Integrand size = 41, antiderivative size = 156 \[ \int (a+a \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {1}{2} a^3 (6 A+7 B+5 C) x+\frac {a^3 (3 A+B) \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^3 (B+C) \sin (c+d x)}{2 d}-\frac {(3 A-C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{3 a d}-\frac {(6 A-3 B-5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {A (a+a \cos (c+d x))^3 \tan (c+d x)}{d} \]
1/2*a^3*(6*A+7*B+5*C)*x+a^3*(3*A+B)*arctanh(sin(d*x+c))/d+5/2*a^3*(B+C)*si n(d*x+c)/d-1/3*(3*A-C)*(a^2+a^2*cos(d*x+c))^2*sin(d*x+c)/a/d-1/6*(6*A-3*B- 5*C)*(a^3+a^3*cos(d*x+c))*sin(d*x+c)/d+A*(a+a*cos(d*x+c))^3*tan(d*x+c)/d
Time = 3.64 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.46 \[ \int (a+a \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {a^3 (1+\cos (c+d x))^3 \sec ^6\left (\frac {1}{2} (c+d x)\right ) \left (6 (6 A+7 B+5 C) (c+d x)-12 (3 A+B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 (3 A+B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {12 A \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {12 A \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}+3 (4 A+12 B+15 C) \sin (c+d x)+3 (B+3 C) \sin (2 (c+d x))+C \sin (3 (c+d x))\right )}{96 d} \]
(a^3*(1 + Cos[c + d*x])^3*Sec[(c + d*x)/2]^6*(6*(6*A + 7*B + 5*C)*(c + d*x ) - 12*(3*A + B)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 12*(3*A + B)*L og[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (12*A*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) + (12*A*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 3*(4*A + 12*B + 15*C)*Sin[c + d*x] + 3*(B + 3*C)*Sin [2*(c + d*x)] + C*Sin[3*(c + d*x)]))/(96*d)
Time = 1.33 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.05, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.366, Rules used = {3042, 3522, 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.
| \(\displaystyle \int \sec ^2(c+d x) (a \cos (c+d x)+a)^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
| \(\Big \downarrow \) 3522 |
| \(\displaystyle \frac {\int (\cos (c+d x) a+a)^3 (a (3 A+B)-a (3 A-C) \cos (c+d x)) \sec (c+d x)dx}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3 \left (a (3 A+B)-a (3 A-C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {\frac {1}{3} \int (\cos (c+d x) a+a)^2 \left (3 a^2 (3 A+B)-a^2 (6 A-3 B-5 C) \cos (c+d x)\right ) \sec (c+d x)dx-\frac {(3 A-C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (3 a^2 (3 A+B)-a^2 (6 A-3 B-5 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {(3 A-C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {1}{2} \int 3 (\cos (c+d x) a+a) \left (2 (3 A+B) a^3+5 (B+C) \cos (c+d x) a^3\right ) \sec (c+d x)dx-\frac {(6 A-3 B-5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )-\frac {(3 A-C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \int (\cos (c+d x) a+a) \left (2 (3 A+B) a^3+5 (B+C) \cos (c+d x) a^3\right ) \sec (c+d x)dx-\frac {(6 A-3 B-5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )-\frac {(3 A-C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (2 (3 A+B) a^3+5 (B+C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {(6 A-3 B-5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )-\frac {(3 A-C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \int \left (5 (B+C) \cos ^2(c+d x) a^4+2 (3 A+B) a^4+\left (2 (3 A+B) a^4+5 (B+C) a^4\right ) \cos (c+d x)\right ) \sec (c+d x)dx-\frac {(6 A-3 B-5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )-\frac {(3 A-C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \int \frac {5 (B+C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^4+2 (3 A+B) a^4+\left (2 (3 A+B) a^4+5 (B+C) a^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {(6 A-3 B-5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )-\frac {(3 A-C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \left (\int \left (2 (3 A+B) a^4+(6 A+7 B+5 C) \cos (c+d x) a^4\right ) \sec (c+d x)dx+\frac {5 a^4 (B+C) \sin (c+d x)}{d}\right )-\frac {(6 A-3 B-5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )-\frac {(3 A-C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \left (\int \frac {2 (3 A+B) a^4+(6 A+7 B+5 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^4}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {5 a^4 (B+C) \sin (c+d x)}{d}\right )-\frac {(6 A-3 B-5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )-\frac {(3 