Integrand size = 33, antiderivative size = 140 \[ \int \frac {(a+b \cos (c+d x))^2 (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\frac {2 \left (3 b^2 B+5 a (2 A b+a B)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (3 a^2 A+A b^2+2 a b B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 b (5 A b+7 a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 b B \sqrt {\cos (c+d x)} (a+b \cos (c+d x)) \sin (c+d x)}{5 d} \]
2/5*(3*b^2*B+5*a*(2*A*b+B*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))/d+2/3*(3*A*a^2+A*b^2+2*B*a*b)*(c os(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))/d+2/15*b*(5*A*b+7*B*a)*sin(d*x+c)*cos(d*x+c)^(1/2)/d+2/5*b*B*(a+ b*cos(d*x+c))*sin(d*x+c)*cos(d*x+c)^(1/2)/d
Time = 1.17 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.76 \[ \int \frac {(a+b \cos (c+d x))^2 (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\frac {2 \left (3 \left (10 a A b+5 a^2 B+3 b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+5 \left (3 a^2 A+A b^2+2 a b B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+b \sqrt {\cos (c+d x)} (5 A b+10 a B+3 b B \cos (c+d x)) \sin (c+d x)\right )}{15 d} \]
(2*(3*(10*a*A*b + 5*a^2*B + 3*b^2*B)*EllipticE[(c + d*x)/2, 2] + 5*(3*a^2* A + A*b^2 + 2*a*b*B)*EllipticF[(c + d*x)/2, 2] + b*Sqrt[Cos[c + d*x]]*(5*A *b + 10*a*B + 3*b*B*Cos[c + d*x])*Sin[c + d*x]))/(15*d)
Time = 0.79 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.04, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3469, 27, 3042, 3502, 27, 3042, 3227, 3042, 3119, 3120}
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.
\(\displaystyle \int \frac {(a+b \cos (c+d x))^2 (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 3469 |
\(\displaystyle \frac {2}{5} \int \frac {b (5 A b+7 a B) \cos ^2(c+d x)+\left (3 B b^2+5 a (2 A b+a B)\right ) \cos (c+d x)+a (5 a A+b B)}{2 \sqrt {\cos (c+d x)}}dx+\frac {2 b B \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}{5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int \frac {b (5 A b+7 a B) \cos ^2(c+d x)+\left (3 B b^2+5 a (2 A b+a B)\right ) \cos (c+d x)+a (5 a A+b B)}{\sqrt {\cos (c+d x)}}dx+\frac {2 b B \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \int \frac {b (5 A b+7 a B) \sin \left (c+d x+\frac {\pi }{2}\right )^2+\left (3 B b^2+5 a (2 A b+a B)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+a (5 a A+b B)}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b B \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}{5 d}\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int \frac {5 \left (3 A a^2+2 b B a+A b^2\right )+3 \left (3 B b^2+5 a (2 A b+a B)\right ) \cos (c+d x)}{2 \sqrt {\cos (c+d x)}}dx+\frac {2 b (7 a B+5 A b) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 b B \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}{5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {5 \left (3 A a^2+2 b B a+A b^2\right )+3 \left (3 B b^2+5 a (2 A b+a B)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx+\frac {2 b (7 a B+5 A b) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 b B \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {5 \left (3 A a^2+2 b B a+A b^2\right )+3 \left (3 B b^2+5 a (2 A b+a B)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b (7 a B+5 A b) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 b B \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}{5 d}\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (5 \left (3 a^2 A+2 a b B+A b^2\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx+3 \left (5 a (a B+2 A b)+3 b^2 B\right ) \int \sqrt {\cos (c+d x)}dx\right )+\frac {2 b (7 a B+5 A b) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 b B \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (5 \left (3 a^2 A+2 a b B+A b^2\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 \left (5 a (a B+2 A b)+3 b^2 B\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )+\frac {2 b (7 a B+5 A b) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 b B \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}{5 d}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (5 \left (3 a^2 A+2 a b B+A b^2\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {6 \left (5 a (a B+2 A b)+3 b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {2 b (7 a B+5 A b) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 b B \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}{5 d}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {10 \left (3 a^2 A+2 a b B+A b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {6 \left (5 a (a B+2 A b)+3 b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {2 b (7 a B+5 A b) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 b B \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}{5 d}\) |
(2*b*B*Sqrt[Cos[c + d*x]]*(a + b*Cos[c + d*x])*Sin[c + d*x])/(5*d) + (((6* (3*b^2*B + 5*a*(2*A*b + a*B))*EllipticE[(c + d*x)/2, 2])/d + (10*(3*a^2*A + A*b^2 + 2*a*b*B)*EllipticF[(c + d*x)/2, 2])/d)/3 + (2*b*(5*A*b + 7*a*B)* Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d))/5
3.4.55.3.1 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[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
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]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[(-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] + Simp[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] && !(IGt Q[n, 1] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
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. \(486\) vs. \(2(180)=360\).
