Integrand size = 40, antiderivative size = 116 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {\left (2 a b B+a^2 C+2 b^2 C\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {\left (2 a^2 B+3 b^2 B+6 a b C\right ) \tan (c+d x)}{3 d}+\frac {a (2 b B+a C) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a^2 B \sec ^2(c+d x) \tan (c+d x)}{3 d} \] Output:
1/2*(2*B*a*b+C*a^2+2*C*b^2)*arctanh(sin(d*x+c))/d+1/3*(2*B*a^2+3*B*b^2+6*C *a*b)*tan(d*x+c)/d+1/2*a*(2*B*b+C*a)*sec(d*x+c)*tan(d*x+c)/d+1/3*a^2*B*sec (d*x+c)^2*tan(d*x+c)/d
Time = 0.67 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.84 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {6 b^2 C \coth ^{-1}(\sin (c+d x))+3 a (2 b B+a C) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (3 a (2 b B+a C) \sec (c+d x)+2 \left (3 a^2 B+3 b^2 B+6 a b C+a^2 B \tan ^2(c+d x)\right )\right )}{6 d} \] Input:
Integrate[(a + b*Cos[c + d*x])^2*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^5,x]
Output:
(6*b^2*C*ArcCoth[Sin[c + d*x]] + 3*a*(2*b*B + a*C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(3*a*(2*b*B + a*C)*Sec[c + d*x] + 2*(3*a^2*B + 3*b^2*B + 6*a *b*C + a^2*B*Tan[c + d*x]^2)))/(6*d)
Time = 0.92 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.05, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.325, Rules used = {3042, 3508, 3042, 3467, 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.
| \(\displaystyle \int \sec ^5(c+d x) (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
| \(\Big \downarrow \) 3508 |
| \(\displaystyle \int \sec ^4(c+d x) (a+b \cos (c+d x))^2 (B+C \cos (c+d x))dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (B+C \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
| \(\Big \downarrow \) 3467 |
| \(\displaystyle \frac {a^2 B \tan (c+d x) \sec ^2(c+d x)}{3 d}-\frac {1}{3} \int -\left (\left (3 b^2 C \cos ^2(c+d x)+\left (2 B a^2+6 b C a+3 b^2 B\right ) \cos (c+d x)+3 a (2 b B+a C)\right ) \sec ^3(c+d x)\right )dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{3} \int \left (3 b^2 C \cos ^2(c+d x)+\left (2 B a^2+6 b C a+3 b^2 B\right ) \cos (c+d x)+3 a (2 b B+a C)\right ) \sec ^3(c+d x)dx+\frac {a^2 B \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \int \frac {3 b^2 C \sin \left (c+d x+\frac {\pi }{2}\right )^2+\left (2 B a^2+6 b C a+3 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+3 a (2 b B+a C)}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {a^2 B \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int \left (2 \left (2 B a^2+6 b C a+3 b^2 B\right )+3 \left (C a^2+2 b B a+2 b^2 C\right ) \cos (c+d x)\right ) \sec ^2(c+d x)dx+\frac {3 a (a C+2 b B) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a^2 B \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int \frac {2 \left (2 B a^2+6 b C a+3 b^2 B\right )+3 \left (C a^2+2 b B a+2 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {3 a (a C+2 b B) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a^2 B \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (2 \left (2 a^2 B+6 a b C+3 b^2 B\right ) \int \sec ^2(c+d x)dx+3 \left (a^2 C+2 a b B+2 b^2 C\right ) \int \sec (c+d x)dx\right )+\frac {3 a (a C+2 b B) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a^2 B \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 \left (a^2 C+2 a b B+2 b^2 C\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+2 \left (2 a^2 B+6 a b C+3 b^2 B\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx\right )+\frac {3 a (a C+2 b B) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a^2 B \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 \left (a^2 C+2 a b B+2 b^2 C\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {2 \left (2 a^2 B+6 a b C+3 b^2 B\right ) \int 1d(-\tan (c+d x))}{d}\right )+\frac {3 a (a C+2 b B) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a^2 B \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 \left (a^2 C+2 a b B+2 b^2 C\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {2 \left (2 a^2 B+6 a b C+3 b^2 B\right ) \tan (c+d x)}{d}\right )+\frac {3 a (a C+2 b B) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a^2 B \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\frac {3 \left (a^2 C+2 a b B+2 b^2 C\right ) \text {arctanh}(\sin (c+d x))}{d}+\frac {2 \left (2 a^2 B+6 a b C+3 b^2 B\right ) \tan (c+d x)}{d}\right )+\frac {3 a (a C+2 b B) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a^2 B \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
Input:
Int[(a + b*Cos[c + d*x])^2*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x ]^5,x]
Output:
(a^2*B*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + ((3*a*(2*b*B + a*C)*Sec[c + d* x]*Tan[c + d*x])/(2*d) + ((3*(2*a*b*B + a^2*C + 2*b^2*C)*ArcTanh[Sin[c + d *x]])/d + (2*(2*a^2*B + 3*b^2*B + 6*a*b*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_)])^2*((A_.) + (B_.)*sin[(e_.) + (f _.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[ (B*c - A*d)*(b*c - a*d)^2*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(f*d^2 *(n + 1)*(c^2 - d^2))), x] - Simp[1/(d^2*(n + 1)*(c^2 - d^2)) Int[(c + d* Sin[e + f*x])^(n + 1)*Simp[d*(n + 1)*(B*(b*c - a*d)^2 - A*d*(a^2*c + b^2*c - 2*a*b*d)) - ((B*c - A*d)*(a^2*d^2*(n + 2) + b^2*(c^2 + d^2*(n + 1))) + 2* a*b*d*(A*c*d*(n + 2) - B*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b^2*B*d*(n + 1)*(c^2 - d^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B} , x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[ n, -1]
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[1/b^2 Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
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 = 0.75 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.02
| method | result | size |
| parts | \(\frac {\left (B \,b^{2}+2 a b C \right ) \tan \left (d x +c \right )}{d}+\frac {\left (2 B a b +a^{2} C \right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {B \,a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {b^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(118\) |
| derivativedivides | \(\frac {a^{2} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B \,a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+2 C \tan \left (d x +c \right ) a b +2 B a b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+b^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \tan \left (d x +c \right ) b^{2}}{d}\) | \(143\) |
| default | \(\frac {a^{2} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B \,a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+2 C \tan \left (d x +c \right ) a b +2 B a b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+b^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \tan \left (d x +c \right ) b^{2}}{d}\) | \(143\) |
| parallelrisch | \(\frac {-9 \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \left (B a b +\frac {1}{2} a^{2} C +b^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+9 \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \left (B a b +\frac {1}{2} a^{2} C +b^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (2 B \,a^{2}+3 B \,b^{2}+6 a b C \right ) \sin \left (3 d x +3 c \right )+3 \left (2 B a b +a^{2} C \right ) \sin \left (2 d x +2 c \right )+6 \sin \left (d x +c \right ) \left (B \,a^{2}+\frac {1}{2} B \,b^{2}+a b C \right )}{3 d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\) | \(197\) |
