Integrand size = 41, antiderivative size = 160 \[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\frac {(2 A-4 B+7 C) x}{2 a^2}-\frac {2 (2 A-5 B+8 C) \sin (c+d x)}{3 a^2 d}+\frac {(2 A-4 B+7 C) \cos (c+d x) \sin (c+d x)}{2 a^2 d}-\frac {(2 A-5 B+8 C) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {(A-B+C) \cos ^3(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2} \]
1/2*(2*A-4*B+7*C)*x/a^2-2/3*(2*A-5*B+8*C)*sin(d*x+c)/a^2/d+1/2*(2*A-4*B+7* C)*cos(d*x+c)*sin(d*x+c)/a^2/d-1/3*(2*A-5*B+8*C)*cos(d*x+c)^2*sin(d*x+c)/a ^2/d/(1+cos(d*x+c))-1/3*(A-B+C)*cos(d*x+c)^3*sin(d*x+c)/d/(a+a*cos(d*x+c)) ^2
Leaf count is larger than twice the leaf count of optimal. \(385\) vs. \(2(160)=320\).
Time = 2.53 (sec) , antiderivative size = 385, normalized size of antiderivative = 2.41 \[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (36 (2 A-4 B+7 C) d x \cos \left (\frac {d x}{2}\right )+36 (2 A-4 B+7 C) d x \cos \left (c+\frac {d x}{2}\right )+24 A d x \cos \left (c+\frac {3 d x}{2}\right )-48 B d x \cos \left (c+\frac {3 d x}{2}\right )+84 C d x \cos \left (c+\frac {3 d x}{2}\right )+24 A d x \cos \left (2 c+\frac {3 d x}{2}\right )-48 B d x \cos \left (2 c+\frac {3 d x}{2}\right )+84 C d x \cos \left (2 c+\frac {3 d x}{2}\right )-144 A \sin \left (\frac {d x}{2}\right )+264 B \sin \left (\frac {d x}{2}\right )-381 C \sin \left (\frac {d x}{2}\right )+96 A \sin \left (c+\frac {d x}{2}\right )-120 B \sin \left (c+\frac {d x}{2}\right )+147 C \sin \left (c+\frac {d x}{2}\right )-80 A \sin \left (c+\frac {3 d x}{2}\right )+164 B \sin \left (c+\frac {3 d x}{2}\right )-239 C \sin \left (c+\frac {3 d x}{2}\right )+36 B \sin \left (2 c+\frac {3 d x}{2}\right )-63 C \sin \left (2 c+\frac {3 d x}{2}\right )+12 B \sin \left (2 c+\frac {5 d x}{2}\right )-15 C \sin \left (2 c+\frac {5 d x}{2}\right )+12 B \sin \left (3 c+\frac {5 d x}{2}\right )-15 C \sin \left (3 c+\frac {5 d x}{2}\right )+3 C \sin \left (3 c+\frac {7 d x}{2}\right )+3 C \sin \left (4 c+\frac {7 d x}{2}\right )\right )}{48 a^2 d (1+\cos (c+d x))^2} \]
(Cos[(c + d*x)/2]*Sec[c/2]*(36*(2*A - 4*B + 7*C)*d*x*Cos[(d*x)/2] + 36*(2* A - 4*B + 7*C)*d*x*Cos[c + (d*x)/2] + 24*A*d*x*Cos[c + (3*d*x)/2] - 48*B*d *x*Cos[c + (3*d*x)/2] + 84*C*d*x*Cos[c + (3*d*x)/2] + 24*A*d*x*Cos[2*c + ( 3*d*x)/2] - 48*B*d*x*Cos[2*c + (3*d*x)/2] + 84*C*d*x*Cos[2*c + (3*d*x)/2] - 144*A*Sin[(d*x)/2] + 264*B*Sin[(d*x)/2] - 381*C*Sin[(d*x)/2] + 96*A*Sin[ c + (d*x)/2] - 120*B*Sin[c + (d*x)/2] + 147*C*Sin[c + (d*x)/2] - 80*A*Sin[ c + (3*d*x)/2] + 164*B*Sin[c + (3*d*x)/2] - 239*C*Sin[c + (3*d*x)/2] + 36* B*Sin[2*c + (3*d*x)/2] - 63*C*Sin[2*c + (3*d*x)/2] + 12*B*Sin[2*c + (5*d*x )/2] - 15*C*Sin[2*c + (5*d*x)/2] + 12*B*Sin[3*c + (5*d*x)/2] - 15*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]))/(48 *a^2*d*(1 + Cos[c + d*x])^2)
Time = 0.72 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3042, 3520, 3042, 3456, 25, 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.
