Integrand size = 31, antiderivative size = 175 \[ \int \cos (c+d x) (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx=\frac {64 a^3 (15 A+13 B) \sin (c+d x)}{315 d \sqrt {a+a \cos (c+d x)}}+\frac {16 a^2 (15 A+13 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{315 d}+\frac {2 a (15 A+13 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{105 d}+\frac {2 (9 A-2 B) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{63 d}+\frac {2 B (a+a \cos (c+d x))^{7/2} \sin (c+d x)}{9 a d} \] Output:
64/315*a^3*(15*A+13*B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+16/315*a^2*(15* A+13*B)*(a+a*cos(d*x+c))^(1/2)*sin(d*x+c)/d+2/105*a*(15*A+13*B)*(a+a*cos(d *x+c))^(3/2)*sin(d*x+c)/d+2/63*(9*A-2*B)*(a+a*cos(d*x+c))^(5/2)*sin(d*x+c) /d+2/9*B*(a+a*cos(d*x+c))^(7/2)*sin(d*x+c)/a/d
Time = 0.60 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.60 \[ \int \cos (c+d x) (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} (6240 A+5653 B+(3030 A+3116 B) \cos (c+d x)+8 (90 A+127 B) \cos (2 (c+d x))+90 A \cos (3 (c+d x))+260 B \cos (3 (c+d x))+35 B \cos (4 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )}{1260 d} \] Input:
Integrate[Cos[c + d*x]*(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x]),x]
Output:
(a^2*Sqrt[a*(1 + Cos[c + d*x])]*(6240*A + 5653*B + (3030*A + 3116*B)*Cos[c + d*x] + 8*(90*A + 127*B)*Cos[2*(c + d*x)] + 90*A*Cos[3*(c + d*x)] + 260* B*Cos[3*(c + d*x)] + 35*B*Cos[4*(c + d*x)])*Tan[(c + d*x)/2])/(1260*d)
Time = 0.87 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.03, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.419, Rules used = {3042, 3447, 3042, 3502, 27, 3042, 3230, 3042, 3126, 3042, 3126, 3042, 3125}
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.
| \(\displaystyle \int \cos (c+d x) (a \cos (c+d x)+a)^{5/2} (A+B \cos (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \int (a \cos (c+d x)+a)^{5/2} \left (A \cos (c+d x)+B \cos ^2(c+d x)\right )dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (A \sin \left (c+d x+\frac {\pi }{2}\right )+B \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {2 \int \frac {1}{2} (\cos (c+d x) a+a)^{5/2} (7 a B+a (9 A-2 B) \cos (c+d x))dx}{9 a}+\frac {2 B \sin (c+d x) (a \cos (c+d x)+a)^{7/2}}{9 a d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int (\cos (c+d x) a+a)^{5/2} (7 a B+a (9 A-2 B) \cos (c+d x))dx}{9 a}+\frac {2 B \sin (c+d x) (a \cos (c+d x)+a)^{7/2}}{9 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (7 a B+a (9 A-2 B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{9 a}+\frac {2 B \sin (c+d x) (a \cos (c+d x)+a)^{7/2}}{9 a d}\) |
| \(\Big \downarrow \) 3230 |
| \(\displaystyle \frac {\frac {3}{7} a (15 A+13 B) \int (\cos (c+d x) a+a)^{5/2}dx+\frac {2 a (9 A-2 B) \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {2 B \sin (c+d x) (a \cos (c+d x)+a)^{7/2}}{9 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3}{7} a (15 A+13 B) \int \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2}dx+\frac {2 a (9 A-2 B) \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {2 B \sin (c+d x) (a \cos (c+d x)+a)^{7/2}}{9 a d}\) |
| \(\Big \downarrow \) 3126 |
| \(\displaystyle \frac {\frac {3}{7} a (15 A+13 B) \left (\frac {8}{5} a \int (\cos (c+d x) a+a)^{3/2}dx+\frac {2 a \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 a (9 A-2 B) \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {2 B \sin (c+d x) (a \cos (c+d x)+a)^{7/2}}{9 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3}{7} a (15 A+13 B) \left (\frac {8}{5} a \int \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}dx+\frac {2 a \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 a (9 A-2 B) \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {2 B \sin (c+d x) (a \cos (c+d x)+a)^{7/2}}{9 a d}\) |
| \(\Big \downarrow \) 3126 |
| \(\displaystyle \frac {\frac {3}{7} a (15 A+13 B) \left (\frac {8}{5} a \left (\frac {4}{3} a \int \sqrt {\cos (c+d x) a+a}dx+\frac {2 a \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {2 a \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 a (9 A-2 B) \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {2 B \sin (c+d x) (a \cos (c+d x)+a)^{7/2}}{9 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3}{7} a (15 A+13 B) \left (\frac {8}{5} a \left (\frac {4}{3} a \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {2 a \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 a (9 A-2 B) \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {2 B \sin (c+d x) (a \cos (c+d x)+a)^{7/2}}{9 a d}\) |
