Integrand size = 40, antiderivative size = 170 \[ \int \frac {\cos ^3(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\frac {(7 B-10 C) x}{2 a^2}-\frac {4 (2 B-3 C) \sin (c+d x)}{a^2 d}+\frac {(7 B-10 C) \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac {(7 B-10 C) \cos ^3(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}+\frac {(B-C) \cos ^4(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {4 (2 B-3 C) \sin ^3(c+d x)}{3 a^2 d} \] Output:
1/2*(7*B-10*C)*x/a^2-4*(2*B-3*C)*sin(d*x+c)/a^2/d+1/2*(7*B-10*C)*cos(d*x+c )*sin(d*x+c)/a^2/d+1/3*(7*B-10*C)*cos(d*x+c)^3*sin(d*x+c)/a^2/d/(1+cos(d*x +c))+1/3*(B-C)*cos(d*x+c)^4*sin(d*x+c)/d/(a+a*cos(d*x+c))^2+4/3*(2*B-3*C)* sin(d*x+c)^3/a^2/d
Leaf count is larger than twice the leaf count of optimal. \(369\) vs. \(2(170)=340\).
Time = 1.99 (sec) , antiderivative size = 369, normalized size of antiderivative = 2.17 \[ \int \frac {\cos ^3(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (36 (7 B-10 C) d x \cos \left (\frac {d x}{2}\right )+36 (7 B-10 C) d x \cos \left (c+\frac {d x}{2}\right )+84 B d x \cos \left (c+\frac {3 d x}{2}\right )-120 C d x \cos \left (c+\frac {3 d x}{2}\right )+84 B d x \cos \left (2 c+\frac {3 d x}{2}\right )-120 C d x \cos \left (2 c+\frac {3 d x}{2}\right )-381 B \sin \left (\frac {d x}{2}\right )+516 C \sin \left (\frac {d x}{2}\right )+147 B \sin \left (c+\frac {d x}{2}\right )-156 C \sin \left (c+\frac {d x}{2}\right )-239 B \sin \left (c+\frac {3 d x}{2}\right )+342 C \sin \left (c+\frac {3 d x}{2}\right )-63 B \sin \left (2 c+\frac {3 d x}{2}\right )+118 C \sin \left (2 c+\frac {3 d x}{2}\right )-15 B \sin \left (2 c+\frac {5 d x}{2}\right )+30 C \sin \left (2 c+\frac {5 d x}{2}\right )-15 B \sin \left (3 c+\frac {5 d x}{2}\right )+30 C \sin \left (3 c+\frac {5 d x}{2}\right )+3 B \sin \left (3 c+\frac {7 d x}{2}\right )-3 C \sin \left (3 c+\frac {7 d x}{2}\right )+3 B \sin \left (4 c+\frac {7 d x}{2}\right )-3 C \sin \left (4 c+\frac {7 d x}{2}\right )+C \sin \left (4 c+\frac {9 d x}{2}\right )+C \sin \left (5 c+\frac {9 d x}{2}\right )\right )}{48 a^2 d (1+\cos (c+d x))^2} \] Input:
Integrate[(Cos[c + d*x]^3*(B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(a + a*Cos[ c + d*x])^2,x]
Output:
(Cos[(c + d*x)/2]*Sec[c/2]*(36*(7*B - 10*C)*d*x*Cos[(d*x)/2] + 36*(7*B - 1 0*C)*d*x*Cos[c + (d*x)/2] + 84*B*d*x*Cos[c + (3*d*x)/2] - 120*C*d*x*Cos[c + (3*d*x)/2] + 84*B*d*x*Cos[2*c + (3*d*x)/2] - 120*C*d*x*Cos[2*c + (3*d*x) /2] - 381*B*Sin[(d*x)/2] + 516*C*Sin[(d*x)/2] + 147*B*Sin[c + (d*x)/2] - 1 56*C*Sin[c + (d*x)/2] - 239*B*Sin[c + (3*d*x)/2] + 342*C*Sin[c + (3*d*x)/2 ] - 63*B*Sin[2*c + (3*d*x)/2] + 118*C*Sin[2*c + (3*d*x)/2] - 15*B*Sin[2*c + (5*d*x)/2] + 30*C*Sin[2*c + (5*d*x)/2] - 15*B*Sin[3*c + (5*d*x)/2] + 30* C*Sin[3*c + (5*d*x)/2] + 3*B*Sin[3*c + (7*d*x)/2] - 3*C*Sin[3*c + (7*d*x)/ 2] + 3*B*Sin[4*c + (7*d*x)/2] - 3*C*Sin[4*c + (7*d*x)/2] + C*Sin[4*c + (9* d*x)/2] + C*Sin[5*c + (9*d*x)/2]))/(48*a^2*d*(1 + Cos[c + d*x])^2)
Time = 1.03 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.94, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {3042, 3508, 3042, 3456, 3042, 3456, 27, 3042, 3227, 3042, 3113, 2009, 3115, 24}
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.
