Integrand size = 31, antiderivative size = 147 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {1}{8} a^3 (28 A+15 C) x+\frac {a^3 A \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^3 (4 A+3 C) \sin (c+d x)}{8 d}+\frac {C (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac {C \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{4 a d}+\frac {(4 A+5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{8 d} \]
1/8*a^3*(28*A+15*C)*x+a^3*A*arctanh(sin(d*x+c))/d+5/8*a^3*(4*A+3*C)*sin(d* x+c)/d+1/4*C*(a+a*cos(d*x+c))^3*sin(d*x+c)/d+1/4*C*(a^2+a^2*cos(d*x+c))^2* sin(d*x+c)/a/d+1/8*(4*A+5*C)*(a^3+a^3*cos(d*x+c))*sin(d*x+c)/d
Time = 1.61 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.84 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {a^3 \left (112 A d x+60 C d x-32 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+32 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+8 (12 A+13 C) \sin (c+d x)+8 (A+4 C) \sin (2 (c+d x))+8 C \sin (3 (c+d x))+C \sin (4 (c+d x))\right )}{32 d} \]
(a^3*(112*A*d*x + 60*C*d*x - 32*A*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 32*A*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 8*(12*A + 13*C)*Sin[c + d*x] + 8*(A + 4*C)*Sin[2*(c + d*x)] + 8*C*Sin[3*(c + d*x)] + C*Sin[4*(c + d*x)]))/(32*d)
Time = 1.17 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.02, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.484, Rules used = {3042, 3525, 3042, 3455, 27, 3042, 3455, 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.
| \(\displaystyle \int \sec (c+d x) (a \cos (c+d x)+a)^3 \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3 \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 3525 |
| \(\displaystyle \frac {\int (\cos (c+d x) a+a)^3 (4 a A+3 a C \cos (c+d x)) \sec (c+d x)dx}{4 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3 \left (4 a A+3 a C \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{4 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {\frac {1}{3} \int 3 (\cos (c+d x) a+a)^2 \left (4 A a^2+(4 A+5 C) \cos (c+d x) a^2\right ) \sec (c+d x)dx+\frac {C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}}{4 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int (\cos (c+d x) a+a)^2 \left (4 A a^2+(4 A+5 C) \cos (c+d x) a^2\right ) \sec (c+d x)dx+\frac {C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}}{4 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (4 A a^2+(4 A+5 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}}{4 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {\frac {1}{2} \int (\cos (c+d x) a+a) \left (8 A a^3+5 (4 A+3 C) \cos (c+d x) a^3\right ) \sec (c+d x)dx+\frac {(4 A+5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}+\frac {C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}}{4 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (8 A a^3+5 (4 A+3 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {(4 A+5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}+\frac {C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}}{4 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {\frac {1}{2} \int \left (5 (4 A+3 C) \cos ^2(c+d x) a^4+8 A a^4+\left (8 A a^4+5 (4 A+3 C) a^4\right ) \cos (c+d x)\right ) \sec (c+d x)dx+\frac {(4 A+5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}+\frac {C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}}{4 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \int \frac {5 (4 A+3 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^4+8 A a^4+\left (8 A a^4+5 (4 A+3 C) a^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {(4 A+5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}+\frac {C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}}{4 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {\frac {1}{2} \left (\int \left (8 A a^4+(28 A+15 C) \cos (c+d x) a^4\right ) \sec (c+d x)dx+\frac {5 a^4 (4 A+3 C) \sin (c+d x)}{d}\right )+\frac {(4 A+5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}+\frac {C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}}{4 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \left (\int \frac {8 A a^4+(28 A+15 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^4}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {5 a^4 (4 A+3 C) \sin (c+d x)}{d}\right )+\frac {(4 A+5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}+\frac {C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}}{4 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {\frac {1}{2} \left (8 a^4 A \int \sec (c+d x)dx+\frac {5 a^4 (4 A+3 C) \sin (c+d x)}{d}+a^4 x (28 A+15 C)\right )+\frac {(4 A+5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}+\frac {C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}}{4 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \left (8 a^4 A \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {5 a^4 (4 A+3 C) \sin (c+d x)}{d}+a^4 x (28 A+15 C)\right )+\frac {(4 A+5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}+\frac {C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}}{4 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {8 a^4 A \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^4 (4 A+3 C) \sin (c+d x)}{d}+a^4 x (28 A+15 C)\right )+\frac {(4 A+5 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}+\frac {C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}}{4 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
