Integrand size = 31, antiderivative size = 138 \[ \int \cos (c+d x) (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx=\frac {8 a^2 (21 A+19 B) \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a (21 A+19 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{105 d}+\frac {2 (7 A-2 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac {2 B (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{7 a d} \]
2/35*(7*A-2*B)*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/d+2/7*B*(a+a*cos(d*x+c))^ (5/2)*sin(d*x+c)/a/d+8/105*a^2*(21*A+19*B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^( 1/2)+2/105*a*(21*A+19*B)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d
Time = 0.21 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.59 \[ \int \cos (c+d x) (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx=\frac {a \sqrt {a (1+\cos (c+d x))} (546 A+494 B+(252 A+253 B) \cos (c+d x)+6 (7 A+13 B) \cos (2 (c+d x))+15 B \cos (3 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )}{210 d} \]
(a*Sqrt[a*(1 + Cos[c + d*x])]*(546*A + 494*B + (252*A + 253*B)*Cos[c + d*x ] + 6*(7*A + 13*B)*Cos[2*(c + d*x)] + 15*B*Cos[3*(c + d*x)])*Tan[(c + d*x) /2])/(210*d)
Time = 0.76 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.06, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {3042, 3447, 3042, 3502, 27, 3042, 3230, 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.
| \(\displaystyle \int \cos (c+d x) (a \cos (c+d x)+a)^{3/2} (A+B \cos (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \int (a \cos (c+d x)+a)^{3/2} \left (A \cos (c+d x)+B \cos ^2(c+d x)\right )dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2} \left (A \sin \left (c+d x+\frac {\pi }{2}\right )+B \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {2 \int \frac {1}{2} (\cos (c+d x) a+a)^{3/2} (5 a B+a (7 A-2 B) \cos (c+d x))dx}{7 a}+\frac {2 B \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 a d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int (\cos (c+d x) a+a)^{3/2} (5 a B+a (7 A-2 B) \cos (c+d x))dx}{7 a}+\frac {2 B \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (5 a B+a (7 A-2 B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{7 a}+\frac {2 B \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 a d}\) |
| \(\Big \downarrow \) 3230 |
| \(\displaystyle \frac {\frac {1}{5} a (21 A+19 B) \int (\cos (c+d x) a+a)^{3/2}dx+\frac {2 a (7 A-2 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 B \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} a (21 A+19 B) \int \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}dx+\frac {2 a (7 A-2 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 B \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 a d}\) |
| \(\Big \downarrow \) 3126 |
| \(\displaystyle \frac {\frac {1}{5} a (21 A+19 B) \left (\frac {4}{3} a \int \sqrt {\cos (c+d x) a+a}dx+\frac {2 a \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {2 a (7 A-2 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 B \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} a (21 A+19 B) \left (\frac {4}{3} a \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {2 a (7 A-2 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 B \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 a d}\) |
| \(\Big \downarrow \) 3125 |
| \(\displaystyle \frac {\frac {1}{5} a (21 A+19 B) \left (\frac {8 a^2 \sin (c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {2 a (7 A-2 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 B \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 a d}\) |
(2*B*(a + a*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(7*a*d) + ((2*a*(7*A - 2*B)* (a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*d) + (a*(21*A + 19*B)*((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*a)
3.1.84.3.1 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*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 = 2.60 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.75
| method | result | size |
| default | \(\frac {4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (-60 B \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (42 A +168 B \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-105 A -175 B \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+105 A +105 B \right ) \sqrt {2}}{105 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) | \(104\) |
| parts | \(\frac {4 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (2 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+2\right ) \sqrt {2}}{5 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}+\frac {4 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (60 \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+19 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+38\right ) \sqrt {2}}{105 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) | \(159\) |
4/105*cos(1/2*d*x+1/2*c)*a^2*sin(1/2*d*x+1/2*c)*(-60*B*sin(1/2*d*x+1/2*c)^ 6+(42*A+168*B)*sin(1/2*d*x+1/2*c)^4+(-105*A-175*B)*sin(1/2*d*x+1/2*c)^2+10 5*A+105*B)*2^(1/2)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d
Time = 0.27 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.64 \[ \int \cos (c+d x) (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx=\frac {2 \, {\left (15 \, B a \cos \left (d x + c\right )^{3} + 3 \, {\left (7 \, A + 13 \, B\right )} a \cos \left (d x + c\right )^{2} + {\left (63 \, A + 52 \, B\right )} a \cos \left (d x + c\right ) + 2 \, {\left (63 \, A + 52 \, B\right )} a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]
2/105*(15*B*a*cos(d*x + c)^3 + 3*(7*A + 13*B)*a*cos(d*x + c)^2 + (63*A + 5 2*B)*a*cos(d*x + c) + 2*(63*A + 52*B)*a)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/(d*cos(d*x + c) + d)
Timed out. \[ \int \cos (c+d x) (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx=\text {Timed out} \]
Time = 0.37 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.89 \[ \int \cos (c+d x) (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx=\frac {42 \, {\left (\sqrt {2} a \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, \sqrt {2} a \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 20 \, \sqrt {2} a \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} A \sqrt {a} + {\left (15 \, \sqrt {2} a \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 63 \, \sqrt {2} a \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 175 \, \sqrt {2} a \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 735 \, \sqrt {2} a \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} B \sqrt {a}}{420 \, d} \]
1/420*(42*(sqrt(2)*a*sin(5/2*d*x + 5/2*c) + 5*sqrt(2)*a*sin(3/2*d*x + 3/2* c) + 20*sqrt(2)*a*sin(1/2*d*x + 1/2*c))*A*sqrt(a) + (15*sqrt(2)*a*sin(7/2* d*x + 7/2*c) + 63*sqrt(2)*a*sin(5/2*d*x + 5/2*c) + 175*sqrt(2)*a*sin(3/2*d *x + 3/2*c) + 735*sqrt(2)*a*sin(1/2*d*x + 1/2*c))*B*sqrt(a))/d
Time = 0.51 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.12 \[ \int \cos (c+d x) (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx=\frac {\sqrt {2} {\left (15 \, B a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 21 \, {\left (2 \, A a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 3 \, B a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 35 \, {\left (6 \, A a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 5 \, B a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 105 \, {\left (8 \, A a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 7 \, B a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{420 \, d} \]
1/420*sqrt(2)*(15*B*a*sgn(cos(1/2*d*x + 1/2*c))*sin(7/2*d*x + 7/2*c) + 21* (2*A*a*sgn(cos(1/2*d*x + 1/2*c)) + 3*B*a*sgn(cos(1/2*d*x + 1/2*c)))*sin(5/ 2*d*x + 5/2*c) + 35*(6*A*a*sgn(cos(1/2*d*x + 1/2*c)) + 5*B*a*sgn(cos(1/2*d *x + 1/2*c)))*sin(3/2*d*x + 3/2*c) + 105*(8*A*a*sgn(cos(1/2*d*x + 1/2*c)) + 7*B*a*sgn(cos(1/2*d*x + 1/2*c)))*sin(1/2*d*x + 1/2*c))*sqrt(a)/d
Timed out. \[ \int \cos (c+d x) (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx=\int \cos \left (c+d\,x\right )\,\left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2} \,d x \]