 2832, 2831, 2742, 2740, 2734, 2732} \[ \int \sqrt {a+b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=-\frac {2 \left (a^2-b^2\right ) (5 b B-2 a C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{15 b^2 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (a (5 b B-2 a C)+3 b^2 (5 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 (5 b B-2 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{15 b d}+\frac {2 C \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 b d} \]

[In]

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

[Out]

(2*(3*b^2*(5*A + 3*C) + a*(5*b*B - 2*a*C))*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(15
*b^2*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) - (2*(a^2 - b^2)*(5*b*B - 2*a*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]
*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(15*b^2*d*Sqrt[a + b*Cos[c + d*x]]) + (2*(5*b*B - 2*a*C)*Sqrt[a + b*Co
s[c + d*x]]*Sin[c + d*x])/(15*b*d) + (2*C*(a + b*Cos[c + d*x])^(3/2)*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 2832

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

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]

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

Mathematica [A] (verified)

Time = 1.95 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.80 \[ \int \sqrt {a+b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \left (b^2 (15 a A+5 b B+7 a C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )+\left (15 A b^2+5 a b B-2 a^2 C+9 b^2 C\right ) \left ((a+b) E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-a \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )\right )\right )+2 b (a+b \cos (c+d x)) (5 b B+a C+3 b C \cos (c+d x)) \sin (c+d x)}{15 b^2 d \sqrt {a+b \cos (c+d x)}} \]

[In]

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

[Out]

(2*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*(b^2*(15*a*A + 5*b*B + 7*a*C)*EllipticF[(c + d*x)/2, (2*b)/(a + b)] + (1
5*A*b^2 + 5*a*b*B - 2*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 + a*C + 3*b*C*Cos[c + d*x])*Sin[c + d*x])/(15*b^2*d*Sqrt
[a + b*Cos[c + d*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1186\) vs. \(2(275)=550\).

Time = 10.16 (sec) , antiderivative size = 1187, normalized size of antiderivative = 5.01

method	result	size
default	\(\text {Expression too large to display}\)	\(1187\)
parts	\(\text {Expression too large to display}\)	\(1289\)

[In]

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

[Out]

-2/15*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*C*(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)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3-9*C*(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)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^3+
16*C*cos(1/2*d*x+1/2*c)^5*a*b^2+10*B*cos(1/2*d*x+1/2*c)^3*a*b^2+2*C*cos(1/2*d*x+1/2*c)^3*a^2*b-10*B*a*b^2*cos(
1/2*d*x+1/2*c)+8*C*a*b^2*cos(1/2*d*x+1/2*c)-2*C*a^2*b*cos(1/2*d*x+1/2*c)-24*C*a*b^2*cos(1/2*d*x+1/2*c)^3+9*C*(
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)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a
-b))^(1/2))*a*b^2-5*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)*EllipticE(cos(
1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b^2+15*A*(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)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b^2-5*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)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b+5*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)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b
-2*C*a*(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)*EllipticF(cos(1/2*d*x+1/2*c),
(-2*b/(a-b))^(1/2))*b^2+2*C*(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)*Elliptic
E(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b-15*A*(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)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^3+5*B*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)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+2*C*(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)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3
+24*C*cos(1/2*d*x+1/2*c)^7*b^3+20*B*cos(1/2*d*x+1/2*c)^5*b^3+10*B*b^3*cos(1/2*d*x+1/2*c)-6*C*b^3*cos(1/2*d*x+1
/2*c)-30*B*b^3*cos(1/2*d*x+1/2*c)^3+30*C*b^3*cos(1/2*d*x+1/2*c)^3-48*C*cos(1/2*d*x+1/2*c)^5*b^3)/b^2/(-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 = 0.13 (sec) , antiderivative size = 512, normalized size of antiderivative = 2.16 \[ \int \sqrt {a+b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {\sqrt {2} {\left (-4 i \, C a^{3} + 10 i \, B a^{2} b - 3 i \, {\left (5 \, A + C\right )} 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 (4 i \, C a^{3} - 10 i \, B a^{2} b + 3 i \, {\left (5 \, A + C\right )} 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 (2 i \, C a^{2} b - 5 i \, B a b^{2} - 3 i \, {\left (5 \, A + 3 \, C\right )} 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 (-2 i \, C a^{2} b + 5 i \, B a b^{2} + 3 i \, {\left (5 \, A + 3 \, C\right )} 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 ) + 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^{3} d} \]

[In]

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

[Out]

1/45*(sqrt(2)*(-4*I*C*a^3 + 10*I*B*a^2*b - 3*I*(5*A + 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
)*(4*I*C*a^3 - 10*I*B*a^2*b + 3*I*(5*A + 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)*(2*I*C*a^
2*b - 5*I*B*a*b^2 - 3*I*(5*A + 3*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)) - 3*sqrt(2)*(-2*I*C*a^2*b + 5*I*B*a*b^2 + 3*I*(5*A + 3*C)*b^3)*sqrt(b)*weierstrassZ
eta(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) + C*a*b^
2 + 5*B*b^3)*sqrt(b*cos(d*x + c) + a)*sin(d*x + c))/(b^3*d)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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