Integrand size = 41, antiderivative size = 109 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx=-\frac {(2 A-B) \text {arctanh}(\sin (c+d x))}{a^2 d}+\frac {(10 A-4 B+C) \tan (c+d x)}{3 a^2 d}-\frac {(2 A-B) \tan (c+d x)}{a^2 d (1+\cos (c+d x))}-\frac {(A-B+C) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2} \] Output:
-(2*A-B)*arctanh(sin(d*x+c))/a^2/d+1/3*(10*A-4*B+C)*tan(d*x+c)/a^2/d-(2*A- B)*tan(d*x+c)/a^2/d/(1+cos(d*x+c))-1/3*(A-B+C)*tan(d*x+c)/d/(a+a*cos(d*x+c ))^2
Leaf count is larger than twice the leaf count of optimal. \(321\) vs. \(2(109)=218\).
Time = 3.33 (sec) , antiderivative size = 321, normalized size of antiderivative = 2.94 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {4 \cos \left (\frac {1}{2} (c+d x)\right ) \cos ^2(c+d x) \left (C+B \sec (c+d x)+A \sec ^2(c+d x)\right ) \left ((A-B+C) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+2 (7 A-4 B+C) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+6 \cos ^3\left (\frac {1}{2} (c+d x)\right ) \left ((2 A-B) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\frac {A \sin (d x)}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )+(A-B+C) \cos \left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {c}{2}\right )\right )}{3 a^2 d (1+\cos (c+d x))^2 (2 A+C+2 B \cos (c+d x)+C \cos (2 (c+d x)))} \] Input:
Integrate[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^2)/(a + a* Cos[c + d*x])^2,x]
Output:
(4*Cos[(c + d*x)/2]*Cos[c + d*x]^2*(C + B*Sec[c + d*x] + A*Sec[c + d*x]^2) *((A - B + C)*Sec[c/2]*Sin[(d*x)/2] + 2*(7*A - 4*B + C)*Cos[(c + d*x)/2]^2 *Sec[c/2]*Sin[(d*x)/2] + 6*Cos[(c + d*x)/2]^3*((2*A - B)*(Log[Cos[(c + d*x )/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + (A* Sin[d*x])/((Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))) + (A - B + C)*C os[(c + d*x)/2]*Tan[c/2]))/(3*a^2*d*(1 + Cos[c + d*x])^2*(2*A + C + 2*B*Co s[c + d*x] + C*Cos[2*(c + d*x)]))
Time = 0.88 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.06, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.244, Rules used = {3042, 3520, 3042, 3457, 3042, 3227, 3042, 4254, 24, 4257}
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.
| \(\displaystyle \int \frac {\sec ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 3520 |
| \(\displaystyle \frac {\int \frac {(a (4 A-B+C)-a (2 A-2 B-C) \cos (c+d x)) \sec ^2(c+d x)}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {(A-B+C) \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (4 A-B+C)-a (2 A-2 B-C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}-\frac {(A-B+C) \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\int \left (a^2 (10 A-4 B+C)-3 a^2 (2 A-B) \cos (c+d x)\right ) \sec ^2(c+d x)dx}{a^2}-\frac {3 (2 A-B) \tan (c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {a^2 (10 A-4 B+C)-3 a^2 (2 A-B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx}{a^2}-\frac {3 (2 A-B) \tan (c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {\frac {a^2 (10 A-4 B+C) \int \sec ^2(c+d x)dx-3 a^2 (2 A-B) \int \sec (c+d x)dx}{a^2}-\frac {3 (2 A-B) \tan (c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a^2 (10 A-4 B+C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx-3 a^2 (2 A-B) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {3 (2 A-B) \tan (c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {\frac {-\frac {a^2 (10 A-4 B+C) \int 1d(-\tan (c+d x))}{d}-3 a^2 (2 A-B) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {3 (2 A-B) \tan (c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\frac {\frac {a^2 (10 A-4 B+C) \tan (c+d x)}{d}-3 a^2 (2 A-B) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {3 (2 A-B) \tan (c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {\frac {a^2 (10 A-4 B+C) \tan (c+d x)}{d}-\frac {3 a^2 (2 A-B) \text {arctanh}(\sin (c+d x))}{d}}{a^2}-\frac {3 (2 A-B) \tan (c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
Input:
Int[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^2)/(a + a*Cos[c + d*x])^2,x]
Output:
-1/3*((A - B + C)*Tan[c + d*x])/(d*(a + a*Cos[c + d*x])^2) + ((-3*(2*A - B )*Tan[c + d*x])/(d*(1 + Cos[c + d*x])) + ((-3*a^2*(2*A - B)*ArcTanh[Sin[c + d*x]])/d + (a^2*(10*A - 4*B + C)*Tan[c + d*x])/d)/a^2)/(3*a^2)
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[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b *d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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] && (IntegerQ[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)]
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.51 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.24
| method | result | size |
| parallelrisch | \(\frac {24 \left (-\frac {B}{2}+A \right ) \cos \left (d x +c \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-24 \left (-\frac {B}{2}+A \right ) \cos \left (d x +c \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+28 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (\left (\frac {5 A}{14}-\frac {B}{7}+\frac {C}{28}\right ) \cos \left (2 d x +2 c \right )+\left (A -\frac {5 B}{14}+\frac {C}{7}\right ) \cos \left (d x +c \right )+\frac {4 A}{7}-\frac {B}{7}+\frac {C}{28}\right )}{12 d \,a^{2} \cos \left (d x +c \right )}\) | \(135\) |
| derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+5 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-4 A +2 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (4 A -2 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {2 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}}{2 d \,a^{2}}\) | \(159\) |
| default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+5 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-4 A +2 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (4 A -2 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {2 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}}{2 d \,a^{2}}\) | \(159\) |
| norman | \(\frac {\frac {\left (A -4 B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 a d}+\frac {\left (A -B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{6 a d}+\frac {\left (8 A -5 B +2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 a d}-\frac {\left (9 A -3 B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}-\frac {\left (20 A -5 B +2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) a}+\frac {\left (2 A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{2} d}-\frac {\left (2 A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{2} d}\) | \(223\) |
| risch | \(\frac {2 i \left (6 A \,{\mathrm e}^{4 i \left (d x +c \right )}-3 B \,{\mathrm e}^{4 i \left (d x +c \right )}+18 A \,{\mathrm e}^{3 i \left (d x +c \right )}-9 B \,{\mathrm e}^{3 i \left (d x +c \right )}+3 C \,{\mathrm e}^{3 i \left (d x +c \right )}+22 A \,{\mathrm e}^{2 i \left (d x +c \right )}-7 B \,{\mathrm e}^{2 i \left (d x +c \right )}+C \,{\mathrm e}^{2 i \left (d x +c \right )}+24 A \,{\mathrm e}^{i \left (d x +c \right )}-9 B \,{\mathrm e}^{i \left (d x +c \right )}+3 C \,{\mathrm e}^{i \left (d x +c \right )}+10 A -4 B +C \right )}{3 a^{2} d \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {2 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{2} d}-\frac {B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{2} d}-\frac {2 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{2} d}+\frac {B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{2} d}\) | \(263\) |
Input:
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+a*cos(d*x+c))^2,x,meth od=_RETURNVERBOSE)
Output:
1/12*(24*(-1/2*B+A)*cos(d*x+c)*ln(tan(1/2*d*x+1/2*c)-1)-24*(-1/2*B+A)*cos( d*x+c)*ln(tan(1/2*d*x+1/2*c)+1)+28*tan(1/2*d*x+1/2*c)*sec(1/2*d*x+1/2*c)^2 *((5/14*A-1/7*B+1/28*C)*cos(2*d*x+2*c)+(A-5/14*B+1/7*C)*cos(d*x+c)+4/7*A-1 /7*B+1/28*C))/d/a^2/cos(d*x+c)
Time = 0.09 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.93 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx=-\frac {3 \, {\left ({\left (2 \, A - B\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (2 \, A - B\right )} \cos \left (d x + c\right )^{2} + {\left (2 \, A - B\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left ({\left (2 \, A - B\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (2 \, A - B\right )} \cos \left (d x + c\right )^{2} + {\left (2 \, A - B\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left ({\left (10 \, A - 4 \, B + C\right )} \cos \left (d x + c\right )^{2} + {\left (14 \, A - 5 \, B + 2 \, C\right )} \cos \left (d x + c\right ) + 3 \, A\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}} \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+a*cos(d*x+c))^2, x, algorithm="fricas")
Output:
-1/6*(3*((2*A - B)*cos(d*x + c)^3 + 2*(2*A - B)*cos(d*x + c)^2 + (2*A - B) *cos(d*x + c))*log(sin(d*x + c) + 1) - 3*((2*A - B)*cos(d*x + c)^3 + 2*(2* A - B)*cos(d*x + c)^2 + (2*A - B)*cos(d*x + c))*log(-sin(d*x + c) + 1) - 2 *((10*A - 4*B + C)*cos(d*x + c)^2 + (14*A - 5*B + 2*C)*cos(d*x + c) + 3*A) *sin(d*x + c))/(a^2*d*cos(d*x + c)^3 + 2*a^2*d*cos(d*x + c)^2 + a^2*d*cos( d*x + c))
\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {\int \frac {A \sec ^{2}{\left (c + d x \right )}}{\cos ^{2}{\left (c + d x \right )} + 2 \cos {\left (c + d x \right )} + 1}\, dx + \int \frac {B \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\cos ^{2}{\left (c + d x \right )} + 2 \cos {\left (c + d x \right )} + 1}\, dx + \int \frac {C \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\cos ^{2}{\left (c + d x \right )} + 2 \cos {\left (c + d x \right )} + 1}\, dx}{a^{2}} \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**2/(a+a*cos(d*x+c))* *2,x)
Output:
(Integral(A*sec(c + d*x)**2/(cos(c + d*x)**2 + 2*cos(c + d*x) + 1), x) + I ntegral(B*cos(c + d*x)*sec(c + d*x)**2/(cos(c + d*x)**2 + 2*cos(c + d*x) + 1), x) + Integral(C*cos(c + d*x)**2*sec(c + d*x)**2/(cos(c + d*x)**2 + 2* cos(c + d*x) + 1), x))/a**2
Leaf count of result is larger than twice the leaf count of optimal. 287 vs. \(2 (105) = 210\).
