Integrand size = 35, antiderivative size = 180 \[ \int \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {2 a (21 A+16 C) \sin (c+d x)}{45 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a C \cos ^3(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \cos (c+d x)}}-\frac {4 (21 A+16 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{315 d}+\frac {2 C \cos ^3(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{9 d}+\frac {2 (21 A+16 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{105 a d} \]
2/105*(21*A+16*C)*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/a/d+2/45*a*(21*A+16*C) *sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/63*a*C*cos(d*x+c)^3*sin(d*x+c)/d/(a +a*cos(d*x+c))^(1/2)-4/315*(21*A+16*C)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d +2/9*C*cos(d*x+c)^3*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d
Time = 0.27 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.51 \[ \int \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {\sqrt {a (1+\cos (c+d x))} (1596 A+1321 C+16 (42 A+47 C) \cos (c+d x)+4 (63 A+83 C) \cos (2 (c+d x))+80 C \cos (3 (c+d x))+35 C \cos (4 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )}{1260 d} \]
(Sqrt[a*(1 + Cos[c + d*x])]*(1596*A + 1321*C + 16*(42*A + 47*C)*Cos[c + d* x] + 4*(63*A + 83*C)*Cos[2*(c + d*x)] + 80*C*Cos[3*(c + d*x)] + 35*C*Cos[4 *(c + d*x)])*Tan[(c + d*x)/2])/(1260*d)
Time = 0.98 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.07, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.343, Rules used = {3042, 3525, 27, 3042, 3460, 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.
\(\displaystyle \int \cos ^2(c+d x) \sqrt {a \cos (c+d x)+a} \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {a \sin \left (c+d x+\frac {\pi }{2}\right )+a} \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
\(\Big \downarrow \) 3525 |
\(\displaystyle \frac {2 \int \frac {1}{2} \cos ^2(c+d x) \sqrt {\cos (c+d x) a+a} (3 a (3 A+2 C)+a C \cos (c+d x))dx}{9 a}+\frac {2 C \sin (c+d x) \cos ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{9 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \cos ^2(c+d x) \sqrt {\cos (c+d x) a+a} (3 a (3 A+2 C)+a C \cos (c+d x))dx}{9 a}+\frac {2 C \sin (c+d x) \cos ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{9 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \sin \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (3 a (3 A+2 C)+a C \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{9 a}+\frac {2 C \sin (c+d x) \cos ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{9 d}\) |
\(\Big \downarrow \) 3460 |
\(\displaystyle \frac {\frac {3}{7} a (21 A+16 C) \int \cos ^2(c+d x) \sqrt {\cos (c+d x) a+a}dx+\frac {2 a^2 C \sin (c+d x) \cos ^3(c+d x)}{7 d \sqrt {a \cos (c+d x)+a}}}{9 a}+\frac {2 C \sin (c+d x) \cos ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{9 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {3}{7} a (21 A+16 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a^2 C \sin (c+d x) \cos ^3(c+d x)}{7 d \sqrt {a \cos (c+d x)+a}}}{9 a}+\frac {2 C \sin (c+d x) \cos ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{9 d}\) |
\(\Big \downarrow \) 3238 |
\(\displaystyle \frac {\frac {3}{7} a (21 A+16 C) \left (\frac {2 \int \frac {1}{2} (3 a-2 a \cos (c+d x)) \sqrt {\cos (c+d x) a+a}dx}{5 a}+\frac {2 \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a^2 C \sin (c+d x) \cos ^3(c+d x)}{7 d \sqrt {a \cos (c+d x)+a}}}{9 a}+\frac {2 C \sin (c+d x) \cos ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{9 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {3}{7} a (21 A+16 C) \left (\frac {\int (3 a-2 a \cos (c+d x)) \sqrt {\cos (c+d x) a+a}dx}{5 a}+\frac {2 \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a^2 C \sin (c+d x) \cos ^3(c+d x)}{7 d \sqrt {a \cos (c+d x)+a}}}{9 a}+\frac {2 C \sin (c+d x) \cos ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{9 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {3}{7} a (21 A+16 C) \left (\frac {\int \left (3 a-2 a \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{5 a}+\frac {2 \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a^2 C \sin (c+d x) \cos ^3(c+d x)}{7 d \sqrt {a \cos (c+d x)+a}}}{9 a}+\frac {2 C \sin (c+d x) \cos ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{9 d}\) |
\(\Big \downarrow \) 3230 |
\(\displaystyle \frac {\frac {3}{7} a (21 A+16 C) \left (\frac {\frac {7}{3} a \int \sqrt {\cos (c+d x) a+a}dx-\frac {4 a \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}+\frac {2 \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a^2 C \sin (c+d x) \cos ^3(c+d x)}{7 d \sqrt {a \cos (c+d x)+a}}}{9 a}+\frac {2 C \sin (c+d x) \cos ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{9 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {3}{7} a (21 A+16 C) \left (\frac {\frac {7}{3} a \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-\frac {4 a \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}+\frac {2 \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a^2 C \sin (c+d x) \cos ^3(c+d x)}{7 d \sqrt {a \cos (c+d x)+a}}}{9 a}+\frac {2 C \sin (c+d x) \cos ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{9 d}\) |
\(\Big \downarrow \) 3125 |
