Integrand size = 31, antiderivative size = 114 \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=a^3 (A+3 B) x+\frac {a^3 (7 A+6 B) \text {arctanh}(\sin (c+d x))}{2 d}-\frac {5 a^3 A \sin (c+d x)}{2 d}+\frac {(2 A+B) \left (a^3+a^3 \cos (c+d x)\right ) \tan (c+d x)}{d}+\frac {a A (a+a \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d} \] Output:
a^3*(A+3*B)*x+1/2*a^3*(7*A+6*B)*arctanh(sin(d*x+c))/d-5/2*a^3*A*sin(d*x+c) /d+(2*A+B)*(a^3+a^3*cos(d*x+c))*tan(d*x+c)/d+1/2*a*A*(a+a*cos(d*x+c))^2*se c(d*x+c)*tan(d*x+c)/d
Time = 4.94 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.82 \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {a^3 \left (4 A c+12 B c+4 A d x+12 B d x-14 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-12 B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+14 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {A}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {A}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+4 B \sin (c+d x)+4 (3 A+B) \tan (c+d x)\right )}{4 d} \] Input:
Integrate[(a + a*Cos[c + d*x])^3*(A + B*Cos[c + d*x])*Sec[c + d*x]^3,x]
Output:
(a^3*(4*A*c + 12*B*c + 4*A*d*x + 12*B*d*x - 14*A*Log[Cos[(c + d*x)/2] - Si n[(c + d*x)/2]] - 12*B*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 14*A*Log [Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 12*B*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + A/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 - A/(Cos[(c + d*x) /2] + Sin[(c + d*x)/2])^2 + 4*B*Sin[c + d*x] + 4*(3*A + B)*Tan[c + d*x]))/ (4*d)
Time = 0.96 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.02, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.387, Rules used = {3042, 3454, 3042, 3454, 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.
| \(\displaystyle \int \sec ^3(c+d x) (a \cos (c+d x)+a)^3 (A+B \cos (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {1}{2} \int (\cos (c+d x) a+a)^2 (2 a (2 A+B)-a (A-2 B) \cos (c+d x)) \sec ^2(c+d x)dx+\frac {a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (2 a (2 A+B)-a (A-2 B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {1}{2} \left (\int (\cos (c+d x) a+a) \left (a^2 (7 A+6 B)-5 a^2 A \cos (c+d x)\right ) \sec (c+d x)dx+\frac {2 (2 A+B) \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d}\right )+\frac {a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (a^2 (7 A+6 B)-5 a^2 A \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 (2 A+B) \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d}\right )+\frac {a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {1}{2} \left (\int \left (-5 A \cos ^2(c+d x) a^3+(7 A+6 B) a^3+\left (a^3 (7 A+6 B)-5 a^3 A\right ) \cos (c+d x)\right ) \sec (c+d x)dx+\frac {2 (2 A+B) \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d}\right )+\frac {a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\int \frac {-5 A \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^3+(7 A+6 B) a^3+\left (a^3 (7 A+6 B)-5 a^3 A\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 (2 A+B) \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d}\right )+\frac {a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {1}{2} \left (\int \left ((7 A+6 B) a^3+2 (A+3 B) \cos (c+d x) a^3\right ) \sec (c+d x)dx+\frac {2 (2 A+B) \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d}-\frac {5 a^3 A \sin (c+d x)}{d}\right )+\frac {a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\int \frac {(7 A+6 B) a^3+2 (A+3 B) \sin \left (c+d x+\frac {\pi }{2}\right ) a^3}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 (2 A+B) \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d}-\frac {5 a^3 A \sin (c+d x)}{d}\right )+\frac {a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {1}{2} \left (a^3 (7 A+6 B) \int \sec (c+d x)dx+\frac {2 (2 A+B) \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d}+2 a^3 x (A+3 B)-\frac {5 a^3 A \sin (c+d x)}{d}\right )+\frac {a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (a^3 (7 A+6 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {2 (2 A+B) \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d}+2 a^3 x (A+3 B)-\frac {5 a^3 A \sin (c+d x)}{d}\right )+\frac {a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{2} \left (\frac {a^3 (7 A+6 B) \text {arctanh}(\sin (c+d x))}{d}+\frac {2 (2 A+B) \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d}+2 a^3 x (A+3 B)-\frac {5 a^3 A \sin (c+d x)}{d}\right )+\frac {a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
Input:
Int[(a + a*Cos[c + d*x])^3*(A + B*Cos[c + d*x])*Sec[c + d*x]^3,x]
Output:
(a*A*(a + a*Cos[c + d*x])^2*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (2*a^3*(A + 3*B)*x + (a^3*(7*A + 6*B)*ArcTanh[Sin[c + d*x]])/d - (5*a^3*A*Sin[c + d*x ])/d + (2*(2*A + B)*(a^3 + a^3*Cos[c + d*x])*Tan[c + d*x])/d)/2
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^2)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[ e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a*d))), x] - Simp[b/(d*(n + 1)*(b*c + a*d)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp [a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n + 1) - B *(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f , A, B}, 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[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 9.49 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.13
