Integrand size = 39, antiderivative size = 162 \[ \int (a+a \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {1}{8} a^3 (28 A+20 B+15 C) x+\frac {a^3 A \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^3 (4 A+4 B+3 C) \sin (c+d x)}{8 d}+\frac {C (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac {(4 B+3 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{12 a d}+\frac {(12 A+20 B+15 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{24 d} \]
1/8*a^3*(28*A+20*B+15*C)*x+a^3*A*arctanh(sin(d*x+c))/d+5/8*a^3*(4*A+4*B+3* C)*sin(d*x+c)/d+1/4*C*(a+a*cos(d*x+c))^3*sin(d*x+c)/d+1/12*(4*B+3*C)*(a^2+ a^2*cos(d*x+c))^2*sin(d*x+c)/a/d+1/24*(12*A+20*B+15*C)*(a^3+a^3*cos(d*x+c) )*sin(d*x+c)/d
Time = 2.25 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.91 \[ \int (a+a \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {a^3 \left (336 A d x+240 B d x+180 C d x-96 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+96 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+24 (12 A+15 B+13 C) \sin (c+d x)+24 (A+3 B+4 C) \sin (2 (c+d x))+8 B \sin (3 (c+d x))+24 C \sin (3 (c+d x))+3 C \sin (4 (c+d x))\right )}{96 d} \]
(a^3*(336*A*d*x + 240*B*d*x + 180*C*d*x - 96*A*Log[Cos[(c + d*x)/2] - Sin[ (c + d*x)/2]] + 96*A*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 24*(12*A + 15*B + 13*C)*Sin[c + d*x] + 24*(A + 3*B + 4*C)*Sin[2*(c + d*x)] + 8*B*Sin [3*(c + d*x)] + 24*C*Sin[3*(c + d*x)] + 3*C*Sin[4*(c + d*x)]))/(96*d)
Time = 1.28 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.07, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3042, 3524, 3042, 3455, 3042, 3455, 27, 3042, 3447, 3042, 3502, 3042, 3214, 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.
| \(\displaystyle \int \sec (c+d x) (a \cos (c+d x)+a)^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 3524 |
| \(\displaystyle \frac {\int (\cos (c+d x) a+a)^3 (4 a A+a (4 B+3 C) \cos (c+d x)) \sec (c+d x)dx}{4 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3 \left (4 a A+a (4 B+3 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{4 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {\frac {1}{3} \int (\cos (c+d x) a+a)^2 \left (12 A a^2+(12 A+20 B+15 C) \cos (c+d x) a^2\right ) \sec (c+d x)dx+\frac {(4 B+3 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}}{4 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (12 A a^2+(12 A+20 B+15 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {(4 B+3 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}}{4 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {1}{2} \int 3 (\cos (c+d x) a+a) \left (8 A a^3+5 (4 A+4 B+3 C) \cos (c+d x) a^3\right ) \sec (c+d x)dx+\frac {(12 A+20 B+15 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )+\frac {(4 B+3 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}}{4 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \int (\cos (c+d x) a+a) \left (8 A a^3+5 (4 A+4 B+3 C) \cos (c+d x) a^3\right ) \sec (c+d x)dx+\frac {(12 A+20 B+15 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )+\frac {(4 B+3 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}}{4 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (8 A a^3+5 (4 A+4 B+3 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {(12 A+20 B+15 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )+\frac {(4 B+3 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}}{4 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \int \left (5 (4 A+4 B+3 C) \cos ^2(c+d x) a^4+8 A a^4+\left (8 A a^4+5 (4 A+4 B+3 C) a^4\right ) \cos (c+d x)\right ) \sec (c+d x)dx+\frac {(12 A+20 B+15 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )+\frac {(4 B+3 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}}{4 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \int \frac {5 (4 A+4 B+3 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^4+8 A a^4+\left (8 A a^4+5 (4 A+4 B+3 C) a^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {(12 A+20 B+15 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )+\frac {(4 B+3 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}}{4 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \left (\int \left (8 A a^4+(28 A+20 B+15 C) \cos (c+d x) a^4\right ) \sec (c+d x)dx+\frac {5 a^4 (4 A+4 B+3 C) \sin (c+d x)}{d}\right )+\frac {(12 A+20 B+15 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )+\frac {(4 B+3 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}}{4 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \left (\int \frac {8 A a^4+(28 A+20 B+15 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^4}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {5 a^4 (4 A+4 B+3 C) \sin (c+d x)}{d}\right )+\frac {(12 A+20 B+15 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )+\frac {(4 B+3 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}}{4 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \left (8 a^4 A \int \sec (c+d x)dx+\frac {5 a^4 (4 A+4 B+3 C) \sin (c+d x)}{d}+a^4 x (28 A+20 B+15 C)\right )+\frac {(12 A+20 B+15 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )+\frac {(4 B+3 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}}{4 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \left (8 a^4 A \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {5 a^4 (4 A+4 B+3 C) \sin (c+d x)}{d}+a^4 x (28 A+20 B+15 C)\right )+\frac {(12 A+20 B+15 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )+\frac {(4 B+3 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}}{4 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \left (\frac {8 a^4 A \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^4 (4 A+4 B+3 C) \sin (c+d x)}{d}+a^4 x (28 A+20 B+15 C)\right )+\frac {(12 A+20 B+15 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )+\frac {(4 B+3 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}}{4 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
(C*(a + a*Cos[c + d*x])^3*Sin[c + d*x])/(4*d) + (((4*B + 3*C)*(a^2 + a^2*C os[c + d*x])^2*Sin[c + d*x])/(3*d) + (((12*A + 20*B + 15*C)*(a^4 + a^4*Cos [c + d*x])*Sin[c + d*x])/(2*d) + (3*(a^4*(28*A + 20*B + 15*C)*x + (8*a^4*A *ArcTanh[Sin[c + d*x]])/d + (5*a^4*(4*A + 4*B + 3*C)*Sin[c + d*x])/d))/2)/ 3)/(4*a)
3.4.21.3.1 Defintions of rubi rules used
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_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
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)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1 ) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*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] && GtQ[m, 1/2] && !LtQ[n, -1 ] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
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[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x] )^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(b*d*(m + n + 2)) Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n*Simp[A*b*d*(m + n + 2) + C*(a*c*m + b*d*(n + 1)) + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, n} , x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !Lt Q[m, -2^(-1)] && NeQ[m + n + 2, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 5.04 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.72
| method | result | size |
| parallelrisch | \(-\frac {\left (A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-\frac {A}{4}-\frac {3 B}{4}-C \right ) \sin \left (2 d x +2 c \right )+\left (-\frac {B}{12}-\frac {C}{4}\right ) \sin \left (3 d x +3 c \right )-\frac {\sin \left (4 d x +4 c \right ) C}{32}+\left (-3 A -\frac {15 B}{4}-\frac {13 C}{4}\right ) \sin \left (d x +c \right )-\frac {7 x \left (A +\frac {5 B}{7}+\frac {15 C}{28}\right ) d}{2}\right ) a^{3}}{d}\) | \(117\) |
| parts | \(\frac {A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {\left (3 A \,a^{3}+B \,a^{3}\right ) \left (d x +c \right )}{d}+\frac {\left (B \,a^{3}+3 C \,a^{3}\right ) \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}+\frac {\left (A \,a^{3}+3 B \,a^{3}+3 C \,a^{3}\right ) \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (3 A \,a^{3}+3 B \,a^{3}+C \,a^{3}\right ) \sin \left (d x +c \right )}{d}+\frac {C \,a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(191\) |
| derivativedivides | \(\frac {A \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {B \,a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+C \,a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+3 A \,a^{3} \sin \left (d x +c \right )+3 B \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C \,a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 A \,a^{3} \left (d x +c \right )+3 B \sin \left (d x +c \right ) a^{3}+3 C \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{3} \left (d x +c \right )+C \,a^{3} \sin \left (d x +c \right )}{d}\) | \(245\) |
| default | \(\frac {A \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {B \,a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+C \,a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+3 A \,a^{3} \sin \left (d x +c \right )+3 B \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C \,a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 A \,a^{3} \left (d x +c \right )+3 B \sin \left (d x +c \right ) a^{3}+3 C \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{3} \left (d x +c \right )+C \,a^{3} \sin \left (d x +c \right )}{d}\) | \(245\) |
