Integrand size = 41, antiderivative size = 210 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {(13 A-6 B+2 C) \text {arctanh}(\sin (c+d x))}{2 a^3 d}-\frac {2 (76 A-36 B+11 C) \tan (c+d x)}{15 a^3 d}+\frac {(13 A-6 B+2 C) \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac {(A-B+C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(11 A-6 B+C) \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(76 A-36 B+11 C) \sec (c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )} \] Output:
1/2*(13*A-6*B+2*C)*arctanh(sin(d*x+c))/a^3/d-2/15*(76*A-36*B+11*C)*tan(d*x +c)/a^3/d+1/2*(13*A-6*B+2*C)*sec(d*x+c)*tan(d*x+c)/a^3/d-1/5*(A-B+C)*sec(d *x+c)*tan(d*x+c)/d/(a+a*cos(d*x+c))^3-1/15*(11*A-6*B+C)*sec(d*x+c)*tan(d*x +c)/a/d/(a+a*cos(d*x+c))^2-1/15*(76*A-36*B+11*C)*sec(d*x+c)*tan(d*x+c)/d/( a^3+a^3*cos(d*x+c))
Time = 4.17 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.98 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {2 \cos ^6\left (\frac {1}{2} (c+d x)\right ) \left (-30 (13 A-6 B+2 C) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )-16 (17 A-12 B+7 C) \csc ^3(c+d x) \sin ^4\left (\frac {1}{2} (c+d x)\right )-96 (A-B+C) \csc ^5(c+d x) \sin ^6\left (\frac {1}{2} (c+d x)\right )-4 (107 A-57 B+22 C) \tan \left (\frac {1}{2} (c+d x)\right )-60 (3 A-B) \tan (c+d x)+30 A \sec (c+d x) \tan (c+d x)\right )}{15 a^3 d (1+\cos (c+d x))^3} \] Input:
Integrate[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^3)/(a + a* Cos[c + d*x])^3,x]
Output:
(2*Cos[(c + d*x)/2]^6*(-30*(13*A - 6*B + 2*C)*(Log[Cos[(c + d*x)/2] - Sin[ (c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - 16*(17*A - 12* B + 7*C)*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4 - 96*(A - B + C)*Csc[c + d*x]^5 *Sin[(c + d*x)/2]^6 - 4*(107*A - 57*B + 22*C)*Tan[(c + d*x)/2] - 60*(3*A - B)*Tan[c + d*x] + 30*A*Sec[c + d*x]*Tan[c + d*x]))/(15*a^3*d*(1 + Cos[c + d*x])^3)
Time = 1.42 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.02, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.341, Rules used = {3042, 3520, 3042, 3457, 3042, 3457, 3042, 3227, 3042, 4254, 24, 4255, 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.
| \(\displaystyle \int \frac {\sec ^3(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 3520 |
| \(\displaystyle \frac {\int \frac {(a (7 A-2 B+2 C)-a (4 A-4 B-C) \cos (c+d x)) \sec ^3(c+d x)}{(\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (7 A-2 B+2 C)-a (4 A-4 B-C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\int \frac {\left (a^2 (43 A-18 B+8 C)-3 a^2 (11 A-6 B+C) \cos (c+d x)\right ) \sec ^3(c+d x)}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {a (11 A-6 B+C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {a^2 (43 A-18 B+8 C)-3 a^2 (11 A-6 B+C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}-\frac {a (11 A-6 B+C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\frac {\int \left (15 a^3 (13 A-6 B+2 C)-2 a^3 (76 A-36 B+11 C) \cos (c+d x)\right ) \sec ^3(c+d x)dx}{a^2}-\frac {a^2 (76 A-36 B+11 C) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (11 A-6 B+C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {15 a^3 (13 A-6 B+2 C)-2 a^3 (76 A-36 B+11 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx}{a^2}-\frac {a^2 (76 A-36 B+11 C) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (11 A-6 B+C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {\frac {\frac {15 a^3 (13 A-6 B+2 C) \int \sec ^3(c+d x)dx-2 a^3 (76 A-36 B+11 C) \int \sec ^2(c+d x)dx}{a^2}-\frac {a^2 (76 A-36 B+11 C) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (11 A-6 B+C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {15 a^3 (13 A-6 B+2 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-2 a^3 (76 A-36 B+11 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx}{a^2}-\frac {a^2 (76 A-36 B+11 C) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (11 A-6 B+C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {\frac {\frac {\frac {2 a^3 (76 A-36 B+11 C) \int 1d(-\tan (c+d x))}{d}+15 a^3 (13 A-6 B+2 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}-\frac {a^2 (76 A-36 B+11 C) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (11 A-6 B+C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\frac {\frac {15 a^3 (13 A-6 B+2 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {2 a^3 (76 A-36 B+11 C) \tan (c+d x)}{d}}{a^2}-\frac {a^2 (76 A-36 B+11 C) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (11 A-6 B+C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {\frac {\frac {15 a^3 (13 A-6 B+2 C) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {2 a^3 (76 A-36 B+11 C) \tan (c+d x)}{d}}{a^2}-\frac {a^2 (76 A-36 B+11 C) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (11 A-6 B+C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {15 a^3 (13 A-6 B+2 C) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {2 a^3 (76 A-36 B+11 C) \tan (c+d x)}{d}}{a^2}-\frac {a^2 (76 A-36 B+11 C) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (11 A-6 B+C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {\frac {15 a^3 (13 A-6 B+2 C) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {2 a^3 (76 A-36 B+11 C) \tan (c+d x)}{d}}{a^2}-\frac {a^2 (76 A-36 B+11 C) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (11 A-6 B+C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
Input:
Int[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^3)/(a + a*Cos[c + d*x])^3,x]
Output:
-1/5*((A - B + C)*Sec[c + d*x]*Tan[c + d*x])/(d*(a + a*Cos[c + d*x])^3) + (-1/3*(a*(11*A - 6*B + C)*Sec[c + d*x]*Tan[c + d*x])/(d*(a + a*Cos[c + d*x ])^2) + (-((a^2*(76*A - 36*B + 11*C)*Sec[c + d*x]*Tan[c + d*x])/(d*(a + a* Cos[c + d*x]))) + ((-2*a^3*(76*A - 36*B + 11*C)*Tan[c + d*x])/d + 15*a^3*( 13*A - 6*B + 2*C)*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x ])/(2*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_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.79 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.95
| method | result | size |
| parallelrisch | \(\frac {-520 \left (\cos \left (2 d x +2 c \right )+1\right ) \left (A -\frac {6 B}{13}+\frac {2 C}{13}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+520 \left (\cos \left (2 d x +2 c \right )+1\right ) \left (A -\frac {6 B}{13}+\frac {2 C}{13}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-239 \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \left (\frac {18 \left (29 A -14 B +4 C \right ) \cos \left (2 d x +2 c \right )}{239}+\left (A -\frac {114 B}{239}+\frac {34 C}{239}\right ) \cos \left (3 d x +3 c \right )+\frac {2 \left (\frac {76 A}{3}-12 B +\frac {11 C}{3}\right ) \cos \left (4 d x +4 c \right )}{239}+\frac {\left (777 A -382 B +102 C \right ) \cos \left (d x +c \right )}{239}+\frac {1354 A}{717}-\frac {228 B}{239}+\frac {194 C}{717}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{80 d \,a^{3} \left (\cos \left (2 d x +2 c \right )+1\right )}\) | \(199\) |
| derivativedivides | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}-\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B -\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}-31 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -7 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {-14 A +4 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (26 A -12 B +4 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2 A}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-14 A +4 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-26 A +12 B -4 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {2 A}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}}{4 d \,a^{3}}\) | \(252\) |
| default | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}-\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B -\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}-31 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -7 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {-14 A +4 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (26 A -12 B +4 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2 A}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-14 A +4 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-26 A +12 B -4 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {2 A}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}}{4 d \,a^{3}}\) | \(252\) |
| norman | \(\frac {-\frac {\left (A -B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{20 a d}-\frac {\left (4 A -3 B +2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{6 d a}-\frac {\left (22 A -15 B +2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 d a}+\frac {\left (25 A -9 B +2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d a}-\frac {\left (51 A -25 B +7 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d a}-\frac {\left (153 A -83 B +33 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{20 d a}+\frac {\left (489 A -209 B +69 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{20 d a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2} a^{2}}-\frac {\left (13 A -6 B +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{3} d}+\frac {\left (13 A -6 B +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{3} d}\) | \(294\) |
