Integrand size = 41, antiderivative size = 152 \[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\frac {(B-3 C) x}{a^3}+\frac {(2 A-7 B+27 C) \sin (c+d x)}{15 a^3 d}-\frac {(A-B+C) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(A+4 B-9 C) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(B-3 C) \sin (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )} \]
(B-3*C)*x/a^3+1/15*(2*A-7*B+27*C)*sin(d*x+c)/a^3/d-1/5*(A-B+C)*cos(d*x+c)^ 3*sin(d*x+c)/d/(a+a*cos(d*x+c))^3+1/15*(A+4*B-9*C)*cos(d*x+c)^2*sin(d*x+c) /a/d/(a+a*cos(d*x+c))^2-(B-3*C)*sin(d*x+c)/d/(a^3+a^3*cos(d*x+c))
Leaf count is larger than twice the leaf count of optimal. \(423\) vs. \(2(152)=304\).
Time = 3.02 (sec) , antiderivative size = 423, normalized size of antiderivative = 2.78 \[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (300 (B-3 C) d x \cos \left (\frac {d x}{2}\right )+300 (B-3 C) d x \cos \left (c+\frac {d x}{2}\right )+150 B d x \cos \left (c+\frac {3 d x}{2}\right )-450 C d x \cos \left (c+\frac {3 d x}{2}\right )+150 B d x \cos \left (2 c+\frac {3 d x}{2}\right )-450 C d x \cos \left (2 c+\frac {3 d x}{2}\right )+30 B d x \cos \left (2 c+\frac {5 d x}{2}\right )-90 C d x \cos \left (2 c+\frac {5 d x}{2}\right )+30 B d x \cos \left (3 c+\frac {5 d x}{2}\right )-90 C d x \cos \left (3 c+\frac {5 d x}{2}\right )+160 A \sin \left (\frac {d x}{2}\right )-740 B \sin \left (\frac {d x}{2}\right )+1755 C \sin \left (\frac {d x}{2}\right )-120 A \sin \left (c+\frac {d x}{2}\right )+540 B \sin \left (c+\frac {d x}{2}\right )-1125 C \sin \left (c+\frac {d x}{2}\right )+80 A \sin \left (c+\frac {3 d x}{2}\right )-460 B \sin \left (c+\frac {3 d x}{2}\right )+1215 C \sin \left (c+\frac {3 d x}{2}\right )-60 A \sin \left (2 c+\frac {3 d x}{2}\right )+180 B \sin \left (2 c+\frac {3 d x}{2}\right )-225 C \sin \left (2 c+\frac {3 d x}{2}\right )+28 A \sin \left (2 c+\frac {5 d x}{2}\right )-128 B \sin \left (2 c+\frac {5 d x}{2}\right )+363 C \sin \left (2 c+\frac {5 d x}{2}\right )+75 C \sin \left (3 c+\frac {5 d x}{2}\right )+15 C \sin \left (3 c+\frac {7 d x}{2}\right )+15 C \sin \left (4 c+\frac {7 d x}{2}\right )\right )}{120 a^3 d (1+\cos (c+d x))^3} \]
(Cos[(c + d*x)/2]*Sec[c/2]*(300*(B - 3*C)*d*x*Cos[(d*x)/2] + 300*(B - 3*C) *d*x*Cos[c + (d*x)/2] + 150*B*d*x*Cos[c + (3*d*x)/2] - 450*C*d*x*Cos[c + ( 3*d*x)/2] + 150*B*d*x*Cos[2*c + (3*d*x)/2] - 450*C*d*x*Cos[2*c + (3*d*x)/2 ] + 30*B*d*x*Cos[2*c + (5*d*x)/2] - 90*C*d*x*Cos[2*c + (5*d*x)/2] + 30*B*d *x*Cos[3*c + (5*d*x)/2] - 90*C*d*x*Cos[3*c + (5*d*x)/2] + 160*A*Sin[(d*x)/ 2] - 740*B*Sin[(d*x)/2] + 1755*C*Sin[(d*x)/2] - 120*A*Sin[c + (d*x)/2] + 5 40*B*Sin[c + (d*x)/2] - 1125*C*Sin[c + (d*x)/2] + 80*A*Sin[c + (3*d*x)/2] - 460*B*Sin[c + (3*d*x)/2] + 1215*C*Sin[c + (3*d*x)/2] - 60*A*Sin[2*c + (3 *d*x)/2] + 180*B*Sin[2*c + (3*d*x)/2] - 225*C*Sin[2*c + (3*d*x)/2] + 28*A* Sin[2*c + (5*d*x)/2] - 128*B*Sin[2*c + (5*d*x)/2] + 363*C*Sin[2*c + (5*d*x )/2] + 75*C*Sin[3*c + (5*d*x)/2] + 15*C*Sin[3*c + (7*d*x)/2] + 15*C*Sin[4* c + (7*d*x)/2]))/(120*a^3*d*(1 + Cos[c + d*x])^3)
Time = 1.16 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.04, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.317, Rules used = {3042, 3520, 3042, 3456, 3042, 3447, 3042, 3502, 27, 3042, 3214, 3042, 3127}
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.