| risch | \(-\frac {i \left (6 B a b \,{\mathrm e}^{5 i \left (d x +c \right )}+3 C \,a^{2} {\mathrm e}^{5 i \left (d x +c \right )}-6 B \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-12 C a b \,{\mathrm e}^{4 i \left (d x +c \right )}-12 B \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-12 B \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-24 C a b \,{\mathrm e}^{2 i \left (d x +c \right )}-6 B a b \,{\mathrm e}^{i \left (d x +c \right )}-3 C \,a^{2} {\mathrm e}^{i \left (d x +c \right )}-4 B \,a^{2}-6 B \,b^{2}-12 a b C \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B b}{d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2} C}{d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B b}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2} C}{d}\) | \(298\) |
| norman | \(\frac {\frac {\left (2 B \,a^{2}+2 B a b -6 B \,b^{2}+a^{2} C -12 a b C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}+\frac {\left (2 B \,a^{2}+2 B a b +2 B \,b^{2}+a^{2} C +4 a b C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (2 B \,a^{2}-30 B a b +18 B \,b^{2}-15 a^{2} C +36 a b C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d}-\frac {\left (2 B \,a^{2}-2 B a b -6 B \,b^{2}-a^{2} C -12 a b C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d}-\frac {\left (2 B \,a^{2}-2 B a b +2 B \,b^{2}-a^{2} C +4 a b C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{d}-\frac {\left (2 B \,a^{2}+30 B a b +18 B \,b^{2}+15 a^{2} C +36 a b C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}-\frac {\left (14 B \,a^{2}-18 B a b +6 B \,b^{2}-9 a^{2} C +12 a b C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{3 d}+\frac {\left (14 B \,a^{2}+18 B a b +6 B \,b^{2}+9 a^{2} C +12 a b C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}-\frac {\left (2 B a b +a^{2} C +2 b^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (2 B a b +a^{2} C +2 b^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(456\) |
Input:
int((a+b*cos(d*x+c))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x,method =_RETURNVERBOSE)
Output:
(B*b^2+2*C*a*b)/d*tan(d*x+c)+(2*B*a*b+C*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/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+ b^2*C/d*ln(sec(d*x+c)+tan(d*x+c))
Time = 0.11 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.29 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {3 \, {\left (C a^{2} + 2 \, B a b + 2 \, C b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (C a^{2} + 2 \, B a b + 2 \, C b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, B a^{2} + 2 \, {\left (2 \, B a^{2} + 6 \, C a b + 3 \, B b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (C a^{2} + 2 \, B a b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \] Input:
integrate((a+b*cos(d*x+c))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="fricas")
Output:
1/12*(3*(C*a^2 + 2*B*a*b + 2*C*b^2)*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(C*a^2 + 2*B*a*b + 2*C*b^2)*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*( 2*B*a^2 + 2*(2*B*a^2 + 6*C*a*b + 3*B*b^2)*cos(d*x + c)^2 + 3*(C*a^2 + 2*B* a*b)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^3)
Timed out. \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+b*cos(d*x+c))**2*(B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**5 ,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.48 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} - 3 \, C a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 6 \, B a b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, C b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, C a b \tan \left (d x + c\right ) + 12 \, B b^{2} \tan \left (d x + c\right )}{12 \, d} \] Input:
integrate((a+b*cos(d*x+c))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="maxima")
Output:
1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^2 - 3*C*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*b*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(s in(d*x + c) - 1)) + 6*C*b^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1) ) + 24*C*a*b*tan(d*x + c) + 12*B*b^2*tan(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 294 vs. \(2 (108) = 216\).
