Integrand size = 41, antiderivative size = 134 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=a^2 C x+\frac {a^2 (2 A+3 B+4 C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a^2 (2 A+3 B+2 C) \tan (c+d x)}{2 d}+\frac {(2 A+3 B) \left (a^2+a^2 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {A (a+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d} \]
a^2*C*x+1/2*a^2*(2*A+3*B+4*C)*arctanh(sin(d*x+c))/d+1/2*a^2*(2*A+3*B+2*C)* tan(d*x+c)/d+1/6*(2*A+3*B)*(a^2+a^2*cos(d*x+c))*sec(d*x+c)*tan(d*x+c)/d+1/ 3*A*(a+a*cos(d*x+c))^2*sec(d*x+c)^2*tan(d*x+c)/d
Time = 1.69 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.57 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {a^2 \left (6 C d x+3 (2 A+3 B+4 C) \text {arctanh}(\sin (c+d x))+3 (2 (2 A+2 B+C)+(2 A+B) \sec (c+d x)) \tan (c+d x)+2 A \tan ^3(c+d x)\right )}{6 d} \]
(a^2*(6*C*d*x + 3*(2*A + 3*B + 4*C)*ArcTanh[Sin[c + d*x]] + 3*(2*(2*A + 2* B + C) + (2*A + B)*Sec[c + d*x])*Tan[c + d*x] + 2*A*Tan[c + d*x]^3))/(6*d)
Time = 1.08 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.06, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.317, Rules used = {3042, 3522, 3042, 3454, 27, 3042, 3447, 3042, 3500, 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.
| \(\displaystyle \int \sec ^4(c+d x) (a \cos (c+d x)+a)^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
| \(\Big \downarrow \) 3522 |
| \(\displaystyle \frac {\int (\cos (c+d x) a+a)^2 (a (2 A+3 B)+3 a C \cos (c+d x)) \sec ^3(c+d x)dx}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^2}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (a (2 A+3 B)+3 a C \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^2}{3 d}\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {\frac {1}{2} \int 3 (\cos (c+d x) a+a) \left ((2 A+3 B+2 C) a^2+2 C \cos (c+d x) a^2\right ) \sec ^2(c+d x)dx+\frac {(2 A+3 B) \tan (c+d x) \sec (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^2}{3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {3}{2} \int (\cos (c+d x) a+a) \left ((2 A+3 B+2 C) a^2+2 C \cos (c+d x) a^2\right ) \sec ^2(c+d x)dx+\frac {(2 A+3 B) \tan (c+d x) \sec (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^2}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3}{2} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left ((2 A+3 B+2 C) a^2+2 C \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {(2 A+3 B) \tan (c+d x) \sec (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^2}{3 d}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {\frac {3}{2} \int \left (2 C \cos ^2(c+d x) a^3+(2 A+3 B+2 C) a^3+\left (2 C a^3+(2 A+3 B+2 C) a^3\right ) \cos (c+d x)\right ) \sec ^2(c+d x)dx+\frac {(2 A+3 B) \tan (c+d x) \sec (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^2}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3}{2} \int \frac {2 C \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^3+(2 A+3 B+2 C) a^3+\left (2 C a^3+(2 A+3 B+2 C) a^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {(2 A+3 B) \tan (c+d x) \sec (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^2}{3 d}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {\frac {3}{2} \left (\int \left ((2 A+3 B+4 C) a^3+2 C \cos (c+d x) a^3\right ) \sec (c+d x)dx+\frac {a^3 (2 A+3 B+2 C) \tan (c+d x)}{d}\right )+\frac {(2 A+3 B) \tan (c+d x) \sec (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^2}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3}{2} \left (\int \frac {(2 A+3 B+4 C) a^3+2 C \sin \left (c+d x+\frac {\pi }{2}\right ) a^3}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a^3 (2 A+3 B+2 C) \tan (c+d x)}{d}\right )+\frac {(2 A+3 B) \tan (c+d x) \sec (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^2}{3 d}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {\frac {3}{2} \left (a^3 (2 A+3 B+4 C) \int \sec (c+d x)dx+\frac {a^3 (2 A+3 B+2 C) \tan (c+d x)}{d}+2 a^3 C x\right )+\frac {(2 A+3 B) \tan (c+d x) \sec (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^2}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3}{2} \left (a^3 (2 A+3 B+4 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {a^3 (2 A+3 B+2 C) \tan (c+d x)}{d}+2 a^3 C x\right )+\frac {(2 A+3 B) \tan (c+d x) \sec (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^2}{3 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {3}{2} \left (\frac {a^3 (2 A+3 B+4 C) \text {arctanh}(\sin (c+d x))}{d}+\frac {a^3 (2 A+3 B+2 C) \tan (c+d x)}{d}+2 a^3 C x\right )+\frac {(2 A+3 B) \tan (c+d x) \sec (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^2}{3 d}\) |
(A*(a + a*Cos[c + d*x])^2*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + (((2*A + 3* B)*(a^3 + a^3*Cos[c + d*x])*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (3*(2*a^3*C *x + (a^3*(2*A + 3*B + 4*C)*ArcTanh[Sin[c + d*x]])/d + (a^3*(2*A + 3*B + 2 *C)*Tan[c + d*x])/d))/2)/(3*a)
3.4.15.3.1 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^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[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* (a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x ])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A *b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
