Integrand size = 39, antiderivative size = 165 \[ \int (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {(3 A b+3 a B+4 b C) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {(4 a A+5 b B+5 a C) \tan (c+d x)}{5 d}+\frac {(3 A b+3 a B+4 b C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {(A b+a B) \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a A \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {(4 a A+5 b B+5 a C) \tan ^3(c+d x)}{15 d} \] Output:
1/8*(3*A*b+3*B*a+4*C*b)*arctanh(sin(d*x+c))/d+1/5*(4*A*a+5*B*b+5*C*a)*tan( d*x+c)/d+1/8*(3*A*b+3*B*a+4*C*b)*sec(d*x+c)*tan(d*x+c)/d+1/4*(A*b+B*a)*sec (d*x+c)^3*tan(d*x+c)/d+1/5*a*A*sec(d*x+c)^4*tan(d*x+c)/d+1/15*(4*A*a+5*B*b +5*C*a)*tan(d*x+c)^3/d
Time = 1.29 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.75 \[ \int (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {15 (3 A b+3 a B+4 b C) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (15 (3 A b+3 a B+4 b C) \sec (c+d x)+30 (A b+a B) \sec ^3(c+d x)+8 \left (15 (b B+a (A+C))+5 (b B+a (2 A+C)) \tan ^2(c+d x)+3 a A \tan ^4(c+d x)\right )\right )}{120 d} \] Input:
Integrate[(a + b*Cos[c + d*x])*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec [c + d*x]^6,x]
Output:
(15*(3*A*b + 3*a*B + 4*b*C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(15*(3*A* b + 3*a*B + 4*b*C)*Sec[c + d*x] + 30*(A*b + a*B)*Sec[c + d*x]^3 + 8*(15*(b *B + a*(A + C)) + 5*(b*B + a*(2*A + C))*Tan[c + d*x]^2 + 3*a*A*Tan[c + d*x ]^4)))/(120*d)
Time = 0.88 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.91, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3510, 25, 3042, 3500, 3042, 3227, 3042, 4254, 2009, 4255, 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.
| \(\displaystyle \int \sec ^6(c+d x) (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^6}dx\) |
| \(\Big \downarrow \) 3510 |
| \(\displaystyle \frac {a A \tan (c+d x) \sec ^4(c+d x)}{5 d}-\frac {1}{5} \int -\left (\left (5 b C \cos ^2(c+d x)+(4 a A+5 b B+5 a C) \cos (c+d x)+5 (A b+a B)\right ) \sec ^5(c+d x)\right )dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{5} \int \left (5 b C \cos ^2(c+d x)+(4 a A+5 b B+5 a C) \cos (c+d x)+5 (A b+a B)\right ) \sec ^5(c+d x)dx+\frac {a A \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {5 b C \sin \left (c+d x+\frac {\pi }{2}\right )^2+(4 a A+5 b B+5 a C) \sin \left (c+d x+\frac {\pi }{2}\right )+5 (A b+a B)}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx+\frac {a A \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \int (4 (4 a A+5 b B+5 a C)+5 (3 A b+4 C b+3 a B) \cos (c+d x)) \sec ^4(c+d x)dx+\frac {5 (a B+A b) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {a A \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \int \frac {4 (4 a A+5 b B+5 a C)+5 (3 A b+4 C b+3 a B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {5 (a B+A b) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {a A \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (4 (4 a A+5 a C+5 b B) \int \sec ^4(c+d x)dx+5 (3 a B+3 A b+4 b C) \int \sec ^3(c+d x)dx\right )+\frac {5 (a B+A b) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {a A \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 (3 a B+3 A b+4 b C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+4 (4 a A+5 a C+5 b B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx\right )+\frac {5 (a B+A b) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {a A \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 (3 a B+3 A b+4 b C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {4 (4 a A+5 a C+5 b B) \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{d}\right )+\frac {5 (a B+A b) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {a A \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 (3 a B+3 A b+4 b C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {4 \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right ) (4 a A+5 a C+5 b B)}{d}\right )+\frac {5 (a B+A b) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {a A \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 (3 a B+3 A b+4 b C) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {4 \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right ) (4 a A+5 a C+5 b B)}{d}\right )+\frac {5 (a B+A b) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {a A \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 (3 a B+3 A b+4 b C) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {4 \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right ) (4 a A+5 a C+5 b B)}{d}\right )+\frac {5 (a B+A b) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {a A \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 (3 a B+3 A b+4 b C) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {4 \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right ) (4 a A+5 a C+5 b B)}{d}\right )+\frac {5 (a B+A b) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {a A \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