| \(\Big \downarrow \) 3125 |
| \(\displaystyle \frac {\frac {3}{7} a (15 A+13 B) \left (\frac {8}{5} a \left (\frac {8 a^2 \sin (c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {2 a \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 a (9 A-2 B) \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {2 B \sin (c+d x) (a \cos (c+d x)+a)^{7/2}}{9 a d}\) |
Input:
Int[Cos[c + d*x]*(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x]),x]
Output:
(2*B*(a + a*Cos[c + d*x])^(7/2)*Sin[c + d*x])/(9*a*d) + ((2*a*(9*A - 2*B)* (a + a*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(7*d) + (3*a*(15*A + 13*B)*((2*a* (a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*d) + (8*a*((8*a^2*Sin[c + d*x] )/(3*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d *x])/(3*d)))/5))/7)/(9*a)
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*b*(Cos [c + d*x]/(d*Sqrt[a + b*Sin[c + d*x]])), x] /; FreeQ[{a, b, c, d}, x] && Eq Q[a^2 - b^2, 0]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos [c + d*x]*((a + b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[a*((2*n - 1)/n) Int[(a + b*Sin[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[ a^2 - b^2, 0] && IGtQ[n - 1/2, 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[(a*d*m + b*c*(m + 1))/(b*(m + 1)) Int[(a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)]
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_.)*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]
Time = 5.80 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.70
| method | result | size |
| default | \(\frac {8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (140 B \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (-90 A -540 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (315 A +819 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-420 A -630 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+315 A +315 B \right ) \sqrt {2}}{315 \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\) | \(123\) |
| parts | \(\frac {8 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (6 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+3 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+8\right ) \sqrt {2}}{21 \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}+\frac {8 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (140 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-20 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+39 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+52 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+104\right ) \sqrt {2}}{315 \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\) | \(187\) |
Input:
int(cos(d*x+c)*(a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)),x,method=_RETURNVER BOSE)
Output:
8/315*cos(1/2*d*x+1/2*c)*a^3*sin(1/2*d*x+1/2*c)*(140*B*sin(1/2*d*x+1/2*c)^ 8+(-90*A-540*B)*sin(1/2*d*x+1/2*c)^6+(315*A+819*B)*sin(1/2*d*x+1/2*c)^4+(- 420*A-630*B)*sin(1/2*d*x+1/2*c)^2+315*A+315*B)*2^(1/2)/(a*cos(1/2*d*x+1/2* c)^2)^(1/2)/d
Time = 0.08 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.66 \[ \int \cos (c+d x) (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx=\frac {2 \, {\left (35 \, B a^{2} \cos \left (d x + c\right )^{4} + 5 \, {\left (9 \, A + 26 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (60 \, A + 73 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (345 \, A + 292 \, B\right )} a^{2} \cos \left (d x + c\right ) + 2 \, {\left (345 \, A + 292 \, B\right )} a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \] Input:
integrate(cos(d*x+c)*(a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)),x, algorithm= "fricas")
Output:
2/315*(35*B*a^2*cos(d*x + c)^4 + 5*(9*A + 26*B)*a^2*cos(d*x + c)^3 + 3*(60 *A + 73*B)*a^2*cos(d*x + c)^2 + (345*A + 292*B)*a^2*cos(d*x + c) + 2*(345* A + 292*B)*a^2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/(d*cos(d*x + c) + d)
Timed out. \[ \int \cos (c+d x) (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)*(a+a*cos(d*x+c))**(5/2)*(A+B*cos(d*x+c)),x)