| \(\displaystyle \int \frac {\cos ^3(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 3508 |
| \(\displaystyle \int \frac {\cos ^4(c+d x) (B+C \cos (c+d x))}{(a \cos (c+d x)+a)^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^4 \left (B+C \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 3456 |
| \(\displaystyle \frac {\int \frac {\cos ^3(c+d x) (4 a (B-C)-3 a (B-2 C) \cos (c+d x))}{\cos (c+d x) a+a}dx}{3 a^2}+\frac {(B-C) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (4 a (B-C)-3 a (B-2 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}+\frac {(B-C) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3456 |
| \(\displaystyle \frac {\frac {\int 3 \cos ^2(c+d x) \left (a^2 (7 B-10 C)-4 a^2 (2 B-3 C) \cos (c+d x)\right )dx}{a^2}+\frac {(7 B-10 C) \sin (c+d x) \cos ^3(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}+\frac {(B-C) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {3 \int \cos ^2(c+d x) \left (a^2 (7 B-10 C)-4 a^2 (2 B-3 C) \cos (c+d x)\right )dx}{a^2}+\frac {(7 B-10 C) \sin (c+d x) \cos ^3(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}+\frac {(B-C) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 \int \sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a^2 (7 B-10 C)-4 a^2 (2 B-3 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}+\frac {(7 B-10 C) \sin (c+d x) \cos ^3(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}+\frac {(B-C) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {\frac {3 \left (a^2 (7 B-10 C) \int \cos ^2(c+d x)dx-4 a^2 (2 B-3 C) \int \cos ^3(c+d x)dx\right )}{a^2}+\frac {(7 B-10 C) \sin (c+d x) \cos ^3(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}+\frac {(B-C) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 \left (a^2 (7 B-10 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-4 a^2 (2 B-3 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^3dx\right )}{a^2}+\frac {(7 B-10 C) \sin (c+d x) \cos ^3(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}+\frac {(B-C) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3113 |
| \(\displaystyle \frac {\frac {3 \left (\frac {4 a^2 (2 B-3 C) \int \left (1-\sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}+a^2 (7 B-10 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx\right )}{a^2}+\frac {(7 B-10 C) \sin (c+d x) \cos ^3(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}+\frac {(B-C) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {3 \left (a^2 (7 B-10 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {4 a^2 (2 B-3 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )}{a^2}+\frac {(7 B-10 C) \sin (c+d x) \cos ^3(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}+\frac {(B-C) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {\frac {3 \left (a^2 (7 B-10 C) \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {4 a^2 (2 B-3 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )}{a^2}+\frac {(7 B-10 C) \sin (c+d x) \cos ^3(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}+\frac {(B-C) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\frac {3 \left (\frac {4 a^2 (2 B-3 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+a^2 (7 B-10 C) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )}{a^2}+\frac {(7 B-10 C) \sin (c+d x) \cos ^3(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}+\frac {(B-C) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
Input:
Int[(Cos[c + d*x]^3*(B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(a + a*Cos[c + d* x])^2,x]
Output:
((B - C)*Cos[c + d*x]^4*Sin[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2) + (((7* B - 10*C)*Cos[c + d*x]^3*Sin[c + d*x])/(d*(1 + Cos[c + d*x])) + (3*(a^2*(7 *B - 10*C)*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)) + (4*a^2*(2*B - 3*C)* (-Sin[c + d*x] + Sin[c + d*x]^3/3))/d))/a^2)/(3*a^2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp and[(1 - x^2)^((n - 1)/2), x], x], x, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/( a*f*(2*m + 1))), x] - Simp[1/(a*b*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x], x] /; Fre eQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] & & NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (In tegerQ[2*n] || EqQ[c, 0])
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_ .) + (f_.)*(x_)]^2), x_Symbol] :> Simp[1/b^2 Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Time = 0.50 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.64
| method | result | size |
| parallelrisch | \(\frac {-163 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\frac {4 \left (3 B -7 C \right ) \cos \left (2 d x +2 c \right )}{163}+\frac {\left (-3 B +2 C \right ) \cos \left (3 d x +3 c \right )}{163}-\frac {C \cos \left (4 d x +4 c \right )}{163}+\left (B -\frac {258 C}{163}\right ) \cos \left (d x +c \right )+\frac {140 B}{163}-\frac {219 C}{163}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+168 \left (B -\frac {10 C}{7}\right ) x d}{48 a^{2} d}\) | \(108\) |
| derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +9 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {8 \left (-\frac {5 B}{4}+\frac {5 C}{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+8 \left (-2 B +\frac {10 C}{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+8 \left (-\frac {3 B}{4}+\frac {3 C}{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+2 \left (7 B -10 C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\) | \(154\) |
| default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +9 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {8 \left (-\frac {5 B}{4}+\frac {5 C}{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+8 \left (-2 B +\frac {10 C}{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+8 \left (-\frac {3 B}{4}+\frac {3 C}{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+2 \left (7 B -10 C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\) | \(154\) |
| risch | \(\frac {7 x B}{2 a^{2}}-\frac {5 C x}{a^{2}}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )} B}{8 a^{2} d}+\frac {i C \,{\mathrm e}^{2 i \left (d x +c \right )}}{4 a^{2} d}+\frac {i {\mathrm e}^{i \left (d x +c \right )} B}{a^{2} d}-\frac {15 i {\mathrm e}^{i \left (d x +c \right )} C}{8 a^{2} d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} B}{a^{2} d}+\frac {15 i {\mathrm e}^{-i \left (d x +c \right )} C}{8 a^{2} d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} B}{8 a^{2} d}-\frac {i C \,{\mathrm e}^{-2 i \left (d x +c \right )}}{4 a^{2} d}-\frac {2 i \left (12 B \,{\mathrm e}^{2 i \left (d x +c \right )}-15 C \,{\mathrm e}^{2 i \left (d x +c \right )}+21 B \,{\mathrm e}^{i \left (d x +c \right )}-27 C \,{\mathrm e}^{i \left (d x +c \right )}+11 B -14 C \right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}+\frac {C \sin \left (3 d x +3 c \right )}{12 a^{2} d}\) | \(263\) |