\(\displaystyle \frac {\frac {3}{7} a (21 A+16 C) \left (\frac {\frac {14 a^2 \sin (c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {4 a \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}+\frac {2 \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a^2 C \sin (c+d x) \cos ^3(c+d x)}{7 d \sqrt {a \cos (c+d x)+a}}}{9 a}+\frac {2 C \sin (c+d x) \cos ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{9 d}\) |
(2*C*Cos[c + d*x]^3*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(9*d) + ((2*a^2 *C*Cos[c + d*x]^3*Sin[c + d*x])/(7*d*Sqrt[a + a*Cos[c + d*x]]) + (3*a*(21* A + 16*C)*((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)))/7)/(9*a)
3.1.75.3.1 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[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)]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp [-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b *d*(2*n + 3)) Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ[n, -1]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (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))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ[m, -2^(-1)] && NeQ[m + n + 2, 0]
Time = 5.12 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.64
method | result | size |
default | \(\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (560 C \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1440 C \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (252 A +1512 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-420 A -840 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+315 A +315 C \right ) \sqrt {2}}{315 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) | \(116\) |
parts | \(\frac {2 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (12 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7\right ) \sqrt {2}}{15 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}+\frac {2 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (560 \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-800 \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+552 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-104 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+107\right ) \sqrt {2}}{315 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) | \(170\) |
2/315*cos(1/2*d*x+1/2*c)*a*sin(1/2*d*x+1/2*c)*(560*C*sin(1/2*d*x+1/2*c)^8- 1440*C*sin(1/2*d*x+1/2*c)^6+(252*A+1512*C)*sin(1/2*d*x+1/2*c)^4+(-420*A-84 0*C)*sin(1/2*d*x+1/2*c)^2+315*A+315*C)*2^(1/2)/(a*cos(1/2*d*x+1/2*c)^2)^(1 /2)/d
Time = 0.27 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.52 \[ \int \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (35 \, C \cos \left (d x + c\right )^{4} + 40 \, C \cos \left (d x + c\right )^{3} + 3 \, {\left (21 \, A + 16 \, C\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (21 \, A + 16 \, C\right )} \cos \left (d x + c\right ) + 168 \, A + 128 \, C\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]
2/315*(35*C*cos(d*x + c)^4 + 40*C*cos(d*x + c)^3 + 3*(21*A + 16*C)*cos(d*x + c)^2 + 4*(21*A + 16*C)*cos(d*x + c) + 168*A + 128*C)*sqrt(a*cos(d*x + c ) + a)*sin(d*x + c)/(d*cos(d*x + c) + d)
Timed out. \[ \int \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\text {Timed out} \]
Time = 0.40 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.73 \[ \int \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {84 \, {\left (3 \, \sqrt {2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, \sqrt {2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 30 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} A \sqrt {a} + {\left (35 \, \sqrt {2} \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 45 \, \sqrt {2} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 252 \, \sqrt {2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 420 \, \sqrt {2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 1890 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} C \sqrt {a}}{2520 \, d} \]
1/2520*(84*(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))*A*sqrt(a) + (35*sqrt(2)*sin(9/2*d*x + 9/2*c) + 45*sqrt(2)*sin(7/2*d*x + 7/2*c) + 252*sqrt(2)*sin(5/2*d*x + 5/2* c) + 420*sqrt(2)*sin(3/2*d*x + 3/2*c) + 1890*sqrt(2)*sin(1/2*d*x + 1/2*c)) *C*sqrt(a))/d
Time = 0.55 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.92 \[ \int \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {\sqrt {2} {\left (35 \, C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 45 \, C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 252 \, {\left (A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 420 \, {\left (A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 630 \, {\left (4 \, A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 3 \, C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{2520 \, d} \]
1/2520*sqrt(2)*(35*C*sgn(cos(1/2*d*x + 1/2*c))*sin(9/2*d*x + 9/2*c) + 45*C *sgn(cos(1/2*d*x + 1/2*c))*sin(7/2*d*x + 7/2*c) + 252*(A*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) + 420*(A*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) + 630*(4*A*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. \[ \int \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\int {\cos \left (c+d\,x\right )}^2\,\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,\sqrt {a+a\,\cos \left (c+d\,x\right )} \,d x \]