Integrand size = 23, antiderivative size = 86 \[ \int \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} \, dx=\frac {14 a \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}-\frac {4 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 a d} \]
2/5*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/a/d+14/15*a*sin(d*x+c)/d/(a+a*cos(d* x+c))^(1/2)-4/15*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d
Time = 0.06 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.79 \[ \int \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} \, dx=\frac {\sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \left (30 \sin \left (\frac {1}{2} (c+d x)\right )+5 \sin \left (\frac {3}{2} (c+d x)\right )+3 \sin \left (\frac {5}{2} (c+d x)\right )\right )}{30 d} \]
(Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(30*Sin[(c + d*x)/2] + 5*Sin[ (3*(c + d*x))/2] + 3*Sin[(5*(c + d*x))/2]))/(30*d)
Time = 0.44 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.13, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 3238, 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.
\(\displaystyle \int \cos ^2(c+d x) \sqrt {a \cos (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {a \sin \left (c+d x+\frac {\pi }{2}\right )+a}dx\) |
\(\Big \downarrow \) 3238 |
\(\displaystyle \frac {2 \int \frac {1}{2} (3 a-2 a \cos (c+d x)) \sqrt {\cos (c+d x) a+a}dx}{5 a}+\frac {2 \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int (3 a-2 a \cos (c+d x)) \sqrt {\cos (c+d x) a+a}dx}{5 a}+\frac {2 \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \left (3 a-2 a \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{5 a}+\frac {2 \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d}\) |
\(\Big \downarrow \) 3230 |
\(\displaystyle \frac {\frac {7}{3} a \int \sqrt {\cos (c+d x) a+a}dx-\frac {4 a \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}+\frac {2 \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {7}{3} a \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-\frac {4 a \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}+\frac {2 \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d}\) |
\(\Big \downarrow \) 3125 |
\(\displaystyle \frac {\frac {14 a^2 \sin (c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {4 a \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}+\frac {2 \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d}\) |
(2*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*a*d) + ((14*a^2*Sin[c + d*x ])/(3*d*Sqrt[a + a*Cos[c + d*x]]) - (4*a*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3*d))/(5*a)
3.1.97.3.1 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[sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-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*(b*(m + 1) - a*Si n[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] && ! LtQ[m, -2^(-1)]
Time = 0.92 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.83
method | result | size |
default | \(\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (12 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7\right ) \sqrt {2}}{15 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) | \(71\) |
2/15*cos(1/2*d*x+1/2*c)*a*sin(1/2*d*x+1/2*c)*(12*cos(1/2*d*x+1/2*c)^4-4*co s(1/2*d*x+1/2*c)^2+7)*2^(1/2)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d
Time = 0.30 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.60 \[ \int \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} \, dx=\frac {2 \, \sqrt {a \cos \left (d x + c\right ) + a} {\left (3 \, \cos \left (d x + c\right )^{2} + 4 \, \cos \left (d x + c\right ) + 8\right )} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]
2/15*sqrt(a*cos(d*x + c) + a)*(3*cos(d*x + c)^2 + 4*cos(d*x + c) + 8)*sin( d*x + c)/(d*cos(d*x + c) + d)
\[ \int \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} \, dx=\int \sqrt {a \left (\cos {\left (c + d x \right )} + 1\right )} \cos ^{2}{\left (c + d x \right )}\, dx \]
Time = 0.44 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.59 \[ \int \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} \, dx=\frac {{\left (3 \, \sqrt {2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, \sqrt {2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 30 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{30 \, d} \]
1/30*(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))*sqrt(a)/d
Time = 0.48 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.87 \[ \int \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} \, dx=\frac {\sqrt {2} {\left (3 \, \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 30 \, \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{30 \, d} \]
1/30*sqrt(2)*(3*sgn(cos(1/2*d*x + 1/2*c))*sin(5/2*d*x + 5/2*c) + 5*sgn(cos (1/2*d*x + 1/2*c))*sin(3/2*d*x + 3/2*c) + 30*sgn(cos(1/2*d*x + 1/2*c))*sin (1/2*d*x + 1/2*c))*sqrt(a)/d
Timed out. \[ \int \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} \, dx=\int {\cos \left (c+d\,x\right )}^2\,\sqrt {a+a\,\cos \left (c+d\,x\right )} \,d x \]