Integrand size = 41, antiderivative size = 150 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {(3 A-B) \text {arctanh}(\sin (c+d x))}{a^3 d}+\frac {2 (36 A-11 B+C) \tan (c+d x)}{15 a^3 d}-\frac {(A-B+C) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(9 A-4 B-C) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(3 A-B) \tan (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )} \] Output:
-(3*A-B)*arctanh(sin(d*x+c))/a^3/d+2/15*(36*A-11*B+C)*tan(d*x+c)/a^3/d-1/5 *(A-B+C)*tan(d*x+c)/d/(a+a*cos(d*x+c))^3-1/15*(9*A-4*B-C)*tan(d*x+c)/a/d/( a+a*cos(d*x+c))^2-(3*A-B)*tan(d*x+c)/d/(a^3+a^3*cos(d*x+c))
Leaf count is larger than twice the leaf count of optimal. \(839\) vs. \(2(150)=300\).
Time = 7.62 (sec) , antiderivative size = 839, normalized size of antiderivative = 5.59 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx =\text {Too large to display} \] 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])^3,x]
Output:
((16*(3*A - B)*Cos[c/2 + (d*x)/2]^6*Cos[c + d*x]^2*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*(C + B*Sec[c + d*x] + A*Sec[c + d*x]^2))/(d*(1 + Cos [c + d*x])^3*(2*A + C + 2*B*Cos[c + d*x] + C*Cos[2*c + 2*d*x])) - (16*(3*A - B)*Cos[c/2 + (d*x)/2]^6*Cos[c + d*x]^2*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*(C + B*Sec[c + d*x] + A*Sec[c + d*x]^2))/(d*(1 + Cos[c + d*x] )^3*(2*A + C + 2*B*Cos[c + d*x] + C*Cos[2*c + 2*d*x])) + (Cos[c/2 + (d*x)/ 2]*Cos[c + d*x]*Sec[c/2]*Sec[c]*(C + B*Sec[c + d*x] + A*Sec[c + d*x]^2)*(- 255*A*Sin[(d*x)/2] + 160*B*Sin[(d*x)/2] - 20*C*Sin[(d*x)/2] + 567*A*Sin[(3 *d*x)/2] - 167*B*Sin[(3*d*x)/2] + 22*C*Sin[(3*d*x)/2] - 600*A*Sin[c - (d*x )/2] + 170*B*Sin[c - (d*x)/2] - 10*C*Sin[c - (d*x)/2] + 375*A*Sin[c + (d*x )/2] - 170*B*Sin[c + (d*x)/2] + 10*C*Sin[c + (d*x)/2] - 480*A*Sin[2*c + (d *x)/2] + 160*B*Sin[2*c + (d*x)/2] - 20*C*Sin[2*c + (d*x)/2] - 60*A*Sin[c + (3*d*x)/2] + 75*B*Sin[c + (3*d*x)/2] + 402*A*Sin[2*c + (3*d*x)/2] - 167*B *Sin[2*c + (3*d*x)/2] + 22*C*Sin[2*c + (3*d*x)/2] - 225*A*Sin[3*c + (3*d*x )/2] + 75*B*Sin[3*c + (3*d*x)/2] + 315*A*Sin[c + (5*d*x)/2] - 95*B*Sin[c + (5*d*x)/2] + 10*C*Sin[c + (5*d*x)/2] + 30*A*Sin[2*c + (5*d*x)/2] + 15*B*S in[2*c + (5*d*x)/2] + 240*A*Sin[3*c + (5*d*x)/2] - 95*B*Sin[3*c + (5*d*x)/ 2] + 10*C*Sin[3*c + (5*d*x)/2] - 45*A*Sin[4*c + (5*d*x)/2] + 15*B*Sin[4*c + (5*d*x)/2] + 72*A*Sin[2*c + (7*d*x)/2] - 22*B*Sin[2*c + (7*d*x)/2] + 2*C *Sin[2*c + (7*d*x)/2] + 15*A*Sin[3*c + (7*d*x)/2] + 57*A*Sin[4*c + (7*d...
