Integrand size = 33, antiderivative size = 136 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=-\frac {3 C x}{a^3}+\frac {(2 A+27 C) \sin (c+d x)}{15 a^3 d}-\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(A-9 C) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {3 C \sin (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )} \] Output:
-3*C*x/a^3+1/15*(2*A+27*C)*sin(d*x+c)/a^3/d-1/5*(A+C)*cos(d*x+c)^3*sin(d*x +c)/d/(a+a*cos(d*x+c))^3+1/15*(A-9*C)*cos(d*x+c)^2*sin(d*x+c)/a/d/(a+a*cos (d*x+c))^2+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. \(283\) vs. \(2(136)=272\).
Time = 2.86 (sec) , antiderivative size = 283, normalized size of antiderivative = 2.08 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=-\frac {\sec \left (\frac {c}{2}\right ) \sec ^5\left (\frac {1}{2} (c+d x)\right ) \left (900 C d x \cos \left (\frac {d x}{2}\right )+900 C d x \cos \left (c+\frac {d x}{2}\right )+450 C d x \cos \left (c+\frac {3 d x}{2}\right )+450 C d x \cos \left (2 c+\frac {3 d x}{2}\right )+90 C d x \cos \left (2 c+\frac {5 d x}{2}\right )+90 C d x \cos \left (3 c+\frac {5 d x}{2}\right )-160 A \sin \left (\frac {d x}{2}\right )-1755 C \sin \left (\frac {d x}{2}\right )+120 A \sin \left (c+\frac {d x}{2}\right )+1125 C \sin \left (c+\frac {d x}{2}\right )-80 A \sin \left (c+\frac {3 d x}{2}\right )-1215 C \sin \left (c+\frac {3 d x}{2}\right )+60 A \sin \left (2 c+\frac {3 d x}{2}\right )+225 C \sin \left (2 c+\frac {3 d x}{2}\right )-28 A \sin \left (2 c+\frac {5 d x}{2}\right )-363 C \sin \left (2 c+\frac {5 d x}{2}\right )-75 C \sin \left (3 c+\frac {5 d x}{2}\right )-15 C \sin \left (3 c+\frac {7 d x}{2}\right )-15 C \sin \left (4 c+\frac {7 d x}{2}\right )\right )}{960 a^3 d} \] Input:
Integrate[(Cos[c + d*x]^2*(A + C*Cos[c + d*x]^2))/(a + a*Cos[c + d*x])^3,x ]
Output:
-1/960*(Sec[c/2]*Sec[(c + d*x)/2]^5*(900*C*d*x*Cos[(d*x)/2] + 900*C*d*x*Co s[c + (d*x)/2] + 450*C*d*x*Cos[c + (3*d*x)/2] + 450*C*d*x*Cos[2*c + (3*d*x )/2] + 90*C*d*x*Cos[2*c + (5*d*x)/2] + 90*C*d*x*Cos[3*c + (5*d*x)/2] - 160 *A*Sin[(d*x)/2] - 1755*C*Sin[(d*x)/2] + 120*A*Sin[c + (d*x)/2] + 1125*C*Si n[c + (d*x)/2] - 80*A*Sin[c + (3*d*x)/2] - 1215*C*Sin[c + (3*d*x)/2] + 60* A*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] - 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]))/(a^3*d)
Time = 1.06 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.07, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.394, Rules used = {3042, 3521, 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.
