Integrand size = 41, antiderivative size = 165 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {(7 A-4 B+2 C) \text {arctanh}(\sin (c+d x))}{2 a^2 d}-\frac {2 (8 A-5 B+2 C) \tan (c+d x)}{3 a^2 d}+\frac {(7 A-4 B+2 C) \sec (c+d x) \tan (c+d x)}{2 a^2 d}-\frac {(8 A-5 B+2 C) \sec (c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {(A-B+C) \sec (c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2} \]
1/2*(7*A-4*B+2*C)*arctanh(sin(d*x+c))/a^2/d-2/3*(8*A-5*B+2*C)*tan(d*x+c)/a ^2/d+1/2*(7*A-4*B+2*C)*sec(d*x+c)*tan(d*x+c)/a^2/d-1/3*(8*A-5*B+2*C)*sec(d *x+c)*tan(d*x+c)/a^2/d/(1+cos(d*x+c))-1/3*(A-B+C)*sec(d*x+c)*tan(d*x+c)/d/ (a+a*cos(d*x+c))^2
Leaf count is larger than twice the leaf count of optimal. \(578\) vs. \(2(165)=330\).
Time = 8.46 (sec) , antiderivative size = 578, normalized size of antiderivative = 3.50 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^2} \, dx=-\frac {2 (7 A-4 B+2 C) \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d (a+a \cos (c+d x))^2}+\frac {2 (7 A-4 B+2 C) \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d (a+a \cos (c+d x))^2}+\frac {A \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right )}{d (a+a \cos (c+d x))^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {A \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right )}{d (a+a \cos (c+d x))^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {4 \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (2 A \sin \left (\frac {1}{2} (c+d x)\right )-B \sin \left (\frac {1}{2} (c+d x)\right )\right )}{d (a+a \cos (c+d x))^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {4 \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (2 A \sin \left (\frac {1}{2} (c+d x)\right )-B \sin \left (\frac {1}{2} (c+d x)\right )\right )}{d (a+a \cos (c+d x))^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {2 \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \left (A \sin \left (\frac {1}{2} (c+d x)\right )-B \sin \left (\frac {1}{2} (c+d x)\right )+C \sin \left (\frac {1}{2} (c+d x)\right )\right )}{3 d (a+a \cos (c+d x))^2}-\frac {4 \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \left (10 A \sin \left (\frac {1}{2} (c+d x)\right )-7 B \sin \left (\frac {1}{2} (c+d x)\right )+4 C \sin \left (\frac {1}{2} (c+d x)\right )\right )}{3 d (a+a \cos (c+d x))^2} \]
(-2*(7*A - 4*B + 2*C)*Cos[c/2 + (d*x)/2]^4*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/(d*(a + a*Cos[c + d*x])^2) + (2*(7*A - 4*B + 2*C)*Cos[c/2 + (d* x)/2]^4*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/(d*(a + a*Cos[c + d*x])^ 2) + (A*Cos[c/2 + (d*x)/2]^4)/(d*(a + a*Cos[c + d*x])^2*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) - (A*Cos[c/2 + (d*x)/2]^4)/(d*(a + a*Cos[c + d*x])^ 2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) - (4*Cos[c/2 + (d*x)/2]^4*(2*A* Sin[(c + d*x)/2] - B*Sin[(c + d*x)/2]))/(d*(a + a*Cos[c + d*x])^2*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) - (4*Cos[c/2 + (d*x)/2]^4*(2*A*Sin[(c + d*x )/2] - B*Sin[(c + d*x)/2]))/(d*(a + a*Cos[c + d*x])^2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) - (2*Cos[c/2 + (d*x)/2]^4*Sec[(c + d*x)/2]^3*(A*Sin[(c + d*x)/2] - B*Sin[(c + d*x)/2] + C*Sin[(c + d*x)/2]))/(3*d*(a + a*Cos[c + d*x])^2) - (4*Cos[c/2 + (d*x)/2]^4*Sec[(c + d*x)/2]*(10*A*Sin[(c + d*x)/2] - 7*B*Sin[(c + d*x)/2] + 4*C*Sin[(c + d*x)/2]))/(3*d*(a + a*Cos[c + d*x]) ^2)
Time = 1.03 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.98, 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, 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.