A-C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \left (2 a^4 (3 A+B) \int \sec (c+d x)dx+a^4 x (6 A+7 B+5 C)+\frac {5 a^4 (B+C) \sin (c+d x)}{d}\right )-\frac {(6 A-3 B-5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )-\frac {(3 A-C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \left (2 a^4 (3 A+B) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+a^4 x (6 A+7 B+5 C)+\frac {5 a^4 (B+C) \sin (c+d x)}{d}\right )-\frac {(6 A-3 B-5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )-\frac {(3 A-C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \left (\frac {2 a^4 (3 A+B) \text {arctanh}(\sin (c+d x))}{d}+a^4 x (6 A+7 B+5 C)+\frac {5 a^4 (B+C) \sin (c+d x)}{d}\right )-\frac {(6 A-3 B-5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )-\frac {(3 A-C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}\) |
(-1/3*((3*A - C)*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/d + (-1/2*((6*A - 3*B - 5*C)*(a^4 + a^4*Cos[c + d*x])*Sin[c + d*x])/d + (3*(a^4*(6*A + 7*B + 5*C)*x + (2*a^4*(3*A + B)*ArcTanh[Sin[c + d*x]])/d + (5*a^4*(B + C)*Sin [c + d*x])/d))/2)/3)/a + (A*(a + a*Cos[c + d*x])^3*Tan[c + d*x])/d
3.4.22.3.1 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^2*C - B*c*d + A*d^2))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(b*d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^m* (c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + (c*C - B*d)*( a*c*m + b*d*(n + 1)) + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2* (n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ [m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0])
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 6.81 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.95
| method | result | size |
| parallelrisch | \(\frac {a^{3} \left (-6 \cos \left (d x +c \right ) \left (A +\frac {B}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+6 \cos \left (d x +c \right ) \left (A +\frac {B}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (A +3 B +\frac {23 C}{6}\right ) \sin \left (2 d x +2 c \right )+\frac {\left (B +3 C \right ) \sin \left (3 d x +3 c \right )}{4}+\frac {\sin \left (4 d x +4 c \right ) C}{12}+6 x \left (A +\frac {7 B}{6}+\frac {5 C}{6}\right ) d \cos \left (d x +c \right )+2 \left (A +\frac {B}{8}+\frac {3 C}{8}\right ) \sin \left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )}\) | \(148\) |
| parts | \(\frac {A \,a^{3} \tan \left (d x +c \right )}{d}+\frac {\left (3 A \,a^{3}+B \,a^{3}\right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {\left (B \,a^{3}+3 C \,a^{3}\right ) \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (A \,a^{3}+3 B \,a^{3}+3 C \,a^{3}\right ) \sin \left (d x +c \right )}{d}+\frac {\left (3 A \,a^{3}+3 B \,a^{3}+C \,a^{3}\right ) \left (d x +c \right )}{d}+\frac {C \,a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}\) | \(164\) |
| derivativedivides | \(\frac {A \,a^{3} \sin \left (d x +c \right )+B \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {C \,a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+3 A \,a^{3} \left (d x +c \right )+3 B \sin \left (d x +c \right ) a^{3}+3 C \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 B \,a^{3} \left (d x +c \right )+3 C \,a^{3} \sin \left (d x +c \right )+A \,a^{3} \tan \left (d x +c \right )+B \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \,a^{3} \left (d x +c \right )}{d}\) | \(200\) |
| default | \(\frac {A \,a^{3} \sin \left (d x +c \right )+B \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {C \,a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+3 A \,a^{3} \left (d x +c \right )+3 B \sin \left (d x +c \right ) a^{3}+3 C \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 B \,a^{3} \left (d x +c \right )+3 C \,a^{3} \sin \left (d x +c \right )+A \,a^{3} \tan \left (d x +c \right )+B \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \,a^{3} \left (d x +c \right )}{d}\) | \(200\) |
