Integrand size = 39, antiderivative size = 91 \[ \int (a+a \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {a (A+B+2 C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a (2 A+3 (B+C)) \tan (c+d x)}{3 d}+\frac {a (A+B) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a A \sec ^2(c+d x) \tan (c+d x)}{3 d} \]
1/2*a*(A+B+2*C)*arctanh(sin(d*x+c))/d+1/3*a*(2*A+3*B+3*C)*tan(d*x+c)/d+1/2 *a*(A+B)*sec(d*x+c)*tan(d*x+c)/d+1/3*a*A*sec(d*x+c)^2*tan(d*x+c)/d
Time = 0.38 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.66 \[ \int (a+a \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {a \left (3 (A+B+2 C) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (6 (A+B+C)+3 (A+B) \sec (c+d x)+2 A \tan ^2(c+d x)\right )\right )}{6 d} \]
(a*(3*(A + B + 2*C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(6*(A + B + C) + 3*(A + B)*Sec[c + d*x] + 2*A*Tan[c + d*x]^2)))/(6*d)
Time = 0.70 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.07, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.282, Rules used = {3042, 3510, 25, 3042, 3500, 3042, 3227, 3042, 4254, 24, 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.
\(\displaystyle \int \sec ^4(c+d x) (a \cos (c+d x)+a) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right ) \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
\(\Big \downarrow \) 3510 |
\(\displaystyle \frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d}-\frac {1}{3} \int -\left (\left (3 a C \cos ^2(c+d x)+a (2 A+3 (B+C)) \cos (c+d x)+3 a (A+B)\right ) \sec ^3(c+d x)\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{3} \int \left (3 a C \cos ^2(c+d x)+a (2 A+3 (B+C)) \cos (c+d x)+3 a (A+B)\right ) \sec ^3(c+d x)dx+\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} \int \frac {3 a C \sin \left (c+d x+\frac {\pi }{2}\right )^2+a (2 A+3 (B+C)) \sin \left (c+d x+\frac {\pi }{2}\right )+3 a (A+B)}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
\(\Big \downarrow \) 3500 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int (2 a (2 A+3 (B+C))+3 a (A+B+2 C) \cos (c+d x)) \sec ^2(c+d x)dx+\frac {3 a (A+B) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int \frac {2 a (2 A+3 (B+C))+3 a (A+B+2 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {3 a (A+B) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (2 a (2 A+3 (B+C)) \int \sec ^2(c+d x)dx+3 a (A+B+2 C) \int \sec (c+d x)dx\right )+\frac {3 a (A+B) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 a (A+B+2 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+2 a (2 A+3 (B+C)) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx\right )+\frac {3 a (A+B) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 a (A+B+2 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {2 a (2 A+3 (B+C)) \int 1d(-\tan (c+d x))}{d}\right )+\frac {3 a (A+B) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 a (A+B+2 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {2 a (2 A+3 (B+C)) \tan (c+d x)}{d}\right )+\frac {3 a (A+B) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\frac {3 a (A+B+2 C) \text {arctanh}(\sin (c+d x))}{d}+\frac {2 a (2 A+3 (B+C)) \tan (c+d x)}{d}\right )+\frac {3 a (A+B) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
(a*A*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + ((3*a*(A + B)*Sec[c + d*x]*Tan[c + d*x])/(2*d) + ((3*a*(A + B + 2*C)*ArcTanh[Sin[c + d*x]])/d + (2*a*(2*A + 3*(B + C))*Tan[c + d*x])/d)/2)/3
3.4.7.3.1 Defintions of rubi rules used
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
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_)])*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f _.)*(x_)]^2), x_Symbol] :> Simp[(-(b*c - a*d))*(A*b^2 - a*b*B + a^2*C)*Cos[ e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] - S imp[1/(b^2*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*( m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1)) ))*Sin[e + f*x] - b*C*d*(m + 1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; F reeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 8.12 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.16
method | result | size |
parts | \(-\frac {a A \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (a A +B a \right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {\left (B a +a C \right ) \tan \left (d x +c \right )}{d}+\frac {a C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(106\) |
derivativedivides | \(\frac {a A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B a \tan \left (d x +c \right )+a C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-a A \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+B a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a C \tan \left (d x +c \right )}{d}\) | \(131\) |
default | \(\frac {a A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B a \tan \left (d x +c \right )+a C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-a A \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+B a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a C \tan \left (d x +c \right )}{d}\) | \(131\) |
parallelrisch | \(\frac {2 a \left (-\frac {3 \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \left (A +B +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4}+\frac {3 \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \left (A +B +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4}+\left (\frac {A}{3}+\frac {B}{2}+\frac {C}{2}\right ) \sin \left (3 d x +3 c \right )+\frac {\left (A +B \right ) \sin \left (2 d x +2 c \right )}{2}+\sin \left (d x +c \right ) \left (A +\frac {B}{2}+\frac {C}{2}\right )\right )}{d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\) | \(153\) |