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 = 8.60 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.09
| method | result | size |
| parts | \(-\frac {A \,a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (2 A \,a^{2}+B \,a^{2}\right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {\left (B \,a^{2}+2 a^{2} C \right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {\left (A \,a^{2}+2 B \,a^{2}+a^{2} C \right ) \tan \left (d x +c \right )}{d}+\frac {a^{2} C \left (d x +c \right )}{d}\) | \(146\) |
| parallelrisch | \(\frac {3 \left (-\left (A +\frac {3 B}{2}+2 C \right ) \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (A +\frac {3 B}{2}+2 C \right ) \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\frac {d x C \cos \left (3 d x +3 c \right )}{3}+\frac {\left (\frac {5 A}{3}+2 B +C \right ) \sin \left (3 d x +3 c \right )}{3}+\frac {\left (2 A +B \right ) \sin \left (2 d x +2 c \right )}{3}+d x C \cos \left (d x +c \right )+\sin \left (d x +c \right ) \left (A +\frac {2 B}{3}+\frac {C}{3}\right )\right ) a^{2}}{d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\) | \(183\) |
| derivativedivides | \(\frac {A \,a^{2} \tan \left (d x +c \right )+B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{2} C \left (d x +c \right )+2 A \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+2 B \,a^{2} \tan \left (d x +c \right )+2 a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-A \,a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+B \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{2} C \tan \left (d x +c \right )}{d}\) | \(186\) |
| default | \(\frac {A \,a^{2} \tan \left (d x +c \right )+B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{2} C \left (d x +c \right )+2 A \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+2 B \,a^{2} \tan \left (d x +c \right )+2 a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-A \,a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+B \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{2} C \tan \left (d x +c \right )}{d}\) | \(186\) |
| risch | \(a^{2} C x -\frac {i a^{2} \left (6 A \,{\mathrm e}^{5 i \left (d x +c \right )}+3 B \,{\mathrm e}^{5 i \left (d x +c \right )}-6 A \,{\mathrm e}^{4 i \left (d x +c \right )}-12 B \,{\mathrm e}^{4 i \left (d x +c \right )}-6 C \,{\mathrm e}^{4 i \left (d x +c \right )}-24 A \,{\mathrm e}^{2 i \left (d x +c \right )}-24 B \,{\mathrm e}^{2 i \left (d x +c \right )}-12 C \,{\mathrm e}^{2 i \left (d x +c \right )}-6 A \,{\mathrm e}^{i \left (d x +c \right )}-3 B \,{\mathrm e}^{i \left (d x +c \right )}-10 A -12 B -6 C \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {A \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 d}+\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}-\frac {A \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 d}-\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}\) | \(291\) |
| norman | \(\frac {a^{2} C x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a^{2} C x \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a^{2} C x -a^{2} C x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 a^{2} C x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 a^{2} C x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 a^{2} C x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 a^{2} C x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {8 a^{2} \left (2 B +C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{2} \left (2 A +3 B +2 C \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a^{2} \left (2 A +3 B +3 C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {a^{2} \left (6 A +5 B +2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {a^{2} \left (10 A +27 B +6 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {4 a^{2} \left (14 A +9 B +3 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {a^{2} \left (50 A +3 B -6 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {a^{2} \left (2 A +3 B +4 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a^{2} \left (2 A +3 B +4 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(421\) |
-A*a^2/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+(2*A*a^2+B*a^2)/d*(1/2*sec(d*x +c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+(B*a^2+2*C*a^2)/d*ln(sec(d*x +c)+tan(d*x+c))+(A*a^2+2*B*a^2+C*a^2)/d*tan(d*x+c)+a^2*C/d*(d*x+c)
Time = 0.30 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.12 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {12 \, C a^{2} d x \cos \left (d x + c\right )^{3} + 3 \, {\left (2 \, A + 3 \, B + 4 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (2 \, A + 3 \, B + 4 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (5 \, A + 6 \, B + 3 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right ) + 2 \, A a^{2}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]
integrate((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4, x, algorithm="fricas")
