Integrand size = 31, antiderivative size = 101 \[ \int \cos (c+d x) \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {2 a (5 A+7 B) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 (5 A-2 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 B (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 a d} \]
2/5*B*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/a/d+2/15*a*(5*A+7*B)*sin(d*x+c)/d/ (a+a*cos(d*x+c))^(1/2)+2/15*(5*A-2*B)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d
Time = 0.11 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.63 \[ \int \cos (c+d x) \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {\sqrt {a (1+\cos (c+d x))} (20 A+19 B+2 (5 A+4 B) \cos (c+d x)+3 B \cos (2 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )}{15 d} \]
(Sqrt[a*(1 + Cos[c + d*x])]*(20*A + 19*B + 2*(5*A + 4*B)*Cos[c + d*x] + 3* B*Cos[2*(c + d*x)])*Tan[(c + d*x)/2])/(15*d)
Time = 0.60 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.11, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {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.
\(\displaystyle \int \cos (c+d x) \sqrt {a \cos (c+d x)+a} (A+B \cos (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a \sin \left (c+d x+\frac {\pi }{2}\right )+a} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
\(\Big \downarrow \) 3447 |
\(\displaystyle \int \sqrt {a \cos (c+d x)+a} \left (A \cos (c+d x)+B \cos ^2(c+d x)\right )dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt {a \sin \left (c+d x+\frac {\pi }{2}\right )+a} \left (A \sin \left (c+d x+\frac {\pi }{2}\right )+B \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \frac {2 \int \frac {1}{2} \sqrt {\cos (c+d x) a+a} (3 a B+a (5 A-2 B) \cos (c+d x))dx}{5 a}+\frac {2 B \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \sqrt {\cos (c+d x) a+a} (3 a B+a (5 A-2 B) \cos (c+d x))dx}{5 a}+\frac {2 B \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (3 a B+a (5 A-2 B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{5 a}+\frac {2 B \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d}\) |
\(\Big \downarrow \) 3230 |
\(\displaystyle \frac {\frac {1}{3} a (5 A+7 B) \int \sqrt {\cos (c+d x) a+a}dx+\frac {2 a (5 A-2 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}+\frac {2 B \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{3} a (5 A+7 B) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a (5 A-2 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}+\frac {2 B \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d}\) |
\(\Big \downarrow \) 3125 |
\(\displaystyle \frac {\frac {2 a^2 (5 A+7 B) \sin (c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a (5 A-2 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}+\frac {2 B \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d}\) |
(2*B*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*a*d) + ((2*a^2*(5*A + 7*B )*Sin[c + d*x])/(3*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*(5*A - 2*B)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3*d))/(5*a)
3.1.76.3.1 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]
Time = 2.71 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.82
method | result | size |
default | \(\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (12 B \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-10 A -20 B \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 A +15 B \right ) \sqrt {2}}{15 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) | \(83\) |
parts | \(\frac {2 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (1+2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right ) \sqrt {2}}{3 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}+\frac {2 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (12 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7\right ) \sqrt {2}}{15 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) | \(131\) |
2/15*cos(1/2*d*x+1/2*c)*a*sin(1/2*d*x+1/2*c)*(12*B*sin(1/2*d*x+1/2*c)^4+(- 10*A-20*B)*sin(1/2*d*x+1/2*c)^2+15*A+15*B)*2^(1/2)/(a*cos(1/2*d*x+1/2*c)^2 )^(1/2)/d
Time = 0.30 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.63 \[ \int \cos (c+d x) \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {2 \, {\left (3 \, B \cos \left (d x + c\right )^{2} + {\left (5 \, A + 4 \, B\right )} \cos \left (d x + c\right ) + 10 \, A + 8 \, B\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]
2/15*(3*B*cos(d*x + c)^2 + (5*A + 4*B)*cos(d*x + c) + 10*A + 8*B)*sqrt(a*c os(d*x + c) + a)*sin(d*x + c)/(d*cos(d*x + c) + d)
\[ \int \cos (c+d x) \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\int \sqrt {a \left (\cos {\left (c + d x \right )} + 1\right )} \left (A + B \cos {\left (c + d x \right )}\right ) \cos {\left (c + d x \right )}\, dx \]
Time = 0.37 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.87 \[ \int \cos (c+d x) \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {10 \, {\left (\sqrt {2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 3 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} A \sqrt {a} + {\left (3 \, \sqrt {2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, \sqrt {2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 30 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} B \sqrt {a}}{30 \, d} \]
1/30*(10*(sqrt(2)*sin(3/2*d*x + 3/2*c) + 3*sqrt(2)*sin(1/2*d*x + 1/2*c))*A *sqrt(a) + (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))/d
Time = 0.33 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.06 \[ \int \cos (c+d x) \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {\sqrt {2} {\left (3 \, B \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, {\left (2 \, A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + B \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 30 \, {\left (A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + B \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{30 \, d} \]
1/30*sqrt(2)*(3*B*sgn(cos(1/2*d*x + 1/2*c))*sin(5/2*d*x + 5/2*c) + 5*(2*A* sgn(cos(1/2*d*x + 1/2*c)) + B*sgn(cos(1/2*d*x + 1/2*c)))*sin(3/2*d*x + 3/2 *c) + 30*(A*sgn(cos(1/2*d*x + 1/2*c)) + B*sgn(cos(1/2*d*x + 1/2*c)))*sin(1 /2*d*x + 1/2*c))*sqrt(a)/d
Timed out. \[ \int \cos (c+d x) \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\int \cos \left (c+d\,x\right )\,\left (A+B\,\cos \left (c+d\,x\right )\right )\,\sqrt {a+a\,\cos \left (c+d\,x\right )} \,d x \]