Integrand size = 39, antiderivative size = 124 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {A \text {arctanh}(\sin (c+d x))}{a^3 d}-\frac {(A-B+C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(7 A-2 B-3 C) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(22 A-2 B-3 C) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )} \] Output:
A*arctanh(sin(d*x+c))/a^3/d-1/5*(A-B+C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^3-1/ 15*(7*A-2*B-3*C)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^2-1/15*(22*A-2*B-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. \(276\) vs. \(2(124)=248\).
Time = 2.94 (sec) , antiderivative size = 276, normalized size of antiderivative = 2.23 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \left (240 A \cos ^6\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (5 (29 A-4 B-3 C) \sin \left (\frac {d x}{2}\right )+15 (-5 A+C) \sin \left (c+\frac {d x}{2}\right )+95 A \sin \left (c+\frac {3 d x}{2}\right )-10 B \sin \left (c+\frac {3 d x}{2}\right )-15 C \sin \left (c+\frac {3 d x}{2}\right )-15 A \sin \left (2 c+\frac {3 d x}{2}\right )+22 A \sin \left (2 c+\frac {5 d x}{2}\right )-2 B \sin \left (2 c+\frac {5 d x}{2}\right )-3 C \sin \left (2 c+\frac {5 d x}{2}\right )\right )\right )}{15 a^3 d (1+\cos (c+d x))^3 (2 A+C+2 B \cos (c+d x)+C \cos (2 (c+d x)))} \] Input:
Integrate[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x])/(a + a*Co s[c + d*x])^3,x]
Output:
-1/15*((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*(240*A*Cos[(c + d*x)/2]^6*( Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + Cos[(c + d*x)/2]*Sec[c/2]*(5*(29*A - 4*B - 3*C)*Sin[(d*x)/2] + 15*(-5*A + C)*Sin[c + (d*x)/2] + 95*A*Sin[c + (3*d*x)/2] - 10*B*Sin[c + (3*d*x)/2] - 15*C*Sin[c + (3*d*x)/2] - 15*A*Sin[2*c + (3*d*x)/2] + 22*A*Si n[2*c + (5*d*x)/2] - 2*B*Sin[2*c + (5*d*x)/2] - 3*C*Sin[2*c + (5*d*x)/2])) )/(a^3*d*(1 + Cos[c + d*x])^3*(2*A + C + 2*B*Cos[c + d*x] + C*Cos[2*(c + d *x)]))
Time = 0.83 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3042, 3520, 3042, 3457, 3042, 3457, 27, 3042, 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.
| \(\displaystyle \int \frac {\sec (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 3520 |
| \(\displaystyle \frac {\int \frac {(5 a A-a (2 A-2 B-3 C) \cos (c+d x)) \sec (c+d x)}{(\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {5 a A-a (2 A-2 B-3 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\int \frac {\left (15 a^2 A-a^2 (7 A-2 B-3 C) \cos (c+d x)\right ) \sec (c+d x)}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {a (7 A-2 B-3 C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {15 a^2 A-a^2 (7 A-2 B-3 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}-\frac {a (7 A-2 B-3 C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\frac {\int 15 a^3 A \sec (c+d x)dx}{a^2}-\frac {a^2 (22 A-2 B-3 C) \sin (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (7 A-2 B-3 C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {15 a A \int \sec (c+d x)dx-\frac {a^2 (22 A-2 B-3 C) \sin (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (7 A-2 B-3 C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {15 a A \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {a^2 (22 A-2 B-3 C) \sin (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (7 A-2 B-3 C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {\frac {15 a A \text {arctanh}(\sin (c+d x))}{d}-\frac {a^2 (22 A-2 B-3 C) \sin (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (7 A-2 B-3 C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
Input:
Int[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x])/(a + a*Cos[c + d*x])^3,x]
Output:
-1/5*((A - B + C)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^3) + (-1/3*(a*(7*A - 2*B - 3*C)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^2) + ((15*a*A*ArcTanh[ Sin[c + d*x]])/d - (a^2*(22*A - 2*B - 3*C)*Sin[c + d*x])/(d*(a + a*Cos[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[(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_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.40 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.80
| method | result | size |
| parallelrisch | \(\frac {-20 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+20 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\left (\left (A -B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\frac {10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (2 A -B \right )}{3}+35 A -5 B -5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{20 a^{3} d}\) | \(99\) |
| derivativedivides | \(\frac {-4 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-7 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}+4 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d \,a^{3}}\) | \(144\) |
| default | \(\frac {-4 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-7 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}+4 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d \,a^{3}}\) | \(144\) |
| risch | \(-\frac {2 i \left (15 A \,{\mathrm e}^{4 i \left (d x +c \right )}+75 A \,{\mathrm e}^{3 i \left (d x +c \right )}-15 C \,{\mathrm e}^{3 i \left (d x +c \right )}+145 A \,{\mathrm e}^{2 i \left (d x +c \right )}-20 B \,{\mathrm e}^{2 i \left (d x +c \right )}-15 C \,{\mathrm e}^{2 i \left (d x +c \right )}+95 A \,{\mathrm e}^{i \left (d x +c \right )}-10 B \,{\mathrm e}^{i \left (d x +c \right )}-15 C \,{\mathrm e}^{i \left (d x +c \right )}+22 A -2 B -3 C \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}+\frac {A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{3} d}-\frac {A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{3} d}\) | \(185\) |