Output:
Timed out
Time = 0.30 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.98 \[ \int \cos (c+d x) (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx=\frac {30 \, {\left (3 \, \sqrt {2} a^{2} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 21 \, \sqrt {2} a^{2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 77 \, \sqrt {2} a^{2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 315 \, \sqrt {2} a^{2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} A \sqrt {a} + {\left (35 \, \sqrt {2} a^{2} \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 225 \, \sqrt {2} a^{2} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 756 \, \sqrt {2} a^{2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 2100 \, \sqrt {2} a^{2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 8190 \, \sqrt {2} a^{2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} B \sqrt {a}}{2520 \, d} \] Input:
integrate(cos(d*x+c)*(a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)),x, algorithm= "maxima")
Output:
1/2520*(30*(3*sqrt(2)*a^2*sin(7/2*d*x + 7/2*c) + 21*sqrt(2)*a^2*sin(5/2*d* x + 5/2*c) + 77*sqrt(2)*a^2*sin(3/2*d*x + 3/2*c) + 315*sqrt(2)*a^2*sin(1/2 *d*x + 1/2*c))*A*sqrt(a) + (35*sqrt(2)*a^2*sin(9/2*d*x + 9/2*c) + 225*sqrt (2)*a^2*sin(7/2*d*x + 7/2*c) + 756*sqrt(2)*a^2*sin(5/2*d*x + 5/2*c) + 2100 *sqrt(2)*a^2*sin(3/2*d*x + 3/2*c) + 8190*sqrt(2)*a^2*sin(1/2*d*x + 1/2*c)) *B*sqrt(a))/d
Time = 1.21 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.22 \[ \int \cos (c+d x) (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx=\frac {\sqrt {2} {\left (35 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 45 \, {\left (2 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 5 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 126 \, {\left (5 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 6 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 210 \, {\left (11 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 10 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 630 \, {\left (15 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 13 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{2520 \, d} \] Input:
integrate(cos(d*x+c)*(a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)),x, algorithm= "giac")
Output:
1/2520*sqrt(2)*(35*B*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(9/2*d*x + 9/2*c) + 45*(2*A*a^2*sgn(cos(1/2*d*x + 1/2*c)) + 5*B*a^2*sgn(cos(1/2*d*x + 1/2*c))) *sin(7/2*d*x + 7/2*c) + 126*(5*A*a^2*sgn(cos(1/2*d*x + 1/2*c)) + 6*B*a^2*s gn(cos(1/2*d*x + 1/2*c)))*sin(5/2*d*x + 5/2*c) + 210*(11*A*a^2*sgn(cos(1/2 *d*x + 1/2*c)) + 10*B*a^2*sgn(cos(1/2*d*x + 1/2*c)))*sin(3/2*d*x + 3/2*c) + 630*(15*A*a^2*sgn(cos(1/2*d*x + 1/2*c)) + 13*B*a^2*sgn(cos(1/2*d*x + 1/2 *c)))*sin(1/2*d*x + 1/2*c))*sqrt(a)/d
Timed out. \[ \int \cos (c+d x) (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx=\int \cos \left (c+d\,x\right )\,\left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2} \,d x \] Input:
int(cos(c + d*x)*(A + B*cos(c + d*x))*(a + a*cos(c + d*x))^(5/2),x)
Output:
int(cos(c + d*x)*(A + B*cos(c + d*x))*(a + a*cos(c + d*x))^(5/2), x)
\[ \int \cos (c+d x) (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx=\sqrt {a}\, a^{2} \left (\left (\int \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )d x \right ) a +\left (\int \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{4}d x \right ) b +\left (\int \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{3}d x \right ) a +2 \left (\int \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{3}d x \right ) b +2 \left (\int \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2}d x \right ) a +\left (\int \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2}d x \right ) b \right ) \] Input:
int(cos(d*x+c)*(a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)),x)
Output:
sqrt(a)*a**2*(int(sqrt(cos(c + d*x) + 1)*cos(c + d*x),x)*a + int(sqrt(cos( c + d*x) + 1)*cos(c + d*x)**4,x)*b + int(sqrt(cos(c + d*x) + 1)*cos(c + d* x)**3,x)*a + 2*int(sqrt(cos(c + d*x) + 1)*cos(c + d*x)**3,x)*b + 2*int(sqr t(cos(c + d*x) + 1)*cos(c + d*x)**2,x)*a + int(sqrt(cos(c + d*x) + 1)*cos( c + d*x)**2,x)*b)