Integrand size = 40, antiderivative size = 145 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {(3 B-C) \text {arctanh}(\sin (c+d x))}{a^3 d}+\frac {2 (36 B-11 C) \tan (c+d x)}{15 a^3 d}-\frac {(B-C) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(9 B-4 C) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(3 B-C) \tan (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )} \] Output:
-(3*B-C)*arctanh(sin(d*x+c))/a^3/d+2/15*(36*B-11*C)*tan(d*x+c)/a^3/d-1/5*( B-C)*tan(d*x+c)/d/(a+a*cos(d*x+c))^3-1/15*(9*B-4*C)*tan(d*x+c)/a/d/(a+a*co s(d*x+c))^2-(3*B-C)*tan(d*x+c)/d/(a^3+a^3*cos(d*x+c))
Leaf count is larger than twice the leaf count of optimal. \(482\) vs. \(2(145)=290\).
Time = 4.12 (sec) , antiderivative size = 482, normalized size of antiderivative = 3.32 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {960 (3 B-C) \cos ^6\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sec (c) \sec (c+d x) \left (-5 (51 B-32 C) \sin \left (\frac {d x}{2}\right )+(567 B-167 C) \sin \left (\frac {3 d x}{2}\right )-600 B \sin \left (c-\frac {d x}{2}\right )+170 C \sin \left (c-\frac {d x}{2}\right )+375 B \sin \left (c+\frac {d x}{2}\right )-170 C \sin \left (c+\frac {d x}{2}\right )-480 B \sin \left (2 c+\frac {d x}{2}\right )+160 C \sin \left (2 c+\frac {d x}{2}\right )-60 B \sin \left (c+\frac {3 d x}{2}\right )+75 C \sin \left (c+\frac {3 d x}{2}\right )+402 B \sin \left (2 c+\frac {3 d x}{2}\right )-167 C \sin \left (2 c+\frac {3 d x}{2}\right )-225 B \sin \left (3 c+\frac {3 d x}{2}\right )+75 C \sin \left (3 c+\frac {3 d x}{2}\right )+315 B \sin \left (c+\frac {5 d x}{2}\right )-95 C \sin \left (c+\frac {5 d x}{2}\right )+30 B \sin \left (2 c+\frac {5 d x}{2}\right )+15 C \sin \left (2 c+\frac {5 d x}{2}\right )+240 B \sin \left (3 c+\frac {5 d x}{2}\right )-95 C \sin \left (3 c+\frac {5 d x}{2}\right )-45 B \sin \left (4 c+\frac {5 d x}{2}\right )+15 C \sin \left (4 c+\frac {5 d x}{2}\right )+72 B \sin \left (2 c+\frac {7 d x}{2}\right )-22 C \sin \left (2 c+\frac {7 d x}{2}\right )+15 B \sin \left (3 c+\frac {7 d x}{2}\right )+57 B \sin \left (4 c+\frac {7 d x}{2}\right )-22 C \sin \left (4 c+\frac {7 d x}{2}\right )\right )}{120 a^3 d (1+\cos (c+d x))^3} \] Input:
Integrate[((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:
(960*(3*B - C)*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]* Sec[c]*Sec[c + d*x]*(-5*(51*B - 32*C)*Sin[(d*x)/2] + (567*B - 167*C)*Sin[( 3*d*x)/2] - 600*B*Sin[c - (d*x)/2] + 170*C*Sin[c - (d*x)/2] + 375*B*Sin[c + (d*x)/2] - 170*C*Sin[c + (d*x)/2] - 480*B*Sin[2*c + (d*x)/2] + 160*C*Sin [2*c + (d*x)/2] - 60*B*Sin[c + (3*d*x)/2] + 75*C*Sin[c + (3*d*x)/2] + 402* B*Sin[2*c + (3*d*x)/2] - 167*C*Sin[2*c + (3*d*x)/2] - 225*B*Sin[3*c + (3*d *x)/2] + 75*C*Sin[3*c + (3*d*x)/2] + 315*B*Sin[c + (5*d*x)/2] - 95*C*Sin[c + (5*d*x)/2] + 30*B*Sin[2*c + (5*d*x)/2] + 15*C*Sin[2*c + (5*d*x)/2] + 24 0*B*Sin[3*c + (5*d*x)/2] - 95*C*Sin[3*c + (5*d*x)/2] - 45*B*Sin[4*c + (5*d *x)/2] + 15*C*Sin[4*c + (5*d*x)/2] + 72*B*Sin[2*c + (7*d*x)/2] - 22*C*Sin[ 2*c + (7*d*x)/2] + 15*B*Sin[3*c + (7*d*x)/2] + 57*B*Sin[4*c + (7*d*x)/2] - 22*C*Sin[4*c + (7*d*x)/2]))/(120*a^3*d*(1 + Cos[c + d*x])^3)
Time = 1.25 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.11, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {3042, 3508, 3042, 3457, 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.