| method | result | size |
| parts | \(\frac {a^{3} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {\left (a^{3} A +3 a^{3} B \right ) \left (d x +c \right )}{d}+\frac {\left (3 a^{3} A +a^{3} B \right ) \tan \left (d x +c \right )}{d}+\frac {\left (3 a^{3} A +3 a^{3} B \right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {a^{3} B \sin \left (d x +c \right )}{d}\) | \(129\) |
| derivativedivides | \(\frac {a^{3} A \left (d x +c \right )+a^{3} B \sin \left (d x +c \right )+3 a^{3} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 a^{3} B \left (d x +c \right )+3 a^{3} A \tan \left (d x +c \right )+3 a^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{3} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{3} B \tan \left (d x +c \right )}{d}\) | \(137\) |
| default | \(\frac {a^{3} A \left (d x +c \right )+a^{3} B \sin \left (d x +c \right )+3 a^{3} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 a^{3} B \left (d x +c \right )+3 a^{3} A \tan \left (d x +c \right )+3 a^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{3} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{3} B \tan \left (d x +c \right )}{d}\) | \(137\) |
| parallelrisch | \(-\frac {7 \left (\left (A +\frac {6 B}{7}\right ) \left (\cos \left (2 d x +2 c \right )+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\left (A +\frac {6 B}{7}\right ) \left (\cos \left (2 d x +2 c \right )+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2 d x \left (A +3 B \right ) \cos \left (2 d x +2 c \right )}{7}+\frac {2 \left (-3 A -B \right ) \sin \left (2 d x +2 c \right )}{7}-\frac {B \sin \left (3 d x +3 c \right )}{7}+\frac {\left (-2 A -B \right ) \sin \left (d x +c \right )}{7}-\frac {2 d x \left (A +3 B \right )}{7}\right ) a^{3}}{2 d \left (\cos \left (2 d x +2 c \right )+1\right )}\) | \(154\) |
| risch | \(a^{3} A x +3 a^{3} B x -\frac {i {\mathrm e}^{i \left (d x +c \right )} a^{3} B}{2 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} a^{3} B}{2 d}-\frac {i a^{3} \left (A \,{\mathrm e}^{3 i \left (d x +c \right )}-6 A \,{\mathrm e}^{2 i \left (d x +c \right )}-2 B \,{\mathrm e}^{2 i \left (d x +c \right )}-A \,{\mathrm e}^{i \left (d x +c \right )}-6 A -2 B \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {7 a^{3} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}+\frac {7 a^{3} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d}\) | \(217\) |
| norman | \(\frac {\left (a^{3} A +3 a^{3} B \right ) x +\left (-4 a^{3} A -12 a^{3} B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (-a^{3} A -3 a^{3} B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-a^{3} A -3 a^{3} B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (a^{3} A +3 a^{3} B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (2 a^{3} A +6 a^{3} B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (2 a^{3} A +6 a^{3} B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\frac {a^{3} \left (7 A +4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {a^{3} \left (23 A +8 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}+\frac {22 a^{3} A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {5 a^{3} A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d}-\frac {2 a^{3} \left (A +4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}-\frac {a^{3} \left (13 A +4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}-\frac {a^{3} \left (7 A +6 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a^{3} \left (7 A +6 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(395\) |
Input:
int((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c))*sec(d*x+c)^3,x,method=_RETURNVERBO SE)
Output:
a^3*A/d*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+(A*a^3+3 *B*a^3)/d*(d*x+c)+(3*A*a^3+B*a^3)/d*tan(d*x+c)+(3*A*a^3+3*B*a^3)/d*ln(sec( d*x+c)+tan(d*x+c))+a^3*B*sin(d*x+c)/d
Time = 0.10 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.20 \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {4 \, {\left (A + 3 \, B\right )} a^{3} d x \cos \left (d x + c\right )^{2} + {\left (7 \, A + 6 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (7 \, A + 6 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, B a^{3} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right ) + A a^{3}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c))*sec(d*x+c)^3,x, algorithm="f ricas")
Output:
1/4*(4*(A + 3*B)*a^3*d*x*cos(d*x + c)^2 + (7*A + 6*B)*a^3*cos(d*x + c)^2*l og(sin(d*x + c) + 1) - (7*A + 6*B)*a^3*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(2*B*a^3*cos(d*x + c)^2 + 2*(3*A + B)*a^3*cos(d*x + c) + A*a^3)*sin (d*x + c))/(d*cos(d*x + c)^2)
\[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=a^{3} \left (\int A \sec ^{3}{\left (c + d x \right )}\, dx + \int 3 A \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int 3 A \cos ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int A \cos ^{3}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int B \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int 3 B \cos ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int 3 B \cos ^{3}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int B \cos ^{4}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate((a+a*cos(d*x+c))**3*(A+B*cos(d*x+c))*sec(d*x+c)**3,x)
Output:
a**3*(Integral(A*sec(c + d*x)**3, x) + Integral(3*A*cos(c + d*x)*sec(c + d *x)**3, x) + Integral(3*A*cos(c + d*x)**2*sec(c + d*x)**3, x) + Integral(A *cos(c + d*x)**3*sec(c + d*x)**3, x) + Integral(B*cos(c + d*x)*sec(c + d*x )**3, x) + Integral(3*B*cos(c + d*x)**2*sec(c + d*x)**3, x) + Integral(3*B *cos(c + d*x)**3*sec(c + d*x)**3, x) + Integral(B*cos(c + d*x)**4*sec(c + d*x)**3, x))
Time = 0.05 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.45 \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {4 \, {\left (d x + c\right )} A a^{3} + 12 \, {\left (d x + c\right )} B a^{3} - A a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, A a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, B a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, B a^{3} \sin \left (d x + c\right ) + 12 \, A a^{3} \tan \left (d x + c\right ) + 4 \, B a^{3} \tan \left (d x + c\right )}{4 \, d} \] Input:
integrate((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c))*sec(d*x+c)^3,x, algorithm="m axima")
Output:
1/4*(4*(d*x + c)*A*a^3 + 12*(d*x + c)*B*a^3 - A*a^3*(2*sin(d*x + c)/(sin(d *x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 6*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(sin(d*x + c) - 1)) + 4*B*a^3*sin(d*x + c) + 12*A*a^3*tan(d*x + c) + 4*B*a^3*tan(d*x + c))/d
Time = 0.37 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.68 \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {\frac {4 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + 2 \, {\left (A a^{3} + 3 \, B a^{3}\right )} {\left (d x + c\right )} + {\left (7 \, A a^{3} + 6 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (7 \, A a^{3} + 6 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (5 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 7 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \] Input:
integrate((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c))*sec(d*x+c)^3,x, algorithm="g iac")
Output:
1/2*(4*B*a^3*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 + 1) + 2*(A*a^3 + 3*B*a^3)*(d*x + c) + (7*A*a^3 + 6*B*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (7*A*a^3 + 6*B*a^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(5*A*a^3* tan(1/2*d*x + 1/2*c)^3 + 2*B*a^3*tan(1/2*d*x + 1/2*c)^3 - 7*A*a^3*tan(1/2* d*x + 1/2*c) - 2*B*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^ 2)/d
Time = 41.73 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.82 \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {B\,a^3\,\sin \left (c+d\,x\right )}{d}+\frac {2\,A\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {7\,A\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {6\,B\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {6\,B\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {3\,A\,a^3\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {A\,a^3\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {B\,a^3\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )} \] Input:
int(((A + B*cos(c + d*x))*(a + a*cos(c + d*x))^3)/cos(c + d*x)^3,x)
Output:
(B*a^3*sin(c + d*x))/d + (2*A*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/ 2)))/d + (7*A*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (6*B*a ^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (6*B*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (3*A*a^3*sin(c + d*x))/(d*cos(c + d*x )) + (A*a^3*sin(c + d*x))/(2*d*cos(c + d*x)^2) + (B*a^3*sin(c + d*x))/(d*c os(c + d*x))
Time = 0.18 (sec) , antiderivative size = 376, normalized size of antiderivative = 3.30 \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {a^{3} \left (-7 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a -6 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b +7 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a +6 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b +7 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a +6 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b -7 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a -6 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b +2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b +2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a d x +6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b d x -\cos \left (d x +c \right ) \sin \left (d x +c \right ) a -2 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b -2 \cos \left (d x +c \right ) a d x -6 \cos \left (d x +c \right ) b d x +6 \sin \left (d x +c \right )^{3} a +2 \sin \left (d x +c \right )^{3} b -6 \sin \left (d x +c \right ) a -2 \sin \left (d x +c \right ) b \right )}{2 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c))*sec(d*x+c)^3,x)
Output:
(a**3*( - 7*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a - 6*c os(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b + 7*cos(c + d*x)*l og(tan((c + d*x)/2) - 1)*a + 6*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*b + 7*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a + 6*cos(c + d*x )*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b - 7*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a - 6*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*b + 2*cos(c + d*x)*sin(c + d*x)**3*b + 2*cos(c + d*x)*sin(c + d*x)**2*a*d*x + 6*cos(c + d*x)*sin(c + d*x)**2*b*d*x - cos(c + d*x)*sin(c + d*x)*a - 2*cos(c + d*x)* sin(c + d*x)*b - 2*cos(c + d*x)*a*d*x - 6*cos(c + d*x)*b*d*x + 6*sin(c + d *x)**3*a + 2*sin(c + d*x)**3*b - 6*sin(c + d*x)*a - 2*sin(c + d*x)*b))/(2* cos(c + d*x)*d*(sin(c + d*x)**2 - 1))