Input:
Int[(a + b*Cos[c + d*x])*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d *x]^6,x]
Output:
(a*A*Sec[c + d*x]^4*Tan[c + d*x])/(5*d) + ((5*(A*b + a*B)*Sec[c + d*x]^3*T an[c + d*x])/(4*d) + (5*(3*A*b + 3*a*B + 4*b*C)*(ArcTanh[Sin[c + d*x]]/(2* d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)) - (4*(4*a*A + 5*b*B + 5*a*C)*(-Tan [c + d*x] - Tan[c + d*x]^3/3))/d)/4)/5
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_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.96 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.96
| method | result | size |
| parts | \(-\frac {a A \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (A b +B a \right ) \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}-\frac {\left (B b +C a \right ) \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {C b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(159\) |
| derivativedivides | \(\frac {-a A \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )+B a \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-C a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+A b \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-B b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+C b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(210\) |
| default | \(\frac {-a A \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )+B a \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-C a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+A b \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-B b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+C b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(210\) |
| parallelrisch | \(\frac {-225 \left (\left (A +\frac {4 C}{3}\right ) b +B a \right ) \left (\frac {\cos \left (5 d x +5 c \right )}{5}+\cos \left (3 d x +3 c \right )+2 \cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+225 \left (\left (A +\frac {4 C}{3}\right ) b +B a \right ) \left (\frac {\cos \left (5 d x +5 c \right )}{5}+\cos \left (3 d x +3 c \right )+2 \cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (\left (420 A +240 C \right ) b +420 B a \right ) \sin \left (2 d x +2 c \right )+\left (400 B b +320 \left (A +\frac {5 C}{4}\right ) a \right ) \sin \left (3 d x +3 c \right )+\left (\left (90 A +120 C \right ) b +90 B a \right ) \sin \left (4 d x +4 c \right )+\left (80 B b +64 \left (A +\frac {5 C}{4}\right ) a \right ) \sin \left (5 d x +5 c \right )+640 \left (\frac {B b}{2}+a \left (A +\frac {C}{2}\right )\right ) \sin \left (d x +c \right )}{600 \left (\frac {\cos \left (5 d x +5 c \right )}{5}+\cos \left (3 d x +3 c \right )+2 \cos \left (d x +c \right )\right ) d}\) | \(260\) |
| risch | \(-\frac {i \left (45 A b \,{\mathrm e}^{9 i \left (d x +c \right )}+45 B a \,{\mathrm e}^{9 i \left (d x +c \right )}+60 C b \,{\mathrm e}^{9 i \left (d x +c \right )}+210 A b \,{\mathrm e}^{7 i \left (d x +c \right )}+210 B a \,{\mathrm e}^{7 i \left (d x +c \right )}+120 C b \,{\mathrm e}^{7 i \left (d x +c \right )}-240 B b \,{\mathrm e}^{6 i \left (d x +c \right )}-240 C a \,{\mathrm e}^{6 i \left (d x +c \right )}-640 A a \,{\mathrm e}^{4 i \left (d x +c \right )}-560 B b \,{\mathrm e}^{4 i \left (d x +c \right )}-560 C a \,{\mathrm e}^{4 i \left (d x +c \right )}-210 A b \,{\mathrm e}^{3 i \left (d x +c \right )}-210 B a \,{\mathrm e}^{3 i \left (d x +c \right )}-120 C b \,{\mathrm e}^{3 i \left (d x +c \right )}-320 A a \,{\mathrm e}^{2 i \left (d x +c \right )}-400 B b \,{\mathrm e}^{2 i \left (d x +c \right )}-400 C a \,{\mathrm e}^{2 i \left (d x +c \right )}-45 A b \,{\mathrm e}^{i \left (d x +c \right )}-45 B a \,{\mathrm e}^{i \left (d x +c \right )}-60 C b \,{\mathrm e}^{i \left (d x +c \right )}-64 a A -80 B b -80 C a \right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A b}{8 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B a}{8 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C b}{2 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A b}{8 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B a}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C b}{2 d}\) | \(414\) |