| \(\displaystyle \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 3520 |
| \(\displaystyle \frac {\int \frac {\cos ^2(c+d x) (a (2 A+3 B-3 C)+a (A-B+6 C) \cos (c+d x))}{(\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a (2 A+3 B-3 C)+a (A-B+6 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3456 |
| \(\displaystyle \frac {\frac {\int \frac {\cos (c+d x) \left (2 (A+4 B-9 C) a^2+(2 A-7 B+27 C) \cos (c+d x) a^2\right )}{\cos (c+d x) a+a}dx}{3 a^2}+\frac {a (A+4 B-9 C) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \left (2 (A+4 B-9 C) a^2+(2 A-7 B+27 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}+\frac {a (A+4 B-9 C) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {\frac {\int \frac {(2 A-7 B+27 C) \cos ^2(c+d x) a^2+2 (A+4 B-9 C) \cos (c+d x) a^2}{\cos (c+d x) a+a}dx}{3 a^2}+\frac {a (A+4 B-9 C) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {(2 A-7 B+27 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^2+2 (A+4 B-9 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}+\frac {a (A+4 B-9 C) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {15 a^3 (B-3 C) \cos (c+d x)}{\cos (c+d x) a+a}dx}{a}+\frac {a (2 A-7 B+27 C) \sin (c+d x)}{d}}{3 a^2}+\frac {a (A+4 B-9 C) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {15 a^2 (B-3 C) \int \frac {\cos (c+d x)}{\cos (c+d x) a+a}dx+\frac {a (2 A-7 B+27 C) \sin (c+d x)}{d}}{3 a^2}+\frac {a (A+4 B-9 C) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {15 a^2 (B-3 C) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a (2 A-7 B+27 C) \sin (c+d x)}{d}}{3 a^2}+\frac {a (A+4 B-9 C) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {\frac {15 a^2 (B-3 C) \left (\frac {x}{a}-\int \frac {1}{\cos (c+d x) a+a}dx\right )+\frac {a (2 A-7 B+27 C) \sin (c+d x)}{d}}{3 a^2}+\frac {a (A+4 B-9 C) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {15 a^2 (B-3 C) \left (\frac {x}{a}-\int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx\right )+\frac {a (2 A-7 B+27 C) \sin (c+d x)}{d}}{3 a^2}+\frac {a (A+4 B-9 C) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3127 |
| \(\displaystyle \frac {\frac {15 a^2 (B-3 C) \left (\frac {x}{a}-\frac {\sin (c+d x)}{d (a \cos (c+d x)+a)}\right )+\frac {a (2 A-7 B+27 C) \sin (c+d x)}{d}}{3 a^2}+\frac {a (A+4 B-9 C) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
-1/5*((A - B + C)*Cos[c + d*x]^3*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^3) + ((a*(A + 4*B - 9*C)*Cos[c + d*x]^2*Sin[c + d*x])/(3*d*(a + a*Cos[c + d*x ])^2) + ((a*(2*A - 7*B + 27*C)*Sin[c + d*x])/d + 15*a^2*(B - 3*C)*(x/a - S in[c + d*x]/(d*(a + a*Cos[c + d*x]))))/(3*a^2))/(5*a^2)
3.4.57.3.1 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[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b ^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{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_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, 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_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[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*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
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[(a*A - b*B + 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)) Int[(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)) + B*(b*c*m + a *d*(n + 1)) - C*(a*c*m + b*d*(n + 1)) + (d*(a*A - b*B)*(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, B, 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 = 1.83 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.64
| method | result | size |
| parallelrisch | \(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (\frac {7 A}{12}-\frac {8 B}{3}+\frac {39 C}{4}\right ) \cos \left (2 d x +2 c \right )+\frac {5 C \cos \left (3 d x +3 c \right )}{8}+\left (A -\frac {17 B}{2}+\frac {243 C}{8}\right ) \cos \left (d x +c \right )+\frac {11 A}{12}-\frac {19 B}{3}+\frac {87 C}{4}\right ) \left (\sec ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+10 d x \left (B -3 C \right )}{10 a^{3} d}\) | \(97\) |