| risch | \(-\frac {i \left (44 C +304 A -144 B -450 B \,{\mathrm e}^{7 i \left (d x +c \right )}+2275 A \,{\mathrm e}^{6 i \left (d x +c \right )}+350 C \,{\mathrm e}^{6 i \left (d x +c \right )}+4329 A \,{\mathrm e}^{4 i \left (d x +c \right )}+654 C \,{\mathrm e}^{4 i \left (d x +c \right )}+2673 A \,{\mathrm e}^{2 i \left (d x +c \right )}+378 C \,{\mathrm e}^{2 i \left (d x +c \right )}+3575 A \,{\mathrm e}^{5 i \left (d x +c \right )}+490 C \,{\mathrm e}^{5 i \left (d x +c \right )}+3805 A \,{\mathrm e}^{3 i \left (d x +c \right )}+530 C \,{\mathrm e}^{3 i \left (d x +c \right )}+1325 A \,{\mathrm e}^{i \left (d x +c \right )}-1278 B \,{\mathrm e}^{2 i \left (d x +c \right )}+190 C \,{\mathrm e}^{i \left (d x +c \right )}-1650 B \,{\mathrm e}^{5 i \left (d x +c \right )}-1050 B \,{\mathrm e}^{6 i \left (d x +c \right )}-1830 B \,{\mathrm e}^{3 i \left (d x +c \right )}-630 B \,{\mathrm e}^{i \left (d x +c \right )}+195 A \,{\mathrm e}^{8 i \left (d x +c \right )}+975 A \,{\mathrm e}^{7 i \left (d x +c \right )}+150 C \,{\mathrm e}^{7 i \left (d x +c \right )}-90 \,{\mathrm e}^{8 i \left (d x +c \right )} B +30 C \,{\mathrm e}^{8 i \left (d x +c \right )}-2094 B \,{\mathrm e}^{4 i \left (d x +c \right )}\right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {13 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 a^{3} d}-\frac {3 B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{3} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{a^{3} d}-\frac {13 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a^{3} d}+\frac {3 B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{3} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{a^{3} d}\) | \(466\) |
Input:
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^3,x,meth od=_RETURNVERBOSE)
Output:
1/80*(-520*(cos(2*d*x+2*c)+1)*(A-6/13*B+2/13*C)*ln(tan(1/2*d*x+1/2*c)-1)+5 20*(cos(2*d*x+2*c)+1)*(A-6/13*B+2/13*C)*ln(tan(1/2*d*x+1/2*c)+1)-239*sec(1 /2*d*x+1/2*c)^4*(18/239*(29*A-14*B+4*C)*cos(2*d*x+2*c)+(A-114/239*B+34/239 *C)*cos(3*d*x+3*c)+2/239*(76/3*A-12*B+11/3*C)*cos(4*d*x+4*c)+1/239*(777*A- 382*B+102*C)*cos(d*x+c)+1354/717*A-228/239*B+194/717*C)*tan(1/2*d*x+1/2*c) )/d/a^3/(cos(2*d*x+2*c)+1)
Time = 0.10 (sec) , antiderivative size = 328, normalized size of antiderivative = 1.56 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {15 \, {\left ({\left (13 \, A - 6 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (13 \, A - 6 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (13 \, A - 6 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (13 \, A - 6 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left ({\left (13 \, A - 6 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (13 \, A - 6 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (13 \, A - 6 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (13 \, A - 6 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (4 \, {\left (76 \, A - 36 \, B + 11 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (239 \, A - 114 \, B + 34 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (479 \, A - 234 \, B + 64 \, C\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right ) - 15 \, A\right )} \sin \left (d x + c\right )}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}} \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^3, x, algorithm="fricas")
Output:
1/60*(15*((13*A - 6*B + 2*C)*cos(d*x + c)^5 + 3*(13*A - 6*B + 2*C)*cos(d*x + c)^4 + 3*(13*A - 6*B + 2*C)*cos(d*x + c)^3 + (13*A - 6*B + 2*C)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 15*((13*A - 6*B + 2*C)*cos(d*x + c)^5 + 3 *(13*A - 6*B + 2*C)*cos(d*x + c)^4 + 3*(13*A - 6*B + 2*C)*cos(d*x + c)^3 + (13*A - 6*B + 2*C)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(4*(76*A - 36*B + 11*C)*cos(d*x + c)^4 + 3*(239*A - 114*B + 34*C)*cos(d*x + c)^3 + (4 79*A - 234*B + 64*C)*cos(d*x + c)^2 + 15*(3*A - 2*B)*cos(d*x + c) - 15*A)* sin(d*x + c))/(a^3*d*cos(d*x + c)^5 + 3*a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos (d*x + c)^3 + a^3*d*cos(d*x + c)^2)
Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\text {Timed out} \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**3/(a+a*cos(d*x+c))* *3,x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 493 vs. \(2 (198) = 396\).
