Integrand size = 33, antiderivative size = 189 \[ \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\frac {(2 A+13 C) x}{2 a^3}-\frac {2 (11 A+76 C) \sin (c+d x)}{15 a^3 d}+\frac {(2 A+13 C) \cos (c+d x) \sin (c+d x)}{2 a^3 d}-\frac {(A+C) \cos ^4(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(A+11 C) \cos ^3(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(11 A+76 C) \cos ^2(c+d x) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )} \] Output:
1/2*(2*A+13*C)*x/a^3-2/15*(11*A+76*C)*sin(d*x+c)/a^3/d+1/2*(2*A+13*C)*cos( d*x+c)*sin(d*x+c)/a^3/d-1/5*(A+C)*cos(d*x+c)^4*sin(d*x+c)/d/(a+a*cos(d*x+c ))^3-1/15*(A+11*C)*cos(d*x+c)^3*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^2-1/15*(11 *A+76*C)*cos(d*x+c)^2*sin(d*x+c)/d/(a^3+a^3*cos(d*x+c))
Leaf count is larger than twice the leaf count of optimal. \(393\) vs. \(2(189)=378\).
Time = 2.88 (sec) , antiderivative size = 393, normalized size of antiderivative = 2.08 \[ \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (600 (2 A+13 C) d x \cos \left (\frac {d x}{2}\right )+600 (2 A+13 C) d x \cos \left (c+\frac {d x}{2}\right )+600 A d x \cos \left (c+\frac {3 d x}{2}\right )+3900 C d x \cos \left (c+\frac {3 d x}{2}\right )+600 A d x \cos \left (2 c+\frac {3 d x}{2}\right )+3900 C d x \cos \left (2 c+\frac {3 d x}{2}\right )+120 A d x \cos \left (2 c+\frac {5 d x}{2}\right )+780 C d x \cos \left (2 c+\frac {5 d x}{2}\right )+120 A d x \cos \left (3 c+\frac {5 d x}{2}\right )+780 C d x \cos \left (3 c+\frac {5 d x}{2}\right )-2960 A \sin \left (\frac {d x}{2}\right )-12760 C \sin \left (\frac {d x}{2}\right )+2160 A \sin \left (c+\frac {d x}{2}\right )+7560 C \sin \left (c+\frac {d x}{2}\right )-1840 A \sin \left (c+\frac {3 d x}{2}\right )-9230 C \sin \left (c+\frac {3 d x}{2}\right )+720 A \sin \left (2 c+\frac {3 d x}{2}\right )+930 C \sin \left (2 c+\frac {3 d x}{2}\right )-512 A \sin \left (2 c+\frac {5 d x}{2}\right )-2782 C \sin \left (2 c+\frac {5 d x}{2}\right )-750 C \sin \left (3 c+\frac {5 d x}{2}\right )-105 C \sin \left (3 c+\frac {7 d x}{2}\right )-105 C \sin \left (4 c+\frac {7 d x}{2}\right )+15 C \sin \left (4 c+\frac {9 d x}{2}\right )+15 C \sin \left (5 c+\frac {9 d x}{2}\right )\right )}{480 a^3 d (1+\cos (c+d x))^3} \] Input:
Integrate[(Cos[c + d*x]^3*(A + C*Cos[c + d*x]^2))/(a + a*Cos[c + d*x])^3,x ]
Output:
(Cos[(c + d*x)/2]*Sec[c/2]*(600*(2*A + 13*C)*d*x*Cos[(d*x)/2] + 600*(2*A + 13*C)*d*x*Cos[c + (d*x)/2] + 600*A*d*x*Cos[c + (3*d*x)/2] + 3900*C*d*x*Co s[c + (3*d*x)/2] + 600*A*d*x*Cos[2*c + (3*d*x)/2] + 3900*C*d*x*Cos[2*c + ( 3*d*x)/2] + 120*A*d*x*Cos[2*c + (5*d*x)/2] + 780*C*d*x*Cos[2*c + (5*d*x)/2 ] + 120*A*d*x*Cos[3*c + (5*d*x)/2] + 780*C*d*x*Cos[3*c + (5*d*x)/2] - 2960 *A*Sin[(d*x)/2] - 12760*C*Sin[(d*x)/2] + 2160*A*Sin[c + (d*x)/2] + 7560*C* Sin[c + (d*x)/2] - 1840*A*Sin[c + (3*d*x)/2] - 9230*C*Sin[c + (3*d*x)/2] + 720*A*Sin[2*c + (3*d*x)/2] + 930*C*Sin[2*c + (3*d*x)/2] - 512*A*Sin[2*c + (5*d*x)/2] - 2782*C*Sin[2*c + (5*d*x)/2] - 750*C*Sin[3*c + (5*d*x)/2] - 1 05*C*Sin[3*c + (7*d*x)/2] - 105*C*Sin[4*c + (7*d*x)/2] + 15*C*Sin[4*c + (9 *d*x)/2] + 15*C*Sin[5*c + (9*d*x)/2]))/(480*a^3*d*(1 + Cos[c + d*x])^3)
Time = 0.95 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3042, 3521, 3042, 3456, 25, 3042, 3456, 3042, 3213}
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.