| \(\displaystyle \int \frac {\sec ^3(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 3520 |
| \(\displaystyle \frac {\int \frac {(a (5 A-2 B+2 C)-3 a (A-B) \cos (c+d x)) \sec ^3(c+d x)}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (5 A-2 B+2 C)-3 a (A-B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\int \left (3 a^2 (7 A-4 B+2 C)-2 a^2 (8 A-5 B+2 C) \cos (c+d x)\right ) \sec ^3(c+d x)dx}{a^2}-\frac {(8 A-5 B+2 C) \tan (c+d x) \sec (c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {3 a^2 (7 A-4 B+2 C)-2 a^2 (8 A-5 B+2 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx}{a^2}-\frac {(8 A-5 B+2 C) \tan (c+d x) \sec (c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {\frac {3 a^2 (7 A-4 B+2 C) \int \sec ^3(c+d x)dx-2 a^2 (8 A-5 B+2 C) \int \sec ^2(c+d x)dx}{a^2}-\frac {(8 A-5 B+2 C) \tan (c+d x) \sec (c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 a^2 (7 A-4 B+2 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-2 a^2 (8 A-5 B+2 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx}{a^2}-\frac {(8 A-5 B+2 C) \tan (c+d x) \sec (c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (8 A-5 B+2 C) \int 1d(-\tan (c+d x))}{d}+3 a^2 (7 A-4 B+2 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}-\frac {(8 A-5 B+2 C) \tan (c+d x) \sec (c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\frac {3 a^2 (7 A-4 B+2 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {2 a^2 (8 A-5 B+2 C) \tan (c+d x)}{d}}{a^2}-\frac {(8 A-5 B+2 C) \tan (c+d x) \sec (c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {\frac {3 a^2 (7 A-4 B+2 C) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {2 a^2 (8 A-5 B+2 C) \tan (c+d x)}{d}}{a^2}-\frac {(8 A-5 B+2 C) \tan (c+d x) \sec (c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 a^2 (7 A-4 B+2 C) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {2 a^2 (8 A-5 B+2 C) \tan (c+d x)}{d}}{a^2}-\frac {(8 A-5 B+2 C) \tan (c+d x) \sec (c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {3 a^2 (7 A-4 B+2 C) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {2 a^2 (8 A-5 B+2 C) \tan (c+d x)}{d}}{a^2}-\frac {(8 A-5 B+2 C) \tan (c+d x) \sec (c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
-1/3*((A - B + C)*Sec[c + d*x]*Tan[c + d*x])/(d*(a + a*Cos[c + d*x])^2) + (-(((8*A - 5*B + 2*C)*Sec[c + d*x]*Tan[c + d*x])/(d*(1 + Cos[c + d*x]))) + ((-2*a^2*(8*A - 5*B + 2*C)*Tan[c + d*x])/d + 3*a^2*(7*A - 4*B + 2*C)*(Arc Tanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/a^2)/(3*a^2 )
3.4.53.3.1 Defintions of rubi rules used
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 = 2.87 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.07
| method | result | size |
| parallelrisch | \(\frac {-21 \left (A -\frac {4 B}{7}+\frac {2 C}{7}\right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+21 \left (A -\frac {4 B}{7}+\frac {2 C}{7}\right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-8 \left (\frac {\left (\frac {43 A}{4}-7 B +\frac {5 C}{2}\right ) \cos \left (2 d x +2 c \right )}{4}+\left (A -\frac {5 B}{8}+\frac {C}{4}\right ) \cos \left (3 d x +3 c \right )+\frac {3 \left (5 A -\frac {7 B}{2}+C \right ) \cos \left (d x +c \right )}{4}+\frac {37 A}{16}-\frac {7 B}{4}+\frac {5 C}{8}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{2} \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(176\) |
| derivativedivides | \(\frac {-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3}+\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{3}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{3}-7 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+5 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -\frac {-5 A +2 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (7 A -4 B +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {A}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-5 A +2 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-7 A +4 B -2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {A}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}}{2 d \,a^{2}}\) | \(209\) |
| default | \(\frac {-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3}+\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{3}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{3}-7 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+5 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -\frac {-5 A +2 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (7 A -4 B +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {A}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-5 A +2 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-7 A +4 B -2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {A}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}}{2 d \,a^{2}}\) | \(209\) |
