Integrand size = 33, antiderivative size = 163 \[ \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=-\frac {(2 A+5 C) x}{a^2}+\frac {(5 A+12 C) \sin (c+d x)}{a^2 d}-\frac {(2 A+5 C) \cos (c+d x) \sin (c+d x)}{a^2 d}-\frac {2 (2 A+5 C) \cos ^3(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {(A+C) \cos ^4(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac {(5 A+12 C) \sin ^3(c+d x)}{3 a^2 d} \] Output:
-(2*A+5*C)*x/a^2+(5*A+12*C)*sin(d*x+c)/a^2/d-(2*A+5*C)*cos(d*x+c)*sin(d*x+ c)/a^2/d-2/3*(2*A+5*C)*cos(d*x+c)^3*sin(d*x+c)/a^2/d/(1+cos(d*x+c))-1/3*(A +C)*cos(d*x+c)^4*sin(d*x+c)/d/(a+a*cos(d*x+c))^2-1/3*(5*A+12*C)*sin(d*x+c) ^3/a^2/d
Leaf count is larger than twice the leaf count of optimal. \(341\) vs. \(2(163)=326\).
Time = 2.51 (sec) , antiderivative size = 341, normalized size of antiderivative = 2.09 \[ \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (-72 (2 A+5 C) d x \cos \left (\frac {d x}{2}\right )-72 (2 A+5 C) d x \cos \left (c+\frac {d x}{2}\right )-48 A d x \cos \left (c+\frac {3 d x}{2}\right )-120 C d x \cos \left (c+\frac {3 d x}{2}\right )-48 A d x \cos \left (2 c+\frac {3 d x}{2}\right )-120 C d x \cos \left (2 c+\frac {3 d x}{2}\right )+264 A \sin \left (\frac {d x}{2}\right )+516 C \sin \left (\frac {d x}{2}\right )-120 A \sin \left (c+\frac {d x}{2}\right )-156 C \sin \left (c+\frac {d x}{2}\right )+164 A \sin \left (c+\frac {3 d x}{2}\right )+342 C \sin \left (c+\frac {3 d x}{2}\right )+36 A \sin \left (2 c+\frac {3 d x}{2}\right )+118 C \sin \left (2 c+\frac {3 d x}{2}\right )+12 A \sin \left (2 c+\frac {5 d x}{2}\right )+30 C \sin \left (2 c+\frac {5 d x}{2}\right )+12 A \sin \left (3 c+\frac {5 d x}{2}\right )+30 C \sin \left (3 c+\frac {5 d x}{2}\right )-3 C \sin \left (3 c+\frac {7 d x}{2}\right )-3 C \sin \left (4 c+\frac {7 d x}{2}\right )+C \sin \left (4 c+\frac {9 d x}{2}\right )+C \sin \left (5 c+\frac {9 d x}{2}\right )\right )}{48 a^2 d (1+\cos (c+d x))^2} \] Input:
Integrate[(Cos[c + d*x]^3*(A + C*Cos[c + d*x]^2))/(a + a*Cos[c + d*x])^2,x ]
Output:
(Cos[(c + d*x)/2]*Sec[c/2]*(-72*(2*A + 5*C)*d*x*Cos[(d*x)/2] - 72*(2*A + 5 *C)*d*x*Cos[c + (d*x)/2] - 48*A*d*x*Cos[c + (3*d*x)/2] - 120*C*d*x*Cos[c + (3*d*x)/2] - 48*A*d*x*Cos[2*c + (3*d*x)/2] - 120*C*d*x*Cos[2*c + (3*d*x)/ 2] + 264*A*Sin[(d*x)/2] + 516*C*Sin[(d*x)/2] - 120*A*Sin[c + (d*x)/2] - 15 6*C*Sin[c + (d*x)/2] + 164*A*Sin[c + (3*d*x)/2] + 342*C*Sin[c + (3*d*x)/2] + 36*A*Sin[2*c + (3*d*x)/2] + 118*C*Sin[2*c + (3*d*x)/2] + 12*A*Sin[2*c + (5*d*x)/2] + 30*C*Sin[2*c + (5*d*x)/2] + 12*A*Sin[3*c + (5*d*x)/2] + 30*C *Sin[3*c + (5*d*x)/2] - 3*C*Sin[3*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 = 0.84 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.98, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.394, Rules used = {3042, 3521, 25, 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.