Time = 0.16 (sec) , antiderivative size = 294, normalized size of antiderivative = 2.53 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {3 \, {\left (C a^{2} + 2 \, B a b + 2 \, C b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (C a^{2} + 2 \, B a b + 2 \, C b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \] Input:
integrate((a+b*cos(d*x+c))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="giac")
Output:
1/6*(3*(C*a^2 + 2*B*a*b + 2*C*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3* (C*a^2 + 2*B*a*b + 2*C*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(6*B*a^ 2*tan(1/2*d*x + 1/2*c)^5 - 3*C*a^2*tan(1/2*d*x + 1/2*c)^5 - 6*B*a*b*tan(1/ 2*d*x + 1/2*c)^5 + 12*C*a*b*tan(1/2*d*x + 1/2*c)^5 + 6*B*b^2*tan(1/2*d*x + 1/2*c)^5 - 4*B*a^2*tan(1/2*d*x + 1/2*c)^3 - 24*C*a*b*tan(1/2*d*x + 1/2*c) ^3 - 12*B*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*B*a^2*tan(1/2*d*x + 1/2*c) + 3*C* a^2*tan(1/2*d*x + 1/2*c) + 6*B*a*b*tan(1/2*d*x + 1/2*c) + 12*C*a*b*tan(1/2 *d*x + 1/2*c) + 6*B*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1) ^3)/d
Time = 2.58 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.96 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {C\,a^2}{2}+B\,a\,b+C\,b^2\right )}{2\,C\,a^2+4\,B\,a\,b+4\,C\,b^2}\right )\,\left (C\,a^2+2\,B\,a\,b+2\,C\,b^2\right )}{d}-\frac {\left (2\,B\,a^2+2\,B\,b^2-C\,a^2-2\,B\,a\,b+4\,C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {4\,B\,a^2}{3}-8\,C\,a\,b-4\,B\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,B\,a^2+2\,B\,b^2+C\,a^2+2\,B\,a\,b+4\,C\,a\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \] Input:
int(((B*cos(c + d*x) + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^2)/cos(c + d *x)^5,x)
Output:
(atanh((4*tan(c/2 + (d*x)/2)*((C*a^2)/2 + C*b^2 + B*a*b))/(2*C*a^2 + 4*C*b ^2 + 4*B*a*b))*(C*a^2 + 2*C*b^2 + 2*B*a*b))/d - (tan(c/2 + (d*x)/2)*(2*B*a ^2 + 2*B*b^2 + C*a^2 + 2*B*a*b + 4*C*a*b) - tan(c/2 + (d*x)/2)^3*((4*B*a^2 )/3 + 4*B*b^2 + 8*C*a*b) + tan(c/2 + (d*x)/2)^5*(2*B*a^2 + 2*B*b^2 - C*a^2 - 2*B*a*b + 4*C*a*b))/(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 = 0.15 (sec) , antiderivative size = 472, normalized size of antiderivative = 4.07 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(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^{2} c -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} a \,b^{2}-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^{2} c +3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2} c +6 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a \,b^{2}+6 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{2} 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^{2} c +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} a \,b^{2}+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^{2} c -3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2} c -6 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a \,b^{2}-6 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{2} c -3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} c -6 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a \,b^{2}+4 \sin \left (d x +c \right )^{3} a^{2} b +12 \sin \left (d x +c \right )^{3} a b c +6 \sin \left (d x +c \right )^{3} b^{3}-6 \sin \left (d x +c \right ) a^{2} b -12 \sin \left (d x +c \right ) a b c -6 \sin \left (d x +c \right ) b^{3}}{6 \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))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x)
Output:
( - 3*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2*c - 6*co s(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a*b**2 - 6*cos(c + d* x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b**2*c + 3*cos(c + d*x)*log(t an((c + d*x)/2) - 1)*a**2*c + 6*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a*b **2 + 6*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*b**2*c + 3*cos(c + d*x)*log (tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2*c + 6*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a*b**2 + 6*cos(c + d*x)*log(tan((c + d*x)/2 ) + 1)*sin(c + d*x)**2*b**2*c - 3*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a **2*c - 6*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a*b**2 - 6*cos(c + d*x)*l og(tan((c + d*x)/2) + 1)*b**2*c - 3*cos(c + d*x)*sin(c + d*x)*a**2*c - 6*c os(c + d*x)*sin(c + d*x)*a*b**2 + 4*sin(c + d*x)**3*a**2*b + 12*sin(c + d* x)**3*a*b*c + 6*sin(c + d*x)**3*b**3 - 6*sin(c + d*x)*a**2*b - 12*sin(c + d*x)*a*b*c - 6*sin(c + d*x)*b**3)/(6*cos(c + d*x)*d*(sin(c + d*x)**2 - 1))