Integrand size = 35, antiderivative size = 104 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {2 a (15 A+5 B+7 C) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 (5 B-2 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 C (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 a d} \]
2/5*C*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/a/d+2/15*a*(15*A+5*B+7*C)*sin(d*x+ c)/d/(a+a*cos(d*x+c))^(1/2)+2/15*(5*B-2*C)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/ 2)/d
Time = 0.20 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.64 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {\sqrt {a (1+\cos (c+d x))} (30 A+20 B+19 C+2 (5 B+4 C) \cos (c+d x)+3 C \cos (2 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )}{15 d} \]
(Sqrt[a*(1 + Cos[c + d*x])]*(30*A + 20*B + 19*C + 2*(5*B + 4*C)*Cos[c + d* x] + 3*C*Cos[2*(c + d*x)])*Tan[(c + d*x)/2])/(15*d)
Time = 0.50 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {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.
\(\displaystyle \int \sqrt {a \cos (c+d x)+a} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt {a \sin \left (c+d x+\frac {\pi }{2}\right )+a} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \frac {2 \int \frac {1}{2} \sqrt {\cos (c+d x) a+a} (a (5 A+3 C)+a (5 B-2 C) \cos (c+d x))dx}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \sqrt {\cos (c+d x) a+a} (a (5 A+3 C)+a (5 B-2 C) \cos (c+d x))dx}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a (5 A+3 C)+a (5 B-2 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d}\) |
\(\Big \downarrow \) 3230 |
\(\displaystyle \frac {\frac {1}{3} a (15 A+5 B+7 C) \int \sqrt {\cos (c+d x) a+a}dx+\frac {2 a (5 B-2 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{3} a (15 A+5 B+7 C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a (5 B-2 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d}\) |
\(\Big \downarrow \) 3125 |
\(\displaystyle \frac {\frac {2 a^2 (15 A+5 B+7 C) \sin (c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a (5 B-2 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d}\) |
(2*C*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*a*d) + ((2*a^2*(15*A + 5* B + 7*C)*Sin[c + d*x])/(3*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*(5*B - 2*C)*S qrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3*d))/(5*a)
3.4.76.3.1 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_.)*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 = 6.44 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.83
method | result | size |
default | \(\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (12 C \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-10 B -20 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 A +15 B +15 C \right ) \sqrt {2}}{15 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) | \(86\) |
parts | \(\frac {2 A a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2}}{\sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}+\frac {2 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (1+2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right ) \sqrt {2}}{3 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}+\frac {2 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (12 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7\right ) \sqrt {2}}{15 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) | \(174\) |
2/15*cos(1/2*d*x+1/2*c)*a*sin(1/2*d*x+1/2*c)*(12*C*sin(1/2*d*x+1/2*c)^4+(- 10*B-20*C)*sin(1/2*d*x+1/2*c)^2+15*A+15*B+15*C)*2^(1/2)/(a*cos(1/2*d*x+1/2 *c)^2)^(1/2)/d
Time = 0.27 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.64 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (3 \, C \cos \left (d x + c\right )^{2} + {\left (5 \, B + 4 \, C\right )} \cos \left (d x + c\right ) + 15 \, A + 10 \, B + 8 \, C\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]
2/15*(3*C*cos(d*x + c)^2 + (5*B + 4*C)*cos(d*x + c) + 15*A + 10*B + 8*C)*s qrt(a*cos(d*x + c) + a)*sin(d*x + c)/(d*cos(d*x + c) + d)
\[ \int \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int \sqrt {a \left (\cos {\left (c + d x \right )} + 1\right )} \left (A + B \cos {\left (c + d x \right )} + C \cos ^{2}{\left (c + d x \right )}\right )\, dx \]
Time = 0.39 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.02 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {60 \, \sqrt {2} A \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 10 \, {\left (\sqrt {2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 3 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} B \sqrt {a} + {\left (3 \, \sqrt {2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, \sqrt {2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 30 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} C \sqrt {a}}{30 \, d} \]
1/30*(60*sqrt(2)*A*sqrt(a)*sin(1/2*d*x + 1/2*c) + 10*(sqrt(2)*sin(3/2*d*x + 3/2*c) + 3*sqrt(2)*sin(1/2*d*x + 1/2*c))*B*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))*C*sqrt(a))/d
Time = 0.37 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.15 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {\sqrt {2} {\left (3 \, C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, {\left (2 \, B \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 30 \, {\left (2 \, A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + B \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{30 \, d} \]
1/30*sqrt(2)*(3*C*sgn(cos(1/2*d*x + 1/2*c))*sin(5/2*d*x + 5/2*c) + 5*(2*B* sgn(cos(1/2*d*x + 1/2*c)) + C*sgn(cos(1/2*d*x + 1/2*c)))*sin(3/2*d*x + 3/2 *c) + 30*(2*A*sgn(cos(1/2*d*x + 1/2*c)) + B*sgn(cos(1/2*d*x + 1/2*c)) + C* sgn(cos(1/2*d*x + 1/2*c)))*sin(1/2*d*x + 1/2*c))*sqrt(a)/d
Timed out. \[ \int \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int \sqrt {a+a\,\cos \left (c+d\,x\right )}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right ) \,d x \]