| risch | \(3 a^{3} A x +\frac {7 a^{3} B x}{2}+\frac {5 a^{3} C x}{2}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} B \,a^{3}}{2 d}+\frac {15 i {\mathrm e}^{-i \left (d x +c \right )} C \,a^{3}}{8 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} A \,a^{3}}{2 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} A \,a^{3}}{2 d}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )} B \,a^{3}}{8 d}-\frac {3 i C \,a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {3 i C \,a^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} B \,a^{3}}{8 d}-\frac {15 i {\mathrm e}^{i \left (d x +c \right )} C \,a^{3}}{8 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} B \,a^{3}}{2 d}+\frac {2 i A \,a^{3}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {3 A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}+\frac {B \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {3 A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {B \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {\sin \left (3 d x +3 c \right ) C \,a^{3}}{12 d}\) | \(341\) |
| norman | \(\frac {\left (-\frac {7}{2} B \,a^{3}-\frac {5}{2} C \,a^{3}-3 A \,a^{3}\right ) x +\left (-\frac {35}{2} B \,a^{3}-\frac {25}{2} C \,a^{3}-15 A \,a^{3}\right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {7}{2} B \,a^{3}+\frac {5}{2} C \,a^{3}+3 A \,a^{3}\right ) x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {35}{2} B \,a^{3}+\frac {25}{2} C \,a^{3}+15 A \,a^{3}\right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-12 A \,a^{3}-14 B \,a^{3}-10 C \,a^{3}\right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (12 A \,a^{3}+14 B \,a^{3}+10 C \,a^{3}\right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {5 a^{3} \left (B +C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{3} \left (4 A +7 B +11 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 a^{3} \left (36 A +15 B +11 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {a^{3} \left (48 A +57 B +73 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {a^{3} \left (-51 B -55 C +12 A \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 a^{3} \left (-21 B -29 C +24 A \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {a^{3} \left (3 A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a^{3} \left (3 A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\) | \(429\) |
1/2*a^3*(-6*cos(d*x+c)*(A+1/3*B)*ln(tan(1/2*d*x+1/2*c)-1)+6*cos(d*x+c)*(A+ 1/3*B)*ln(tan(1/2*d*x+1/2*c)+1)+(A+3*B+23/6*C)*sin(2*d*x+2*c)+1/4*(B+3*C)* sin(3*d*x+3*c)+1/12*sin(4*d*x+4*c)*C+6*x*(A+7/6*B+5/6*C)*d*cos(d*x+c)+2*(A +1/8*B+3/8*C)*sin(d*x+c))/d/cos(d*x+c)
Time = 0.29 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.00 \[ \int (a+a \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {3 \, {\left (6 \, A + 7 \, B + 5 \, C\right )} a^{3} d x \cos \left (d x + c\right ) + 3 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, C a^{3} \cos \left (d x + c\right )^{3} + 3 \, {\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, A + 9 \, B + 11 \, C\right )} a^{3} \cos \left (d x + c\right ) + 6 \, A a^{3}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )} \]
integrate((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2, x, algorithm="fricas")
1/6*(3*(6*A + 7*B + 5*C)*a^3*d*x*cos(d*x + c) + 3*(3*A + B)*a^3*cos(d*x + c)*log(sin(d*x + c) + 1) - 3*(3*A + B)*a^3*cos(d*x + c)*log(-sin(d*x + c) + 1) + (2*C*a^3*cos(d*x + c)^3 + 3*(B + 3*C)*a^3*cos(d*x + c)^2 + 2*(3*A + 9*B + 11*C)*a^3*cos(d*x + c) + 6*A*a^3)*sin(d*x + c))/(d*cos(d*x + c))
\[ \int (a+a \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=a^{3} \left (\int A \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 A \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 A \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int A \cos ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int B \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 B \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 B \cos ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int B \cos ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int C \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 C \cos ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 C \cos ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int C \cos ^{5}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx\right ) \]
a**3*(Integral(A*sec(c + d*x)**2, x) + Integral(3*A*cos(c + d*x)*sec(c + d *x)**2, x) + Integral(3*A*cos(c + d*x)**2*sec(c + d*x)**2, x) + Integral(A *cos(c + d*x)**3*sec(c + d*x)**2, x) + Integral(B*cos(c + d*x)*sec(c + d*x )**2, x) + Integral(3*B*cos(c + d*x)**2*sec(c + d*x)**2, x) + Integral(3*B *cos(c + d*x)**3*sec(c + d*x)**2, x) + Integral(B*cos(c + d*x)**4*sec(c + d*x)**2, x) + Integral(C*cos(c + d*x)**2*sec(c + d*x)**2, x) + Integral(3* C*cos(c + d*x)**3*sec(c + d*x)**2, x) + Integral(3*C*cos(c + d*x)**4*sec(c + d*x)**2, x) + Integral(C*cos(c + d*x)**5*sec(c + d*x)**2, x))