| risch | \(\frac {7 a^{3} A x}{2}+\frac {5 a^{3} B x}{2}+\frac {15 a^{3} C x}{8}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} A \,a^{3}}{2 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} A \,a^{3}}{2 d}+\frac {15 i {\mathrm e}^{-i \left (d x +c \right )} B \,a^{3}}{8 d}-\frac {13 i {\mathrm e}^{i \left (d x +c \right )} C \,a^{3}}{8 d}+\frac {13 i {\mathrm e}^{-i \left (d x +c \right )} C \,a^{3}}{8 d}-\frac {15 i {\mathrm e}^{i \left (d x +c \right )} B \,a^{3}}{8 d}+\frac {A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {\sin \left (4 d x +4 c \right ) C \,a^{3}}{32 d}+\frac {\sin \left (3 d x +3 c \right ) B \,a^{3}}{12 d}+\frac {\sin \left (3 d x +3 c \right ) C \,a^{3}}{4 d}+\frac {\sin \left (2 d x +2 c \right ) A \,a^{3}}{4 d}+\frac {3 \sin \left (2 d x +2 c \right ) B \,a^{3}}{4 d}+\frac {\sin \left (2 d x +2 c \right ) C \,a^{3}}{d}\) | \(287\) |
| norman | \(\frac {\left (\frac {7}{2} A \,a^{3}+\frac {5}{2} B \,a^{3}+\frac {15}{8} C \,a^{3}\right ) x +\left (35 A \,a^{3}+25 B \,a^{3}+\frac {75}{4} C \,a^{3}\right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (35 A \,a^{3}+25 B \,a^{3}+\frac {75}{4} C \,a^{3}\right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {7}{2} A \,a^{3}+\frac {5}{2} B \,a^{3}+\frac {15}{8} C \,a^{3}\right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {35}{2} A \,a^{3}+\frac {25}{2} B \,a^{3}+\frac {75}{8} C \,a^{3}\right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {35}{2} A \,a^{3}+\frac {25}{2} B \,a^{3}+\frac {75}{8} C \,a^{3}\right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {5 a^{3} \left (4 A +4 B +3 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {4 a^{3} \left (27 A +32 B +24 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {a^{3} \left (28 A +44 B +49 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {a^{3} \left (132 A +140 B +105 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {a^{3} \left (156 A +212 B +183 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {A \,a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {A \,a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\) | \(384\) |
-(A*ln(tan(1/2*d*x+1/2*c)-1)-A*ln(tan(1/2*d*x+1/2*c)+1)+(-1/4*A-3/4*B-C)*s in(2*d*x+2*c)+(-1/12*B-1/4*C)*sin(3*d*x+3*c)-1/32*sin(4*d*x+4*c)*C+(-3*A-1 5/4*B-13/4*C)*sin(d*x+c)-7/2*x*(A+5/7*B+15/28*C)*d)*a^3/d
Time = 0.29 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.81 \[ \int (a+a \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {3 \, {\left (28 \, A + 20 \, B + 15 \, C\right )} a^{3} d x + 12 \, A a^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 12 \, A a^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (6 \, C a^{3} \cos \left (d x + c\right )^{3} + 8 \, {\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 3 \, {\left (4 \, A + 12 \, B + 15 \, C\right )} a^{3} \cos \left (d x + c\right ) + 8 \, {\left (9 \, A + 11 \, B + 9 \, C\right )} a^{3}\right )} \sin \left (d x + c\right )}{24 \, d} \]
integrate((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c),x, algorithm="fricas")
1/24*(3*(28*A + 20*B + 15*C)*a^3*d*x + 12*A*a^3*log(sin(d*x + c) + 1) - 12 *A*a^3*log(-sin(d*x + c) + 1) + (6*C*a^3*cos(d*x + c)^3 + 8*(B + 3*C)*a^3* cos(d*x + c)^2 + 3*(4*A + 12*B + 15*C)*a^3*cos(d*x + c) + 8*(9*A + 11*B + 9*C)*a^3)*sin(d*x + c))/d
\[ \int (a+a \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=a^{3} \left (\int A \sec {\left (c + d x \right )}\, dx + \int 3 A \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 3 A \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int A \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int B \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 3 B \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 3 B \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int B \cos ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int C \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 3 C \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 3 C \cos ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int C \cos ^{5}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx\right ) \]
a**3*(Integral(A*sec(c + d*x), x) + Integral(3*A*cos(c + d*x)*sec(c + d*x) , x) + Integral(3*A*cos(c + d*x)**2*sec(c + d*x), x) + Integral(A*cos(c + d*x)**3*sec(c + d*x), x) + Integral(B*cos(c + d*x)*sec(c + d*x), x) + Inte gral(3*B*cos(c + d*x)**2*sec(c + d*x), x) + Integral(3*B*cos(c + d*x)**3*s ec(c + d*x), x) + Integral(B*cos(c + d*x)**4*sec(c + d*x), x) + Integral(C *cos(c + d*x)**2*sec(c + d*x), x) + Integral(3*C*cos(c + d*x)**3*sec(c + d *x), x) + Integral(3*C*cos(c + d*x)**4*sec(c + d*x), x) + Integral(C*cos(c + d*x)**5*sec(c + d*x), x))