| \(\displaystyle \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 3521 |
| \(\displaystyle \frac {\int \frac {\cos ^2(c+d x) (a (2 A-3 C)+a (A+6 C) \cos (c+d x))}{(\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a (2 A-3 C)+a (A+6 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3456 |
| \(\displaystyle \frac {\frac {\int \frac {\cos (c+d x) \left (2 (A-9 C) a^2+(2 A+27 C) \cos (c+d x) a^2\right )}{\cos (c+d x) a+a}dx}{3 a^2}+\frac {a (A-9 C) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \left (2 (A-9 C) a^2+(2 A+27 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}+\frac {a (A-9 C) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {\frac {\int \frac {(2 A+27 C) \cos ^2(c+d x) a^2+2 (A-9 C) \cos (c+d x) a^2}{\cos (c+d x) a+a}dx}{3 a^2}+\frac {a (A-9 C) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {(2 A+27 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^2+2 (A-9 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}+\frac {a (A-9 C) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {\frac {\frac {\int -\frac {45 a^3 C \cos (c+d x)}{\cos (c+d x) a+a}dx}{a}+\frac {a (2 A+27 C) \sin (c+d x)}{d}}{3 a^2}+\frac {a (A-9 C) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {a (2 A+27 C) \sin (c+d x)}{d}-45 a^2 C \int \frac {\cos (c+d x)}{\cos (c+d x) a+a}dx}{3 a^2}+\frac {a (A-9 C) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {a (2 A+27 C) \sin (c+d x)}{d}-45 a^2 C \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}+\frac {a (A-9 C) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {\frac {\frac {a (2 A+27 C) \sin (c+d x)}{d}-45 a^2 C \left (\frac {x}{a}-\int \frac {1}{\cos (c+d x) a+a}dx\right )}{3 a^2}+\frac {a (A-9 C) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {a (2 A+27 C) \sin (c+d x)}{d}-45 a^2 C \left (\frac {x}{a}-\int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx\right )}{3 a^2}+\frac {a (A-9 C) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3127 |
| \(\displaystyle \frac {\frac {\frac {a (2 A+27 C) \sin (c+d x)}{d}-45 a^2 C \left (\frac {x}{a}-\frac {\sin (c+d x)}{d (a \cos (c+d x)+a)}\right )}{3 a^2}+\frac {a (A-9 C) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
Input:
Int[(Cos[c + d*x]^2*(A + C*Cos[c + d*x]^2))/(a + a*Cos[c + d*x])^3,x]
Output:
-1/5*((A + C)*Cos[c + d*x]^3*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^3) + (( a*(A - 9*C)*Cos[c + d*x]^2*Sin[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2) + (( a*(2*A + 27*C)*Sin[c + d*x])/d - 45*a^2*C*(x/a - Sin[c + d*x]/(d*(a + a*Co s[c + d*x]))))/(3*a^2))/(5*a^2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((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_.) + (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 = 0.31 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.62
| method | result | size |
| parallelrisch | \(\frac {\left (\frac {\left (\frac {7 A}{3}+39 C \right ) \cos \left (2 d x +2 c \right )}{4}+\frac {5 \cos \left (3 d x +3 c \right ) C}{8}+\left (A +\frac {243 C}{8}\right ) \cos \left (d x +c \right )+\frac {11 A}{12}+\frac {87 C}{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-30 d x C}{10 a^{3} d}\) | \(85\) |
| derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C +A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+17 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-16 C \left (-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}+\frac {3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{4 d \,a^{3}}\) | \(131\) |
| default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C +A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+17 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-16 C \left (-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}+\frac {3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{4 d \,a^{3}}\) | \(131\) |
| risch | \(-\frac {3 C x}{a^{3}}-\frac {i {\mathrm e}^{i \left (d x +c \right )} C}{2 a^{3} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} C}{2 a^{3} d}+\frac {2 i \left (15 A \,{\mathrm e}^{4 i \left (d x +c \right )}+90 C \,{\mathrm e}^{4 i \left (d x +c \right )}+30 A \,{\mathrm e}^{3 i \left (d x +c \right )}+300 C \,{\mathrm e}^{3 i \left (d x +c \right )}+40 A \,{\mathrm e}^{2 i \left (d x +c \right )}+420 C \,{\mathrm e}^{2 i \left (d x +c \right )}+20 A \,{\mathrm e}^{i \left (d x +c \right )}+270 C \,{\mathrm e}^{i \left (d x +c \right )}+7 A +72 C \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\) | \(172\) |
| norman | \(\frac {-\frac {3 C x}{a}-\frac {12 C x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a}-\frac {18 C x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{a}-\frac {12 C x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{a}-\frac {3 C x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{a}+\frac {\left (A -9 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{30 a d}+\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{20 a d}+\frac {\left (A +25 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {5 \left (A +27 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 a d}+\frac {\left (A +81 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{5 a d}-\frac {\left (7 A -153 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{60 a d}+\frac {\left (53 A +1773 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{60 a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} a^{2}}\) | \(268\) |
Input:
int(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x,method=_RETURNVER BOSE)
Output:
1/10*((1/4*(7/3*A+39*C)*cos(2*d*x+2*c)+5/8*cos(3*d*x+3*c)*C+(A+243/8*C)*co s(d*x+c)+11/12*A+87/4*C)*tan(1/2*d*x+1/2*c)*sec(1/2*d*x+1/2*c)^4-30*d*x*C) /a^3/d
Time = 0.08 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.10 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=-\frac {45 \, C d x \cos \left (d x + c\right )^{3} + 135 \, C d x \cos \left (d x + c\right )^{2} + 135 \, C d x \cos \left (d x + c\right ) + 45 \, C d x - {\left (15 \, C \cos \left (d x + c\right )^{3} + {\left (7 \, A + 117 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, A + 57 \, C\right )} \cos \left (d x + c\right ) + 2 \, A + 72 \, C\right )} \sin \left (d x + c\right )}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \] Input:
integrate(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x, algorithm= "fricas")
Output:
-1/15*(45*C*d*x*cos(d*x + c)^3 + 135*C*d*x*cos(d*x + c)^2 + 135*C*d*x*cos( d*x + c) + 45*C*d*x - (15*C*cos(d*x + c)^3 + (7*A + 117*C)*cos(d*x + c)^2 + 3*(2*A + 57*C)*cos(d*x + c) + 2*A + 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. 422 vs. \(2 (129) = 258\).