| \(\displaystyle \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 3521 |
| \(\displaystyle \frac {\int -\frac {\cos ^3(c+d x) (a (A+4 C)-3 a (A+2 C) \cos (c+d x))}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {(A+C) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int \frac {\cos ^3(c+d x) (a (A+4 C)-3 a (A+2 C) \cos (c+d x))}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {(A+C) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a (A+4 C)-3 a (A+2 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {(A+C) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3456 |
| \(\displaystyle -\frac {\frac {\int 3 \cos ^2(c+d x) \left (2 a^2 (2 A+5 C)-a^2 (5 A+12 C) \cos (c+d x)\right )dx}{a^2}+\frac {2 (2 A+5 C) \sin (c+d x) \cos ^3(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A+C) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\frac {3 \int \cos ^2(c+d x) \left (2 a^2 (2 A+5 C)-a^2 (5 A+12 C) \cos (c+d x)\right )dx}{a^2}+\frac {2 (2 A+5 C) \sin (c+d x) \cos ^3(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A+C) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {3 \int \sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (2 a^2 (2 A+5 C)-a^2 (5 A+12 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}+\frac {2 (2 A+5 C) \sin (c+d x) \cos ^3(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A+C) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle -\frac {\frac {3 \left (2 a^2 (2 A+5 C) \int \cos ^2(c+d x)dx-a^2 (5 A+12 C) \int \cos ^3(c+d x)dx\right )}{a^2}+\frac {2 (2 A+5 C) \sin (c+d x) \cos ^3(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A+C) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {3 \left (2 a^2 (2 A+5 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-a^2 (5 A+12 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^3dx\right )}{a^2}+\frac {2 (2 A+5 C) \sin (c+d x) \cos ^3(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A+C) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3113 |
| \(\displaystyle -\frac {\frac {3 \left (\frac {a^2 (5 A+12 C) \int \left (1-\sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}+2 a^2 (2 A+5 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx\right )}{a^2}+\frac {2 (2 A+5 C) \sin (c+d x) \cos ^3(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A+C) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\frac {3 \left (2 a^2 (2 A+5 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {a^2 (5 A+12 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )}{a^2}+\frac {2 (2 A+5 C) \sin (c+d x) \cos ^3(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A+C) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle -\frac {\frac {3 \left (2 a^2 (2 A+5 C) \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {a^2 (5 A+12 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )}{a^2}+\frac {2 (2 A+5 C) \sin (c+d x) \cos ^3(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A+C) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle -\frac {\frac {3 \left (\frac {a^2 (5 A+12 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+2 a^2 (2 A+5 C) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )}{a^2}+\frac {2 (2 A+5 C) \sin (c+d x) \cos ^3(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A+C) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
Input:
Int[(Cos[c + d*x]^3*(A + C*Cos[c + d*x]^2))/(a + a*Cos[c + d*x])^2,x]
Output:
-1/3*((A + C)*Cos[c + d*x]^4*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^2) - (( 2*(2*A + 5*C)*Cos[c + d*x]^3*Sin[c + d*x])/(d*(1 + Cos[c + d*x])) + (3*(2* a^2*(2*A + 5*C)*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)) + (a^2*(5*A + 12 *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_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[a*(A + C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(f*(b*c - a*d)*(2*m + 1))), x] + Simp[1/(b*(b*c - a*d)*(2*m + 1)) I nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(m + n + 2) + C*(b* c*(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)]
Time = 0.59 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.61
| method | result | size |
| parallelrisch | \(\frac {7 \left (\frac {\left (\frac {3 A}{7}+C \right ) \cos \left (2 d x +2 c \right )}{4}-\frac {\cos \left (3 d x +3 c \right ) C}{56}+\frac {C \cos \left (4 d x +4 c \right )}{112}+\left (A +\frac {129 C}{56}\right ) \cos \left (d x +c \right )+\frac {23 A}{28}+\frac {219 C}{112}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-6 x \left (A +\frac {5 C}{2}\right ) d}{3 a^{2} d}\) | \(100\) |
| derivativedivides | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+5 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+9 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {4 \left (\left (-A -5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-2 A -\frac {20 C}{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-A -3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}-4 \left (2 A +5 C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\) | \(154\) |