Time = 1.22 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.11, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.293, Rules used = {3042, 3520, 3042, 3457, 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.
| \(\displaystyle \int \frac {\sec ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 3520 |
| \(\displaystyle \frac {\int \frac {(a (6 A-B+C)-a (3 A-3 B-2 C) \cos (c+d x)) \sec ^2(c+d x)}{(\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A-B+C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (6 A-B+C)-a (3 A-3 B-2 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {(A-B+C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\int \frac {\left (a^2 (27 A-7 B+2 C)-2 a^2 (9 A-4 B-C) \cos (c+d x)\right ) \sec ^2(c+d x)}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {a (9 A-4 B-C) \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {a^2 (27 A-7 B+2 C)-2 a^2 (9 A-4 B-C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}-\frac {a (9 A-4 B-C) \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\frac {\int \left (2 a^3 (36 A-11 B+C)-15 a^3 (3 A-B) \cos (c+d x)\right ) \sec ^2(c+d x)dx}{a^2}-\frac {15 a^2 (3 A-B) \tan (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (9 A-4 B-C) \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {2 a^3 (36 A-11 B+C)-15 a^3 (3 A-B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx}{a^2}-\frac {15 a^2 (3 A-B) \tan (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (9 A-4 B-C) \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {\frac {\frac {2 a^3 (36 A-11 B+C) \int \sec ^2(c+d x)dx-15 a^3 (3 A-B) \int \sec (c+d x)dx}{a^2}-\frac {15 a^2 (3 A-B) \tan (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (9 A-4 B-C) \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {2 a^3 (36 A-11 B+C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx-15 a^3 (3 A-B) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {15 a^2 (3 A-B) \tan (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (9 A-4 B-C) \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {\frac {\frac {-\frac {2 a^3 (36 A-11 B+C) \int 1d(-\tan (c+d x))}{d}-15 a^3 (3 A-B) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {15 a^2 (3 A-B) \tan (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (9 A-4 B-C) \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\frac {\frac {\frac {2 a^3 (36 A-11 B+C) \tan (c+d x)}{d}-15 a^3 (3 A-B) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {15 a^2 (3 A-B) \tan (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (9 A-4 B-C) \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {\frac {\frac {2 a^3 (36 A-11 B+C) \tan (c+d x)}{d}-\frac {15 a^3 (3 A-B) \text {arctanh}(\sin (c+d x))}{d}}{a^2}-\frac {15 a^2 (3 A-B) \tan (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (9 A-4 B-C) \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
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])^3,x]
Output:
-1/5*((A - B + C)*Tan[c + d*x])/(d*(a + a*Cos[c + d*x])^3) + (-1/3*(a*(9*A - 4*B - C)*Tan[c + d*x])/(d*(a + a*Cos[c + d*x])^2) + ((-15*a^2*(3*A - B) *Tan[c + d*x])/(d*(a + a*Cos[c + d*x])) + ((-15*a^3*(3*A - B)*ArcTanh[Sin[ c + d*x]])/d + (2*a^3*(36*A - 11*B + C)*Tan[c + d*x])/d)/a^2)/(3*a^2))/(5* 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 = 0.59 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.03
| method | result | size |
| parallelrisch | \(\frac {360 \left (A -\frac {B}{3}\right ) \cos \left (d x +c \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-360 \left (A -\frac {B}{3}\right ) \cos \left (d x +c \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+36 \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (\frac {19 A}{4}-\frac {17 B}{12}+\frac {C}{6}\right ) \cos \left (2 d x +2 c \right )+\left (A -\frac {11 B}{36}+\frac {C}{36}\right ) \cos \left (3 d x +3 c \right )+\left (\frac {19 A}{2}-\frac {97 B}{36}+\frac {17 C}{36}\right ) \cos \left (d x +c \right )+\frac {67 A}{12}-\frac {17 B}{12}+\frac {C}{6}\right )}{120 d \,a^{3} \cos \left (d x +c \right )}\) | \(155\) |
| derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A -\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+17 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (12 A -4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {4 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-12 A +4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {4 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{4 d \,a^{3}}\) | \(201\) |
| default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A -\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+17 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (12 A -4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {4 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-12 A +4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {4 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{4 d \,a^{3}}\) | \(201\) |