Time = 0.21 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.44 \[ \int (a+a \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} + 288 \, {\left (d x + c\right )} A a^{3} - 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{3} + 72 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} + 96 \, {\left (d x + c\right )} B a^{3} - 96 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{3} + 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} + 72 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} + 96 \, A a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 288 \, A a^{3} \sin \left (d x + c\right ) + 288 \, B a^{3} \sin \left (d x + c\right ) + 96 \, C a^{3} \sin \left (d x + c\right )}{96 \, d} \]
integrate((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c),x, algorithm="maxima")
1/96*(24*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^3 + 288*(d*x + c)*A*a^3 - 32 *(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^3 + 72*(2*d*x + 2*c + sin(2*d*x + 2 *c))*B*a^3 + 96*(d*x + c)*B*a^3 - 96*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a ^3 + 3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*C*a^3 + 72* (2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^3 + 96*A*a^3*log(sec(d*x + c) + tan(d *x + c)) + 288*A*a^3*sin(d*x + c) + 288*B*a^3*sin(d*x + c) + 96*C*a^3*sin( d*x + c))/d
Time = 0.40 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.77 \[ \int (a+a \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {24 \, A a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 24 \, A a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 3 \, {\left (28 \, A a^{3} + 20 \, B a^{3} + 15 \, C a^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (60 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 60 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 45 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 204 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 220 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 165 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 228 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 292 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 219 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 84 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 132 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 147 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \]
1/24*(24*A*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 24*A*a^3*log(abs(tan(1 /2*d*x + 1/2*c) - 1)) + 3*(28*A*a^3 + 20*B*a^3 + 15*C*a^3)*(d*x + c) + 2*( 60*A*a^3*tan(1/2*d*x + 1/2*c)^7 + 60*B*a^3*tan(1/2*d*x + 1/2*c)^7 + 45*C*a ^3*tan(1/2*d*x + 1/2*c)^7 + 204*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 220*B*a^3*t an(1/2*d*x + 1/2*c)^5 + 165*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 228*A*a^3*tan(1 /2*d*x + 1/2*c)^3 + 292*B*a^3*tan(1/2*d*x + 1/2*c)^3 + 219*C*a^3*tan(1/2*d *x + 1/2*c)^3 + 84*A*a^3*tan(1/2*d*x + 1/2*c) + 132*B*a^3*tan(1/2*d*x + 1/ 2*c) + 147*C*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d
Time = 2.19 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.49 \[ \int (a+a \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {7\,A\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+2\,A\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+5\,B\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+\frac {15\,C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{4}+\frac {A\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{4}+\frac {3\,B\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{4}+\frac {B\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{12}+C\,a^3\,\sin \left (2\,c+2\,d\,x\right )+\frac {C\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {C\,a^3\,\sin \left (4\,c+4\,d\,x\right )}{32}+3\,A\,a^3\,\sin \left (c+d\,x\right )+\frac {15\,B\,a^3\,\sin \left (c+d\,x\right )}{4}+\frac {13\,C\,a^3\,\sin \left (c+d\,x\right )}{4}}{d} \]
(7*A*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + 2*A*a^3*atanh(sin(c /2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + 5*B*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/ 2 + (d*x)/2)) + (15*C*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/4 + (A*a^3*sin(2*c + 2*d*x))/4 + (3*B*a^3*sin(2*c + 2*d*x))/4 + (B*a^3*sin(3* c + 3*d*x))/12 + C*a^3*sin(2*c + 2*d*x) + (C*a^3*sin(3*c + 3*d*x))/4 + (C* a^3*sin(4*c + 4*d*x))/32 + 3*A*a^3*sin(c + d*x) + (15*B*a^3*sin(c + d*x))/ 4 + (13*C*a^3*sin(c + d*x))/4)/d