| \(\displaystyle \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 3521 |
| \(\displaystyle \frac {\int \frac {\cos ^3(c+d x) (a (A-4 C)+a (2 A+7 C) \cos (c+d x))}{(\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A+C) \sin (c+d x) \cos ^4(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a (A-4 C)+a (2 A+7 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {(A+C) \sin (c+d x) \cos ^4(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3456 |
| \(\displaystyle \frac {\frac {\int -\frac {\cos ^2(c+d x) \left (3 a^2 (A+11 C)-a^2 (8 A+43 C) \cos (c+d x)\right )}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {a (A+11 C) \sin (c+d x) \cos ^3(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \sin (c+d x) \cos ^4(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\frac {\int \frac {\cos ^2(c+d x) \left (3 a^2 (A+11 C)-a^2 (8 A+43 C) \cos (c+d x)\right )}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {a (A+11 C) \sin (c+d x) \cos ^3(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \sin (c+d x) \cos ^4(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (3 a^2 (A+11 C)-a^2 (8 A+43 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {a (A+11 C) \sin (c+d x) \cos ^3(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \sin (c+d x) \cos ^4(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3456 |
| \(\displaystyle \frac {-\frac {\frac {\int \cos (c+d x) \left (2 a^3 (11 A+76 C)-15 a^3 (2 A+13 C) \cos (c+d x)\right )dx}{a^2}+\frac {a^2 (11 A+76 C) \sin (c+d x) \cos ^2(c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (A+11 C) \sin (c+d x) \cos ^3(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \sin (c+d x) \cos ^4(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\frac {\int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (2 a^3 (11 A+76 C)-15 a^3 (2 A+13 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}+\frac {a^2 (11 A+76 C) \sin (c+d x) \cos ^2(c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (A+11 C) \sin (c+d x) \cos ^3(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \sin (c+d x) \cos ^4(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3213 |
| \(\displaystyle \frac {-\frac {\frac {a^2 (11 A+76 C) \sin (c+d x) \cos ^2(c+d x)}{d (a \cos (c+d x)+a)}+\frac {\frac {2 a^3 (11 A+76 C) \sin (c+d x)}{d}-\frac {15 a^3 (2 A+13 C) \sin (c+d x) \cos (c+d x)}{2 d}-\frac {15}{2} a^3 x (2 A+13 C)}{a^2}}{3 a^2}-\frac {a (A+11 C) \sin (c+d x) \cos ^3(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \sin (c+d x) \cos ^4(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
Input:
Int[(Cos[c + d*x]^3*(A + C*Cos[c + d*x]^2))/(a + a*Cos[c + d*x])^3,x]
Output:
-1/5*((A + C)*Cos[c + d*x]^4*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^3) + (- 1/3*(a*(A + 11*C)*Cos[c + d*x]^3*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^2) - ((a^2*(11*A + 76*C)*Cos[c + d*x]^2*Sin[c + d*x])/(d*(a + a*Cos[c + d*x]) ) + ((-15*a^3*(2*A + 13*C)*x)/2 + (2*a^3*(11*A + 76*C)*Sin[c + d*x])/d - ( 15*a^3*(2*A + 13*C)*Cos[c + d*x]*Sin[c + d*x])/(2*d))/a^2)/(3*a^2))/(5*a^2 )
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.) *(x_)]), x_Symbol] :> Simp[(2*a*c + b*d)*(x/2), x] + (-Simp[(b*c + a*d)*(Co s[e + f*x]/f), x] - Simp[b*d*Cos[e + f*x]*(Sin[e + f*x]/(2*f)), x]) /; Free Q[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