Time = 2.30 (sec) , antiderivative size = 422, normalized size of antiderivative = 3.10 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\begin {cases} \frac {3 A \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} - \frac {7 A \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} + \frac {5 A \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} + \frac {15 A \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} - \frac {180 C d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} - \frac {180 C d x}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} + \frac {3 C \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} - \frac {27 C \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} + \frac {225 C \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} + \frac {375 C \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \left (A + C \cos ^{2}{\left (c \right )}\right ) \cos ^{2}{\left (c \right )}}{\left (a \cos {\left (c \right )} + a\right )^{3}} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**2*(A+C*cos(d*x+c)**2)/(a+a*cos(d*x+c))**3,x)
Output:
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) - 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/(6 0*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 + C*cos(c)**2)*cos(c) **2/(a*cos(c) + a)**3, True))
Time = 0.14 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.51 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\frac {3 \, C {\left (\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} + \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {120 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} + \frac {A {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \] Input:
integrate(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x, algorithm= "maxima")
Output:
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) + 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 = 151, normalized size of antiderivative = 1.11 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=-\frac {\frac {180 \, {\left (d x + c\right )} C}{a^{3}} - \frac {120 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 10 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 255 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \] Input:
integrate(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x, algorithm= "giac")
Output:
-1/60*(180*(d*x + c)*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*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 - 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) + 255*C*a^12*tan(1/2*d*x + 1/2*c) )/a^15)/d
Time = 0.19 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.12 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\frac {\left (\frac {7\,A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{15}+\frac {24\,C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{5}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (-\frac {4\,A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{15}-\frac {3\,C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{5}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{20}+\frac {C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{20}}{a^3\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}-\frac {3\,C\,x}{a^3}+\frac {2\,C\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^3\,d} \] Input:
int((cos(c + d*x)^2*(A + C*cos(c + d*x)^2))/(a + a*cos(c + d*x))^3,x)
Output:
((A*sin(c/2 + (d*x)/2))/20 + (C*sin(c/2 + (d*x)/2))/20 - cos(c/2 + (d*x)/2 )^2*((4*A*sin(c/2 + (d*x)/2))/15 + (3*C*sin(c/2 + (d*x)/2))/5) + cos(c/2 + (d*x)/2)^4*((7*A*sin(c/2 + (d*x)/2))/15 + (24*C*sin(c/2 + (d*x)/2))/5))/( a^3*d*cos(c/2 + (d*x)/2)^5) - (3*C*x)/a^3 + (2*C*cos(c/2 + (d*x)/2)*sin(c/ 2 + (d*x)/2))/(a^3*d)
Time = 0.18 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.12 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} a +3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} c -7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} a -27 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} c +5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a +225 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} c -180 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} c d x +15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +375 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) c -180 c d x}{60 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )} \] Input:
int(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x)
Output:
(3*tan((c + d*x)/2)**7*a + 3*tan((c + d*x)/2)**7*c - 7*tan((c + d*x)/2)**5 *a - 27*tan((c + d*x)/2)**5*c + 5*tan((c + d*x)/2)**3*a + 225*tan((c + d*x )/2)**3*c - 180*tan((c + d*x)/2)**2*c*d*x + 15*tan((c + d*x)/2)*a + 375*ta n((c + d*x)/2)*c - 180*c*d*x)/(60*a**3*d*(tan((c + d*x)/2)**2 + 1))