Integrand size = 41, antiderivative size = 185 \[ \int \frac {\cos ^3(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=-\frac {(4 A-7 B+10 C) x}{2 a^2}+\frac {(5 A-8 B+12 C) \sin (c+d x)}{a^2 d}-\frac {(4 A-7 B+10 C) \cos (c+d x) \sin (c+d x)}{2 a^2 d}-\frac {(4 A-7 B+10 C) \cos ^3(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {(A-B+C) \cos ^4(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac {(5 A-8 B+12 C) \sin ^3(c+d x)}{3 a^2 d} \]
-1/2*(4*A-7*B+10*C)*x/a^2+(5*A-8*B+12*C)*sin(d*x+c)/a^2/d-1/2*(4*A-7*B+10* C)*cos(d*x+c)*sin(d*x+c)/a^2/d-1/3*(4*A-7*B+10*C)*cos(d*x+c)^3*sin(d*x+c)/ a^2/d/(1+cos(d*x+c))-1/3*(A-B+C)*cos(d*x+c)^4*sin(d*x+c)/d/(a+a*cos(d*x+c) )^2-1/3*(5*A-8*B+12*C)*sin(d*x+c)^3/a^2/d
Leaf count is larger than twice the leaf count of optimal. \(481\) vs. \(2(185)=370\).
Time = 2.78 (sec) , antiderivative size = 481, normalized size of antiderivative = 2.60 \[ \int \frac {\cos ^3(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (-36 (4 A-7 B+10 C) d x \cos \left (\frac {d x}{2}\right )-36 (4 A-7 B+10 C) d x \cos \left (c+\frac {d x}{2}\right )-48 A d x \cos \left (c+\frac {3 d x}{2}\right )+84 B d x \cos \left (c+\frac {3 d x}{2}\right )-120 C d x \cos \left (c+\frac {3 d x}{2}\right )-48 A d x \cos \left (2 c+\frac {3 d x}{2}\right )+84 B d x \cos \left (2 c+\frac {3 d x}{2}\right )-120 C d x \cos \left (2 c+\frac {3 d x}{2}\right )+264 A \sin \left (\frac {d x}{2}\right )-381 B \sin \left (\frac {d x}{2}\right )+516 C \sin \left (\frac {d x}{2}\right )-120 A \sin \left (c+\frac {d x}{2}\right )+147 B \sin \left (c+\frac {d x}{2}\right )-156 C \sin \left (c+\frac {d x}{2}\right )+164 A \sin \left (c+\frac {3 d x}{2}\right )-239 B \sin \left (c+\frac {3 d x}{2}\right )+342 C \sin \left (c+\frac {3 d x}{2}\right )+36 A \sin \left (2 c+\frac {3 d x}{2}\right )-63 B \sin \left (2 c+\frac {3 d x}{2}\right )+118 C \sin \left (2 c+\frac {3 d x}{2}\right )+12 A \sin \left (2 c+\frac {5 d x}{2}\right )-15 B \sin \left (2 c+\frac {5 d x}{2}\right )+30 C \sin \left (2 c+\frac {5 d x}{2}\right )+12 A \sin \left (3 c+\frac {5 d x}{2}\right )-15 B \sin \left (3 c+\frac {5 d x}{2}\right )+30 C \sin \left (3 c+\frac {5 d x}{2}\right )+3 B \sin \left (3 c+\frac {7 d x}{2}\right )-3 C \sin \left (3 c+\frac {7 d x}{2}\right )+3 B \sin \left (4 c+\frac {7 d x}{2}\right )-3 C \sin \left (4 c+\frac {7 d x}{2}\right )+C \sin \left (4 c+\frac {9 d x}{2}\right )+C \sin \left (5 c+\frac {9 d x}{2}\right )\right )}{48 a^2 d (1+\cos (c+d x))^2} \]
(Cos[(c + d*x)/2]*Sec[c/2]*(-36*(4*A - 7*B + 10*C)*d*x*Cos[(d*x)/2] - 36*( 4*A - 7*B + 10*C)*d*x*Cos[c + (d*x)/2] - 48*A*d*x*Cos[c + (3*d*x)/2] + 84* B*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] + 84*B*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] - 381*B*Sin[(d*x)/2] + 516*C*Sin[(d*x)/2] - 120* A*Sin[c + (d*x)/2] + 147*B*Sin[c + (d*x)/2] - 156*C*Sin[c + (d*x)/2] + 164 *A*Sin[c + (3*d*x)/2] - 239*B*Sin[c + (3*d*x)/2] + 342*C*Sin[c + (3*d*x)/2 ] + 36*A*Sin[2*c + (3*d*x)/2] - 63*B*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] - 15*B*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] - 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 = 0.92 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.91, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.317, Rules used = {3042, 3520, 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.