| norman | \(\frac {\frac {\left (14 A -7 B +3 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {\left (A -B +C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}-\frac {\left (7 A -13 B +C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}-\frac {\left (7 A -5 B +3 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a d}-\frac {\left (13 A -9 B +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}+\frac {\left (16 A -7 B +C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2} a}-\frac {\left (7 A -4 B +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{2} d}+\frac {\left (7 A -4 B +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{2} d}\) | \(260\) |
| risch | \(-\frac {i \left (21 A \,{\mathrm e}^{6 i \left (d x +c \right )}-12 B \,{\mathrm e}^{6 i \left (d x +c \right )}+6 C \,{\mathrm e}^{6 i \left (d x +c \right )}+63 A \,{\mathrm e}^{5 i \left (d x +c \right )}-36 B \,{\mathrm e}^{5 i \left (d x +c \right )}+18 C \,{\mathrm e}^{5 i \left (d x +c \right )}+98 A \,{\mathrm e}^{4 i \left (d x +c \right )}-56 B \,{\mathrm e}^{4 i \left (d x +c \right )}+20 C \,{\mathrm e}^{4 i \left (d x +c \right )}+126 A \,{\mathrm e}^{3 i \left (d x +c \right )}-84 B \,{\mathrm e}^{3 i \left (d x +c \right )}+36 C \,{\mathrm e}^{3 i \left (d x +c \right )}+97 A \,{\mathrm e}^{2 i \left (d x +c \right )}-64 B \,{\mathrm e}^{2 i \left (d x +c \right )}+22 C \,{\mathrm e}^{2 i \left (d x +c \right )}+75 A \,{\mathrm e}^{i \left (d x +c \right )}-48 B \,{\mathrm e}^{i \left (d x +c \right )}+18 C \,{\mathrm e}^{i \left (d x +c \right )}+32 A -20 B +8 C \right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {7 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a^{2} d}+\frac {2 B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{2} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{a^{2} d}+\frac {7 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 a^{2} d}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{a^{2} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{a^{2} d}\) | \(394\) |
1/6*(-21*(A-4/7*B+2/7*C)*(1+cos(2*d*x+2*c))*ln(tan(1/2*d*x+1/2*c)-1)+21*(A -4/7*B+2/7*C)*(1+cos(2*d*x+2*c))*ln(tan(1/2*d*x+1/2*c)+1)-8*(1/4*(43/4*A-7 *B+5/2*C)*cos(2*d*x+2*c)+(A-5/8*B+1/4*C)*cos(3*d*x+3*c)+3/4*(5*A-7/2*B+C)* cos(d*x+c)+37/16*A-7/4*B+5/8*C)*tan(1/2*d*x+1/2*c)*sec(1/2*d*x+1/2*c)^2)/d /a^2/(1+cos(2*d*x+2*c))
Time = 0.27 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.53 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {3 \, {\left ({\left (7 \, A - 4 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (7 \, A - 4 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (7 \, A - 4 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left ({\left (7 \, A - 4 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (7 \, A - 4 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (7 \, A - 4 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (4 \, {\left (8 \, A - 5 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (43 \, A - 28 \, B + 10 \, C\right )} \cos \left (d x + c\right )^{2} + 6 \, {\left (A - B\right )} \cos \left (d x + c\right ) - 3 \, A\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^2, x, algorithm="fricas")
1/12*(3*((7*A - 4*B + 2*C)*cos(d*x + c)^4 + 2*(7*A - 4*B + 2*C)*cos(d*x + c)^3 + (7*A - 4*B + 2*C)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 3*((7*A - 4*B + 2*C)*cos(d*x + c)^4 + 2*(7*A - 4*B + 2*C)*cos(d*x + c)^3 + (7*A - 4 *B + 2*C)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(4*(8*A - 5*B + 2*C)* cos(d*x + c)^3 + (43*A - 28*B + 10*C)*cos(d*x + c)^2 + 6*(A - B)*cos(d*x + c) - 3*A)*sin(d*x + c))/(a^2*d*cos(d*x + c)^4 + 2*a^2*d*cos(d*x + c)^3 + a^2*d*cos(d*x + c)^2)
\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {\int \frac {A \sec ^{3}{\left (c + d x \right )}}{\cos ^{2}{\left (c + d x \right )} + 2 \cos {\left (c + d x \right )} + 1}\, dx + \int \frac {B \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}}{\cos ^{2}{\left (c + d x \right )} + 2 \cos {\left (c + d x \right )} + 1}\, dx + \int \frac {C \cos ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}}{\cos ^{2}{\left (c + d x \right )} + 2 \cos {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]
(Integral(A*sec(c + d*x)**3/(cos(c + d*x)**2 + 2*cos(c + d*x) + 1), x) + I ntegral(B*cos(c + d*x)*sec(c + d*x)**3/(cos(c + d*x)**2 + 2*cos(c + d*x) + 1), x) + Integral(C*cos(c + d*x)**2*sec(c + d*x)**3/(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. 431 vs. \(2 (155) = 310\).