1/12*(12*C*a^2*d*x*cos(d*x + c)^3 + 3*(2*A + 3*B + 4*C)*a^2*cos(d*x + c)^3 *log(sin(d*x + c) + 1) - 3*(2*A + 3*B + 4*C)*a^2*cos(d*x + c)^3*log(-sin(d *x + c) + 1) + 2*(2*(5*A + 6*B + 3*C)*a^2*cos(d*x + c)^2 + 3*(2*A + B)*a^2 *cos(d*x + c) + 2*A*a^2)*sin(d*x + c))/(d*cos(d*x + c)^3)
Timed out. \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\text {Timed out} \]
Time = 0.21 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.67 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} + 12 \, {\left (d x + c\right )} C a^{2} - 6 \, A a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 3 \, B a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, B a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, C a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, A a^{2} \tan \left (d x + c\right ) + 24 \, B a^{2} \tan \left (d x + c\right ) + 12 \, C a^{2} \tan \left (d x + c\right )}{12 \, d} \]
integrate((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4, x, algorithm="maxima")
1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^2 + 12*(d*x + c)*C*a^2 - 6*A *a^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(si n(d*x + c) - 1)) - 3*B*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)) + 6*B*a^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 12*C*a^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c ) - 1)) + 12*A*a^2*tan(d*x + c) + 24*B*a^2*tan(d*x + c) + 12*C*a^2*tan(d*x + c))/d
Time = 0.35 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.87 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {6 \, {\left (d x + c\right )} C a^{2} + 3 \, {\left (2 \, A a^{2} + 3 \, B a^{2} + 4 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (2 \, A a^{2} + 3 \, B a^{2} + 4 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 16 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 18 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]
integrate((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4, x, algorithm="giac")
1/6*(6*(d*x + c)*C*a^2 + 3*(2*A*a^2 + 3*B*a^2 + 4*C*a^2)*log(abs(tan(1/2*d *x + 1/2*c) + 1)) - 3*(2*A*a^2 + 3*B*a^2 + 4*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(6*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 9*B*a^2*tan(1/2*d*x + 1 /2*c)^5 + 6*C*a^2*tan(1/2*d*x + 1/2*c)^5 - 16*A*a^2*tan(1/2*d*x + 1/2*c)^3 - 24*B*a^2*tan(1/2*d*x + 1/2*c)^3 - 12*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 18* A*a^2*tan(1/2*d*x + 1/2*c) + 15*B*a^2*tan(1/2*d*x + 1/2*c) + 6*C*a^2*tan(1 /2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d
Time = 2.41 (sec) , antiderivative size = 440, normalized size of antiderivative = 3.28 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {\frac {A\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {5\,A\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{12}+\frac {B\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{4}+\frac {B\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{2}+\frac {C\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {3\,A\,a^2\,\sin \left (c+d\,x\right )}{4}+\frac {B\,a^2\,\sin \left (c+d\,x\right )}{2}+\frac {C\,a^2\,\sin \left (c+d\,x\right )}{4}-\frac {A\,a^2\,\cos \left (c+d\,x\right )\,\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 )\,3{}\mathrm {i}}{2}-\frac {B\,a^2\,\cos \left (c+d\,x\right )\,\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 )\,9{}\mathrm {i}}{4}+\frac {3\,C\,a^2\,\cos \left (c+d\,x\right )\,\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}-C\,a^2\,\cos \left (c+d\,x\right )\,\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 )\,3{}\mathrm {i}-\frac {A\,a^2\,\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 )\,\cos \left (3\,c+3\,d\,x\right )\,1{}\mathrm {i}}{2}-\frac {B\,a^2\,\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 )\,\cos \left (3\,c+3\,d\,x\right )\,3{}\mathrm {i}}{4}+\frac {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 )\,\cos \left (3\,c+3\,d\,x\right )}{2}-C\,a^2\,\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 )\,\cos \left (3\,c+3\,d\,x\right )\,1{}\mathrm {i}}{d\,\left (\frac {3\,\cos \left (c+d\,x\right )}{4}+\frac {\cos \left (3\,c+3\,d\,x\right )}{4}\right )} \]
((A*a^2*sin(2*c + 2*d*x))/2 + (5*A*a^2*sin(3*c + 3*d*x))/12 + (B*a^2*sin(2 *c + 2*d*x))/4 + (B*a^2*sin(3*c + 3*d*x))/2 + (C*a^2*sin(3*c + 3*d*x))/4 + (3*A*a^2*sin(c + d*x))/4 + (B*a^2*sin(c + d*x))/2 + (C*a^2*sin(c + d*x))/ 4 - (A*a^2*cos(c + d*x)*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*3 i)/2 - (B*a^2*cos(c + d*x)*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2) )*9i)/4 + (3*C*a^2*cos(c + d*x)*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2) ))/2 - C*a^2*cos(c + d*x)*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2)) *3i - (A*a^2*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*cos(3*c + 3* d*x)*1i)/2 - (B*a^2*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*cos(3 *c + 3*d*x)*3i)/4 + (C*a^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos (3*c + 3*d*x))/2 - C*a^2*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))* cos(3*c + 3*d*x)*1i)/(d*((3*cos(c + d*x))/4 + cos(3*c + 3*d*x)/4))