Integrand size = 33, antiderivative size = 126 \[ \int \frac {(a+a \cos (c+d x))^2 (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\frac {4 a^2 (5 A+4 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^2 (2 A+B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 a^2 (5 A+7 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 B \sqrt {\cos (c+d x)} \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d} \]
4/5*a^2*(5*A+4*B)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Elliptic E(sin(1/2*d*x+1/2*c),2^(1/2))/d+4/3*a^2*(2*A+B)*(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))/d+2/15*a^2*(5* A+7*B)*sin(d*x+c)*cos(d*x+c)^(1/2)/d+2/5*B*(a^2+a^2*cos(d*x+c))*sin(d*x+c) *cos(d*x+c)^(1/2)/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 6.14 (sec) , antiderivative size = 369, normalized size of antiderivative = 2.93 \[ \int \frac {(a+a \cos (c+d x))^2 (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\frac {a^2 (1+\cos (c+d x))^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (-6 (5 A+4 B) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sec (c) \sin (d x+\arctan (\tan (c)))+\left (9 (5 A+4 B) \cos (c-d x-\arctan (\tan (c))) \csc (c) \sec (c)+15 A \cos (c+d x+\arctan (\tan (c))) \csc (c) \sec (c)+12 B \cos (c+d x+\arctan (\tan (c))) \csc (c) \sec (c)-60 A \cos (c+d x) \cot (c) \sqrt {\sec ^2(c)}-48 B \cos (c+d x) \cot (c) \sqrt {\sec ^2(c)}-20 (2 A+B) \cos (c+d x) \sqrt {\cos ^2(d x-\arctan (\cot (c)))} \sqrt {\csc ^2(c)} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sqrt {\sec ^2(c)} \sec (d x-\arctan (\cot (c))) \sin (c)+10 A \cos (c+d x) \sqrt {\sec ^2(c)} \sin (c+d x)+20 B \cos (c+d x) \sqrt {\sec ^2(c)} \sin (c+d x)+6 B \cos ^2(c+d x) \sqrt {\sec ^2(c)} \sin (c+d x)\right ) \sqrt {\sin ^2(d x+\arctan (\tan (c)))}\right )}{60 d \sqrt {\cos (c+d x)} \sqrt {\sec ^2(c)} \sqrt {\sin ^2(d x+\arctan (\tan (c)))}} \]
(a^2*(1 + Cos[c + d*x])^2*Sec[(c + d*x)/2]^4*(-6*(5*A + 4*B)*Hypergeometri cPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sec[c]*Sin[d*x + Ar cTan[Tan[c]]] + (9*(5*A + 4*B)*Cos[c - d*x - ArcTan[Tan[c]]]*Csc[c]*Sec[c] + 15*A*Cos[c + d*x + ArcTan[Tan[c]]]*Csc[c]*Sec[c] + 12*B*Cos[c + d*x + A rcTan[Tan[c]]]*Csc[c]*Sec[c] - 60*A*Cos[c + d*x]*Cot[c]*Sqrt[Sec[c]^2] - 4 8*B*Cos[c + d*x]*Cot[c]*Sqrt[Sec[c]^2] - 20*(2*A + B)*Cos[c + d*x]*Sqrt[Co s[d*x - ArcTan[Cot[c]]]^2]*Sqrt[Csc[c]^2]*HypergeometricPFQ[{1/4, 1/2}, {5 /4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sqrt[Sec[c]^2]*Sec[d*x - ArcTan[Cot[c]]] *Sin[c] + 10*A*Cos[c + d*x]*Sqrt[Sec[c]^2]*Sin[c + d*x] + 20*B*Cos[c + d*x ]*Sqrt[Sec[c]^2]*Sin[c + d*x] + 6*B*Cos[c + d*x]^2*Sqrt[Sec[c]^2]*Sin[c + d*x])*Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2]))/(60*d*Sqrt[Cos[c + d*x]]*Sqrt[Se c[c]^2]*Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2])
Time = 0.84 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.05, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3042, 3455, 27, 3042, 3447, 3042, 3502, 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.