Time = 0.06 (sec) , antiderivative size = 493, normalized size of antiderivative = 2.35 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^3, x, algorithm="maxima")
Output:
-1/60*(A*(60*(5*sin(d*x + c)/(cos(d*x + c) + 1) - 7*sin(d*x + c)^3/(cos(d* x + c) + 1)^3)/(a^3 - 2*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a^3*sin( d*x + c)^4/(cos(d*x + c) + 1)^4) + (465*sin(d*x + c)/(cos(d*x + c) + 1) + 40*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 - 390*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 390*log(sin (d*x + c)/(cos(d*x + c) + 1) - 1)/a^3) - 3*B*(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) + C*((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))/d
Time = 0.18 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.37 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\frac {30 \, {\left (13 \, A - 6 \, B + 2 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {30 \, {\left (13 \, A - 6 \, B + 2 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {60 \, {\left (7 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 20 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 465 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 255 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 105 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^3, x, algorithm="giac")
Output:
1/60*(30*(13*A - 6*B + 2*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 30*(1 3*A - 6*B + 2*C)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3 + 60*(7*A*tan(1/2* d*x + 1/2*c)^3 - 2*B*tan(1/2*d*x + 1/2*c)^3 - 5*A*tan(1/2*d*x + 1/2*c) + 2 *B*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*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 + 40*A*a^12*tan(1/2*d*x + 1/2*c)^3 - 30*B*a^12*tan(1/2*d* x + 1/2*c)^3 + 20*C*a^12*tan(1/2*d*x + 1/2*c)^3 + 465*A*a^12*tan(1/2*d*x + 1/2*c) - 255*B*a^12*tan(1/2*d*x + 1/2*c) + 105*C*a^12*tan(1/2*d*x + 1/2*c ))/a^15)/d
Time = 0.33 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.18 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (7\,A-2\,B\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (5\,A-2\,B\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (5\,A-3\,B+C\right )}{4\,a^3}-\frac {2\,B-10\,A+2\,C}{4\,a^3}+\frac {3\,\left (A-B+C\right )}{2\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {5\,A-3\,B+C}{12\,a^3}+\frac {A-B+C}{4\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A-B+C\right )}{20\,a^3\,d}+\frac {2\,\mathrm {atanh}\left (\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {13\,A}{2}-3\,B+C\right )}{13\,A-6\,B+2\,C}\right )\,\left (\frac {13\,A}{2}-3\,B+C\right )}{a^3\,d} \] Input:
int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^3*(a + a*cos(c + d*x))^3),x)
Output:
(tan(c/2 + (d*x)/2)^3*(7*A - 2*B) - tan(c/2 + (d*x)/2)*(5*A - 2*B))/(d*(a^ 3*tan(c/2 + (d*x)/2)^4 - 2*a^3*tan(c/2 + (d*x)/2)^2 + a^3)) - (tan(c/2 + ( d*x)/2)*((3*(5*A - 3*B + C))/(4*a^3) - (2*B - 10*A + 2*C)/(4*a^3) + (3*(A - B + C))/(2*a^3)))/d - (tan(c/2 + (d*x)/2)^3*((5*A - 3*B + C)/(12*a^3) + (A - B + C)/(4*a^3)))/d - (tan(c/2 + (d*x)/2)^5*(A - B + C))/(20*a^3*d) + (2*atanh((2*tan(c/2 + (d*x)/2)*((13*A)/2 - 3*B + C))/(13*A - 6*B + 2*C))*( (13*A)/2 - 3*B + C))/(a^3*d)
Time = 0.17 (sec) , antiderivative size = 643, normalized size of antiderivative = 3.06 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^3,x)
Output:
( - 390*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**4*a + 180*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**4*b - 60*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**4*c + 780*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**2*a - 360 *log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**2*b + 120*log(tan((c + d*x)/2 ) - 1)*tan((c + d*x)/2)**2*c - 390*log(tan((c + d*x)/2) - 1)*a + 180*log(t an((c + d*x)/2) - 1)*b - 60*log(tan((c + d*x)/2) - 1)*c + 390*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**4*a - 180*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**4*b + 60*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**4*c - 780 *log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**2*a + 360*log(tan((c + d*x)/2 ) + 1)*tan((c + d*x)/2)**2*b - 120*log(tan((c + d*x)/2) + 1)*tan((c + d*x) /2)**2*c + 390*log(tan((c + d*x)/2) + 1)*a - 180*log(tan((c + d*x)/2) + 1) *b + 60*log(tan((c + d*x)/2) + 1)*c - 3*tan((c + d*x)/2)**9*a + 3*tan((c + d*x)/2)**9*b - 3*tan((c + d*x)/2)**9*c - 34*tan((c + d*x)/2)**7*a + 24*ta n((c + d*x)/2)**7*b - 14*tan((c + d*x)/2)**7*c - 388*tan((c + d*x)/2)**5*a + 198*tan((c + d*x)/2)**5*b - 68*tan((c + d*x)/2)**5*c + 1310*tan((c + d* x)/2)**3*a - 600*tan((c + d*x)/2)**3*b + 190*tan((c + d*x)/2)**3*c - 765*t an((c + d*x)/2)*a + 375*tan((c + d*x)/2)*b - 105*tan((c + d*x)/2)*c)/(60*a **3*d*(tan((c + d*x)/2)**4 - 2*tan((c + d*x)/2)**2 + 1))