Integrand size = 41, antiderivative size = 147 \[ \int \cos (c+d x) \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {2 a (35 A+49 B+27 C) \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}+\frac {2 (35 A-14 B+18 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{105 d}+\frac {2 C \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{7 d}+\frac {2 (7 B+C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{35 a d} \]
2/35*(7*B+C)*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/a/d+2/105*a*(35*A+49*B+27*C )*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/105*(35*A-14*B+18*C)*sin(d*x+c)*(a +a*cos(d*x+c))^(1/2)/d+2/7*C*cos(d*x+c)^2*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2 )/d
Time = 0.35 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.59 \[ \int \cos (c+d x) \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {\sqrt {a (1+\cos (c+d x))} (280 A+266 B+228 C+(140 A+112 B+141 C) \cos (c+d x)+6 (7 B+6 C) \cos (2 (c+d x))+15 C \cos (3 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )}{210 d} \]
(Sqrt[a*(1 + Cos[c + d*x])]*(280*A + 266*B + 228*C + (140*A + 112*B + 141* C)*Cos[c + d*x] + 6*(7*B + 6*C)*Cos[2*(c + d*x)] + 15*C*Cos[3*(c + d*x)])* Tan[(c + d*x)/2])/(210*d)
Time = 0.89 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.12, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.293, Rules used = {3042, 3524, 27, 3042, 3447, 3042, 3502, 27, 3042, 3230, 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.
| \(\displaystyle \int \cos (c+d x) \sqrt {a \cos (c+d x)+a} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a \sin \left (c+d x+\frac {\pi }{2}\right )+a} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 3524 |
| \(\displaystyle \frac {2 \int \frac {1}{2} \cos (c+d x) \sqrt {\cos (c+d x) a+a} (a (7 A+4 C)+a (7 B+C) \cos (c+d x))dx}{7 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{7 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \cos (c+d x) \sqrt {\cos (c+d x) a+a} (a (7 A+4 C)+a (7 B+C) \cos (c+d x))dx}{7 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a (7 A+4 C)+a (7 B+C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{7 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{7 d}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {\int \sqrt {\cos (c+d x) a+a} \left (a (7 B+C) \cos ^2(c+d x)+a (7 A+4 C) \cos (c+d x)\right )dx}{7 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a (7 B+C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+a (7 A+4 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{7 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{7 d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {\frac {2 \int \frac {1}{2} \sqrt {\cos (c+d x) a+a} \left (3 (7 B+C) a^2+(35 A-14 B+18 C) \cos (c+d x) a^2\right )dx}{5 a}+\frac {2 (7 B+C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{7 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \sqrt {\cos (c+d x) a+a} \left (3 (7 B+C) a^2+(35 A-14 B+18 C) \cos (c+d x) a^2\right )dx}{5 a}+\frac {2 (7 B+C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (3 (7 B+C) a^2+(35 A-14 B+18 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx}{5 a}+\frac {2 (7 B+C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{7 d}\) |
| \(\Big \downarrow \) 3230 |
| \(\displaystyle \frac {\frac {\frac {1}{3} a^2 (35 A+49 B+27 C) \int \sqrt {\cos (c+d x) a+a}dx+\frac {2 a^2 (35 A-14 B+18 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}+\frac {2 (7 B+C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {1}{3} a^2 (35 A+49 B+27 C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a^2 (35 A-14 B+18 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}+\frac {2 (7 B+C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{7 d}\) |
| \(\Big \downarrow \) 3125 |
| \(\displaystyle \frac {\frac {\frac {2 a^3 (35 A+49 B+27 C) \sin (c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a^2 (35 A-14 B+18 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}+\frac {2 (7 B+C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{7 d}\) |
(2*C*Cos[c + d*x]^2*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(7*d) + ((2*(7* B + C)*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*d) + ((2*a^3*(35*A + 49 *B + 27*C)*Sin[c + d*x])/(3*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(35*A - 1 4*B + 18*C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3*d))/(5*a))/(7*a)
3.4.75.3.1 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[(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]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_. ) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x] )^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(b*d*(m + n + 2)) Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n*Simp[A*b*d*(m + n + 2) + C*(a*c*m + b*d*(n + 1)) + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, n} , x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !Lt Q[m, -2^(-1)] && NeQ[m + n + 2, 0]
Time = 6.62 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.73
