Integrand size = 34, antiderivative size = 231 \[ \int \sqrt {a+b \cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {2 \left (5 a b B-2 a^2 C+9 b^2 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 \left (a^2-b^2\right ) (5 b B-2 a C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{15 b^2 d \sqrt {a+b \cos (c+d x)}}+\frac {2 (5 b B-2 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 b d}+\frac {2 C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 b d} \]
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*(5*B*a*b-2*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^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*cos(d*x+c))/( a+b))^(1/2)/b^2/d/(a+b*cos(d*x+c))^(1/2)
Time = 3.10 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.77 \[ \int \sqrt {a+b \cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \left (b^2 (5 b B+7 a C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )+\left (5 a b B-2 a^2 C+9 b^2 C\right ) \left ((a+b) E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-a \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )\right )\right )+2 b (a+b \cos (c+d x)) (5 b B+a C+3 b C \cos (c+d x)) \sin (c+d x)}{15 b^2 d \sqrt {a+b \cos (c+d x)}} \]
(2*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*(b^2*(5*b*B + 7*a*C)*EllipticF[(c + d*x)/2, (2*b)/(a + b)] + (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]])
Time = 1.12 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.03, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.441, Rules used = {3042, 3502, 27, 3042, 3232, 27, 3042, 3231, 3042, 3134, 3042, 3132, 3142, 3042, 3140}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \sqrt {a+b \cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )} \left (B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {2 \int \frac {1}{2} \sqrt {a+b \cos (c+d x)} (3 b C+(5 b B-2 a C) \cos (c+d x))dx}{5 b}+\frac {2 C \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \sqrt {a+b \cos (c+d x)} (3 b C+(5 b B-2 a C) \cos (c+d x))dx}{5 b}+\frac {2 C \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )} \left (3 b C+(5 b B-2 a C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{5 b}+\frac {2 C \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 b d}\) |
| \(\Big \downarrow \) 3232 |
| \(\displaystyle \frac {\frac {2}{3} \int \frac {b (5 b B+7 a C)+\left (-2 C a^2+5 b B a+9 b^2 C\right ) \cos (c+d x)}{2 \sqrt {a+b \cos (c+d x)}}dx+\frac {2 (5 b B-2 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}}{5 b}+\frac {2 C \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{3} \int \frac {b (5 b B+7 a C)+\left (-2 C a^2+5 b B a+9 b^2 C\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx+\frac {2 (5 b B-2 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}}{5 b}+\frac {2 C \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \int \frac {b (5 b B+7 a C)+\left (-2 C a^2+5 b B a+9 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (5 b B-2 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}}{5 b}+\frac {2 C \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 b d}\) |
| \(\Big \downarrow \) 3231 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {\left (-2 a^2 C+5 a b B+9 b^2 C\right ) \int \sqrt {a+b \cos (c+d x)}dx}{b}-\frac {\left (a^2-b^2\right ) (5 b B-2 a C) \int \frac {1}{\sqrt {a+b \cos (c+d x)}}dx}{b}\right )+\frac {2 (5 b B-2 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}}{5 b}+\frac {2 C \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {\left (-2 a^2 C+5 a b B+9 b^2 C\right ) \int \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {\left (a^2-b^2\right ) (5 b B-2 a C) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}\right )+\frac {2 (5 b B-2 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}}{5 b}+\frac {2 C \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 b d}\) |
| \(\Big \downarrow \) 3134 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {\left (-2 a^2 C+5 a b B+9 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}dx}{b \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (a^2-b^2\right ) (5 b B-2 a C) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}\right )+\frac {2 (5 b B-2 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}}{5 b}+\frac {2 C \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {\left (-2 a^2 C+5 a b B+9 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{b \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (a^2-b^2\right ) (5 b B-2 a C) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}\right )+\frac {2 (5 b B-2 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}}{5 b}+\frac {2 C \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 b d}\) |
| \(\Big \downarrow \) 3132 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {2 \left (-2 a^2 C+5 a b B+9 b^2 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (a^2-b^2\right ) (5 b B-2 a C) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}\right )+\frac {2 (5 b B-2 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}}{5 b}+\frac {2 C \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 b d}\) |
