Integrand size = 33, antiderivative size = 211 \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=\frac {4 a^3 (5 A+9 B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^3 (5 A+3 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {4 a^3 (5 A-6 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 a B (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 (5 A+9 B) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}} \]
2/5*a*B*(a+a*sec(d*x+c))^2*sin(d*x+c)/d/sec(d*x+c)^(3/2)+2/15*(5*A+9*B)*(a ^3+a^3*sec(d*x+c))*sin(d*x+c)/d/sec(d*x+c)^(1/2)+4/15*a^3*(5*A-6*B)*sin(d* x+c)*sec(d*x+c)^(1/2)/d+4/5*a^3*(5*A+9*B)*(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))*cos(d*x+c)^(1/2)*sec (d*x+c)^(1/2)/d+4/3*a^3*(5*A+3*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))*cos(d*x+c)^(1/2)*sec(d*x+c)^ (1/2)/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 2.14 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.98 \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=\frac {a^3 e^{-i d x} \sqrt {\sec (c+d x)} (\cos (d x)+i \sin (d x)) \left (120 i A \cos (c+d x)+216 i B \cos (c+d x)+40 (5 A+3 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-8 i (5 A+9 B) e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )+60 A \sin (c+d x)+3 B \sin (c+d x)+10 A \sin (2 (c+d x))+30 B \sin (2 (c+d x))+3 B \sin (3 (c+d x))\right )}{30 d} \]
(a^3*Sqrt[Sec[c + d*x]]*(Cos[d*x] + I*Sin[d*x])*((120*I)*A*Cos[c + d*x] + (216*I)*B*Cos[c + d*x] + 40*(5*A + 3*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] - (8*I)*(5*A + 9*B)*E^(I*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x) )]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))] + 60*A*Sin[c + d *x] + 3*B*Sin[c + d*x] + 10*A*Sin[2*(c + d*x)] + 30*B*Sin[2*(c + d*x)] + 3 *B*Sin[3*(c + d*x)]))/(30*d*E^(I*d*x))
Time = 1.35 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.02, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.515, Rules used = {3042, 3439, 3042, 4505, 27, 3042, 4505, 3042, 4485, 27, 3042, 4274, 3042, 4258, 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.
| \(\displaystyle \int \sec ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3 (A+B \cos (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 3439 |
| \(\displaystyle \int \frac {(a \sec (c+d x)+a)^3 (A \sec (c+d x)+B)}{\sec ^{\frac {5}{2}}(c+d x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3 \left (A \csc \left (c+d x+\frac {\pi }{2}\right )+B\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4505 |
| \(\displaystyle \frac {2}{5} \int \frac {(\sec (c+d x) a+a)^2 (a (5 A+9 B)+a (5 A-B) \sec (c+d x))}{2 \sec ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a B \sin (c+d x) (a \sec (c+d x)+a)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \frac {(\sec (c+d x) a+a)^2 (a (5 A+9 B)+a (5 A-B) \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a B \sin (c+d x) (a \sec (c+d x)+a)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (a (5 A+9 B)+a (5 A-B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a B \sin (c+d x) (a \sec (c+d x)+a)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4505 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int \frac {(\sec (c+d x) a+a) \left ((20 A+21 B) a^2+(5 A-6 B) \sec (c+d x) a^2\right )}{\sqrt {\sec (c+d x)}}dx+\frac {2 (5 A+9 B) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a B \sin (c+d x) (a \sec (c+d x)+a)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left ((20 A+21 B) a^2+(5 A-6 B) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (5 A+9 B) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a B \sin (c+d x) (a \sec (c+d x)+a)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4485 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \left (2 \int \frac {3 (5 A+9 B) a^3+5 (5 A+3 B) \sec (c+d x) a^3}{2 \sqrt {\sec (c+d x)}}dx+\frac {2 a^3 (5 A-6 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {2 (5 A+9 B) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a B \sin (c+d x) (a \sec (c+d x)+a)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \left (\int \frac {3 (5 A+9 B) a^3+5 (5 A+3 B) \sec (c+d x) a^3}{\sqrt {\sec (c+d x)}}dx+\frac {2 a^3 (5 A-6 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {2 (5 A+9 B) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a B \sin (c+d x) (a \sec (c+d x)+a)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \left (\int \frac {3 (5 A+9 B) a^3+5 (5 A+3 B) \csc \left (c+d x+\frac {\pi }{2}\right ) a^3}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a^3 (5 A-6 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {2 (5 A+9 B) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a B \sin (c+d x) (a \sec (c+d x)+a)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \left (3 a^3 (5 A+9 B) \int \frac {1}{\sqrt {\sec (c+d x)}}dx+5 a^3 (5 A+3 B) \int \sqrt {\sec (c+d x)}dx+\frac {2 a^3 (5 A-6 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {2 (5 A+9 B) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a B \sin (c+d x) (a \sec (c+d x)+a)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \left (3 a^3 (5 A+9 B) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+5 a^3 (5 A+3 B) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 a^3 (5 A-6 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {2 (5 A+9 B) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a B \sin (c+d x) (a \sec (c+d x)+a)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \left (5 a^3 (5 A+3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+3 a^3 (5 A+9 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx+\frac {2 a^3 (5 A-6 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {2 (5 A+9 B) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a B \sin (c+d x) (a \sec (c+d x)+a)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \left (5 a^3 (5 A+3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 a^3 (5 A+9 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 a^3 (5 A-6 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {2 (5 A+9 B) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a B \sin (c+d x) (a \sec (c+d x)+a)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \left (5 a^3 (5 A+3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a^3 (5 A-6 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {6 a^3 (5 A+9 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {2 (5 A+9 B) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a B \sin (c+d x) (a \sec (c+d x)+a)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {1}{5} \left (\frac {2 (5 A+9 B) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{3 d \sqrt {\sec (c+d x)}}+\frac {2}{3} \left (\frac {2 a^3 (5 A-6 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {10 a^3 (5 A+3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {6 a^3 (5 A+9 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {2 a B \sin (c+d x) (a \sec (c+d x)+a)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
(2*a*B*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + ((2 *(5*A + 9*B)*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x] ]) + (2*((6*a^3*(5*A + 9*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*S qrt[Sec[c + d*x]])/d + (10*a^3*(5*A + 3*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (2*a^3*(5*A - 6*B)*Sqrt[Sec[c + d*x] ]*Sin[c + d*x])/d))/3)/5
3.5.73.3.1 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[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)* (x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Sim p[g^(m + n) Int[(g*Csc[e + f*x])^(p - m - n)*(b + a*Csc[e + f*x])^m*(d + c*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] && !IntegerQ[p] && IntegerQ[m] && IntegerQ[n]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] )^n*Sin[c + d*x]^n Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B*Cot[ e + f*x]*((d*Csc[e + f*x])^n/(f*(n + 1))), x] + Simp[1/(n + 1) Int[(d*Csc [e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x ], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && !LeQ[ n, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[a*A*Cot [e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*n)), x] - Sim p[b/(a*d*n) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Sim p[a*A*(m - n - 1) - b*B*n - (a*B*n + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0 ] && GtQ[m, 1/2] && LtQ[n, -1]
Time = 11.50 (sec) , antiderivative size = 337, normalized size of antiderivative = 1.60
| method | result | size |
| default | \(-\frac {4 a^{3} \left (-12 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+10 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+42 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-20 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+25 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-15 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-18 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-27 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{15 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\) | \(337\) |
| parts | \(\text {Expression too large to display}\) | \(871\) |
-4/15*a^3*(-12*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+10*A*cos(1/2*d*x+ 1/2*c)*sin(1/2*d*x+1/2*c)^4+42*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-2 0*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+25*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(cos(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)*Ellipt icE(cos(1/2*d*x+1/2*c),2^(1/2))-18*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c) ^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)*Elli pticF(cos(1/2*d*x+1/2*c),2^(1/2))-27*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)))/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 = 0.10 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.93 \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (5 \, A + 3 \, B\right )} a^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (5 \, A + 3 \, B\right )} a^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 i \, \sqrt {2} {\left (5 \, A + 9 \, B\right )} a^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 i \, \sqrt {2} {\left (5 \, A + 9 \, B\right )} a^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (3 \, B a^{3} \cos \left (d x + c\right )^{2} + 5 \, {\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right ) + 15 \, A a^{3}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{15 \, d} \]
-2/15*(5*I*sqrt(2)*(5*A + 3*B)*a^3*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 5*I*sqrt(2)*(5*A + 3*B)*a^3*weierstrassPInverse(-4, 0 , cos(d*x + c) - I*sin(d*x + c)) - 3*I*sqrt(2)*(5*A + 9*B)*a^3*weierstrass Zeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3 *I*sqrt(2)*(5*A + 9*B)*a^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (3*B*a^3*cos(d*x + c)^2 + 5*(A + 3*B) *a^3*cos(d*x + c) + 15*A*a^3)*sin(d*x + c)/sqrt(cos(d*x + c)))/d
Timed out. \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=\text {Timed out} \]
\[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{3} \sec \left (d x + c\right )^{\frac {3}{2}} \,d x } \]
\[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{3} \sec \left (d x + c\right )^{\frac {3}{2}} \,d x } \]
Timed out. \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=\int \left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^3 \,d x \]