| \(\displaystyle \int \frac {\sec ^3(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 3508 |
| \(\displaystyle \int \frac {\sec ^2(c+d x) (B+C \cos (c+d x))}{(a \cos (c+d x)+a)^3}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {B+C \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\int \frac {(a (6 B-C)-3 a (B-C) \cos (c+d x)) \sec ^2(c+d x)}{(\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {(B-C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (6 B-C)-3 a (B-C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {(B-C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\int \frac {\left (a^2 (27 B-7 C)-2 a^2 (9 B-4 C) \cos (c+d x)\right ) \sec ^2(c+d x)}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {a (9 B-4 C) \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(B-C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {a^2 (27 B-7 C)-2 a^2 (9 B-4 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}-\frac {a (9 B-4 C) \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(B-C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\frac {\int \left (2 a^3 (36 B-11 C)-15 a^3 (3 B-C) \cos (c+d x)\right ) \sec ^2(c+d x)dx}{a^2}-\frac {15 a^2 (3 B-C) \tan (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (9 B-4 C) \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(B-C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {2 a^3 (36 B-11 C)-15 a^3 (3 B-C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx}{a^2}-\frac {15 a^2 (3 B-C) \tan (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (9 B-4 C) \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(B-C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {\frac {\frac {2 a^3 (36 B-11 C) \int \sec ^2(c+d x)dx-15 a^3 (3 B-C) \int \sec (c+d x)dx}{a^2}-\frac {15 a^2 (3 B-C) \tan (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (9 B-4 C) \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(B-C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {2 a^3 (36 B-11 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx-15 a^3 (3 B-C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {15 a^2 (3 B-C) \tan (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (9 B-4 C) \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(B-C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {\frac {\frac {-\frac {2 a^3 (36 B-11 C) \int 1d(-\tan (c+d x))}{d}-15 a^3 (3 B-C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {15 a^2 (3 B-C) \tan (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (9 B-4 C) \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(B-C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\frac {\frac {\frac {2 a^3 (36 B-11 C) \tan (c+d x)}{d}-15 a^3 (3 B-C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {15 a^2 (3 B-C) \tan (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (9 B-4 C) \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(B-C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {\frac {\frac {2 a^3 (36 B-11 C) \tan (c+d x)}{d}-\frac {15 a^3 (3 B-C) \text {arctanh}(\sin (c+d x))}{d}}{a^2}-\frac {15 a^2 (3 B-C) \tan (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (9 B-4 C) \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(B-C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
Input:
Int[((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*((B - C)*Tan[c + d*x])/(d*(a + a*Cos[c + d*x])^3) + (-1/3*(a*(9*B - 4 *C)*Tan[c + d*x])/(d*(a + a*Cos[c + d*x])^2) + ((-15*a^2*(3*B - C)*Tan[c + d*x])/(d*(a + a*Cos[c + d*x])) + ((-15*a^3*(3*B - C)*ArcTanh[Sin[c + d*x] ])/d + (2*a^3*(36*B - 11*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[1/b^2 Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
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 = 0.52 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.99
| method | result | size |
| parallelrisch | \(\frac {3 \cos \left (d x +c \right ) \left (B -\frac {C}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-3 \cos \left (d x +c \right ) \left (B -\frac {C}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\frac {57 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\frac {\left (B -\frac {17 C}{57}\right ) \cos \left (2 d x +2 c \right )}{2}+\frac {\left (2 B -\frac {11 C}{18}\right ) \cos \left (3 d x +3 c \right )}{19}+\left (B -\frac {97 C}{342}\right ) \cos \left (d x +c \right )+\frac {67 B}{114}-\frac {17 C}{114}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{20}}{\cos \left (d x +c \right ) a^{3} d}\) | \(143\) |
| derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B -\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -7 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-12 B +4 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {4 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (12 B -4 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {4 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}}{4 d \,a^{3}}\) | \(162\) |
| default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B -\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -7 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-12 B +4 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {4 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (12 B -4 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {4 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}}{4 d \,a^{3}}\) | \(162\) |
| norman | \(\frac {\frac {\left (B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{20 a d}+\frac {\left (3 B -2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{6 a d}-\frac {\left (9 B -2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d a}+\frac {\left (15 B -2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 d a}+\frac {\left (25 B -7 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d a}+\frac {\left (83 B -33 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{20 d a}-\frac {\left (209 B -69 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{20 d a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2} a^{2}}+\frac {\left (3 B -C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3} d}-\frac {\left (3 B -C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3} d}\) | \(268\) |
| risch | \(\frac {2 i \left (45 B \,{\mathrm e}^{6 i \left (d x +c \right )}-15 C \,{\mathrm e}^{6 i \left (d x +c \right )}+225 B \,{\mathrm e}^{5 i \left (d x +c \right )}-75 C \,{\mathrm e}^{5 i \left (d x +c \right )}+480 B \,{\mathrm e}^{4 i \left (d x +c \right )}-160 C \,{\mathrm e}^{4 i \left (d x +c \right )}+600 B \,{\mathrm e}^{3 i \left (d x +c \right )}-170 C \,{\mathrm e}^{3 i \left (d x +c \right )}+567 B \,{\mathrm e}^{2 i \left (d x +c \right )}-167 C \,{\mathrm e}^{2 i \left (d x +c \right )}+315 B \,{\mathrm e}^{i \left (d x +c \right )}-95 C \,{\mathrm e}^{i \left (d x +c \right )}+72 B -22 C \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {3 B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{3} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{a^{3} d}+\frac {3 B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{3} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{a^{3} d}\) | \(275\) |
Input:
int((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^3,x,method =_RETURNVERBOSE)
Output:
3/20*(20*cos(d*x+c)*(B-1/3*C)*ln(tan(1/2*d*x+1/2*c)-1)-20*cos(d*x+c)*(B-1/ 3*C)*ln(tan(1/2*d*x+1/2*c)+1)+19*tan(1/2*d*x+1/2*c)*(1/2*(B-17/57*C)*cos(2 *d*x+2*c)+1/19*(2*B-11/18*C)*cos(3*d*x+3*c)+(B-97/342*C)*cos(d*x+c)+67/114 *B-17/114*C)*sec(1/2*d*x+1/2*c)^4)/d/a^3/cos(d*x+c)
Time = 0.09 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.88 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {15 \, {\left ({\left (3 \, B - C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (3 \, B - C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, B - C\right )} \cos \left (d x + c\right )^{2} + {\left (3 \, B - C\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left ({\left (3 \, B - C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (3 \, B - C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, B - C\right )} \cos \left (d x + c\right )^{2} + {\left (3 \, B - C\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, {\left (36 \, B - 11 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (57 \, B - 17 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (117 \, B - 32 \, C\right )} \cos \left (d x + c\right ) + 15 \, B\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}} \] Input:
integrate((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:
-1/30*(15*((3*B - C)*cos(d*x + c)^4 + 3*(3*B - C)*cos(d*x + c)^3 + 3*(3*B - C)*cos(d*x + c)^2 + (3*B - C)*cos(d*x + c))*log(sin(d*x + c) + 1) - 15*( (3*B - C)*cos(d*x + c)^4 + 3*(3*B - C)*cos(d*x + c)^3 + 3*(3*B - C)*cos(d* x + c)^2 + (3*B - C)*cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*(2*(36*B - 1 1*C)*cos(d*x + c)^3 + 3*(57*B - 17*C)*cos(d*x + c)^2 + (117*B - 32*C)*cos( d*x + c) + 15*B)*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))
Timed out. \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\text {Timed out} \] Input:
integrate((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. 286 vs. \(2 (139) = 278\).