| norman | \(\frac {\frac {\left (7 B -10 C \right ) x}{2 a}+\frac {\left (B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{6 a d}+\frac {5 \left (7 B -10 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 a}+\frac {5 \left (7 B -10 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{a}+\frac {5 \left (7 B -10 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{a}+\frac {5 \left (7 B -10 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{2 a}+\frac {\left (7 B -10 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{2 a}-\frac {\left (8 B -11 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 a d}-\frac {\left (13 B -21 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}-\frac {5 \left (25 B -37 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{6 a d}-\frac {2 \left (77 B -115 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 a d}-\frac {\left (94 B -143 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 a d}-\frac {\left (349 B -521 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{6 a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5} a}\) | \(332\) |
Input:
int(cos(d*x+c)^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2,x,method =_RETURNVERBOSE)
Output:
1/48*(-163*tan(1/2*d*x+1/2*c)*(4/163*(3*B-7*C)*cos(2*d*x+2*c)+1/163*(-3*B+ 2*C)*cos(3*d*x+3*c)-1/163*C*cos(4*d*x+4*c)+(B-258/163*C)*cos(d*x+c)+140/16 3*B-219/163*C)*sec(1/2*d*x+1/2*c)^2+168*(B-10/7*C)*x*d)/a^2/d
Time = 0.08 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.91 \[ \int \frac {\cos ^3(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\frac {3 \, {\left (7 \, B - 10 \, C\right )} d x \cos \left (d x + c\right )^{2} + 6 \, {\left (7 \, B - 10 \, C\right )} d x \cos \left (d x + c\right ) + 3 \, {\left (7 \, B - 10 \, C\right )} d x + {\left (2 \, C \cos \left (d x + c\right )^{4} + {\left (3 \, B - 2 \, C\right )} \cos \left (d x + c\right )^{3} - 6 \, {\left (B - 2 \, C\right )} \cos \left (d x + c\right )^{2} - {\left (43 \, B - 66 \, C\right )} \cos \left (d x + c\right ) - 32 \, B + 48 \, C\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \] Input:
integrate(cos(d*x+c)^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2,x, algorithm="fricas")
Output:
1/6*(3*(7*B - 10*C)*d*x*cos(d*x + c)^2 + 6*(7*B - 10*C)*d*x*cos(d*x + c) + 3*(7*B - 10*C)*d*x + (2*C*cos(d*x + c)^4 + (3*B - 2*C)*cos(d*x + c)^3 - 6 *(B - 2*C)*cos(d*x + c)^2 - (43*B - 66*C)*cos(d*x + c) - 32*B + 48*C)*sin( d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)
Leaf count of result is larger than twice the leaf count of optimal. 1430 vs. \(2 (155) = 310\).
Time = 3.71 (sec) , antiderivative size = 1430, normalized size of antiderivative = 8.41 \[ \int \frac {\cos ^3(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)**3*(B*cos(d*x+c)+C*cos(d*x+c)**2)/(a+a*cos(d*x+c))**2 ,x)
Output:
Piecewise((21*B*d*x*tan(c/2 + d*x/2)**6/(6*a**2*d*tan(c/2 + d*x/2)**6 + 18 *a**2*d*tan(c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 63*B*d*x*tan(c/2 + d*x/2)**4/(6*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan (c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 63*B*d*x*ta n(c/2 + d*x/2)**2/(6*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/ 2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 21*B*d*x/(6*a**2*d*tan (c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x /2)**2 + 6*a**2*d) + B*tan(c/2 + d*x/2)**9/(6*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 18*B*tan(c/2 + d*x/2)**7/(6*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan( c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 90*B*tan(c/2 + d*x/2)**5/(6*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 110*B*tan(c/2 + d*x/2)**3/( 6*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 18*a**2*d*t an(c/2 + d*x/2)**2 + 6*a**2*d) - 39*B*tan(c/2 + d*x/2)/(6*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 30*C*d*x*tan(c/2 + d*x/2)**6/(6*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d ) - 90*C*d*x*tan(c/2 + d*x/2)**4/(6*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d *tan(c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 90*C...
Leaf count of result is larger than twice the leaf count of optimal. 372 vs. \(2 (160) = 320\).