(C*(a + a*Cos[c + d*x])^3*Sin[c + d*x])/(4*d) + ((C*(a^2 + a^2*Cos[c + d*x ])^2*Sin[c + d*x])/d + ((4*A + 5*C)*(a^4 + a^4*Cos[c + d*x])*Sin[c + d*x]) /(2*d) + (a^4*(28*A + 15*C)*x + (8*a^4*A*ArcTanh[Sin[c + d*x]])/d + (5*a^4 *(4*A + 3*C)*Sin[c + d*x])/d)/2)/(4*a)
3.1.21.3.1 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_.) + (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))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ[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 = 5.82 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.70
| method | result | size |
| parallelrisch | \(-\frac {\left (A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-\frac {A}{4}-C \right ) \sin \left (2 d x +2 c \right )-\frac {\sin \left (3 d x +3 c \right ) C}{4}-\frac {\sin \left (4 d x +4 c \right ) C}{32}+\left (-3 A -\frac {13 C}{4}\right ) \sin \left (d x +c \right )-\frac {7 x \left (A +\frac {15 C}{28}\right ) d}{2}\right ) a^{3}}{d}\) | \(103\) |
| parts | \(\frac {A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {\left (A \,a^{3}+3 C \,a^{3}\right ) \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (3 A \,a^{3}+C \,a^{3}\right ) \sin \left (d x +c \right )}{d}+\frac {C \,a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}+\frac {3 A \,a^{3} \left (d x +c \right )}{d}+\frac {C \,a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{d}\) | \(163\) |
| derivativedivides | \(\frac {A \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C \,a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+3 A \,a^{3} \sin \left (d x +c \right )+C \,a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 A \,a^{3} \left (d x +c \right )+3 C \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \,a^{3} \sin \left (d x +c \right )}{d}\) | \(173\) |
| default | \(\frac {A \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C \,a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+3 A \,a^{3} \sin \left (d x +c \right )+C \,a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 A \,a^{3} \left (d x +c \right )+3 C \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \,a^{3} \sin \left (d x +c \right )}{d}\) | \(173\) |
| risch | \(\frac {7 a^{3} A x}{2}+\frac {15 a^{3} C x}{8}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} A \,a^{3}}{2 d}-\frac {13 i {\mathrm e}^{i \left (d x +c \right )} C \,a^{3}}{8 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} A \,a^{3}}{2 d}+\frac {13 i {\mathrm e}^{-i \left (d x +c \right )} C \,a^{3}}{8 d}+\frac {A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {\sin \left (4 d x +4 c \right ) C \,a^{3}}{32 d}+\frac {\sin \left (3 d x +3 c \right ) C \,a^{3}}{4 d}+\frac {\sin \left (2 d x +2 c \right ) A \,a^{3}}{4 d}+\frac {\sin \left (2 d x +2 c \right ) C \,a^{3}}{d}\) | \(206\) |
| norman | \(\frac {\left (\frac {7}{2} A \,a^{3}+\frac {15}{8} C \,a^{3}\right ) x +\left (35 A \,a^{3}+\frac {75}{4} C \,a^{3}\right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (35 A \,a^{3}+\frac {75}{4} C \,a^{3}\right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {7}{2} A \,a^{3}+\frac {15}{8} C \,a^{3}\right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {35}{2} A \,a^{3}+\frac {75}{8} C \,a^{3}\right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {35}{2} A \,a^{3}+\frac {75}{8} C \,a^{3}\right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {5 a^{3} \left (4 A +3 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {7 a^{3} \left (4 A +7 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {4 a^{3} \left (9 A +8 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {a^{3} \left (44 A +35 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {a^{3} \left (52 A +61 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {A \,a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {A \,a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\) | \(333\) |
-(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-C)*sin(2*d *x+2*c)-1/4*sin(3*d*x+3*c)*C-1/32*sin(4*d*x+4*c)*C+(-3*A-13/4*C)*sin(d*x+c )-7/2*x*(A+15/28*C)*d)*a^3/d
