Integrand size = 41, antiderivative size = 141 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=b^2 C x+\frac {\left (a^2 B+2 b^2 B+2 a b (A+2 C)\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {\left (2 A b^2+6 a b B+a^2 (2 A+3 C)\right ) \tan (c+d x)}{3 d}+\frac {a (2 A b+3 a B) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {A (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d} \] Output:
b^2*C*x+1/2*(B*a^2+2*B*b^2+2*a*b*(A+2*C))*arctanh(sin(d*x+c))/d+1/3*(2*A*b ^2+6*B*a*b+a^2*(2*A+3*C))*tan(d*x+c)/d+1/6*a*(2*A*b+3*B*a)*sec(d*x+c)*tan( d*x+c)/d+1/3*A*(a+b*cos(d*x+c))^2*sec(d*x+c)^2*tan(d*x+c)/d
Time = 0.91 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.86 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {6 b^2 C d x+6 b (b B+2 a C) \coth ^{-1}(\sin (c+d x))+3 a (2 A b+a B) \text {arctanh}(\sin (c+d x))+6 \left (A b^2+a (2 b B+a C)\right ) \tan (c+d x)+3 a (2 A b+a B) \sec (c+d x) \tan (c+d x)+2 a^2 A \tan (c+d x) \left (3+\tan ^2(c+d x)\right )}{6 d} \] Input:
Integrate[(a + b*Cos[c + d*x])^2*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*S ec[c + d*x]^4,x]
Output:
(6*b^2*C*d*x + 6*b*(b*B + 2*a*C)*ArcCoth[Sin[c + d*x]] + 3*a*(2*A*b + a*B) *ArcTanh[Sin[c + d*x]] + 6*(A*b^2 + a*(2*b*B + a*C))*Tan[c + d*x] + 3*a*(2 *A*b + a*B)*Sec[c + d*x]*Tan[c + d*x] + 2*a^2*A*Tan[c + d*x]*(3 + Tan[c + d*x]^2))/(6*d)
Time = 1.03 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.06, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.293, Rules used = {3042, 3526, 3042, 3510, 25, 3042, 3500, 27, 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.
| \(\displaystyle \int \sec ^4(c+d x) (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
| \(\Big \downarrow \) 3526 |
| \(\displaystyle \frac {1}{3} \int (a+b \cos (c+d x)) \left (3 b C \cos ^2(c+d x)+(2 a A+3 b B+3 a C) \cos (c+d x)+2 A b+3 a B\right ) \sec ^3(c+d x)dx+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (3 b C \sin \left (c+d x+\frac {\pi }{2}\right )^2+(2 a A+3 b B+3 a C) \sin \left (c+d x+\frac {\pi }{2}\right )+2 A b+3 a B\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 3510 |
| \(\displaystyle \frac {1}{3} \left (\frac {a (3 a B+2 A b) \tan (c+d x) \sec (c+d x)}{2 d}-\frac {1}{2} \int -\left (\left (6 b^2 C \cos ^2(c+d x)+3 \left (B a^2+2 b (A+2 C) a+2 b^2 B\right ) \cos (c+d x)+2 \left ((2 A+3 C) a^2+6 b B a+2 A b^2\right )\right ) \sec ^2(c+d x)\right )dx\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int \left (6 b^2 C \cos ^2(c+d x)+3 \left (B a^2+2 b (A+2 C) a+2 b^2 B\right ) \cos (c+d x)+2 \left ((2 A+3 C) a^2+6 b B a+2 A b^2\right )\right ) \sec ^2(c+d x)dx+\frac {a (3 a B+2 A b) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int \frac {6 b^2 C \sin \left (c+d x+\frac {\pi }{2}\right )^2+3 \left (B a^2+2 b (A+2 C) a+2 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+2 \left ((2 A+3 C) a^2+6 b B a+2 A b^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a (3 a B+2 A b) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\int 3 \left (B a^2+2 b (A+2 C) a+2 b^2 B+2 b^2 C \cos (c+d x)\right ) \sec (c+d x)dx+\frac {2 \tan (c+d x) \left (a^2 (2 A+3 C)+6 a b B+2 A b^2\right )}{d}\right )+\frac {a (3 a B+2 A b) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 \int \left (B a^2+2 b (A+2 C) a+2 b^2 B+2 b^2 C \cos (c+d x)\right ) \sec (c+d x)dx+\frac {2 \tan (c+d x) \left (a^2 (2 A+3 C)+6 a b B+2 A b^2\right )}{d}\right )+\frac {a (3 a B+2 A b) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 \int \frac {B a^2+2 b (A+2 C) a+2 b^2 B+2 b^2 C \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \tan (c+d x) \left (a^2 (2 A+3 C)+6 a b B+2 A b^2\right )}{d}\right )+\frac {a (3 a B+2 A b) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 \left (\left (a^2 B+2 a b (A+2 C)+2 b^2 B\right ) \int \sec (c+d x)dx+2 b^2 C x\right )+\frac {2 \tan (c+d x) \left (a^2 (2 A+3 C)+6 a b B+2 A b^2\right )}{d}\right )+\frac {a (3 a B+2 A b) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 \left (\left (a^2 B+2 a b (A+2 C)+2 b^2 B\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+2 b^2 C x\right )+\frac {2 \tan (c+d x) \left (a^2 (2 A+3 C)+6 a b B+2 A b^2\right )}{d}\right )+\frac {a (3 a B+2 A b) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 \left (\frac {\left (a^2 B+2 a b (A+2 C)+2 b^2 B\right ) \text {arctanh}(\sin (c+d x))}{d}+2 b^2 C x\right )+\frac {2 \tan (c+d x) \left (a^2 (2 A+3 C)+6 a b B+2 A b^2\right )}{d}\right )+\frac {a (3 a B+2 A b) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}\) |