| default | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+5 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+9 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {4 \left (\left (-A -5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-2 A -\frac {20 C}{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-A -3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}-4 \left (2 A +5 C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\) | \(154\) |
| risch | \(-\frac {2 x A}{a^{2}}-\frac {5 C x}{a^{2}}+\frac {i {\mathrm e}^{2 i \left (d x +c \right )} C}{4 a^{2} d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} A}{2 a^{2} d}-\frac {15 i {\mathrm e}^{i \left (d x +c \right )} C}{8 a^{2} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} A}{2 a^{2} d}+\frac {15 i {\mathrm e}^{-i \left (d x +c \right )} C}{8 a^{2} d}-\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} C}{4 a^{2} d}+\frac {2 i \left (9 A \,{\mathrm e}^{2 i \left (d x +c \right )}+15 C \,{\mathrm e}^{2 i \left (d x +c \right )}+15 A \,{\mathrm e}^{i \left (d x +c \right )}+27 C \,{\mathrm e}^{i \left (d x +c \right )}+8 A +14 C \right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}+\frac {C \sin \left (3 d x +3 c \right )}{12 a^{2} d}\) | \(225\) |
| norman | \(\frac {-\frac {\left (2 A +5 C \right ) x}{a}-\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{6 a d}-\frac {5 \left (2 A +5 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a}-\frac {10 \left (2 A +5 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{a}-\frac {10 \left (2 A +5 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{a}-\frac {5 \left (2 A +5 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{a}-\frac {\left (2 A +5 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{a}+\frac {3 \left (3 A +7 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}+\frac {\left (5 A +11 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 a d}+\frac {2 \left (47 A +115 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 a d}+\frac {\left (61 A +143 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 a d}+\frac {\left (77 A +185 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{6 a d}+\frac {\left (217 A +521 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{6 a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5} a}\) | \(330\) |
Input:
int(cos(d*x+c)^3*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2,x,method=_RETURNVER BOSE)
Output:
1/3*(7*(1/4*(3/7*A+C)*cos(2*d*x+2*c)-1/56*cos(3*d*x+3*c)*C+1/112*C*cos(4*d *x+4*c)+(A+129/56*C)*cos(d*x+c)+23/28*A+219/112*C)*tan(1/2*d*x+1/2*c)*sec( 1/2*d*x+1/2*c)^2-6*x*(A+5/2*C)*d)/a^2/d
Time = 0.12 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.91 \[ \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=-\frac {3 \, {\left (2 \, A + 5 \, C\right )} d x \cos \left (d x + c\right )^{2} + 6 \, {\left (2 \, A + 5 \, C\right )} d x \cos \left (d x + c\right ) + 3 \, {\left (2 \, A + 5 \, C\right )} d x - {\left (C \cos \left (d x + c\right )^{4} - C \cos \left (d x + c\right )^{3} + 3 \, {\left (A + 2 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (14 \, A + 33 \, C\right )} \cos \left (d x + c\right ) + 10 \, A + 24 \, C\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \] Input:
integrate(cos(d*x+c)^3*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2,x, algorithm= "fricas")
Output:
-1/3*(3*(2*A + 5*C)*d*x*cos(d*x + c)^2 + 6*(2*A + 5*C)*d*x*cos(d*x + c) + 3*(2*A + 5*C)*d*x - (C*cos(d*x + c)^4 - C*cos(d*x + c)^3 + 3*(A + 2*C)*cos (d*x + c)^2 + (14*A + 33*C)*cos(d*x + c) + 10*A + 24*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. 1426 vs. \(2 (153) = 306\).
Time = 2.97 (sec) , antiderivative size = 1426, normalized size of antiderivative = 8.75 \[ \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)**3*(A+C*cos(d*x+c)**2)/(a+a*cos(d*x+c))**2,x)
Output:
Piecewise((-12*A*d*x*tan(c/2 + d*x/2)**6/(6*a**2*d*tan(c/2 + d*x/2)**6 + 1 8*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) - 36*A*d*x*tan(c/2 + d*x/2)**4/(6*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*ta n(c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 36*A*d*x*t an(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) - 12*A*d*x/(6*a**2*d*ta n(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) - A*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 ) + 12*A*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) + 54*A*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) + 68*A*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) + 27*A*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. 325 vs. \(2 (157) = 314\).