Time = 0.29 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.76 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {{\left (28 \, A + 15 \, C\right )} a^{3} d x + 4 \, A a^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 4 \, A a^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, C a^{3} \cos \left (d x + c\right )^{3} + 8 \, C a^{3} \cos \left (d x + c\right )^{2} + {\left (4 \, A + 15 \, C\right )} a^{3} \cos \left (d x + c\right ) + 24 \, {\left (A + C\right )} a^{3}\right )} \sin \left (d x + c\right )}{8 \, d} \]
1/8*((28*A + 15*C)*a^3*d*x + 4*A*a^3*log(sin(d*x + c) + 1) - 4*A*a^3*log(- sin(d*x + c) + 1) + (2*C*a^3*cos(d*x + c)^3 + 8*C*a^3*cos(d*x + c)^2 + (4* A + 15*C)*a^3*cos(d*x + c) + 24*(A + C)*a^3)*sin(d*x + c))/d
\[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=a^{3} \left (\int A \sec {\left (c + d x \right )}\, dx + \int 3 A \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 3 A \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int A \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int C \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 3 C \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 3 C \cos ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int C \cos ^{5}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx\right ) \]
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(C*cos(c + d*x)**2*sec(c + d*x), x) + I ntegral(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 = 0.22 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.11 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {8 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} + 96 \, {\left (d x + c\right )} A a^{3} - 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{3} + {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} + 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} + 32 \, A a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 96 \, A a^{3} \sin \left (d x + c\right ) + 32 \, C a^{3} \sin \left (d x + c\right )}{32 \, d} \]
1/32*(8*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^3 + 96*(d*x + c)*A*a^3 - 32*( sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^3 + (12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*C*a^3 + 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^3 + 32*A*a^3*log(sec(d*x + c) + tan(d*x + c)) + 96*A*a^3*sin(d*x + c) + 32*C* a^3*sin(d*x + c))/d
Time = 0.30 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.45 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {8 \, A a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 8 \, A a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + {\left (28 \, A a^{3} + 15 \, C a^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (20 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 68 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 55 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 76 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 73 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 28 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 49 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{8 \, d} \]
1/8*(8*A*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 8*A*a^3*log(abs(tan(1/2* d*x + 1/2*c) - 1)) + (28*A*a^3 + 15*C*a^3)*(d*x + c) + 2*(20*A*a^3*tan(1/2 *d*x + 1/2*c)^7 + 15*C*a^3*tan(1/2*d*x + 1/2*c)^7 + 68*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 55*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 76*A*a^3*tan(1/2*d*x + 1/2*c )^3 + 73*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 28*A*a^3*tan(1/2*d*x + 1/2*c) + 49 *C*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d
Time = 1.29 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.33 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {3\,A\,a^3\,\sin \left (c+d\,x\right )}{d}+\frac {13\,C\,a^3\,\sin \left (c+d\,x\right )}{4\,d}+\frac {7\,A\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,A\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {15\,C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{4\,d}+\frac {A\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {C\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{d}+\frac {C\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{4\,d}+\frac {C\,a^3\,\sin \left (4\,c+4\,d\,x\right )}{32\,d} \]
(3*A*a^3*sin(c + d*x))/d + (13*C*a^3*sin(c + d*x))/(4*d) + (7*A*a^3*atan(s in(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*A*a^3*atanh(sin(c/2 + (d*x)/ 2)/cos(c/2 + (d*x)/2)))/d + (15*C*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d *x)/2)))/(4*d) + (A*a^3*sin(2*c + 2*d*x))/(4*d) + (C*a^3*sin(2*c + 2*d*x)) /d + (C*a^3*sin(3*c + 3*d*x))/(4*d) + (C*a^3*sin(4*c + 4*d*x))/(32*d)