Input:
Int[(a + b*Cos[c + d*x])^2*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]
Output:
(A*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + ((a*(2*A*b + 3*a*B)*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (3*(2*b^2*C*x + ((a^2*B + 2*b^ 2*B + 2*a*b*(A + 2*C))*ArcTanh[Sin[c + d*x]])/d) + (2*(2*A*b^2 + 6*a*b*B + a^2*(2*A + 3*C))*Tan[c + d*x])/d)/2)/3
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_.)*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[((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/(d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + (c*C - B* d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x ] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f *x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d , 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.81 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.01
| method | result | size |
| parts | \(-\frac {A \,a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (2 a A b +B \,a^{2}\right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {\left (B \,b^{2}+2 a b C \right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {\left (A \,b^{2}+2 B a b +a^{2} C \right ) \tan \left (d x +c \right )}{d}+\frac {C \,b^{2} \left (d x +c \right )}{d}\) | \(143\) |
| derivativedivides | \(\frac {-A \,a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+B \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C \tan \left (d x +c \right ) a^{2}+2 a A b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+2 B \tan \left (d x +c \right ) a b +2 a b C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+A \tan \left (d x +c \right ) b^{2}+B \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \,b^{2} \left (d x +c \right )}{d}\) | \(183\) |
| default | \(\frac {-A \,a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+B \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C \tan \left (d x +c \right ) a^{2}+2 a A b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+2 B \tan \left (d x +c \right ) a b +2 a b C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+A \tan \left (d x +c \right ) b^{2}+B \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \,b^{2} \left (d x +c \right )}{d}\) | \(183\) |
| parallelrisch | \(\frac {-3 \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right ) \left (\frac {B \,a^{2}}{2}+a b \left (A +2 C \right )+B \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+3 \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right ) \left (\frac {B \,a^{2}}{2}+a b \left (A +2 C \right )+B \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+3 C \,b^{2} d x \cos \left (3 d x +3 c \right )+\left (3 A \,b^{2}+6 B a b +a^{2} \left (2 A +3 C \right )\right ) \sin \left (3 d x +3 c \right )+3 \left (2 a A b +B \,a^{2}\right ) \sin \left (2 d x +2 c \right )+9 C \,b^{2} d x \cos \left (d x +c \right )+6 \left (a^{2} \left (A +\frac {C}{2}\right )+B a b +\frac {A \,b^{2}}{2}\right ) \sin \left (d x +c \right )}{3 d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\) | \(245\) |
| risch | \(b^{2} C x -\frac {i \left (6 A a b \,{\mathrm e}^{5 i \left (d x +c \right )}+3 B \,a^{2} {\mathrm e}^{5 i \left (d x +c \right )}-6 A \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-12 B a b \,{\mathrm e}^{4 i \left (d x +c \right )}-6 C \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-12 A \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-12 A \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-24 B a b \,{\mathrm e}^{2 i \left (d x +c \right )}-12 C \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-6 A a b \,{\mathrm e}^{i \left (d x +c \right )}-3 B \,a^{2} {\mathrm e}^{i \left (d x +c \right )}-4 A \,a^{2}-6 A \,b^{2}-12 B a b -6 a^{2} C \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A b}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B \,b^{2}}{d}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a b C}{d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A b}{d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B \,b^{2}}{d}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a b C}{d}\) | \(382\) |
