Integrand size = 33, antiderivative size = 112 \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=2 a^2 C x+\frac {a^2 (3 A+2 C) \text {arctanh}(\sin (c+d x))}{2 d}-\frac {a^2 (3 A-2 C) \sin (c+d x)}{2 d}+\frac {A \left (a^2+a^2 \cos (c+d x)\right ) \tan (c+d x)}{d}+\frac {A (a+a \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d} \]
2*a^2*C*x+1/2*a^2*(3*A+2*C)*arctanh(sin(d*x+c))/d-1/2*a^2*(3*A-2*C)*sin(d* x+c)/d+A*(a^2+a^2*cos(d*x+c))*tan(d*x+c)/d+1/2*A*(a+a*cos(d*x+c))^2*sec(d* x+c)*tan(d*x+c)/d
Leaf count is larger than twice the leaf count of optimal. \(293\) vs. \(2(112)=224\).
Time = 2.79 (sec) , antiderivative size = 293, normalized size of antiderivative = 2.62 \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {1}{16} a^2 (1+\cos (c+d x))^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (8 C x-\frac {2 (3 A+2 C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {2 (3 A+2 C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {4 C \cos (d x) \sin (c)}{d}+\frac {4 C \cos (c) \sin (d x)}{d}+\frac {A}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {8 A \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {A}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {8 A \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right ) \]
(a^2*(1 + Cos[c + d*x])^2*Sec[(c + d*x)/2]^4*(8*C*x - (2*(3*A + 2*C)*Log[C os[(c + d*x)/2] - Sin[(c + d*x)/2]])/d + (2*(3*A + 2*C)*Log[Cos[(c + d*x)/ 2] + Sin[(c + d*x)/2]])/d + (4*C*Cos[d*x]*Sin[c])/d + (4*C*Cos[c]*Sin[d*x] )/d + A/(d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + (8*A*Sin[(d*x)/2])/( d*(Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) - A/(d*(Cos [(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (8*A*Sin[(d*x)/2])/(d*(Cos[c/2] + S in[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))))/16
Time = 0.96 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.04, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3042, 3523, 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.
| \(\displaystyle \int \sec ^3(c+d x) (a \cos (c+d x)+a)^2 \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
| \(\Big \downarrow \) 3523 |
| \(\displaystyle \frac {\int (\cos (c+d x) a+a)^2 (2 a A-a (A-2 C) \cos (c+d x)) \sec ^2(c+d x)dx}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (2 a A-a (A-2 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {\int (\cos (c+d x) a+a) \left (a^2 (3 A+2 C)-a^2 (3 A-2 C) \cos (c+d x)\right ) \sec (c+d x)dx+\frac {2 A \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d}}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (a^2 (3 A+2 C)-a^2 (3 A-2 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 A \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d}}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {\int \left (-\left ((3 A-2 C) \cos ^2(c+d x) a^3\right )+(3 A+2 C) a^3+\left (a^3 (3 A+2 C)-a^3 (3 A-2 C)\right ) \cos (c+d x)\right ) \sec (c+d x)dx+\frac {2 A \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d}}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {-\left ((3 A-2 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^3\right )+(3 A+2 C) a^3+\left (a^3 (3 A+2 C)-a^3 (3 A-2 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 A \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d}}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {\int \left ((3 A+2 C) a^3+4 C \cos (c+d x) a^3\right ) \sec (c+d x)dx-\frac {a^3 (3 A-2 C) \sin (c+d x)}{d}+\frac {2 A \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d}}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {(3 A+2 C) a^3+4 C \sin \left (c+d x+\frac {\pi }{2}\right ) a^3}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {a^3 (3 A-2 C) \sin (c+d x)}{d}+\frac {2 A \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d}}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {a^3 (3 A+2 C) \int \sec (c+d x)dx-\frac {a^3 (3 A-2 C) \sin (c+d x)}{d}+\frac {2 A \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d}+4 a^3 C x}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a^3 (3 A+2 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {a^3 (3 A-2 C) \sin (c+d x)}{d}+\frac {2 A \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d}+4 a^3 C x}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {a^3 (3 A+2 C) \text {arctanh}(\sin (c+d x))}{d}-\frac {a^3 (3 A-2 C) \sin (c+d x)}{d}+\frac {2 A \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d}+4 a^3 C x}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
(A*(a + a*Cos[c + d*x])^2*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (4*a^3*C*x + (a^3*(3*A + 2*C)*ArcTanh[Sin[c + d*x]])/d - (a^3*(3*A - 2*C)*Sin[c + d*x]) /d + (2*A*(a^3 + a^3*Cos[c + d*x])*Tan[c + d*x])/d)/(2*a)
3.1.14.3.1 Defintions of rubi rules used
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[((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^2*C + 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*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(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, 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 = 5.86 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.01
| method | result | size |
| parts | \(\frac {A \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {\left (A \,a^{2}+a^{2} C \right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {\sin \left (d x +c \right ) a^{2} C}{d}+\frac {2 a^{2} A \tan \left (d x +c \right )}{d}+\frac {2 a^{2} C \left (d x +c \right )}{d}\) | \(113\) |
| derivativedivides | \(\frac {A \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{2} C \sin \left (d x +c \right )+2 A \,a^{2} \tan \left (d x +c \right )+2 a^{2} C \left (d x +c \right )+A \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(114\) |