Time = 9.59 (sec) , antiderivative size = 487, normalized size of antiderivative = 3.48
method | result | size |
default | \(-\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-24 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}+\left (20 A \,b^{2}+40 B a b +24 B \,b^{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-10 A \,b^{2}-20 B a b -6 B \,b^{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+15 A \,a^{2} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+5 A \,b^{2} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-30 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a b +10 B a b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-15 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{2}-9 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b^{2}\right )}{15 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\) | \(487\) |
parts | \(-\frac {2 \left (A \,b^{2}+2 B a b \right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (4 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{3 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}+\frac {2 \left (2 A a b +B \,a^{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{\sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}+\frac {2 A \,a^{2} \operatorname {am}^{-1}\left (\frac {d x}{2}+\frac {c}{2}| \sqrt {2}\right )}{d}-\frac {2 B \,b^{2} \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{5 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\) | \(560\) |
-2/15*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-24*B*cos(1 /2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6*b^2+(20*A*b^2+40*B*a*b+24*B*b^2)*sin(1/ 2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-10*A*b^2-20*B*a*b-6*B*b^2)*sin(1/2*d*x +1/2*c)^2*cos(1/2*d*x+1/2*c)+15*A*a^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))+5*A*b^2*(s in(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))-30*A*(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))*a*b+10*B*a*b*(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))-15*B*(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))*a^2-9*B*(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))*b^2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1 /2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.54 \[ \int \frac {(a+b \cos (c+d x))^2 (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\frac {2 \, {\left (3 \, B b^{2} \cos \left (d x + c\right ) + 10 \, B a b + 5 \, A b^{2}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 5 \, \sqrt {2} {\left (3 i \, A a^{2} + 2 i \, B a b + i \, A b^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 \, \sqrt {2} {\left (-3 i \, A a^{2} - 2 i \, B a b - i \, A b^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, \sqrt {2} {\left (-5 i \, B a^{2} - 10 i \, A a b - 3 i \, B b^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, \sqrt {2} {\left (5 i \, B a^{2} + 10 i \, A a b + 3 i \, B b^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{15 \, d} \]
1/15*(2*(3*B*b^2*cos(d*x + c) + 10*B*a*b + 5*A*b^2)*sqrt(cos(d*x + c))*sin (d*x + c) - 5*sqrt(2)*(3*I*A*a^2 + 2*I*B*a*b + I*A*b^2)*weierstrassPInvers e(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 5*sqrt(2)*(-3*I*A*a^2 - 2*I*B*a* b - I*A*b^2)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3 *sqrt(2)*(-5*I*B*a^2 - 10*I*A*a*b - 3*I*B*b^2)*weierstrassZeta(-4, 0, weie rstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*sqrt(2)*(5*I*B* a^2 + 10*I*A*a*b + 3*I*B*b^2)*weierstrassZeta(-4, 0, weierstrassPInverse(- 4, 0, cos(d*x + c) - I*sin(d*x + c))))/d
Timed out. \[ \int \frac {(a+b \cos (c+d x))^2 (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\text {Timed out} \]
\[ \int \frac {(a+b \cos (c+d x))^2 (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\sqrt {\cos \left (d x + c\right )}} \,d x } \]
\[ \int \frac {(a+b \cos (c+d x))^2 (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\sqrt {\cos \left (d x + c\right )}} \,d x } \]
Time = 1.67 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.26 \[ \int \frac {(a+b \cos (c+d x))^2 (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\frac {A\,b^2\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}+\frac {2\,A\,a^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,B\,a^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,B\,a\,b\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}+\frac {4\,A\,a\,b\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}-\frac {2\,B\,b^2\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]
(A*b^2*((2*cos(c + d*x)^(1/2)*sin(c + d*x))/3 + (2*ellipticF(c/2 + (d*x)/2 , 2))/3))/d + (2*A*a^2*ellipticF(c/2 + (d*x)/2, 2))/d + (2*B*a^2*ellipticE (c/2 + (d*x)/2, 2))/d + (2*B*a*b*((2*cos(c + d*x)^(1/2)*sin(c + d*x))/3 + (2*ellipticF(c/2 + (d*x)/2, 2))/3))/d + (4*A*a*b*ellipticE(c/2 + (d*x)/2, 2))/d - (2*B*b^2*cos(c + d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/ 4, cos(c + d*x)^2))/(7*d*(sin(c + d*x)^2)^(1/2))