Time = 0.24 (sec) , antiderivative size = 431, normalized size of antiderivative = 2.61 \[ \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))^2} \, dx=-\frac {A {\left (\frac {6 \, {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} - \frac {2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {21 \, \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 {21 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {21 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )} - B {\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 )} + C {\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 )}}{6 \, d} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^2, x, algorithm="maxima")
-1/6*(A*(6*(3*sin(d*x + c)/(cos(d*x + c) + 1) - 5*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^2 - 2*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a^2*sin(d* x + c)^4/(cos(d*x + c) + 1)^4) + (21*sin(d*x + c)/(cos(d*x + c) + 1) + sin (d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 21*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 21*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^2) - B*((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(sin(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/(c os(d*x + c) + 1)^2)*(cos(d*x + c) + 1))) + C*((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)/(co s(d*x + c) + 1) + 1)/a^2 + 6*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^2) )/d
Time = 0.36 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.42 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {\frac {3 \, {\left (7 \, A - 4 \, B + 2 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {3 \, {\left (7 \, A - 4 \, B + 2 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac {6 \, {\left (5 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{2}} - \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 21 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^2, x, algorithm="giac")
1/6*(3*(7*A - 4*B + 2*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 3*(7*A - 4*B + 2*C)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^2 + 6*(5*A*tan(1/2*d*x + 1/2*c)^3 - 2*B*tan(1/2*d*x + 1/2*c)^3 - 3*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^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 + 21*A*a^4*tan(1/2*d*x + 1/2*c) - 15*B*a^4*tan(1/2*d*x + 1/2*c) + 9*C*a^ 4*tan(1/2*d*x + 1/2*c))/a^6)/d
Time = 1.54 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.16 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (5\,A-2\,B\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,A-2\,B\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A-B+C\right )}{2\,a^2}+\frac {4\,A-2\,B}{2\,a^2}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A-B+C\right )}{6\,a^2\,d}+\frac {2\,\mathrm {atanh}\left (\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {7\,A}{2}-2\,B+C\right )}{7\,A-4\,B+2\,C}\right )\,\left (\frac {7\,A}{2}-2\,B+C\right )}{a^2\,d} \]
(tan(c/2 + (d*x)/2)^3*(5*A - 2*B) - tan(c/2 + (d*x)/2)*(3*A - 2*B))/(d*(a^ 2*tan(c/2 + (d*x)/2)^4 - 2*a^2*tan(c/2 + (d*x)/2)^2 + a^2)) - (tan(c/2 + ( d*x)/2)*((3*(A - B + C))/(2*a^2) + (4*A - 2*B)/(2*a^2)))/d - (tan(c/2 + (d *x)/2)^3*(A - B + C))/(6*a^2*d) + (2*atanh((2*tan(c/2 + (d*x)/2)*((7*A)/2 - 2*B + C))/(7*A - 4*B + 2*C))*((7*A)/2 - 2*B + C))/(a^2*d)