Integrand size = 39, antiderivative size = 101 \[ \int (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {(A b+a B+2 b C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {(2 a A+3 b B+3 a C) \tan (c+d x)}{3 d}+\frac {(A b+a B) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a A \sec ^2(c+d x) \tan (c+d x)}{3 d} \] Output:
1/2*(A*b+B*a+2*C*b)*arctanh(sin(d*x+c))/d+1/3*(2*A*a+3*B*b+3*C*a)*tan(d*x+ c)/d+1/2*(A*b+B*a)*sec(d*x+c)*tan(d*x+c)/d+1/3*a*A*sec(d*x+c)^2*tan(d*x+c) /d
Time = 0.65 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.79 \[ \int (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {6 b C \coth ^{-1}(\sin (c+d x))+3 (A b+a B) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (6 b B+6 a (A+C)+3 (A b+a B) \sec (c+d x)+2 a A \tan ^2(c+d x)\right )}{6 d} \] Input:
Integrate[(a + b*Cos[c + d*x])*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec [c + d*x]^4,x]
Output:
(6*b*C*ArcCoth[Sin[c + d*x]] + 3*(A*b + a*B)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(6*b*B + 6*a*(A + C) + 3*(A*b + a*B)*Sec[c + d*x] + 2*a*A*Tan[c + d*x]^2))/(6*d)
Time = 0.73 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.06, 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.
| \(\displaystyle \int \sec ^4(c+d x) (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
| \(\Big \downarrow \) 3510 |
| \(\displaystyle \frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d}-\frac {1}{3} \int -\left (\left (3 b C \cos ^2(c+d x)+(2 a A+3 b B+3 a C) \cos (c+d x)+3 (A b+a B)\right ) \sec ^3(c+d x)\right )dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{3} \int \left (3 b C \cos ^2(c+d x)+(2 a A+3 b B+3 a C) \cos (c+d x)+3 (A b+a B)\right ) \sec ^3(c+d x)dx+\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \int \frac {3 b C \sin \left (c+d x+\frac {\pi }{2}\right )^2+(2 a A+3 b B+3 a C) \sin \left (c+d x+\frac {\pi }{2}\right )+3 (A b+a B)}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int (2 (2 a A+3 b B+3 a C)+3 (A b+2 C b+a B) \cos (c+d x)) \sec ^2(c+d x)dx+\frac {3 (a B+A b) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int \frac {2 (2 a A+3 b B+3 a C)+3 (A b+2 C b+a B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {3 (a B+A b) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (2 (2 a A+3 a C+3 b B) \int \sec ^2(c+d x)dx+3 (a B+A b+2 b C) \int \sec (c+d x)dx\right )+\frac {3 (a B+A b) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 (a B+A b+2 b C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+2 (2 a A+3 a C+3 b B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx\right )+\frac {3 (a B+A b) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 (a B+A b+2 b C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {2 (2 a A+3 a C+3 b B) \int 1d(-\tan (c+d x))}{d}\right )+\frac {3 (a B+A b) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 (a B+A b+2 b C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {2 \tan (c+d x) (2 a A+3 a C+3 b B)}{d}\right )+\frac {3 (a B+A b) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\frac {3 (a B+A b+2 b C) \text {arctanh}(\sin (c+d x))}{d}+\frac {2 \tan (c+d x) (2 a A+3 a C+3 b B)}{d}\right )+\frac {3 (a B+A b) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
Input:
Int[(a + b*Cos[c + d*x])*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d *x]^4,x]
Output:
(a*A*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + ((3*(A*b + a*B)*Sec[c + d*x]*Tan [c + d*x])/(2*d) + ((3*(A*b + a*B + 2*b*C)*ArcTanh[Sin[c + d*x]])/d + (2*( 2*a*A + 3*b*B + 3*a*C)*Tan[c + d*x])/d)/2)/3
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 = 0.62 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.05
| method | result | size |
| parts | \(-\frac {a A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (A b +B a \right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {\left (B b +C a \right ) \tan \left (d x +c \right )}{d}+\frac {C b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(106\) |
| derivativedivides | \(\frac {-a A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+B a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C a \tan \left (d x +c \right )+A b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B b \tan \left (d x +c \right )+C b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(131\) |
| default | \(\frac {-a A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+B a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C a \tan \left (d x +c \right )+A b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B b \tan \left (d x +c \right )+C b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(131\) |
| parallelrisch | \(\frac {-3 \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right ) \left (\left (A +2 C \right ) b +B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+3 \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right ) \left (\left (A +2 C \right ) b +B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (6 B b +4 \left (A +\frac {3 C}{2}\right ) a \right ) \sin \left (3 d x +3 c \right )+\left (6 A b +6 B a \right ) \sin \left (2 d x +2 c \right )+12 \left (\frac {B b}{2}+a \left (A +\frac {C}{2}\right )\right ) \sin \left (d x +c \right )}{6 d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\) | \(175\) |
| risch | \(-\frac {i \left (3 A b \,{\mathrm e}^{5 i \left (d x +c \right )}+3 B a \,{\mathrm e}^{5 i \left (d x +c \right )}-6 B b \,{\mathrm e}^{4 i \left (d x +c \right )}-6 C a \,{\mathrm e}^{4 i \left (d x +c \right )}-12 A a \,{\mathrm e}^{2 i \left (d x +c \right )}-12 B b \,{\mathrm e}^{2 i \left (d x +c \right )}-12 C a \,{\mathrm e}^{2 i \left (d x +c \right )}-3 A b \,{\mathrm e}^{i \left (d x +c \right )}-3 B a \,{\mathrm e}^{i \left (d x +c \right )}-4 a A -6 B b -6 C a \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A b}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B a}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C b}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A b}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B a}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C b}{d}\) | \(270\) |