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 = 0.44 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.54
| method | result | size |
| parallelrisch | \(\frac {-17 \left (\frac {4 \left (4 A +29 C \right ) \cos \left (2 d x +2 c \right )}{51}+\frac {15 \cos \left (3 d x +3 c \right ) C}{136}-\frac {5 C \cos \left (4 d x +4 c \right )}{272}+\left (A +\frac {1001 C}{136}\right ) \cos \left (d x +c \right )+\frac {38 A}{51}+\frac {4303 C}{816}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+20 x \left (A +\frac {13 C}{2}\right ) d}{20 a^{3} d}\) | \(102\) |
| derivativedivides | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}+\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}-7 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-31 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {-28 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C -20 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+4 \left (2 A +13 C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\) | \(153\) |
| default | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}+\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}-7 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-31 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {-28 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C -20 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+4 \left (2 A +13 C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\) | \(153\) |
| risch | \(\frac {x A}{a^{3}}+\frac {13 C x}{2 a^{3}}-\frac {i C \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 a^{3} d}+\frac {3 i {\mathrm e}^{i \left (d x +c \right )} C}{2 a^{3} d}-\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} C}{2 a^{3} d}+\frac {i C \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 a^{3} d}-\frac {2 i \left (45 A \,{\mathrm e}^{4 i \left (d x +c \right )}+150 C \,{\mathrm e}^{4 i \left (d x +c \right )}+135 A \,{\mathrm e}^{3 i \left (d x +c \right )}+525 C \,{\mathrm e}^{3 i \left (d x +c \right )}+185 A \,{\mathrm e}^{2 i \left (d x +c \right )}+745 C \,{\mathrm e}^{2 i \left (d x +c \right )}+115 A \,{\mathrm e}^{i \left (d x +c \right )}+485 C \,{\mathrm e}^{i \left (d x +c \right )}+32 A +127 C \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\) | \(216\) |
| norman | \(\frac {\frac {\left (2 A +13 C \right ) x}{2 a}-\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{20 a d}+\frac {\left (A +5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{12 a d}+\frac {5 \left (2 A +13 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 a}+\frac {5 \left (2 A +13 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{a}+\frac {5 \left (2 A +13 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{a}+\frac {5 \left (2 A +13 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{2 a}+\frac {\left (2 A +13 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{2 a}-\frac {\left (7 A +51 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}-\frac {\left (7 A +59 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{12 a d}-\frac {\left (71 A +475 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{12 a d}-\frac {\left (101 A +721 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 a d}-\frac {\left (173 A +1165 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{12 a d}-\frac {\left (953 A +6613 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{60 a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5} a^{2}}\) | \(354\) |