| norman | \(\frac {-\frac {\left (A -B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{20 a d}-\frac {\left (7 A -B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d a}-\frac {\left (13 A -8 B +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{30 d a}-\frac {\left (23 A -4 B -3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 d a}-\frac {\left (74 A -19 B -6 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{30 d a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} a^{2}}+\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3} d}-\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3} d}\) | \(202\) |
Input:
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)/(a+a*cos(d*x+c))^3,x,method =_RETURNVERBOSE)
Output:
1/20*(-20*A*ln(tan(1/2*d*x+1/2*c)-1)+20*A*ln(tan(1/2*d*x+1/2*c)+1)-((A-B+C )*tan(1/2*d*x+1/2*c)^4+10/3*tan(1/2*d*x+1/2*c)^2*(2*A-B)+35*A-5*B-5*C)*tan (1/2*d*x+1/2*c))/a^3/d
Time = 0.09 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.56 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {15 \, {\left (A \cos \left (d x + c\right )^{3} + 3 \, A \cos \left (d x + c\right )^{2} + 3 \, A \cos \left (d x + c\right ) + A\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (A \cos \left (d x + c\right )^{3} + 3 \, A \cos \left (d x + c\right )^{2} + 3 \, A \cos \left (d x + c\right ) + A\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left ({\left (22 \, A - 2 \, B - 3 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (17 \, A - 2 \, B - 3 \, C\right )} \cos \left (d x + c\right ) + 32 \, A - 7 \, B - 3 \, C\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)/(a+a*cos(d*x+c))^3,x, algorithm="fricas")
Output:
1/30*(15*(A*cos(d*x + c)^3 + 3*A*cos(d*x + c)^2 + 3*A*cos(d*x + c) + A)*lo g(sin(d*x + c) + 1) - 15*(A*cos(d*x + c)^3 + 3*A*cos(d*x + c)^2 + 3*A*cos( d*x + c) + A)*log(-sin(d*x + c) + 1) - 2*((22*A - 2*B - 3*C)*cos(d*x + c)^ 2 + 3*(17*A - 2*B - 3*C)*cos(d*x + c) + 32*A - 7*B - 3*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)
\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\int \frac {A \sec {\left (c + d x \right )}}{\cos ^{3}{\left (c + d x \right )} + 3 \cos ^{2}{\left (c + d x \right )} + 3 \cos {\left (c + d x \right )} + 1}\, dx + \int \frac {B \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}}{\cos ^{3}{\left (c + d x \right )} + 3 \cos ^{2}{\left (c + d x \right )} + 3 \cos {\left (c + d x \right )} + 1}\, dx + \int \frac {C \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{\cos ^{3}{\left (c + d x \right )} + 3 \cos ^{2}{\left (c + d x \right )} + 3 \cos {\left (c + d x \right )} + 1}\, dx}{a^{3}} \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)/(a+a*cos(d*x+c))**3, x)
Output:
(Integral(A*sec(c + d*x)/(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)/(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)/(cos(c + d*x)**3 + 3*cos(c + d*x)**2 + 3*cos(c + d*x) + 1), x))/ a**3
Time = 0.05 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.87 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {A {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} - \frac {B {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}} - \frac {3 \, C {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)/(a+a*cos(d*x+c))^3,x, algorithm="maxima")
Output:
-1/60*(A*((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) - B*(15*sin(d*x + c)/(cos(d*x + c) + 1) + 10*sin(d*x + c)^3/(c os(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 3*C*(5*s in(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3) /d
Time = 0.16 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.45 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\frac {60 \, A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {60 \, A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 20 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 10 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)/(a+a*cos(d*x+c))^3,x, algorithm="giac")
Output:
1/60*(60*A*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 60*A*log(abs(tan(1/2*d *x + 1/2*c) - 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 + 20*A*a^12*tan(1/2*d*x + 1/2*c)^3 - 10*B*a^12*tan(1/2*d*x + 1/2*c)^3 + 105*A*a^12*tan(1/2*d*x + 1/2*c) - 15*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 = 134, normalized size of antiderivative = 1.08 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {2\,A\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A-B+C}{12\,a^3}-\frac {B-3\,A+C}{12\,a^3}\right )}{d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,A+B-C}{4\,a^3}+\frac {A-B+C}{4\,a^3}-\frac {B-3\,A+C}{4\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A-B+C\right )}{20\,a^3\,d} \] Input:
int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)*(a + a*cos(c + d *x))^3),x)
Output:
(2*A*atanh(tan(c/2 + (d*x)/2)))/(a^3*d) - (tan(c/2 + (d*x)/2)^3*((A - B + C)/(12*a^3) - (B - 3*A + C)/(12*a^3)))/d - (tan(c/2 + (d*x)/2)*((3*A + B - C)/(4*a^3) + (A - B + C)/(4*a^3) - (B - 3*A + C)/(4*a^3)))/d - (tan(c/2 + (d*x)/2)^5*(A - B + C))/(20*a^3*d)
Time = 0.17 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.17 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {-60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a +60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a -3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} a +3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} b -3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} c -20 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a +10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b -105 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) c}{60 a^{3} d} \] Input:
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)/(a+a*cos(d*x+c))^3,x)
Output:
( - 60*log(tan((c + d*x)/2) - 1)*a + 60*log(tan((c + d*x)/2) + 1)*a - 3*ta n((c + d*x)/2)**5*a + 3*tan((c + d*x)/2)**5*b - 3*tan((c + d*x)/2)**5*c - 20*tan((c + d*x)/2)**3*a + 10*tan((c + d*x)/2)**3*b - 105*tan((c + d*x)/2) *a + 15*tan((c + d*x)/2)*b + 15*tan((c + d*x)/2)*c)/(60*a**3*d)