| \(\displaystyle \int \frac {(a \cos (c+d x)+a)^2 (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {2}{5} \int \frac {(\cos (c+d x) a+a) (a (5 A+B)+a (5 A+7 B) \cos (c+d x))}{2 \sqrt {\cos (c+d x)}}dx+\frac {2 B \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^2 \cos (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \frac {(\cos (c+d x) a+a) (a (5 A+B)+a (5 A+7 B) \cos (c+d x))}{\sqrt {\cos (c+d x)}}dx+\frac {2 B \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^2 \cos (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (a (5 A+B)+a (5 A+7 B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 B \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^2 \cos (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {1}{5} \int \frac {(5 A+7 B) \cos ^2(c+d x) a^2+(5 A+B) a^2+\left ((5 A+B) a^2+(5 A+7 B) a^2\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx+\frac {2 B \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^2 \cos (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {(5 A+7 B) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^2+(5 A+B) a^2+\left ((5 A+B) a^2+(5 A+7 B) a^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 B \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^2 \cos (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int \frac {5 (2 A+B) a^2+3 (5 A+4 B) \cos (c+d x) a^2}{\sqrt {\cos (c+d x)}}dx+\frac {2 a^2 (5 A+7 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 B \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^2 \cos (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int \frac {5 (2 A+B) a^2+3 (5 A+4 B) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a^2 (5 A+7 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 B \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^2 \cos (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \left (5 a^2 (2 A+B) \int \frac {1}{\sqrt {\cos (c+d x)}}dx+3 a^2 (5 A+4 B) \int \sqrt {\cos (c+d x)}dx\right )+\frac {2 a^2 (5 A+7 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 B \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^2 \cos (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \left (5 a^2 (2 A+B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 a^2 (5 A+4 B) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )+\frac {2 a^2 (5 A+7 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 B \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^2 \cos (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \left (5 a^2 (2 A+B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {6 a^2 (5 A+4 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {2 a^2 (5 A+7 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 B \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^2 \cos (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {1}{5} \left (\frac {2 a^2 (5 A+7 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}+\frac {2}{3} \left (\frac {10 a^2 (2 A+B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {6 a^2 (5 A+4 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {2 B \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^2 \cos (c+d x)+a^2\right )}{5 d}\) |
(2*B*Sqrt[Cos[c + d*x]]*(a^2 + a^2*Cos[c + d*x])*Sin[c + d*x])/(5*d) + ((2 *((6*a^2*(5*A + 4*B)*EllipticE[(c + d*x)/2, 2])/d + (10*a^2*(2*A + B)*Elli pticF[(c + d*x)/2, 2])/d))/3 + (2*a^2*(5*A + 7*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d))/5
3.2.32.3.1 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_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(-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 - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1 ) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && !LtQ[n, -1 ] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[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. \(356\) vs. \(2(166)=332\).
Time = 8.46 (sec) , antiderivative size = 357, normalized size of antiderivative = 2.83
| method | result | size |
| default | \(-\frac {4 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, a^{2} \left (-12 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (10 A +32 B \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-5 A -13 B \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+10 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-15 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+5 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-12 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{15 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\) | \(357\) |
| parts | \(-\frac {2 \left (A \,a^{2}+2 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 )}\, \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^{2}+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 \,a^{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}\) | \(562\) |
-4/15*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*(-12*B*c os(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+(10*A+32*B)*sin(1/2*d*x+1/2*c)^4*co s(1/2*d*x+1/2*c)+(-5*A-13*B)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+10*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)*EllipticF(co s(1/2*d*x+1/2*c),2^(1/2))-15*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))+5*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))-12*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)))/(-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 = 0.12 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.42 \[ \int \frac {(a+a \cos (c+d x))^2 (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (2 \, A + B\right )} a^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (2 \, A + B\right )} a^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 i \, \sqrt {2} {\left (5 \, A + 4 \, B\right )} a^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 i \, \sqrt {2} {\left (5 \, A + 4 \, B\right )} a^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (3 \, B a^{2} \cos \left (d x + c\right ) + 5 \, {\left (A + 2 \, B\right )} a^{2}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{15 \, d} \]
-2/15*(5*I*sqrt(2)*(2*A + B)*a^2*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 5*I*sqrt(2)*(2*A + B)*a^2*weierstrassPInverse(-4, 0, co s(d*x + c) - I*sin(d*x + c)) - 3*I*sqrt(2)*(5*A + 4*B)*a^2*weierstrassZeta (-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*I*s qrt(2)*(5*A + 4*B)*a^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, c os(d*x + c) - I*sin(d*x + c))) - (3*B*a^2*cos(d*x + c) + 5*(A + 2*B)*a^2)* sqrt(cos(d*x + c))*sin(d*x + c))/d
Timed out. \[ \int \frac {(a+a \cos (c+d x))^2 (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\text {Timed out} \]
\[ \int \frac {(a+a \cos (c+d x))^2 (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{2}}{\sqrt {\cos \left (d x + c\right )}} \,d x } \]
\[ \int \frac {(a+a \cos (c+d x))^2 (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{2}}{\sqrt {\cos \left (d x + c\right )}} \,d x } \]
Time = 1.11 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.21 \[ \int \frac {(a+a \cos (c+d x))^2 (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\frac {2\,B\,a^2\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}+\frac {2\,A\,a^2\,\left (\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )+6\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+4\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{3\,d}+\frac {2\,B\,a^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}-\frac {2\,B\,a^2\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]
(2*B*a^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*(cos(c + d*x)^(1/2)*sin(c + d*x) + 6*ellipticE(c/ 2 + (d*x)/2, 2) + 4*ellipticF(c/2 + (d*x)/2, 2)))/(3*d) + (2*B*a^2*ellipti cE(c/2 + (d*x)/2, 2))/d - (2*B*a^2*cos(c + d*x)^(7/2)*sin(c + d*x)*hyperge om([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d*(sin(c + d*x)^2)^(1/2))