Input:
int(cos(d*x+c)^3*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x,method=_RETURNVER BOSE)
Output:
1/20*(-17*(4/51*(4*A+29*C)*cos(2*d*x+2*c)+15/136*cos(3*d*x+3*c)*C-5/272*C* cos(4*d*x+4*c)+(A+1001/136*C)*cos(d*x+c)+38/51*A+4303/816*C)*tan(1/2*d*x+1 /2*c)*sec(1/2*d*x+1/2*c)^4+20*x*(A+13/2*C)*d)/a^3/d
Time = 0.08 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.97 \[ \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\frac {15 \, {\left (2 \, A + 13 \, C\right )} d x \cos \left (d x + c\right )^{3} + 45 \, {\left (2 \, A + 13 \, C\right )} d x \cos \left (d x + c\right )^{2} + 45 \, {\left (2 \, A + 13 \, C\right )} d x \cos \left (d x + c\right ) + 15 \, {\left (2 \, A + 13 \, C\right )} d x + {\left (15 \, C \cos \left (d x + c\right )^{4} - 45 \, C \cos \left (d x + c\right )^{3} - {\left (64 \, A + 479 \, C\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (34 \, A + 239 \, C\right )} \cos \left (d x + c\right ) - 44 \, A - 304 \, C\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \] Input:
integrate(cos(d*x+c)^3*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x, algorithm= "fricas")
Output:
1/30*(15*(2*A + 13*C)*d*x*cos(d*x + c)^3 + 45*(2*A + 13*C)*d*x*cos(d*x + c )^2 + 45*(2*A + 13*C)*d*x*cos(d*x + c) + 15*(2*A + 13*C)*d*x + (15*C*cos(d *x + c)^4 - 45*C*cos(d*x + c)^3 - (64*A + 479*C)*cos(d*x + c)^2 - 3*(34*A + 239*C)*cos(d*x + c) - 44*A - 304*C)*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)
Leaf count of result is larger than twice the leaf count of optimal. 967 vs. \(2 (173) = 346\).
Time = 3.94 (sec) , antiderivative size = 967, normalized size of antiderivative = 5.12 \[ \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)**3*(A+C*cos(d*x+c)**2)/(a+a*cos(d*x+c))**3,x)
Output:
Piecewise((60*A*d*x*tan(c/2 + d*x/2)**4/(60*a**3*d*tan(c/2 + d*x/2)**4 + 1 20*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 120*A*d*x*tan(c/2 + d*x/2)**2 /(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3 *d) + 60*A*d*x/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2 )**2 + 60*a**3*d) - 3*A*tan(c/2 + d*x/2)**9/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 14*A*tan(c/2 + d*x/2)**7/ (60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3* d) - 68*A*tan(c/2 + d*x/2)**5/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d* tan(c/2 + d*x/2)**2 + 60*a**3*d) - 190*A*tan(c/2 + d*x/2)**3/(60*a**3*d*ta n(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 105*A*ta n(c/2 + d*x/2)/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2 )**2 + 60*a**3*d) + 390*C*d*x*tan(c/2 + d*x/2)**4/(60*a**3*d*tan(c/2 + d*x /2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 780*C*d*x*tan(c/2 + d*x/2)**2/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 390*C*d*x/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan( c/2 + d*x/2)**2 + 60*a**3*d) - 3*C*tan(c/2 + d*x/2)**9/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 34*C*tan(c/2 + d*x/2)**7/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 388*C*tan(c/2 + d*x/2)**5/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 1310*C*tan(c/2 + d*x/2)*...