| norman | \(\frac {\frac {\left (A -B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{20 a d}-\frac {5 \left (21 A -5 B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d a}-\frac {\left (25 A -7 B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d a}+\frac {\left (33 A -23 B +13 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{60 d a}+\frac {\left (51 A -41 B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{30 d a}+\frac {\left (141 A -61 B +11 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{30 d a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) a^{2}}+\frac {\left (3 A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3} d}-\frac {\left (3 A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3} d}\) | \(252\) |
| risch | \(\frac {2 i \left (45 A \,{\mathrm e}^{6 i \left (d x +c \right )}-15 B \,{\mathrm e}^{6 i \left (d x +c \right )}+225 A \,{\mathrm e}^{5 i \left (d x +c \right )}-75 B \,{\mathrm e}^{5 i \left (d x +c \right )}+480 A \,{\mathrm e}^{4 i \left (d x +c \right )}-160 B \,{\mathrm e}^{4 i \left (d x +c \right )}+20 C \,{\mathrm e}^{4 i \left (d x +c \right )}+600 A \,{\mathrm e}^{3 i \left (d x +c \right )}-170 B \,{\mathrm e}^{3 i \left (d x +c \right )}+10 C \,{\mathrm e}^{3 i \left (d x +c \right )}+567 A \,{\mathrm e}^{2 i \left (d x +c \right )}-167 B \,{\mathrm e}^{2 i \left (d x +c \right )}+22 C \,{\mathrm e}^{2 i \left (d x +c \right )}+315 A \,{\mathrm e}^{i \left (d x +c \right )}-95 B \,{\mathrm e}^{i \left (d x +c \right )}+10 C \,{\mathrm e}^{i \left (d x +c \right )}+72 A -22 B +2 C \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {3 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{3} d}-\frac {B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{3} d}-\frac {3 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{3} d}+\frac {B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{3} d}\) | \(326\) |
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))^3,x,meth od=_RETURNVERBOSE)
Output:
1/120*(360*(A-1/3*B)*cos(d*x+c)*ln(tan(1/2*d*x+1/2*c)-1)-360*(A-1/3*B)*cos (d*x+c)*ln(tan(1/2*d*x+1/2*c)+1)+36*sec(1/2*d*x+1/2*c)^4*tan(1/2*d*x+1/2*c )*((19/4*A-17/12*B+1/6*C)*cos(2*d*x+2*c)+(A-11/36*B+1/36*C)*cos(3*d*x+3*c) +(19/2*A-97/36*B+17/36*C)*cos(d*x+c)+67/12*A-17/12*B+1/6*C))/d/a^3/cos(d*x +c)
Time = 0.09 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.86 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {15 \, {\left ({\left (3 \, A - B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (3 \, A - B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A - B\right )} \cos \left (d x + c\right )^{2} + {\left (3 \, A - B\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left ({\left (3 \, A - B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (3 \, A - B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A - B\right )} \cos \left (d x + c\right )^{2} + {\left (3 \, A - B\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, {\left (36 \, A - 11 \, B + C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (57 \, A - 17 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (117 \, A - 32 \, B + 7 \, C\right )} \cos \left (d x + c\right ) + 15 \, A\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}} \] 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))^3, x, algorithm="fricas")
Output:
-1/30*(15*((3*A - B)*cos(d*x + c)^4 + 3*(3*A - B)*cos(d*x + c)^3 + 3*(3*A - B)*cos(d*x + c)^2 + (3*A - B)*cos(d*x + c))*log(sin(d*x + c) + 1) - 15*( (3*A - B)*cos(d*x + c)^4 + 3*(3*A - B)*cos(d*x + c)^3 + 3*(3*A - B)*cos(d* x + c)^2 + (3*A - B)*cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*(2*(36*A - 1 1*B + C)*cos(d*x + c)^3 + 3*(57*A - 17*B + 2*C)*cos(d*x + c)^2 + (117*A - 32*B + 7*C)*cos(d*x + c) + 15*A)*sin(d*x + c))/(a^3*d*cos(d*x + c)^4 + 3*a ^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + a^3*d*cos(d*x + c))
\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\int \frac {A \sec ^{2}{\left (c + d x \right )}}{\cos ^{3}{\left (c + d x \right )} + 3 \cos ^{2}{\left (c + d x \right )} + 3 \cos {\left (c + d x \right )} + 1}\, dx + \int \frac {B \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\cos ^{3}{\left (c + d x \right )} + 3 \cos ^{2}{\left (c + d x \right )} + 3 \cos {\left (c + d x \right )} + 1}\, dx + \int \frac {C \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\cos ^{3}{\left (c + d x \right )} + 3 \cos ^{2}{\left (c + d x \right )} + 3 \cos {\left (c + d x \right )} + 1}\, dx}{a^{3}} \] 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))* *3,x)
Output:
(Integral(A*sec(c + d*x)**2/(cos(c + d*x)**3 + 3*cos(c + d*x)**2 + 3*cos(c + d*x) + 1), x) + Integral(B*cos(c + d*x)*sec(c + d*x)**2/(cos(c + d*x)** 3 + 3*cos(c + d*x)**2 + 3*cos(c + d*x) + 1), x) + Integral(C*cos(c + d*x)* *2*sec(c + d*x)**2/(cos(c + d*x)**3 + 3*cos(c + d*x)**2 + 3*cos(c + d*x) + 1), x))/a**3
Leaf count of result is larger than twice the leaf count of optimal. 350 vs. \(2 (144) = 288\).