| norman | \(\frac {-\frac {2 \left (2 a A -A b -B a -2 B b -2 C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}-\frac {\left (2 a A -A b -B a +2 B b +2 C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d}-\frac {2 \left (2 a A +A b +B a -2 B b -2 C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {\left (2 a A +A b +B a +2 B b +2 C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {\left (14 a A -9 A b -9 B a +6 B b +6 C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d}-\frac {\left (14 a A +9 A b +9 B a +6 B b +6 C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}-\frac {\left (A b +B a +2 C b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (A b +B a +2 C b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(306\) |
Input:
int((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x,method =_RETURNVERBOSE)
Output:
-a*A/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+(A*b+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*b+C*a)*tan(d*x+c)/d+C*b/d*ln(sec( d*x+c)+tan(d*x+c))
Time = 0.09 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.27 \[ \int (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {3 \, {\left (B a + {\left (A + 2 \, C\right )} b\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (B a + {\left (A + 2 \, C\right )} b\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left ({\left (2 \, A + 3 \, C\right )} a + 3 \, B b\right )} \cos \left (d x + c\right )^{2} + 2 \, A a + 3 \, {\left (B a + A b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \] Input:
integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="fricas")
Output:
1/12*(3*(B*a + (A + 2*C)*b)*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(B*a + (A + 2*C)*b)*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(2*((2*A + 3*C)*a + 3*B*b)*cos(d*x + c)^2 + 2*A*a + 3*(B*a + A*b)*cos(d*x + c))*sin(d*x + c ))/(d*cos(d*x + c)^3)
Timed out. \[ \int (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**4, x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.60 \[ \int (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a - 3 \, B a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 3 \, A b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, C b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, C a \tan \left (d x + c\right ) + 12 \, B b \tan \left (d x + c\right )}{12 \, d} \] Input:
integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="maxima")
Output:
1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a - 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)) - 3*A*b* (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*b*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 12*C *a*tan(d*x + c) + 12*B*b*tan(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 261 vs. \(2 (93) = 186\).
Time = 0.15 (sec) , antiderivative size = 261, normalized size of antiderivative = 2.58 \[ \int (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {3 \, {\left (B a + A b + 2 \, C b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (B a + A b + 2 \, C b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \] Input:
integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="giac")
Output:
1/6*(3*(B*a + A*b + 2*C*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(B*a + A *b + 2*C*b)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(6*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 - 3*A *b*tan(1/2*d*x + 1/2*c)^5 + 6*B*b*tan(1/2*d*x + 1/2*c)^5 - 4*A*a*tan(1/2*d *x + 1/2*c)^3 - 12*C*a*tan(1/2*d*x + 1/2*c)^3 - 12*B*b*tan(1/2*d*x + 1/2*c )^3 + 6*A*a*tan(1/2*d*x + 1/2*c) + 3*B*a*tan(1/2*d*x + 1/2*c) + 6*C*a*tan( 1/2*d*x + 1/2*c) + 3*A*b*tan(1/2*d*x + 1/2*c) + 6*B*b*tan(1/2*d*x + 1/2*c) )/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d
Time = 2.70 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.88 \[ \int (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {A\,b}{2}+\frac {B\,a}{2}+C\,b\right )}{2\,A\,b+2\,B\,a+4\,C\,b}\right )\,\left (A\,b+B\,a+2\,C\,b\right )}{d}-\frac {\left (2\,A\,a-A\,b-B\,a+2\,B\,b+2\,C\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {4\,A\,a}{3}-4\,B\,b-4\,C\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a+A\,b+B\,a+2\,B\,b+2\,C\,a\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \] Input:
int(((a + b*cos(c + d*x))*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + d*x)^4,x)
Output:
(atanh((4*tan(c/2 + (d*x)/2)*((A*b)/2 + (B*a)/2 + C*b))/(2*A*b + 2*B*a + 4 *C*b))*(A*b + B*a + 2*C*b))/d - (tan(c/2 + (d*x)/2)*(2*A*a + A*b + B*a + 2 *B*b + 2*C*a) - tan(c/2 + (d*x)/2)^3*((4*A*a)/3 + 4*B*b + 4*C*a) + tan(c/2 + (d*x)/2)^5*(2*A*a - A*b - B*a + 2*B*b + 2*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))
Time = 0.16 (sec) , antiderivative size = 320, normalized size of antiderivative = 3.17 \[ \int (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {-3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a b -3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b c +3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a b +3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b c +3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a b +3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b c -3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a b -3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b c -3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a b +2 \sin \left (d x +c \right )^{3} a^{2}+3 \sin \left (d x +c \right )^{3} a c +3 \sin \left (d x +c \right )^{3} b^{2}-3 \sin \left (d x +c \right ) a^{2}-3 \sin \left (d x +c \right ) a c -3 \sin \left (d x +c \right ) b^{2}}{3 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x)
Output:
( - 3*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a*b - 3*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b*c + 3*cos(c + d*x)*log (tan((c + d*x)/2) - 1)*a*b + 3*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*b*c + 3*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a*b + 3*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b*c - 3*cos(c + d*x)*log(t an((c + d*x)/2) + 1)*a*b - 3*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*b*c - 3*cos(c + d*x)*sin(c + d*x)*a*b + 2*sin(c + d*x)**3*a**2 + 3*sin(c + d*x)* *3*a*c + 3*sin(c + d*x)**3*b**2 - 3*sin(c + d*x)*a**2 - 3*sin(c + d*x)*a*c - 3*sin(c + d*x)*b**2)/(3*cos(c + d*x)*d*(sin(c + d*x)**2 - 1))