| default | \(\frac {A \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{2} C \sin \left (d x +c \right )+2 A \,a^{2} \tan \left (d x +c \right )+2 a^{2} C \left (d x +c \right )+A \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(114\) |
| parallelrisch | \(-\frac {3 a^{2} \left (\left (1+\cos \left (2 d x +2 c \right )\right ) \left (A +\frac {2 C}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\left (1+\cos \left (2 d x +2 c \right )\right ) \left (A +\frac {2 C}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {4 d x C \cos \left (2 d x +2 c \right )}{3}-\frac {4 \sin \left (2 d x +2 c \right ) A}{3}-\frac {\sin \left (3 d x +3 c \right ) C}{3}+\frac {\left (-2 A -C \right ) \sin \left (d x +c \right )}{3}-\frac {4 d x C}{3}\right )}{2 d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(140\) |
| risch | \(2 a^{2} C x -\frac {i {\mathrm e}^{i \left (d x +c \right )} a^{2} C}{2 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} a^{2} C}{2 d}-\frac {i A \,a^{2} \left ({\mathrm e}^{3 i \left (d x +c \right )}-4 \,{\mathrm e}^{2 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}-4\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {3 A \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}-\frac {3 A \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}\) | \(190\) |
| norman | \(\frac {\frac {a^{2} \left (5 A +2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {a^{2} \left (17 A +2 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+2 a^{2} C x +4 a^{2} C x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 a^{2} C x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8 a^{2} C x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 a^{2} C x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 a^{2} C x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 a^{2} C x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {2 a^{2} \left (A -2 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{2} \left (3 A -2 C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{2} \left (7 A -2 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a^{2} \left (9 A -2 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {a^{2} \left (3 A +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a^{2} \left (3 A +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(353\) |
A*a^2/d*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+(A*a^2+C *a^2)/d*ln(sec(d*x+c)+tan(d*x+c))+1/d*sin(d*x+c)*a^2*C+2*a^2*A*tan(d*x+c)/ d+2*a^2*C/d*(d*x+c)
Time = 0.28 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.15 \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {8 \, C a^{2} d x \cos \left (d x + c\right )^{2} + {\left (3 \, A + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, A + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, C a^{2} \cos \left (d x + c\right )^{2} + 4 \, A a^{2} \cos \left (d x + c\right ) + A a^{2}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]
1/4*(8*C*a^2*d*x*cos(d*x + c)^2 + (3*A + 2*C)*a^2*cos(d*x + c)^2*log(sin(d *x + c) + 1) - (3*A + 2*C)*a^2*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*( 2*C*a^2*cos(d*x + c)^2 + 4*A*a^2*cos(d*x + c) + A*a^2)*sin(d*x + c))/(d*co s(d*x + c)^2)
\[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=a^{2} \left (\int A \sec ^{3}{\left (c + d x \right )}\, dx + \int 2 A \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int A \cos ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int C \cos ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int 2 C \cos ^{3}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int C \cos ^{4}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx\right ) \]
a**2*(Integral(A*sec(c + d*x)**3, x) + Integral(2*A*cos(c + d*x)*sec(c + d *x)**3, x) + Integral(A*cos(c + d*x)**2*sec(c + d*x)**3, x) + Integral(C*c os(c + d*x)**2*sec(c + d*x)**3, x) + Integral(2*C*cos(c + d*x)**3*sec(c + d*x)**3, x) + Integral(C*cos(c + d*x)**4*sec(c + d*x)**3, x))
Time = 0.27 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.27 \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {8 \, {\left (d x + c\right )} C a^{2} - A a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, A a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, C a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, C a^{2} \sin \left (d x + c\right ) + 8 \, A a^{2} \tan \left (d x + c\right )}{4 \, d} \]
1/4*(8*(d*x + c)*C*a^2 - A*a^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log( sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 2*A*a^2*(log(sin(d*x + c) + 1 ) - log(sin(d*x + c) - 1)) + 2*C*a^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 4*C*a^2*sin(d*x + c) + 8*A*a^2*tan(d*x + c))/d
Time = 0.30 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.36 \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {4 \, {\left (d x + c\right )} C a^{2} + \frac {4 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + {\left (3 \, A a^{2} + 2 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (3 \, A a^{2} + 2 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]
1/2*(4*(d*x + c)*C*a^2 + 4*C*a^2*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c )^2 + 1) + (3*A*a^2 + 2*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (3*A*a ^2 + 2*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(3*A*a^2*tan(1/2*d*x + 1/2*c)^3 - 5*A*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2) /d
Time = 1.11 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.38 \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {C\,a^2\,\sin \left (c+d\,x\right )}{d}+\frac {3\,A\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {4\,C\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,A\,a^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {A\,a^2\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2} \]