Time = 0.06 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.97 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {3 \, B {\left (\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} - \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} - C {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )}}{60 \, d} \] Input:
integrate((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*(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 = 190, normalized size of antiderivative = 1.31 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {\frac {60 \, {\left (3 \, B - C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {60 \, {\left (3 \, B - C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {120 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{3}} - \frac {3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 30 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 20 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 255 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 105 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \] Input:
integrate((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*(60*(3*B - C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 60*(3*B - C)* log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3 + 120*B*tan(1/2*d*x + 1/2*c)/((tan( 1/2*d*x + 1/2*c)^2 - 1)*a^3) - (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*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 + 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 = 0.28 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.16 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {B-C}{6\,a^3}+\frac {4\,B-2\,C}{12\,a^3}\right )}{d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,B}{2\,a^3}+\frac {3\,\left (B-C\right )}{4\,a^3}+\frac {4\,B-2\,C}{2\,a^3}\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (B-C\right )}{20\,a^3\,d}-\frac {2\,B\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^3\right )}-\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (3\,B-C\right )}{a^3\,d} \] Input:
int((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*((B - C)/(6*a^3) + (4*B - 2*C)/(12*a^3)))/d + (tan(c /2 + (d*x)/2)*((3*B)/(2*a^3) + (3*(B - C))/(4*a^3) + (4*B - 2*C)/(2*a^3))) /d + (tan(c/2 + (d*x)/2)^5*(B - C))/(20*a^3*d) - (2*B*tan(c/2 + (d*x)/2))/ (d*(a^3*tan(c/2 + (d*x)/2)^2 - a^3)) - (2*atanh(tan(c/2 + (d*x)/2))*(3*B - C))/(a^3*d)
Time = 0.17 (sec) , antiderivative size = 296, normalized size of antiderivative = 2.04 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {180 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} c -180 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b +60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) c -180 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} c +180 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b -60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) c +3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} b -3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} c +27 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} b -17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} c +225 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b -85 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} c -375 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +105 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) c}{60 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )} \] Input:
int((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:
(180*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**2*b - 60*log(tan((c + d*x )/2) - 1)*tan((c + d*x)/2)**2*c - 180*log(tan((c + d*x)/2) - 1)*b + 60*log (tan((c + d*x)/2) - 1)*c - 180*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)* *2*b + 60*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**2*c + 180*log(tan((c + d*x)/2) + 1)*b - 60*log(tan((c + d*x)/2) + 1)*c + 3*tan((c + d*x)/2)**7 *b - 3*tan((c + d*x)/2)**7*c + 27*tan((c + d*x)/2)**5*b - 17*tan((c + d*x) /2)**5*c + 225*tan((c + d*x)/2)**3*b - 85*tan((c + d*x)/2)**3*c - 375*tan( (c + d*x)/2)*b + 105*tan((c + d*x)/2)*c)/(60*a**3*d*(tan((c + d*x)/2)**2 - 1))