| \(\displaystyle \int \frac {\cos ^3(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 3520 |
| \(\displaystyle \frac {\int -\frac {\cos ^3(c+d x) (a (A-4 B+4 C)-3 a (A-B+2 C) \cos (c+d x))}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int \frac {\cos ^3(c+d x) (a (A-4 B+4 C)-3 a (A-B+2 C) \cos (c+d x))}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a (A-4 B+4 C)-3 a (A-B+2 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3456 |
| \(\displaystyle -\frac {\frac {\int 3 \cos ^2(c+d x) \left (a^2 (4 A-7 B+10 C)-a^2 (5 A-8 B+12 C) \cos (c+d x)\right )dx}{a^2}+\frac {(4 A-7 B+10 C) \sin (c+d x) \cos ^3(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\frac {3 \int \cos ^2(c+d x) \left (a^2 (4 A-7 B+10 C)-a^2 (5 A-8 B+12 C) \cos (c+d x)\right )dx}{a^2}+\frac {(4 A-7 B+10 C) \sin (c+d x) \cos ^3(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {3 \int \sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a^2 (4 A-7 B+10 C)-a^2 (5 A-8 B+12 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}+\frac {(4 A-7 B+10 C) \sin (c+d x) \cos ^3(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle -\frac {\frac {3 \left (a^2 (4 A-7 B+10 C) \int \cos ^2(c+d x)dx-a^2 (5 A-8 B+12 C) \int \cos ^3(c+d x)dx\right )}{a^2}+\frac {(4 A-7 B+10 C) \sin (c+d x) \cos ^3(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {3 \left (a^2 (4 A-7 B+10 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-a^2 (5 A-8 B+12 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^3dx\right )}{a^2}+\frac {(4 A-7 B+10 C) \sin (c+d x) \cos ^3(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3113 |
| \(\displaystyle -\frac {\frac {3 \left (\frac {a^2 (5 A-8 B+12 C) \int \left (1-\sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}+a^2 (4 A-7 B+10 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx\right )}{a^2}+\frac {(4 A-7 B+10 C) \sin (c+d x) \cos ^3(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\frac {3 \left (a^2 (4 A-7 B+10 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {a^2 (5 A-8 B+12 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )}{a^2}+\frac {(4 A-7 B+10 C) \sin (c+d x) \cos ^3(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle -\frac {\frac {3 \left (a^2 (4 A-7 B+10 C) \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {a^2 (5 A-8 B+12 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )}{a^2}+\frac {(4 A-7 B+10 C) \sin (c+d x) \cos ^3(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle -\frac {\frac {3 \left (\frac {a^2 (5 A-8 B+12 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+a^2 (4 A-7 B+10 C) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )}{a^2}+\frac {(4 A-7 B+10 C) \sin (c+d x) \cos ^3(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
-1/3*((A - B + C)*Cos[c + d*x]^4*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^2) - (((4*A - 7*B + 10*C)*Cos[c + d*x]^3*Sin[c + d*x])/(d*(1 + Cos[c + d*x])) + (3*(a^2*(4*A - 7*B + 10*C)*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)) + (a^2*(5*A - 8*B + 12*C)*(-Sin[c + d*x] + Sin[c + d*x]^3/3))/d))/a^2)/(3*a^ 2)
3.4.47.3.1 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[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[(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 = 1.97 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.64
| method | result | size |
| parallelrisch | \(\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\frac {\left (\frac {3 A}{7}-\frac {3 B}{7}+C \right ) \cos \left (2 d x +2 c \right )}{4}+\frac {\left (\frac {3 B}{2}-C \right ) \cos \left (3 d x +3 c \right )}{56}+\frac {C \cos \left (4 d x +4 c \right )}{112}+\left (A -\frac {163 B}{112}+\frac {129 C}{56}\right ) \cos \left (d x +c \right )+\frac {23 A}{28}-\frac {5 B}{4}+\frac {219 C}{112}\right ) \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6 x d \left (A -\frac {7 B}{4}+\frac {5 C}{2}\right )}{3 a^{2} d}\) | \(118\) |
