Integrand size = 27, antiderivative size = 132 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {8 a^2 (35 A+19 C) \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a (35 A+19 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{105 d}-\frac {4 C (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac {2 C (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{7 a d} \]
-4/35*C*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/d+2/7*C*(a+a*cos(d*x+c))^(5/2)*s in(d*x+c)/a/d+8/105*a^2*(35*A+19*C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/ 105*a*(35*A+19*C)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d
Time = 0.17 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.57 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {a \sqrt {a (1+\cos (c+d x))} (700 A+494 C+(140 A+253 C) \cos (c+d x)+78 C \cos (2 (c+d x))+15 C \cos (3 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )}{210 d} \]
(a*Sqrt[a*(1 + Cos[c + d*x])]*(700*A + 494*C + (140*A + 253*C)*Cos[c + d*x ] + 78*C*Cos[2*(c + d*x)] + 15*C*Cos[3*(c + d*x)])*Tan[(c + d*x)/2])/(210* d)
Time = 0.60 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3503, 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.
| \(\displaystyle \int (a \cos (c+d x)+a)^{3/2} \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2} \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 3503 |
| \(\displaystyle \frac {2 \int \frac {1}{2} (\cos (c+d x) a+a)^{3/2} (a (7 A+5 C)-2 a C \cos (c+d x))dx}{7 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 a d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int (\cos (c+d x) a+a)^{3/2} (a (7 A+5 C)-2 a C \cos (c+d x))dx}{7 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a (7 A+5 C)-2 a C \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{7 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 a d}\) |
| \(\Big \downarrow \) 3230 |
| \(\displaystyle \frac {\frac {1}{5} a (35 A+19 C) \int (\cos (c+d x) a+a)^{3/2}dx-\frac {4 a C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} a (35 A+19 C) \int \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}dx-\frac {4 a C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 a d}\) |
| \(\Big \downarrow \) 3126 |
| \(\displaystyle \frac {\frac {1}{5} a (35 A+19 C) \left (\frac {4}{3} a \int \sqrt {\cos (c+d x) a+a}dx+\frac {2 a \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )-\frac {4 a C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} a (35 A+19 C) \left (\frac {4}{3} a \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )-\frac {4 a C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 a d}\) |
| \(\Big \downarrow \) 3125 |
| \(\displaystyle \frac {\frac {1}{5} a (35 A+19 C) \left (\frac {8 a^2 \sin (c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )-\frac {4 a C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 a d}\) |
(2*C*(a + a*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(7*a*d) + ((-4*a*C*(a + a*Co s[c + d*x])^(3/2)*Sin[c + d*x])/(5*d) + (a*(35*A + 19*C)*((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)
3.1.85.3.1 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_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[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) - a*C*Sin[e + f*x], x], x], x] /; FreeQ[{a , b, e, f, A, C, m}, x] && !LtQ[m, -1]
Time = 4.75 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.82
| method | result | size |
| default | \(\frac {4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (60 C \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 C \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+35 A \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+19 C \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+70 A +38 C \right ) \sqrt {2}}{105 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) | \(108\) |
| parts | \(\frac {4 A \,a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+2\right ) \sqrt {2}}{3 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}+\frac {4 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (60 \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+19 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+38\right ) \sqrt {2}}{105 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) | \(146\) |
4/105*cos(1/2*d*x+1/2*c)*a^2*sin(1/2*d*x+1/2*c)*(60*C*cos(1/2*d*x+1/2*c)^6 -12*C*cos(1/2*d*x+1/2*c)^4+35*A*cos(1/2*d*x+1/2*c)^2+19*C*cos(1/2*d*x+1/2* c)^2+70*A+38*C)*2^(1/2)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d
Time = 0.26 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.61 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (15 \, C a \cos \left (d x + c\right )^{3} + 39 \, C a \cos \left (d x + c\right )^{2} + {\left (35 \, A + 52 \, C\right )} a \cos \left (d x + c\right ) + {\left (175 \, A + 104 \, C\right )} a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]
2/105*(15*C*a*cos(d*x + c)^3 + 39*C*a*cos(d*x + c)^2 + (35*A + 52*C)*a*cos (d*x + c) + (175*A + 104*C)*a)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/(d*co s(d*x + c) + d)
Timed out. \[ \int (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\text {Timed out} \]
Time = 0.43 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.82 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {140 \, {\left (\sqrt {2} a \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 9 \, \sqrt {2} a \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} A \sqrt {a} + {\left (15 \, \sqrt {2} a \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 63 \, \sqrt {2} a \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 175 \, \sqrt {2} a \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 735 \, \sqrt {2} a \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} C \sqrt {a}}{420 \, d} \]
1/420*(140*(sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 9*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))*C*sqrt(a))/d
Time = 0.41 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.05 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {\sqrt {2} {\left (15 \, C a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 63 \, C a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 35 \, {\left (4 \, A a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 5 \, C a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 105 \, {\left (12 \, A a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 7 \, C a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{420 \, d} \]
1/420*sqrt(2)*(15*C*a*sgn(cos(1/2*d*x + 1/2*c))*sin(7/2*d*x + 7/2*c) + 63* C*a*sgn(cos(1/2*d*x + 1/2*c))*sin(5/2*d*x + 5/2*c) + 35*(4*A*a*sgn(cos(1/2 *d*x + 1/2*c)) + 5*C*a*sgn(cos(1/2*d*x + 1/2*c)))*sin(3/2*d*x + 3/2*c) + 1 05*(12*A*a*sgn(cos(1/2*d*x + 1/2*c)) + 7*C*a*sgn(cos(1/2*d*x + 1/2*c)))*si n(1/2*d*x + 1/2*c))*sqrt(a)/d
Timed out. \[ \int (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\int \left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2} \,d x \]