Time = 0.05 (sec) , antiderivative size = 287, normalized size of antiderivative = 2.63 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {A {\left (\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}} + \frac {12 \, \sin \left (d x + c\right )}{{\left (a^{2} - \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - B {\left (\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {6 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {6 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )} + \frac {C {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, d} \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+a*cos(d*x+c))^2, x, algorithm="maxima")
Output:
1/6*(A*((15*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 12*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 12*log(si n(d*x + c)/(cos(d*x + c) + 1) - 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))) - B*((9*sin(d*x + c)/ (cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 6*log(sin( d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 6*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^2) + C*(3*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos( d*x + c) + 1)^3)/a^2)/d
Time = 0.17 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.71 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx=-\frac {\frac {6 \, {\left (2 \, A - B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {6 \, {\left (2 \, A - B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac {12 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{2}} - \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 9 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+a*cos(d*x+c))^2, x, algorithm="giac")
Output:
-1/6*(6*(2*A - B)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 6*(2*A - B)*log (abs(tan(1/2*d*x + 1/2*c) - 1))/a^2 + 12*A*tan(1/2*d*x + 1/2*c)/((tan(1/2* d*x + 1/2*c)^2 - 1)*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) - 9*B*a^4*tan(1/2*d*x + 1/2*c) + 3*C*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d
Time = 0.33 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.14 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {A-B+C}{a^2}-\frac {B-3\,A+C}{2\,a^2}\right )}{d}-\frac {2\,A\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^2\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A-B+C\right )}{6\,a^2\,d}-\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (2\,A-B\right )}{a^2\,d} \] Input:
int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^2*(a + a*cos(c + d*x))^2),x)
Output:
(tan(c/2 + (d*x)/2)*((A - B + C)/a^2 - (B - 3*A + C)/(2*a^2)))/d - (2*A*ta n(c/2 + (d*x)/2))/(d*(a^2*tan(c/2 + (d*x)/2)^2 - a^2)) + (tan(c/2 + (d*x)/ 2)^3*(A - B + C))/(6*a^2*d) - (2*atanh(tan(c/2 + (d*x)/2))*(2*A - B))/(a^2 *d)
Time = 0.19 (sec) , antiderivative size = 306, normalized size of antiderivative = 2.81 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a +6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b -12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a -6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b +\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} b +\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} c +14 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a -8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} c -27 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b -3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) c}{6 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )} \] Input:
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+a*cos(d*x+c))^2,x)
Output:
(12*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**2*a - 6*log(tan((c + d*x)/ 2) - 1)*tan((c + d*x)/2)**2*b - 12*log(tan((c + d*x)/2) - 1)*a + 6*log(tan ((c + d*x)/2) - 1)*b - 12*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**2*a + 6*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**2*b + 12*log(tan((c + d*x) /2) + 1)*a - 6*log(tan((c + d*x)/2) + 1)*b + tan((c + d*x)/2)**5*a - tan(( c + d*x)/2)**5*b + tan((c + d*x)/2)**5*c + 14*tan((c + d*x)/2)**3*a - 8*ta n((c + d*x)/2)**3*b + 2*tan((c + d*x)/2)**3*c - 27*tan((c + d*x)/2)*a + 9* tan((c + d*x)/2)*b - 3*tan((c + d*x)/2)*c)/(6*a**2*d*(tan((c + d*x)/2)**2 - 1))