| \(\displaystyle \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 3520 |
| \(\displaystyle \frac {\int \frac {\cos ^2(c+d x) (3 a (B-C)+a (2 A-2 B+5 C) \cos (c+d x))}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^3(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (3 a (B-C)+a (2 A-2 B+5 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^3(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3456 |
| \(\displaystyle \frac {\frac {\int -\cos (c+d x) \left (2 a^2 (2 A-5 B+8 C)-3 a^2 (2 A-4 B+7 C) \cos (c+d x)\right )dx}{a^2}-\frac {(2 A-5 B+8 C) \sin (c+d x) \cos ^2(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^3(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\frac {\int \cos (c+d x) \left (2 a^2 (2 A-5 B+8 C)-3 a^2 (2 A-4 B+7 C) \cos (c+d x)\right )dx}{a^2}-\frac {(2 A-5 B+8 C) \sin (c+d x) \cos ^2(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^3(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (2 a^2 (2 A-5 B+8 C)-3 a^2 (2 A-4 B+7 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}-\frac {(2 A-5 B+8 C) \sin (c+d x) \cos ^2(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^3(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3213 |
| \(\displaystyle \frac {-\frac {\frac {2 a^2 (2 A-5 B+8 C) \sin (c+d x)}{d}-\frac {3 a^2 (2 A-4 B+7 C) \sin (c+d x) \cos (c+d x)}{2 d}-\frac {3}{2} a^2 x (2 A-4 B+7 C)}{a^2}-\frac {(2 A-5 B+8 C) \sin (c+d x) \cos ^2(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^3(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
-1/3*((A - B + C)*Cos[c + d*x]^3*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^2) + (-(((2*A - 5*B + 8*C)*Cos[c + d*x]^2*Sin[c + d*x])/(d*(1 + Cos[c + d*x]) )) - ((-3*a^2*(2*A - 4*B + 7*C)*x)/2 + (2*a^2*(2*A - 5*B + 8*C)*Sin[c + d* x])/d - (3*a^2*(2*A - 4*B + 7*C)*Cos[c + d*x]*Sin[c + d*x])/(2*d))/a^2)/(3 *a^2)
3.4.48.3.1 Defintions of rubi rules used
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_.) + (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 = 1.70 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.62
| method | result | size |
| parallelrisch | \(\frac {\left (\left (12 \left (B -C \right ) \cos \left (2 d x +2 c \right )+3 C \cos \left (3 d x +3 c \right )+\left (112 B -163 C \right ) \cos \left (d x +c \right )+8 A +92 B -140 C \right ) \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-80 A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+48 x \left (\frac {7 C}{2}-2 B +A \right ) d}{48 a^{2} d}\) | \(100\) |
| derivativedivides | \(\frac {\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{3}+\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{3}-3 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+5 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +\frac {4 \left (-\frac {5 C}{2}+B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 \left (-\frac {3 C}{2}+B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+2 \left (2 A -4 B +7 C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\) | \(160\) |
| default | \(\frac {\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{3}+\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{3}-3 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+5 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +\frac {4 \left (-\frac {5 C}{2}+B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 \left (-\frac {3 C}{2}+B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+2 \left (2 A -4 B +7 C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\) | \(160\) |
| risch | \(\frac {x A}{a^{2}}-\frac {2 B x}{a^{2}}+\frac {7 C x}{2 a^{2}}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )} C}{8 a^{2} d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} B}{2 a^{2} d}+\frac {i C \,{\mathrm e}^{i \left (d x +c \right )}}{a^{2} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B}{2 a^{2} d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} C}{a^{2} d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} C}{8 a^{2} d}-\frac {2 i \left (6 A \,{\mathrm e}^{2 i \left (d x +c \right )}-9 B \,{\mathrm e}^{2 i \left (d x +c \right )}+12 C \,{\mathrm e}^{2 i \left (d x +c \right )}+9 A \,{\mathrm e}^{i \left (d x +c \right )}-15 B \,{\mathrm e}^{i \left (d x +c \right )}+21 C \,{\mathrm e}^{i \left (d x +c \right )}+5 A -8 B +11 C \right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}\) | \(240\) |
| norman | \(\frac {\frac {\left (2 A -4 B +7 C \right ) x}{2 a}+\frac {\left (A -B +C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}+\frac {2 \left (2 A -4 B +7 C \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {3 \left (2 A -4 B +7 C \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {2 \left (2 A -4 B +7 C \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {\left (2 A -4 B +7 C \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}-\frac {\left (3 A -9 B +13 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}-\frac {\left (5 A -11 B +17 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}-\frac {\left (5 A -11 B +18 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {\left (25 A -61 B +100 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}-\frac {\left (35 A -95 B +149 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}}{a \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}\) | \(313\) |
1/48*(((12*(B-C)*cos(2*d*x+2*c)+3*C*cos(3*d*x+3*c)+(112*B-163*C)*cos(d*x+c )+8*A+92*B-140*C)*sec(1/2*d*x+1/2*c)^2-80*A)*tan(1/2*d*x+1/2*c)+48*x*(7/2* C-2*B+A)*d)/a^2/d
Time = 0.27 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.96 \[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\frac {3 \, {\left (2 \, A - 4 \, B + 7 \, C\right )} d x \cos \left (d x + c\right )^{2} + 6 \, {\left (2 \, A - 4 \, B + 7 \, C\right )} d x \cos \left (d x + c\right ) + 3 \, {\left (2 \, A - 4 \, B + 7 \, C\right )} d x + {\left (3 \, C \cos \left (d x + c\right )^{3} + 6 \, {\left (B - C\right )} \cos \left (d x + c\right )^{2} - {\left (10 \, A - 28 \, B + 43 \, C\right )} \cos \left (d x + c\right ) - 8 \, A + 20 \, B - 32 \, C\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]
integrate(cos(d*x+c)^2*(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*(2*A - 4*B + 7*C)*d*x*cos(d*x + c)^2 + 6*(2*A - 4*B + 7*C)*d*x*cos( d*x + c) + 3*(2*A - 4*B + 7*C)*d*x + (3*C*cos(d*x + c)^3 + 6*(B - C)*cos(d *x + c)^2 - (10*A - 28*B + 43*C)*cos(d*x + c) - 8*A + 20*B - 32*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. 1261 vs. \(2 (153) = 306\).