| derivativedivides | \(\frac {\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{5}+\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{5}-\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3}+\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{3}-2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C +A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-7 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +\frac {8 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}+8 \left (B -3 C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\) | \(173\) |
| default | \(\frac {\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{5}+\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{5}-\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3}+\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{3}-2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C +A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-7 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +\frac {8 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}+8 \left (B -3 C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\) | \(173\) |
| risch | \(\frac {B x}{a^{3}}-\frac {3 C x}{a^{3}}-\frac {i {\mathrm e}^{i \left (d x +c \right )} C}{2 a^{3} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} C}{2 a^{3} d}+\frac {2 i \left (15 A \,{\mathrm e}^{4 i \left (d x +c \right )}-45 B \,{\mathrm e}^{4 i \left (d x +c \right )}+90 C \,{\mathrm e}^{4 i \left (d x +c \right )}+30 A \,{\mathrm e}^{3 i \left (d x +c \right )}-135 B \,{\mathrm e}^{3 i \left (d x +c \right )}+300 C \,{\mathrm e}^{3 i \left (d x +c \right )}+40 A \,{\mathrm e}^{2 i \left (d x +c \right )}-185 B \,{\mathrm e}^{2 i \left (d x +c \right )}+420 C \,{\mathrm e}^{2 i \left (d x +c \right )}+20 A \,{\mathrm e}^{i \left (d x +c \right )}-115 B \,{\mathrm e}^{i \left (d x +c \right )}+270 C \,{\mathrm e}^{i \left (d x +c \right )}+7 A -32 B +72 C \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\) | \(229\) |
| norman | \(\frac {\frac {\left (B -3 C \right ) x}{a}+\frac {\left (B -3 C \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {4 \left (B -3 C \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {6 \left (B -3 C \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {4 \left (B -3 C \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {\left (A -26 B +81 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 a d}+\frac {5 \left (A -8 B +27 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}+\frac {\left (A -7 B +25 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {\left (A -B +C \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 a d}+\frac {\left (A +4 B -9 C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{30 a d}-\frac {\left (7 A +43 B -153 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 a d}+\frac {\left (53 A -553 B +1773 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} a^{2}}\) | \(307\) |
1/10*(tan(1/2*d*x+1/2*c)*((7/12*A-8/3*B+39/4*C)*cos(2*d*x+2*c)+5/8*C*cos(3 *d*x+3*c)+(A-17/2*B+243/8*C)*cos(d*x+c)+11/12*A-19/3*B+87/4*C)*sec(1/2*d*x +1/2*c)^4+10*d*x*(B-3*C))/a^3/d
Time = 0.27 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.14 \[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\frac {15 \, {\left (B - 3 \, C\right )} d x \cos \left (d x + c\right )^{3} + 45 \, {\left (B - 3 \, C\right )} d x \cos \left (d x + c\right )^{2} + 45 \, {\left (B - 3 \, C\right )} d x \cos \left (d x + c\right ) + 15 \, {\left (B - 3 \, C\right )} d x + {\left (15 \, C \cos \left (d x + c\right )^{3} + {\left (7 \, A - 32 \, B + 117 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, A - 17 \, B + 57 \, C\right )} \cos \left (d x + c\right ) + 2 \, A - 22 \, B + 72 \, C\right )} \sin \left (d x + c\right )}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]
integrate(cos(d*x+c)^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3, x, algorithm="fricas")
1/15*(15*(B - 3*C)*d*x*cos(d*x + c)^3 + 45*(B - 3*C)*d*x*cos(d*x + c)^2 + 45*(B - 3*C)*d*x*cos(d*x + c) + 15*(B - 3*C)*d*x + (15*C*cos(d*x + c)^3 + (7*A - 32*B + 117*C)*cos(d*x + c)^2 + 3*(2*A - 17*B + 57*C)*cos(d*x + c) + 2*A - 22*B + 72*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. 665 vs. \(2 (146) = 292\).