Time = 0.21 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.35 \[ \int (a+a \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {36 \, {\left (d x + c\right )} A a^{3} + 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} + 36 \, {\left (d x + c\right )} B a^{3} - 4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{3} + 9 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} + 12 \, {\left (d x + c\right )} C a^{3} + 18 \, A a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, B a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, A a^{3} \sin \left (d x + c\right ) + 36 \, B a^{3} \sin \left (d x + c\right ) + 36 \, C a^{3} \sin \left (d x + c\right ) + 12 \, A a^{3} \tan \left (d x + c\right )}{12 \, d} \]
integrate((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2, x, algorithm="maxima")
1/12*(36*(d*x + c)*A*a^3 + 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^3 + 36*( d*x + c)*B*a^3 - 4*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^3 + 9*(2*d*x + 2* c + sin(2*d*x + 2*c))*C*a^3 + 12*(d*x + c)*C*a^3 + 18*A*a^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 6*B*a^3*(log(sin(d*x + c) + 1) - log(s in(d*x + c) - 1)) + 12*A*a^3*sin(d*x + c) + 36*B*a^3*sin(d*x + c) + 36*C*a ^3*sin(d*x + c) + 12*A*a^3*tan(d*x + c))/d
Time = 0.38 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.80 \[ \int (a+a \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=-\frac {\frac {12 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - 3 \, {\left (6 \, A a^{3} + 7 \, B a^{3} + 5 \, C a^{3}\right )} {\left (d x + c\right )} - 6 \, {\left (3 \, A a^{3} + B a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 6 \, {\left (3 \, A a^{3} + B a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 21 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 33 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]
integrate((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2, x, algorithm="giac")
-1/6*(12*A*a^3*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) - 3*(6*A* a^3 + 7*B*a^3 + 5*C*a^3)*(d*x + c) - 6*(3*A*a^3 + B*a^3)*log(abs(tan(1/2*d *x + 1/2*c) + 1)) + 6*(3*A*a^3 + B*a^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(6*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 15*B*a^3*tan(1/2*d*x + 1/2*c)^5 + 1 5*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 12*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 36*B*a^ 3*tan(1/2*d*x + 1/2*c)^3 + 40*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 6*A*a^3*tan(1 /2*d*x + 1/2*c) + 21*B*a^3*tan(1/2*d*x + 1/2*c) + 33*C*a^3*tan(1/2*d*x + 1 /2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d
Time = 2.16 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.86 \[ \int (a+a \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {6\,A\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+7\,B\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+5\,C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-A\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,6{}\mathrm {i}-B\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}}{d}+\frac {\frac {A\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {3\,B\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {B\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{8}+\frac {23\,C\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{12}+\frac {3\,C\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{8}+\frac {C\,a^3\,\sin \left (4\,c+4\,d\,x\right )}{24}+A\,a^3\,\sin \left (c+d\,x\right )+\frac {B\,a^3\,\sin \left (c+d\,x\right )}{8}+\frac {3\,C\,a^3\,\sin \left (c+d\,x\right )}{8}}{d\,\cos \left (c+d\,x\right )} \]
(6*A*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) - A*a^3*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*6i + 7*B*a^3*atan(sin(c/2 + (d*x)/2)/c os(c/2 + (d*x)/2)) - B*a^3*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2) )*2i + 5*C*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + ((A*a^3*si n(2*c + 2*d*x))/2 + (3*B*a^3*sin(2*c + 2*d*x))/2 + (B*a^3*sin(3*c + 3*d*x) )/8 + (23*C*a^3*sin(2*c + 2*d*x))/12 + (3*C*a^3*sin(3*c + 3*d*x))/8 + (C*a ^3*sin(4*c + 4*d*x))/24 + A*a^3*sin(c + d*x) + (B*a^3*sin(c + d*x))/8 + (3 *C*a^3*sin(c + d*x))/8)/(d*cos(c + d*x))