| \(\Big \downarrow \) 3142 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {2 \left (-2 a^2 C+5 a b B+9 b^2 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (a^2-b^2\right ) (5 b B-2 a C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{b \sqrt {a+b \cos (c+d x)}}\right )+\frac {2 (5 b B-2 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}}{5 b}+\frac {2 C \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {2 \left (-2 a^2 C+5 a b B+9 b^2 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (a^2-b^2\right ) (5 b B-2 a C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{b \sqrt {a+b \cos (c+d x)}}\right )+\frac {2 (5 b B-2 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}}{5 b}+\frac {2 C \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 b d}\) |
| \(\Big \downarrow \) 3140 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {2 \left (-2 a^2 C+5 a b B+9 b^2 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 \left (a^2-b^2\right ) (5 b B-2 a C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{b d \sqrt {a+b \cos (c+d x)}}\right )+\frac {2 (5 b B-2 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}}{5 b}+\frac {2 C \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 b d}\) |
(2*C*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*b*d) + (((2*(5*a*b*B - 2* a^2*C + 9*b^2*C)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(b*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)])/(b*d*Sqrt[a + b*Cos[c + d*x]]))/3 + (2*(5*b*B - 2*a*C)*Sqrt[a + b*C os[c + d*x]]*Sin[c + d*x])/(3*d))/(5*b)
3.9.15.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[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]
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*S qrt[a + b]))*EllipticF[(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]
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[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]
Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + ( f_.)*(x_)]], x_Symbol] :> Simp[(b*c - a*d)/b Int[1/Sqrt[a + b*Sin[e + f*x ]], x], x] + Simp[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]
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] + Simp[1/(m + 1) Int[(a + b*Sin[e + f*x])^(m - 1)*Simp[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]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[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]
Leaf count of result is larger than twice the leaf count of optimal. \(992\) vs. \(2(269)=538\).
Time = 10.59 (sec) , antiderivative size = 993, normalized size of antiderivative = 4.30
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(993\) |
| parts | \(\text {Expression too large to display}\) | \(1119\) |
-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*s in(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)*b^3+(20*B*b^3+16*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^3-2*C*a^2*b-8* 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)-5*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)*Ellipt icF(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))+5*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-5*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+2*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^3-2*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^2-2*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+2*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)...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.13 (sec) , antiderivative size = 492, normalized size of antiderivative = 2.13 \[ \int \sqrt {a+b \cos (c+d x)} \left (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 \, 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 (4 i \, C a^{3} - 10 i \, B a^{2} b + 3 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 (2 i \, C a^{2} b - 5 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 (-2 i \, C a^{2} b + 5 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 ) + 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} \]
1/45*(sqrt(2)*(-4*I*C*a^3 + 10*I*B*a^2*b - 3*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)*(4*I*C*a ^3 - 10*I*B*a^2*b + 3*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)*(2*I*C*a^2*b - 5*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)) - 3*s qrt(2)*(-2*I*C*a^2*b + 5*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) + C*a*b^2 + 5*B* b^3)*sqrt(b*cos(d*x + c) + a)*sin(d*x + c))/(b^3*d)
\[ \int \sqrt {a+b \cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int \left (B + C \cos {\left (c + d x \right )}\right ) \sqrt {a + b \cos {\left (c + d x \right )}} \cos {\left (c + d x \right )}\, dx \]
\[ \int \sqrt {a+b \cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \sqrt {b \cos \left (d x + c\right ) + a} \,d x } \]
\[ \int \sqrt {a+b \cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \sqrt {b \cos \left (d x + c\right ) + a} \,d x } \]
Timed out. \[ \int \sqrt {a+b \cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int \left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )\right )\,\sqrt {a+b\,\cos \left (c+d\,x\right )} \,d x \]