| method | result | size |
| default | \(\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (-120 C \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (84 B +252 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-70 A -140 B -210 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+105 A +105 B +105 C \right ) \sqrt {2}}{105 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) | \(108\) |
| parts | \(\frac {2 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (1+2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right ) \sqrt {2}}{3 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}+\frac {2 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (12 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7\right ) \sqrt {2}}{15 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}+\frac {2 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (40 \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-36 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+22 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+9\right ) \sqrt {2}}{35 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) | \(215\) |
int(cos(d*x+c)*(a+cos(d*x+c)*a)^(1/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x,me thod=_RETURNVERBOSE)
2/105*cos(1/2*d*x+1/2*c)*a*sin(1/2*d*x+1/2*c)*(-120*C*sin(1/2*d*x+1/2*c)^6 +(84*B+252*C)*sin(1/2*d*x+1/2*c)^4+(-70*A-140*B-210*C)*sin(1/2*d*x+1/2*c)^ 2+105*A+105*B+105*C)*2^(1/2)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d
Time = 0.29 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.59 \[ \int \cos (c+d x) \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (15 \, C \cos \left (d x + c\right )^{3} + 3 \, {\left (7 \, B + 6 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (35 \, A + 28 \, B + 24 \, C\right )} \cos \left (d x + c\right ) + 70 \, A + 56 \, B + 48 \, C\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]
integrate(cos(d*x+c)*(a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2 ),x, algorithm="fricas")
2/105*(15*C*cos(d*x + c)^3 + 3*(7*B + 6*C)*cos(d*x + c)^2 + (35*A + 28*B + 24*C)*cos(d*x + c) + 70*A + 56*B + 48*C)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/(d*cos(d*x + c) + d)
Timed out. \[ \int \cos (c+d x) \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\text {Timed out} \]
Time = 0.41 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.03 \[ \int \cos (c+d x) \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {140 \, {\left (\sqrt {2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 3 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} A \sqrt {a} + 14 \, {\left (3 \, \sqrt {2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, \sqrt {2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 30 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} B \sqrt {a} + 3 \, {\left (5 \, \sqrt {2} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 7 \, \sqrt {2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 35 \, \sqrt {2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 105 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} C \sqrt {a}}{420 \, d} \]
integrate(cos(d*x+c)*(a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2 ),x, algorithm="maxima")
1/420*(140*(sqrt(2)*sin(3/2*d*x + 3/2*c) + 3*sqrt(2)*sin(1/2*d*x + 1/2*c)) *A*sqrt(a) + 14*(3*sqrt(2)*sin(5/2*d*x + 5/2*c) + 5*sqrt(2)*sin(3/2*d*x + 3/2*c) + 30*sqrt(2)*sin(1/2*d*x + 1/2*c))*B*sqrt(a) + 3*(5*sqrt(2)*sin(7/2 *d*x + 7/2*c) + 7*sqrt(2)*sin(5/2*d*x + 5/2*c) + 35*sqrt(2)*sin(3/2*d*x + 3/2*c) + 105*sqrt(2)*sin(1/2*d*x + 1/2*c))*C*sqrt(a))/d
Time = 0.46 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.18 \[ \int \cos (c+d x) \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {\sqrt {2} {\left (15 \, C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 21 \, {\left (2 \, B \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 35 \, {\left (4 \, A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 2 \, B \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 3 \, C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 105 \, {\left (4 \, A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 4 \, B \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 3 \, C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{420 \, d} \]
integrate(cos(d*x+c)*(a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2 ),x, algorithm="giac")
1/420*sqrt(2)*(15*C*sgn(cos(1/2*d*x + 1/2*c))*sin(7/2*d*x + 7/2*c) + 21*(2 *B*sgn(cos(1/2*d*x + 1/2*c)) + C*sgn(cos(1/2*d*x + 1/2*c)))*sin(5/2*d*x + 5/2*c) + 35*(4*A*sgn(cos(1/2*d*x + 1/2*c)) + 2*B*sgn(cos(1/2*d*x + 1/2*c)) + 3*C*sgn(cos(1/2*d*x + 1/2*c)))*sin(3/2*d*x + 3/2*c) + 105*(4*A*sgn(cos( 1/2*d*x + 1/2*c)) + 4*B*sgn(cos(1/2*d*x + 1/2*c)) + 3*C*sgn(cos(1/2*d*x + 1/2*c)))*sin(1/2*d*x + 1/2*c))*sqrt(a)/d
Timed out. \[ \int \cos (c+d x) \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int \cos \left (c+d\,x\right )\,\sqrt {a+a\,\cos \left (c+d\,x\right )}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right ) \,d x \]