Time = 0.16 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.46 \[ \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=-\frac {C {\left (\frac {60 \, {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{3} + \frac {2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {465 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {40 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {780 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} + A {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {120 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )}}{60 \, d} \] Input:
integrate(cos(d*x+c)^3*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x, algorithm= "maxima")
Output:
-1/60*(C*(60*(5*sin(d*x + c)/(cos(d*x + c) + 1) + 7*sin(d*x + c)^3/(cos(d* x + c) + 1)^3)/(a^3 + 2*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a^3*sin( d*x + c)^4/(cos(d*x + c) + 1)^4) + (465*sin(d*x + c)/(cos(d*x + c) + 1) - 40*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 780*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3) + A*((105*sin (d*x + c)/(cos(d*x + c) + 1) - 20*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3* sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 120*arctan(sin(d*x + c)/(cos(d* x + c) + 1))/a^3))/d
Time = 0.32 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.92 \[ \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\frac {\frac {30 \, {\left (d x + c\right )} {\left (2 \, A + 13 \, C\right )}}{a^{3}} - \frac {60 \, {\left (7 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 20 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 465 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \] Input:
integrate(cos(d*x+c)^3*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x, algorithm= "giac")
Output:
1/60*(30*(d*x + c)*(2*A + 13*C)/a^3 - 60*(7*C*tan(1/2*d*x + 1/2*c)^3 + 5*C *tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*a^3) - (3*A*a^12*ta n(1/2*d*x + 1/2*c)^5 + 3*C*a^12*tan(1/2*d*x + 1/2*c)^5 - 20*A*a^12*tan(1/2 *d*x + 1/2*c)^3 - 40*C*a^12*tan(1/2*d*x + 1/2*c)^3 + 105*A*a^12*tan(1/2*d* x + 1/2*c) + 465*C*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d
Time = 0.16 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.97 \[ \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A+C}{4\,a^3}+\frac {A+5\,C}{12\,a^3}\right )}{d}+\frac {x\,\left (2\,A+13\,C\right )}{2\,a^3}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A+C\right )}{2\,a^3}+\frac {3\,\left (A+5\,C\right )}{4\,a^3}-\frac {2\,A-10\,C}{4\,a^3}\right )}{d}-\frac {7\,C\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+5\,C\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A+C\right )}{20\,a^3\,d} \] Input:
int((cos(c + d*x)^3*(A + C*cos(c + d*x)^2))/(a + a*cos(c + d*x))^3,x)
Output:
(tan(c/2 + (d*x)/2)^3*((A + C)/(4*a^3) + (A + 5*C)/(12*a^3)))/d + (x*(2*A + 13*C))/(2*a^3) - (tan(c/2 + (d*x)/2)*((3*(A + C))/(2*a^3) + (3*(A + 5*C) )/(4*a^3) - (2*A - 10*C)/(4*a^3)))/d - (5*C*tan(c/2 + (d*x)/2) + 7*C*tan(c /2 + (d*x)/2)^3)/(d*(2*a^3*tan(c/2 + (d*x)/2)^2 + a^3*tan(c/2 + (d*x)/2)^4 + a^3)) - (tan(c/2 + (d*x)/2)^5*(A + C))/(20*a^3*d)
Time = 0.16 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.31 \[ \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\frac {-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9} a -3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9} c +14 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} a +34 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} c -68 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} a -388 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} c +60 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a d x +390 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} c d x -190 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a -1310 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} c +120 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a d x +780 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} c d x -105 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -765 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) c +60 a d x +390 c d x}{60 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )} \] Input:
int(cos(d*x+c)^3*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x)
Output:
( - 3*tan((c + d*x)/2)**9*a - 3*tan((c + d*x)/2)**9*c + 14*tan((c + d*x)/2 )**7*a + 34*tan((c + d*x)/2)**7*c - 68*tan((c + d*x)/2)**5*a - 388*tan((c + d*x)/2)**5*c + 60*tan((c + d*x)/2)**4*a*d*x + 390*tan((c + d*x)/2)**4*c* d*x - 190*tan((c + d*x)/2)**3*a - 1310*tan((c + d*x)/2)**3*c + 120*tan((c + d*x)/2)**2*a*d*x + 780*tan((c + d*x)/2)**2*c*d*x - 105*tan((c + d*x)/2)* a - 765*tan((c + d*x)/2)*c + 60*a*d*x + 390*c*d*x)/(60*a**3*d*(tan((c + d* x)/2)**4 + 2*tan((c + d*x)/2)**2 + 1))