| norman | \(\frac {b^{2} C x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+b^{2} C x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}-b^{2} C x -\frac {8 \left (A \,a^{2}-A \,b^{2}-2 B a b -a^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}-\frac {\left (2 A \,a^{2}-2 a A b +2 A \,b^{2}-B \,a^{2}+4 B a b +2 a^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{d}-\frac {\left (2 A \,a^{2}+2 a A b +2 A \,b^{2}+B \,a^{2}+4 B a b +2 a^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {4 \left (5 A \,a^{2}-6 a A b +3 A \,b^{2}-3 B \,a^{2}+6 B a b +3 a^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 d}-\frac {4 \left (5 A \,a^{2}+6 a A b +3 A \,b^{2}+3 B \,a^{2}+6 B a b +3 a^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}-\frac {\left (26 A \,a^{2}-30 a A b -6 A \,b^{2}-15 B \,a^{2}-12 B a b -6 a^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d}-\frac {\left (26 A \,a^{2}+30 a A b -6 A \,b^{2}+15 B \,a^{2}-12 B a b -6 a^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d}-b^{2} C x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+3 b^{2} C x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+3 b^{2} C x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-3 b^{2} C x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-3 b^{2} C x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}-\frac {\left (2 a A b +B \,a^{2}+2 B \,b^{2}+4 a b C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (2 a A b +B \,a^{2}+2 B \,b^{2}+4 a b C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(585\) |
Input:
int((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x,meth od=_RETURNVERBOSE)
Output:
-A*a^2/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+(2*A*a*b+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*b^2+2*C*a*b)/d*ln(sec(d*x +c)+tan(d*x+c))+(A*b^2+2*B*a*b+C*a^2)/d*tan(d*x+c)+C*b^2/d*(d*x+c)
Time = 0.10 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.27 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {12 \, C b^{2} d x \cos \left (d x + c\right )^{3} + 3 \, {\left (B a^{2} + 2 \, {\left (A + 2 \, C\right )} a b + 2 \, B b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (B a^{2} + 2 \, {\left (A + 2 \, C\right )} a b + 2 \, B b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, A a^{2} + 2 \, {\left ({\left (2 \, A + 3 \, C\right )} a^{2} + 6 \, B a b + 3 \, A b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \] Input:
integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4, x, algorithm="fricas")
Output:
1/12*(12*C*b^2*d*x*cos(d*x + c)^3 + 3*(B*a^2 + 2*(A + 2*C)*a*b + 2*B*b^2)* cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(B*a^2 + 2*(A + 2*C)*a*b + 2*B*b^ 2)*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(2*A*a^2 + 2*((2*A + 3*C)*a^2 + 6*B*a*b + 3*A*b^2)*cos(d*x + c)^2 + 3*(B*a^2 + 2*A*a*b)*cos(d*x + c))*s in(d*x + c))/(d*cos(d*x + c)^3)
Timed out. \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+b*cos(d*x+c))**2*(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 = 221, normalized size of antiderivative = 1.57 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} + 12 \, {\left (d x + c\right )} C b^{2} - 3 \, B a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 6 \, A a b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, C a b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, B b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, C a^{2} \tan \left (d x + c\right ) + 24 \, B a b \tan \left (d x + c\right ) + 12 \, A b^{2} \tan \left (d x + c\right )}{12 \, d} \] Input:
integrate((a+b*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")
Output:
1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^2 + 12*(d*x + c)*C*b^2 - 3*B *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)) - 6*A*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)) + 12*C*a*b*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 6*B*b^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c ) - 1)) + 12*C*a^2*tan(d*x + c) + 24*B*a*b*tan(d*x + c) + 12*A*b^2*tan(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 364 vs. \(2 (133) = 266\).