Time = 0.13 (sec) , antiderivative size = 325, normalized size of antiderivative = 1.99 \[ \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\frac {C {\left (\frac {4 \, {\left (\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{2} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac {\frac {27 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {60 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )} + A {\left (\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {24 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {12 \, \sin \left (d x + c\right )}{{\left (a^{2} + \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{6 \, d} \] Input:
integrate(cos(d*x+c)^3*(A+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) + A*((15*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d* x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 24*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 + 12*sin(d*x + c)/((a^2 + a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^ 2)*(cos(d*x + c) + 1))))/d
Time = 0.34 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.17 \[ \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=-\frac {\frac {6 \, {\left (d x + c\right )} {\left (2 \, A + 5 \, C\right )}}{a^{2}} - \frac {4 \, {\left (3 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 20 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} a^{2}} + \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 27 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \] Input:
integrate(cos(d*x+c)^3*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2,x, algorithm= "giac")
Output:
-1/6*(6*(d*x + c)*(2*A + 5*C)/a^2 - 4*(3*A*tan(1/2*d*x + 1/2*c)^5 + 15*C*t an(1/2*d*x + 1/2*c)^5 + 6*A*tan(1/2*d*x + 1/2*c)^3 + 20*C*tan(1/2*d*x + 1/ 2*c)^3 + 3*A*tan(1/2*d*x + 1/2*c) + 9*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d* x + 1/2*c)^2 + 1)^3*a^2) + (A*a^4*tan(1/2*d*x + 1/2*c)^3 + C*a^4*tan(1/2*d *x + 1/2*c)^3 - 15*A*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 = 0.19 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.11 \[ \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\frac {\left (2\,A+10\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (4\,A+\frac {40\,C}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A+6\,C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )}-\frac {x\,\left (2\,A+5\,C\right )}{a^2}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {2\,\left (A+C\right )}{a^2}+\frac {A+5\,C}{2\,a^2}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A+C\right )}{6\,a^2\,d} \] Input:
int((cos(c + d*x)^3*(A + C*cos(c + d*x)^2))/(a + a*cos(c + d*x))^2,x)
Output:
(tan(c/2 + (d*x)/2)^5*(2*A + 10*C) + tan(c/2 + (d*x)/2)^3*(4*A + (40*C)/3) + tan(c/2 + (d*x)/2)*(2*A + 6*C))/(d*(3*a^2*tan(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)) - (x*(2*A + 5*C))/ a^2 + (tan(c/2 + (d*x)/2)*((2*(A + C))/a^2 + (A + 5*C)/(2*a^2)))/d - (tan( c/2 + (d*x)/2)^3*(A + C))/(6*a^2*d)
Time = 0.17 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.16 \[ \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\frac {-\cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} c +3 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a +9 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} c -6 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a d x -15 \cos \left (d x +c \right ) \sin \left (d x +c \right ) c d x +\cos \left (d x +c \right ) a +\cos \left (d x +c \right ) c +2 \sin \left (d x +c \right )^{4} c +11 \sin \left (d x +c \right )^{2} a +23 \sin \left (d x +c \right )^{2} c -6 \sin \left (d x +c \right ) a d x -15 \sin \left (d x +c \right ) c d x -a -c}{3 \sin \left (d x +c \right ) a^{2} d \left (\cos \left (d x +c \right )+1\right )} \] Input:
int(cos(d*x+c)^3*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2,x)
Output:
( - cos(c + d*x)*sin(c + d*x)**4*c + 3*cos(c + d*x)*sin(c + d*x)**2*a + 9* cos(c + d*x)*sin(c + d*x)**2*c - 6*cos(c + d*x)*sin(c + d*x)*a*d*x - 15*co s(c + d*x)*sin(c + d*x)*c*d*x + cos(c + d*x)*a + cos(c + d*x)*c + 2*sin(c + d*x)**4*c + 11*sin(c + d*x)**2*a + 23*sin(c + d*x)**2*c - 6*sin(c + d*x) *a*d*x - 15*sin(c + d*x)*c*d*x - a - c)/(3*sin(c + d*x)*a**2*d*(cos(c + d* x) + 1))