Integrand size = 27, antiderivative size = 95 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {2 a (15 A+7 C) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}-\frac {4 C \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 C (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 a d} \]
2/5*C*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/a/d+2/15*a*(15*A+7*C)*sin(d*x+c)/d /(a+a*cos(d*x+c))^(1/2)-4/15*C*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d
Time = 0.09 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.61 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {\sqrt {a (1+\cos (c+d x))} (30 A+19 C+8 C \cos (c+d x)+3 C \cos (2 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )}{15 d} \]
(Sqrt[a*(1 + Cos[c + d*x])]*(30*A + 19*C + 8*C*Cos[c + d*x] + 3*C*Cos[2*(c + d*x)])*Tan[(c + d*x)/2])/(15*d)
Time = 0.47 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.12, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {3042, 3503, 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.
\(\displaystyle \int \sqrt {a \cos (c+d x)+a} \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt {a \sin \left (c+d x+\frac {\pi }{2}\right )+a} \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
\(\Big \downarrow \) 3503 |
\(\displaystyle \frac {2 \int \frac {1}{2} \sqrt {\cos (c+d x) a+a} (a (5 A+3 C)-2 a C \cos (c+d x))dx}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \sqrt {\cos (c+d x) a+a} (a (5 A+3 C)-2 a C \cos (c+d x))dx}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a (5 A+3 C)-2 a C \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d}\) |
\(\Big \downarrow \) 3230 |
\(\displaystyle \frac {\frac {1}{3} a (15 A+7 C) \int \sqrt {\cos (c+d x) a+a}dx-\frac {4 a C \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{3} a (15 A+7 C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-\frac {4 a C \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d}\) |
\(\Big \downarrow \) 3125 |
\(\displaystyle \frac {\frac {2 a^2 (15 A+7 C) \sin (c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {4 a C \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d}\) |
(2*C*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*a*d) + ((2*a^2*(15*A + 7* C)*Sin[c + d*x])/(3*d*Sqrt[a + a*Cos[c + d*x]]) - (4*a*C*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3*d))/(5*a)
3.1.77.3.1 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_.) + (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 = 4.46 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.82
method | result | size |
default | \(\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (12 C \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 C \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 A +7 C \right ) \sqrt {2}}{15 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) | \(78\) |
parts | \(\frac {2 A a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2}}{\sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}+\frac {2 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (12 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7\right ) \sqrt {2}}{15 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) | \(116\) |
2/15*cos(1/2*d*x+1/2*c)*a*sin(1/2*d*x+1/2*c)*(12*C*cos(1/2*d*x+1/2*c)^4-4* C*cos(1/2*d*x+1/2*c)^2+15*A+7*C)*2^(1/2)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d
Time = 0.27 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.62 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (3 \, C \cos \left (d x + c\right )^{2} + 4 \, C \cos \left (d x + c\right ) + 15 \, A + 8 \, C\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]
2/15*(3*C*cos(d*x + c)^2 + 4*C*cos(d*x + c) + 15*A + 8*C)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/(d*cos(d*x + c) + d)
\[ \int \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\int \sqrt {a \left (\cos {\left (c + d x \right )} + 1\right )} \left (A + C \cos ^{2}{\left (c + d x \right )}\right )\, dx \]
Time = 0.39 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.76 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {60 \, \sqrt {2} A \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + {\left (3 \, \sqrt {2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, \sqrt {2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 30 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} C \sqrt {a}}{30 \, d} \]
1/30*(60*sqrt(2)*A*sqrt(a)*sin(1/2*d*x + 1/2*c) + (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 = 0.29 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.98 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {\sqrt {2} {\left (3 \, C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 30 \, {\left (2 \, A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{30 \, d} \]
1/30*sqrt(2)*(3*C*sgn(cos(1/2*d*x + 1/2*c))*sin(5/2*d*x + 5/2*c) + 5*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)) + C*sgn(cos(1/2*d*x + 1/2*c)))*sin(1/2*d*x + 1/2*c))*sqrt(a)/d
Timed out. \[ \int \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\int \left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,\sqrt {a+a\,\cos \left (c+d\,x\right )} \,d x \]