Time = 2.36 (sec) , antiderivative size = 1261, normalized size of antiderivative = 7.88 \[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\text {Too large to display} \]
Piecewise((6*A*d*x*tan(c/2 + d*x/2)**4/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12* a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 12*A*d*x*tan(c/2 + d*x/2)**2/(6*a **2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 6* A*d*x/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a* *2*d) + A*tan(c/2 + d*x/2)**7/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*ta n(c/2 + d*x/2)**2 + 6*a**2*d) - 7*A*tan(c/2 + d*x/2)**5/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 17*A*tan(c/2 + d *x/2)**3/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6 *a**2*d) - 9*A*tan(c/2 + d*x/2)/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d* tan(c/2 + d*x/2)**2 + 6*a**2*d) - 12*B*d*x*tan(c/2 + d*x/2)**4/(6*a**2*d*t an(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 24*B*d*x* tan(c/2 + d*x/2)**2/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d* x/2)**2 + 6*a**2*d) - 12*B*d*x/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*t an(c/2 + d*x/2)**2 + 6*a**2*d) - B*tan(c/2 + d*x/2)**7/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 13*B*tan(c/2 + d* x/2)**5/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6* a**2*d) + 41*B*tan(c/2 + d*x/2)**3/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2 *d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 27*B*tan(c/2 + d*x/2)/(6*a**2*d*tan(c /2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 21*C*d*x*tan( c/2 + d*x/2)**4/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x...
Leaf count of result is larger than twice the leaf count of optimal. 352 vs. \(2 (150) = 300\).
Time = 0.29 (sec) , antiderivative size = 352, normalized size of antiderivative = 2.20 \[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=-\frac {C {\left (\frac {6 \, {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} + \frac {2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {42 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )} - B {\left (\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {24 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {12 \, \sin \left (d x + c\right )}{{\left (a^{2} + \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + A {\left (\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )}}{6 \, d} \]
integrate(cos(d*x+c)^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2, x, algorithm="maxima")
-1/6*(C*(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)/(cos(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) - B*((15*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(c os(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))) + A*((9*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos (d*x + c) + 1)^3)/a^2 - 12*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2))/d
Time = 0.30 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.24 \[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\frac {\frac {3 \, {\left (d x + c\right )} {\left (2 \, A - 4 \, B + 7 \, C\right )}}{a^{2}} + \frac {6 \, {\left (2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{2}} + \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 21 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]
integrate(cos(d*x+c)^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2, x, algorithm="giac")
1/6*(3*(d*x + c)*(2*A - 4*B + 7*C)/a^2 + 6*(2*B*tan(1/2*d*x + 1/2*c)^3 - 5 *C*tan(1/2*d*x + 1/2*c)^3 + 2*B*tan(1/2*d*x + 1/2*c) - 3*C*tan(1/2*d*x + 1 /2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*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 - 9*A*a^4* tan(1/2*d*x + 1/2*c) + 15*B*a^4*tan(1/2*d*x + 1/2*c) - 21*C*a^4*tan(1/2*d* x + 1/2*c))/a^6)/d
Time = 1.54 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.99 \[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\frac {\left (2\,B-5\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,B-3\,C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A-B+C\right )}{2\,a^2}-\frac {2\,B-4\,C}{2\,a^2}\right )}{d}+\frac {x\,\left (2\,A-4\,B+7\,C\right )}{2\,a^2}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A-B+C\right )}{6\,a^2\,d} \]