Time = 0.05 (sec) , antiderivative size = 350, normalized size of antiderivative = 2.33 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {3 \, A {\left (\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} - \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} - B {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} + \frac {C {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+a*cos(d*x+c))^3, x, algorithm="maxima")
Output:
1/60*(3*A*(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 - 60 *log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 60*log(sin(d*x + c)/(cos(d *x + c) + 1) - 1)/a^3) - B*((105*sin(d*x + c)/(cos(d*x + c) + 1) + 20*sin( d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a ^3 - 60*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 60*log(sin(d*x + c) /(cos(d*x + c) + 1) - 1)/a^3) + C*(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 = 0.16 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.59 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {\frac {60 \, {\left (3 \, A - B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {60 \, {\left (3 \, A - B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {120 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 30 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 20 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 10 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 255 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 105 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \] 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))^3, x, algorithm="giac")
Output:
-1/60*(60*(3*A - B)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 60*(3*A - B)* log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3 + 120*A*tan(1/2*d*x + 1/2*c)/((tan( 1/2*d*x + 1/2*c)^2 - 1)*a^3) - (3*A*a^12*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^12 *tan(1/2*d*x + 1/2*c)^5 + 3*C*a^12*tan(1/2*d*x + 1/2*c)^5 + 30*A*a^12*tan( 1/2*d*x + 1/2*c)^3 - 20*B*a^12*tan(1/2*d*x + 1/2*c)^3 + 10*C*a^12*tan(1/2* d*x + 1/2*c)^3 + 255*A*a^12*tan(1/2*d*x + 1/2*c) - 105*B*a^12*tan(1/2*d*x + 1/2*c) + 15*C*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d
Time = 0.32 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.18 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A-B+C}{6\,a^3}+\frac {4\,A-2\,B}{12\,a^3}\right )}{d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A-B+C\right )}{4\,a^3}+\frac {4\,A-2\,B}{2\,a^3}+\frac {6\,A-2\,C}{4\,a^3}\right )}{d}-\frac {2\,A\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^3\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A-B+C\right )}{20\,a^3\,d}-\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (3\,A-B\right )}{a^3\,d} \] 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))^3),x)
Output:
(tan(c/2 + (d*x)/2)^3*((A - B + C)/(6*a^3) + (4*A - 2*B)/(12*a^3)))/d + (t an(c/2 + (d*x)/2)*((3*(A - B + C))/(4*a^3) + (4*A - 2*B)/(2*a^3) + (6*A - 2*C)/(4*a^3)))/d - (2*A*tan(c/2 + (d*x)/2))/(d*(a^3*tan(c/2 + (d*x)/2)^2 - a^3)) + (tan(c/2 + (d*x)/2)^5*(A - B + C))/(20*a^3*d) - (2*atanh(tan(c/2 + (d*x)/2))*(3*A - B))/(a^3*d)
Time = 0.17 (sec) , antiderivative size = 350, normalized size of antiderivative = 2.33 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {180 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -180 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a +60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b -180 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +180 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a -60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b +3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} a -3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} b +3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} c +27 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} a -17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} b +7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} c +225 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a -85 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b +5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} c -375 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +105 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b -15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) c}{60 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )} \] 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))^3,x)
Output:
(180*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**2*a - 60*log(tan((c + d*x )/2) - 1)*tan((c + d*x)/2)**2*b - 180*log(tan((c + d*x)/2) - 1)*a + 60*log (tan((c + d*x)/2) - 1)*b - 180*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)* *2*a + 60*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**2*b + 180*log(tan((c + d*x)/2) + 1)*a - 60*log(tan((c + d*x)/2) + 1)*b + 3*tan((c + d*x)/2)**7 *a - 3*tan((c + d*x)/2)**7*b + 3*tan((c + d*x)/2)**7*c + 27*tan((c + d*x)/ 2)**5*a - 17*tan((c + d*x)/2)**5*b + 7*tan((c + d*x)/2)**5*c + 225*tan((c + d*x)/2)**3*a - 85*tan((c + d*x)/2)**3*b + 5*tan((c + d*x)/2)**3*c - 375* tan((c + d*x)/2)*a + 105*tan((c + d*x)/2)*b - 15*tan((c + d*x)/2)*c)/(60*a **3*d*(tan((c + d*x)/2)**2 - 1))