Time = 0.16 (sec) , antiderivative size = 364, normalized size of antiderivative = 2.58 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {6 \, {\left (d x + c\right )} C b^{2} + 3 \, {\left (B a^{2} + 2 \, A a b + 4 \, C a b + 2 \, B b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (B a^{2} + 2 \, A a b + 4 \, C a b + 2 \, B b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \] Input:
integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4, x, algorithm="giac")
Output:
1/6*(6*(d*x + c)*C*b^2 + 3*(B*a^2 + 2*A*a*b + 4*C*a*b + 2*B*b^2)*log(abs(t an(1/2*d*x + 1/2*c) + 1)) - 3*(B*a^2 + 2*A*a*b + 4*C*a*b + 2*B*b^2)*log(ab s(tan(1/2*d*x + 1/2*c) - 1)) - 2*(6*A*a^2*tan(1/2*d*x + 1/2*c)^5 - 3*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 - 6*A*a*b*tan(1/2 *d*x + 1/2*c)^5 + 12*B*a*b*tan(1/2*d*x + 1/2*c)^5 + 6*A*b^2*tan(1/2*d*x + 1/2*c)^5 - 4*A*a^2*tan(1/2*d*x + 1/2*c)^3 - 12*C*a^2*tan(1/2*d*x + 1/2*c)^ 3 - 24*B*a*b*tan(1/2*d*x + 1/2*c)^3 - 12*A*b^2*tan(1/2*d*x + 1/2*c)^3 + 6* A*a^2*tan(1/2*d*x + 1/2*c) + 3*B*a^2*tan(1/2*d*x + 1/2*c) + 6*C*a^2*tan(1/ 2*d*x + 1/2*c) + 6*A*a*b*tan(1/2*d*x + 1/2*c) + 12*B*a*b*tan(1/2*d*x + 1/2 *c) + 6*A*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d
Time = 1.62 (sec) , antiderivative size = 512, normalized size of antiderivative = 3.63 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx =\text {Too large to display} \] Input:
int(((a + b*cos(c + d*x))^2*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + d*x)^4,x)
Output:
((A*a^2*sin(3*c + 3*d*x))/6 + (B*a^2*sin(2*c + 2*d*x))/4 + (A*b^2*sin(3*c + 3*d*x))/4 + (C*a^2*sin(3*c + 3*d*x))/4 + (A*a^2*sin(c + d*x))/2 + (A*b^2 *sin(c + d*x))/4 + (C*a^2*sin(c + d*x))/4 + (A*a*b*sin(2*c + 2*d*x))/2 + ( B*a*b*sin(3*c + 3*d*x))/2 + (3*B*a^2*cos(c + d*x)*atanh(sin(c/2 + (d*x)/2) /cos(c/2 + (d*x)/2)))/4 + (3*B*b^2*cos(c + d*x)*atanh(sin(c/2 + (d*x)/2)/c os(c/2 + (d*x)/2)))/2 + (3*C*b^2*cos(c + d*x)*atan(sin(c/2 + (d*x)/2)/cos( c/2 + (d*x)/2)))/2 + (B*a^2*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*c os(3*c + 3*d*x))/4 + (B*b^2*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*c os(3*c + 3*d*x))/2 + (B*a*b*sin(c + d*x))/2 + (C*b^2*atan(sin(c/2 + (d*x)/ 2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/2 + (3*A*a*b*cos(c + d*x)*atanh(s in(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/2 + 3*C*a*b*cos(c + d*x)*atanh(sin( c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + (A*a*b*atanh(sin(c/2 + (d*x)/2)/cos(c /2 + (d*x)/2))*cos(3*c + 3*d*x))/2 + C*a*b*atanh(sin(c/2 + (d*x)/2)/cos(c/ 2 + (d*x)/2))*cos(3*c + 3*d*x))/(d*((3*cos(c + d*x))/4 + cos(3*c + 3*d*x)/ 4))
Time = 0.18 (sec) , antiderivative size = 484, normalized size of antiderivative = 3.43 \[ \int (a+b \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 {-9 \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} b -12 \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 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} b^{3}+9 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2} b +12 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a b c +6 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{3}+9 \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} b +12 \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 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} b^{3}-9 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2} b -12 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a b c -6 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{3}+6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b^{2} c d x -9 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b -6 \cos \left (d x +c \right ) b^{2} c d x +4 \sin \left (d x +c \right )^{3} a^{3}+6 \sin \left (d x +c \right )^{3} a^{2} c +18 \sin \left (d x +c \right )^{3} a \,b^{2}-6 \sin \left (d x +c \right ) a^{3}-6 \sin \left (d x +c \right ) a^{2} c -18 \sin \left (d x +c \right ) a \,b^{2}}{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*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x)
Output:
( - 9*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2*b - 12*c os(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a*b*c - 6*cos(c + d* x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b**3 + 9*cos(c + d*x)*log(tan ((c + d*x)/2) - 1)*a**2*b + 12*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a*b* c + 6*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*b**3 + 9*cos(c + d*x)*log(tan ((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2*b + 12*cos(c + d*x)*log(tan((c + d *x)/2) + 1)*sin(c + d*x)**2*a*b*c + 6*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b**3 - 9*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a**2*b - 12*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a*b*c - 6*cos(c + d*x)*log(tan ((c + d*x)/2) + 1)*b**3 + 6*cos(c + d*x)*sin(c + d*x)**2*b**2*c*d*x - 9*co s(c + d*x)*sin(c + d*x)*a**2*b - 6*cos(c + d*x)*b**2*c*d*x + 4*sin(c + d*x )**3*a**3 + 6*sin(c + d*x)**3*a**2*c + 18*sin(c + d*x)**3*a*b**2 - 6*sin(c + d*x)*a**3 - 6*sin(c + d*x)*a**2*c - 18*sin(c + d*x)*a*b**2)/(6*cos(c + d*x)*d*(sin(c + d*x)**2 - 1))