| derivativedivides | \(\frac {-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3}+\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{3}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{3}+5 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-7 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -\frac {4 \left (\left (-A +\frac {5 B}{2}-5 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-2 A +4 B -\frac {20 C}{3}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-A +\frac {3 B}{2}-3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-2 \left (4 A -7 B +10 C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\) | \(192\) |
| default | \(\frac {-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3}+\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{3}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{3}+5 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-7 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -\frac {4 \left (\left (-A +\frac {5 B}{2}-5 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-2 A +4 B -\frac {20 C}{3}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-A +\frac {3 B}{2}-3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-2 \left (4 A -7 B +10 C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\) | \(192\) |
| risch | \(-\frac {2 x A}{a^{2}}+\frac {7 B x}{2 a^{2}}-\frac {5 C x}{a^{2}}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )} B}{8 a^{2} d}+\frac {i {\mathrm e}^{2 i \left (d x +c \right )} C}{4 a^{2} d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} A}{2 a^{2} d}+\frac {i {\mathrm e}^{i \left (d x +c \right )} B}{a^{2} d}-\frac {15 i C \,{\mathrm e}^{i \left (d x +c \right )}}{8 a^{2} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} A}{2 a^{2} d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} B}{a^{2} d}+\frac {15 i {\mathrm e}^{-i \left (d x +c \right )} C}{8 a^{2} d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} B}{8 a^{2} d}-\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} C}{4 a^{2} d}+\frac {2 i \left (9 A \,{\mathrm e}^{2 i \left (d x +c \right )}-12 B \,{\mathrm e}^{2 i \left (d x +c \right )}+15 C \,{\mathrm e}^{2 i \left (d x +c \right )}+15 A \,{\mathrm e}^{i \left (d x +c \right )}-21 B \,{\mathrm e}^{i \left (d x +c \right )}+27 C \,{\mathrm e}^{i \left (d x +c \right )}+8 A -11 B +14 C \right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}+\frac {C \sin \left (3 d x +3 c \right )}{12 a^{2} d}\) | \(335\) |
| norman | \(\frac {-\frac {\left (4 A -7 B +10 C \right ) x}{2 a}-\frac {\left (A -B +C \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}-\frac {5 \left (4 A -7 B +10 C \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}-\frac {5 \left (4 A -7 B +10 C \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {5 \left (4 A -7 B +10 C \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {5 \left (4 A -7 B +10 C \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}-\frac {\left (4 A -7 B +10 C \right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}+\frac {\left (5 A -8 B +11 C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}+\frac {\left (9 A -13 B +21 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}+\frac {2 \left (47 A -77 B +115 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}+\frac {\left (61 A -94 B +143 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}+\frac {\left (77 A -125 B +185 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}+\frac {\left (217 A -349 B +521 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5} a}\) | \(369\) |
1/3*(7*tan(1/2*d*x+1/2*c)*(1/4*(3/7*A-3/7*B+C)*cos(2*d*x+2*c)+1/56*(3/2*B- C)*cos(3*d*x+3*c)+1/112*C*cos(4*d*x+4*c)+(A-163/112*B+129/56*C)*cos(d*x+c) +23/28*A-5/4*B+219/112*C)*sec(1/2*d*x+1/2*c)^2-6*x*d*(A-7/4*B+5/2*C))/a^2/ d