Time = 3.17 (sec) , antiderivative size = 665, normalized size of antiderivative = 4.38 \[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\begin {cases} \frac {3 A \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} - \frac {7 A \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} + \frac {5 A \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} + \frac {15 A \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} + \frac {60 B d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} + \frac {60 B d x}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} - \frac {3 B \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} + \frac {17 B \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} - \frac {85 B \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} - \frac {105 B \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} - \frac {180 C d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} - \frac {180 C d x}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} + \frac {3 C \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} - \frac {27 C \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} + \frac {225 C \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} + \frac {375 C \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \left (A + B \cos {\left (c \right )} + C \cos ^{2}{\left (c \right )}\right ) \cos ^{2}{\left (c \right )}}{\left (a \cos {\left (c \right )} + a\right )^{3}} & \text {otherwise} \end {cases} \]
Piecewise((3*A*tan(c/2 + d*x/2)**7/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60*a** 3*d) - 7*A*tan(c/2 + d*x/2)**5/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 5*A*tan(c/2 + d*x/2)**3/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 1 5*A*tan(c/2 + d*x/2)/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 60*B*d* x*tan(c/2 + d*x/2)**2/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 60*B*d *x/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 3*B*tan(c/2 + d*x/2)**7/( 60*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 17*B*tan(c/2 + d*x/2)**5/(60* a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 85*B*tan(c/2 + d*x/2)**3/(60*a** 3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 105*B*tan(c/2 + d*x/2)/(60*a**3*d*t an(c/2 + d*x/2)**2 + 60*a**3*d) - 180*C*d*x*tan(c/2 + d*x/2)**2/(60*a**3*d *tan(c/2 + d*x/2)**2 + 60*a**3*d) - 180*C*d*x/(60*a**3*d*tan(c/2 + d*x/2)* *2 + 60*a**3*d) + 3*C*tan(c/2 + d*x/2)**7/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 27*C*tan(c/2 + d*x/2)**5/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60 *a**3*d) + 225*C*tan(c/2 + d*x/2)**3/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60*a **3*d) + 375*C*tan(c/2 + d*x/2)/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d ), Ne(d, 0)), (x*(A + B*cos(c) + C*cos(c)**2)*cos(c)**2/(a*cos(c) + a)**3, True))
Leaf count of result is larger than twice the leaf count of optimal. 295 vs. \(2 (146) = 292\).
Time = 0.30 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.94 \[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\frac {3 \, C {\left (\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} + \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {120 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} - B {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {120 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} + \frac {A {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \]
integrate(cos(d*x+c)^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3, x, algorithm="maxima")
1/60*(3*C*(40*sin(d*x + c)/((a^3 + a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 )*(cos(d*x + c) + 1)) + (85*sin(d*x + c)/(cos(d*x + c) + 1) - 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 12 0*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3) - B*((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) + A*(15*sin(d*x + c)/(cos(d*x + c) + 1) - 10*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)/d
Time = 0.33 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.34 \[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\frac {\frac {60 \, {\left (d x + c\right )} {\left (B - 3 \, C\right )}}{a^{3}} + \frac {120 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{3}} + \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 10 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 20 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 105 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 255 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]
integrate(cos(d*x+c)^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3, x, algorithm="giac")
1/60*(60*(d*x + c)*(B - 3*C)/a^3 + 120*C*tan(1/2*d*x + 1/2*c)/((tan(1/2*d* x + 1/2*c)^2 + 1)*a^3) + (3*A*a^12*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^12*tan(1 /2*d*x + 1/2*c)^5 + 3*C*a^12*tan(1/2*d*x + 1/2*c)^5 - 10*A*a^12*tan(1/2*d* x + 1/2*c)^3 + 20*B*a^12*tan(1/2*d*x + 1/2*c)^3 - 30*C*a^12*tan(1/2*d*x + 1/2*c)^3 + 15*A*a^12*tan(1/2*d*x + 1/2*c) - 105*B*a^12*tan(1/2*d*x + 1/2*c ) + 255*C*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d
Time = 1.53 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.07 \[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\frac {x\,\left (B-3\,C\right )}{a^3}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {2\,A-6\,C}{4\,a^3}-\frac {3\,\left (A-B+C\right )}{4\,a^3}+\frac {2\,B-4\,C}{2\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A-B+C}{6\,a^3}-\frac {2\,B-4\,C}{12\,a^3}\right )}{d}+\frac {2\,C\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A-B+C\right )}{20\,a^3\,d} \]
(x*(B - 3*C))/a^3 - (tan(c/2 + (d*x)/2)*((2*A - 6*C)/(4*a^3) - (3*(A - B + C))/(4*a^3) + (2*B - 4*C)/(2*a^3)))/d - (tan(c/2 + (d*x)/2)^3*((A - B + C )/(6*a^3) - (2*B - 4*C)/(12*a^3)))/d + (2*C*tan(c/2 + (d*x)/2))/(d*(a^3*ta n(c/2 + (d*x)/2)^2 + a^3)) + (tan(c/2 + (d*x)/2)^5*(A - B + C))/(20*a^3*d)