Time = 0.13 (sec) , antiderivative size = 372, normalized size of antiderivative = 2.19 \[ \int \frac {\cos ^3(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\frac {C {\left (\frac {4 \, {\left (\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{2} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac {\frac {27 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {60 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )} - B {\left (\frac {6 \, {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} + \frac {2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {42 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )}}{6 \, d} \] Input:
integrate(cos(d*x+c)^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2,x, algorithm="maxima")
Output:
1/6*(C*(4*(9*sin(d*x + c)/(cos(d*x + c) + 1) + 20*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/(a^2 + 3*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + (27*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 60*arctan(sin(d*x + c)/ (cos(d*x + c) + 1))/a^2) - B*(6*(3*sin(d*x + c)/(cos(d*x + c) + 1) + 5*sin (d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^2 + 2*a^2*sin(d*x + c)^2/(cos(d*x + c ) + 1)^2 + a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (21*sin(d*x + c)/(co s(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 42*arctan(sin (d*x + c)/(cos(d*x + c) + 1))/a^2))/d
Time = 0.17 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.13 \[ \int \frac {\cos ^3(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\frac {\frac {3 \, {\left (d x + c\right )} {\left (7 \, B - 10 \, C\right )}}{a^{2}} - \frac {2 \, {\left (15 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 30 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 24 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 18 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} a^{2}} + \frac {B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 21 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \] Input:
integrate(cos(d*x+c)^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2,x, algorithm="giac")
Output:
1/6*(3*(d*x + c)*(7*B - 10*C)/a^2 - 2*(15*B*tan(1/2*d*x + 1/2*c)^5 - 30*C* tan(1/2*d*x + 1/2*c)^5 + 24*B*tan(1/2*d*x + 1/2*c)^3 - 40*C*tan(1/2*d*x + 1/2*c)^3 + 9*B*tan(1/2*d*x + 1/2*c) - 18*C*tan(1/2*d*x + 1/2*c))/((tan(1/2 *d*x + 1/2*c)^2 + 1)^3*a^2) + (B*a^4*tan(1/2*d*x + 1/2*c)^3 - C*a^4*tan(1/ 2*d*x + 1/2*c)^3 - 21*B*a^4*tan(1/2*d*x + 1/2*c) + 27*C*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d
Time = 0.29 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.11 \[ \int \frac {\cos ^3(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\frac {x\,\left (7\,B-10\,C\right )}{2\,a^2}-\frac {\left (5\,B-10\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (8\,B-\frac {40\,C}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (3\,B-6\,C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {2\,\left (B-C\right )}{a^2}+\frac {3\,B-5\,C}{2\,a^2}\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (B-C\right )}{6\,a^2\,d} \] Input:
int((cos(c + d*x)^3*(B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + a*cos(c + d* x))^2,x)
Output:
(x*(7*B - 10*C))/(2*a^2) - (tan(c/2 + (d*x)/2)^5*(5*B - 10*C) + tan(c/2 + (d*x)/2)^3*(8*B - (40*C)/3) + tan(c/2 + (d*x)/2)*(3*B - 6*C))/(d*(3*a^2*ta n(c/2 + (d*x)/2)^2 + 3*a^2*tan(c/2 + (d*x)/2)^4 + a^2*tan(c/2 + (d*x)/2)^6 + a^2)) - (tan(c/2 + (d*x)/2)*((2*(B - C))/a^2 + (3*B - 5*C)/(2*a^2)))/d + (tan(c/2 + (d*x)/2)^3*(B - C))/(6*a^2*d)
Time = 0.17 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.19 \[ \int \frac {\cos ^3(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\frac {-2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} c -9 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b +18 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} c +21 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b d x -30 \cos \left (d x +c \right ) \sin \left (d x +c \right ) c d x -2 \cos \left (d x +c \right ) b +2 \cos \left (d x +c \right ) c -3 \sin \left (d x +c \right )^{4} b +4 \sin \left (d x +c \right )^{4} c -31 \sin \left (d x +c \right )^{2} b +46 \sin \left (d x +c \right )^{2} c +21 \sin \left (d x +c \right ) b d x -30 \sin \left (d x +c \right ) c d x +2 b -2 c}{6 \sin \left (d x +c \right ) a^{2} d \left (\cos \left (d x +c \right )+1\right )} \] Input:
int(cos(d*x+c)^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2,x)
Output:
( - 2*cos(c + d*x)*sin(c + d*x)**4*c - 9*cos(c + d*x)*sin(c + d*x)**2*b + 18*cos(c + d*x)*sin(c + d*x)**2*c + 21*cos(c + d*x)*sin(c + d*x)*b*d*x - 3 0*cos(c + d*x)*sin(c + d*x)*c*d*x - 2*cos(c + d*x)*b + 2*cos(c + d*x)*c - 3*sin(c + d*x)**4*b + 4*sin(c + d*x)**4*c - 31*sin(c + d*x)**2*b + 46*sin( c + d*x)**2*c + 21*sin(c + d*x)*b*d*x - 30*sin(c + d*x)*c*d*x + 2*b - 2*c) /(6*sin(c + d*x)*a**2*d*(cos(c + d*x) + 1))