Time = 0.25 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.93 \[ \int \frac {\cos ^3(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=-\frac {3 \, {\left (4 \, A - 7 \, B + 10 \, C\right )} d x \cos \left (d x + c\right )^{2} + 6 \, {\left (4 \, A - 7 \, B + 10 \, C\right )} d x \cos \left (d x + c\right ) + 3 \, {\left (4 \, A - 7 \, B + 10 \, C\right )} d x - {\left (2 \, C \cos \left (d x + c\right )^{4} + {\left (3 \, B - 2 \, C\right )} \cos \left (d x + c\right )^{3} + 6 \, {\left (A - B + 2 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (28 \, A - 43 \, B + 66 \, C\right )} \cos \left (d x + c\right ) + 20 \, A - 32 \, B + 48 \, C\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]
integrate(cos(d*x+c)^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2, x, algorithm="fricas")
-1/6*(3*(4*A - 7*B + 10*C)*d*x*cos(d*x + c)^2 + 6*(4*A - 7*B + 10*C)*d*x*c os(d*x + c) + 3*(4*A - 7*B + 10*C)*d*x - (2*C*cos(d*x + c)^4 + (3*B - 2*C) *cos(d*x + c)^3 + 6*(A - B + 2*C)*cos(d*x + c)^2 + (28*A - 43*B + 66*C)*co s(d*x + c) + 20*A - 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. 2134 vs. \(2 (178) = 356\).
Time = 3.72 (sec) , antiderivative size = 2134, normalized size of antiderivative = 11.54 \[ \int \frac {\cos ^3(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\text {Too large to display} \]
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) + 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...
Leaf count of result is larger than twice the leaf count of optimal. 487 vs. \(2 (175) = 350\).
Time = 0.31 (sec) , antiderivative size = 487, normalized size of antiderivative = 2.63 \[ \int \frac {\cos ^3(c+d x) \left (A+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 )} + 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} \]
integrate(cos(d*x+c)^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2, x, algorithm="maxima")
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) + 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 = 0.38 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.44 \[ \int \frac {\cos ^3(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=-\frac {\frac {3 \, {\left (d x + c\right )} {\left (4 \, A - 7 \, B + 10 \, C\right )}}{a^{2}} - \frac {2 \, {\left (6 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 15 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 30 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 9 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} a^{2}} + \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 21 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 27 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]
integrate(cos(d*x+c)^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2, x, algorithm="giac")
-1/6*(3*(d*x + c)*(4*A - 7*B + 10*C)/a^2 - 2*(6*A*tan(1/2*d*x + 1/2*c)^5 - 15*B*tan(1/2*d*x + 1/2*c)^5 + 30*C*tan(1/2*d*x + 1/2*c)^5 + 12*A*tan(1/2* d*x + 1/2*c)^3 - 24*B*tan(1/2*d*x + 1/2*c)^3 + 40*C*tan(1/2*d*x + 1/2*c)^3 + 6*A*tan(1/2*d*x + 1/2*c) - 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) + (A*a^4*tan(1/2*d*x + 1/2* c)^3 - B*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) + 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 = 1.72 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.09 \[ \int \frac {\cos ^3(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\frac {\left (2\,A-5\,B+10\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (4\,A-8\,B+\frac {40\,C}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A-3\,B+6\,C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )}-\frac {x\,\left (4\,A-7\,B+10\,C\right )}{2\,a^2}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {A-3\,B+5\,C}{2\,a^2}+\frac {2\,\left (A-B+C\right )}{a^2}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A-B+C\right )}{6\,a^2\,d} \]
(tan(c/2 + (d*x)/2)*(2*A - 3*B + 6*C) + tan(c/2 + (d*x)/2)^5*(2*A - 5*B + 10*C) + tan(c/2 + (d*x)/2)^3*(4*A - 8*B + (40*C)/3))/(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*(4*A - 7*B + 10*C))/(2*a^2) + (tan(c/2 + (d*x)/2)*((A - 3*B + 5*C)/(2 *a^2) + (2*(A - B + C))/a^2))/d - (tan(c/2 + (d*x)/2)^3*(A - B + C))/(6*a^ 2*d)