norman | \(\frac {-\frac {2 a \left (A -3 B -2 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a \left (A +B +2 C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a \left (3 A -B -2 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a \left (3 A +3 B +2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {a \left (5 A -3 B +6 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {a \left (23 A +15 B +6 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {a \left (A +B +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a \left (A +B +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(236\) |
risch | \(-\frac {i a \left (3 A \,{\mathrm e}^{5 i \left (d x +c \right )}+3 B \,{\mathrm e}^{5 i \left (d x +c \right )}-6 B \,{\mathrm e}^{4 i \left (d x +c \right )}-6 C \,{\mathrm e}^{4 i \left (d x +c \right )}-12 A \,{\mathrm e}^{2 i \left (d x +c \right )}-12 B \,{\mathrm e}^{2 i \left (d x +c \right )}-12 C \,{\mathrm e}^{2 i \left (d x +c \right )}-3 A \,{\mathrm e}^{i \left (d x +c \right )}-3 B \,{\mathrm e}^{i \left (d x +c \right )}-4 A -6 B -6 C \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {a A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {B a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}-\frac {a A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {B a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}\) | \(259\) |
-a*A/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+(A*a+B*a)/d*(1/2*sec(d*x+c)*tan( d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+(B*a+C*a)/d*tan(d*x+c)+a*C/d*ln(sec( d*x+c)+tan(d*x+c))
Time = 0.28 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.25 \[ \int (a+a \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {3 \, {\left (A + B + 2 \, C\right )} a \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (A + B + 2 \, C\right )} a \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (2 \, A + 3 \, B + 3 \, C\right )} a \cos \left (d x + c\right )^{2} + 3 \, {\left (A + B\right )} a \cos \left (d x + c\right ) + 2 \, A a\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]
integrate((a+a*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="fricas")
1/12*(3*(A + B + 2*C)*a*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(A + B + 2*C)*a*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(2*(2*A + 3*B + 3*C)*a*co s(d*x + c)^2 + 3*(A + B)*a*cos(d*x + c) + 2*A*a)*sin(d*x + c))/(d*cos(d*x + c)^3)
\[ \int (a+a \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=a \left (\int A \sec ^{4}{\left (c + d x \right )}\, dx + \int A \cos {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int B \cos {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int B \cos ^{2}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int C \cos ^{2}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int C \cos ^{3}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx\right ) \]
a*(Integral(A*sec(c + d*x)**4, x) + Integral(A*cos(c + d*x)*sec(c + d*x)** 4, x) + Integral(B*cos(c + d*x)*sec(c + d*x)**4, x) + Integral(B*cos(c + d *x)**2*sec(c + d*x)**4, x) + Integral(C*cos(c + d*x)**2*sec(c + d*x)**4, x ) + Integral(C*cos(c + d*x)**3*sec(c + d*x)**4, x))
Time = 0.22 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.78 \[ \int (a+a \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a - 3 \, A a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 3 \, B a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, C a {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, B a \tan \left (d x + c\right ) + 12 \, C a \tan \left (d x + c\right )}{12 \, d} \]
integrate((a+a*cos(d*x+c))*(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 - 3*A*a*(2*sin(d*x + c)/(sin (d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 3*B*a* (2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 6*C*a*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 12*B *a*tan(d*x + c) + 12*C*a*tan(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 205 vs. \(2 (84) = 168\).
Time = 0.35 (sec) , antiderivative size = 205, normalized size of antiderivative = 2.25 \[ \int (a+a \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {3 \, {\left (A a + B a + 2 \, C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (A a + B a + 2 \, C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]
1/6*(3*(A*a + B*a + 2*C*a)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(A*a + B *a + 2*C*a)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(3*A*a*tan(1/2*d*x + 1/ 2*c)^5 + 3*B*a*tan(1/2*d*x + 1/2*c)^5 + 6*C*a*tan(1/2*d*x + 1/2*c)^5 - 4*A *a*tan(1/2*d*x + 1/2*c)^3 - 12*B*a*tan(1/2*d*x + 1/2*c)^3 - 12*C*a*tan(1/2 *d*x + 1/2*c)^3 + 9*A*a*tan(1/2*d*x + 1/2*c) + 9*B*a*tan(1/2*d*x + 1/2*c) + 6*C*a*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d
Time = 4.37 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.81 \[ \int (a+a \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {a\,\mathrm {atanh}\left (\frac {2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A+B+2\,C\right )}{2\,A\,a+2\,B\,a+4\,C\,a}\right )\,\left (A+B+2\,C\right )}{d}-\frac {\left (A\,a+B\,a+2\,C\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {4\,A\,a}{3}-4\,B\,a-4\,C\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (3\,A\,a+3\,B\,a+2\,C\,a\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]
(a*atanh((2*a*tan(c/2 + (d*x)/2)*(A + B + 2*C))/(2*A*a + 2*B*a + 4*C*a))*( A + B + 2*C))/d - (tan(c/2 + (d*x)/2)*(3*A*a + 3*B*a + 2*C*a) + tan(c/2 + (d*x)/2)^5*(A*a + B*a + 2*C*a) - tan(c/2 + (d*x)/2)^3*((4*A*a)/3 + 4*B*a + 4*C*a))/(d*(3*tan(c/2 + (d*x)/2)^2 - 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + ( d*x)/2)^6 - 1))