| norman | \(\frac {-\frac {\left (8 a A -5 A b -5 B a +8 B b +8 C a -4 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{4 d}-\frac {\left (8 a A +5 A b +5 B a +8 B b +8 C a +4 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {\left (40 a A -39 A b -39 B a +8 B b +8 C a -12 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{12 d}-\frac {\left (40 a A +39 A b +39 B a +8 B b +8 C a +12 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}-\frac {\left (344 a A -135 A b -135 B a -200 B b -200 C a +180 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{60 d}-\frac {\left (344 a A +135 A b +135 B a -200 B b -200 C a -180 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{60 d}-\frac {\left (872 a A -15 A b -15 B a +40 B b +40 C a +180 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{60 d}-\frac {\left (872 a A +15 A b +15 B a +40 B b +40 C a -180 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{60 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{5}}-\frac {\left (3 A b +3 B a +4 C b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {\left (3 A b +3 B a +4 C b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(420\) |
Input:
int((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x,method =_RETURNVERBOSE)
Output:
-a*A/d*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c)+(A*b+B*a)/d*( -(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c )))-(B*b+C*a)/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+C*b/d*(1/2*sec(d*x+c)*t an(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))
Time = 0.10 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.10 \[ \int (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {15 \, {\left (3 \, B a + {\left (3 \, A + 4 \, C\right )} b\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (3 \, B a + {\left (3 \, A + 4 \, C\right )} b\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left ({\left (4 \, A + 5 \, C\right )} a + 5 \, B b\right )} \cos \left (d x + c\right )^{4} + 15 \, {\left (3 \, B a + {\left (3 \, A + 4 \, C\right )} b\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left ({\left (4 \, A + 5 \, C\right )} a + 5 \, B b\right )} \cos \left (d x + c\right )^{2} + 24 \, A a + 30 \, {\left (B a + A b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \] Input:
integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="fricas")
Output:
1/240*(15*(3*B*a + (3*A + 4*C)*b)*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 1 5*(3*B*a + (3*A + 4*C)*b)*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(16*(( 4*A + 5*C)*a + 5*B*b)*cos(d*x + c)^4 + 15*(3*B*a + (3*A + 4*C)*b)*cos(d*x + c)^3 + 8*((4*A + 5*C)*a + 5*B*b)*cos(d*x + c)^2 + 24*A*a + 30*(B*a + A*b )*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^5)
Timed out. \[ \int (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**6, x)
Output:
Timed out
Time = 0.05 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.61 \[ \int (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a + 80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a + 80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B b - 15 \, B a {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 15 \, A b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, C b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{240 \, d} \] Input:
integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="maxima")
Output:
1/240*(16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*A*a + 8 0*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a + 80*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*b - 15*B*a*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1) ) - 15*A*b*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin( d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 60* C*b*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin (d*x + c) - 1)))/d
Leaf count of result is larger than twice the leaf count of optimal. 473 vs. \(2 (153) = 306\).
Time = 0.16 (sec) , antiderivative size = 473, normalized size of antiderivative = 2.87 \[ \int (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx =\text {Too large to display} \] Input:
integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="giac")
Output:
1/120*(15*(3*B*a + 3*A*b + 4*C*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15* (3*B*a + 3*A*b + 4*C*b)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(120*A*a*ta n(1/2*d*x + 1/2*c)^9 - 75*B*a*tan(1/2*d*x + 1/2*c)^9 + 120*C*a*tan(1/2*d*x + 1/2*c)^9 - 75*A*b*tan(1/2*d*x + 1/2*c)^9 + 120*B*b*tan(1/2*d*x + 1/2*c) ^9 - 60*C*b*tan(1/2*d*x + 1/2*c)^9 - 160*A*a*tan(1/2*d*x + 1/2*c)^7 + 30*B *a*tan(1/2*d*x + 1/2*c)^7 - 320*C*a*tan(1/2*d*x + 1/2*c)^7 + 30*A*b*tan(1/ 2*d*x + 1/2*c)^7 - 320*B*b*tan(1/2*d*x + 1/2*c)^7 + 120*C*b*tan(1/2*d*x + 1/2*c)^7 + 464*A*a*tan(1/2*d*x + 1/2*c)^5 + 400*C*a*tan(1/2*d*x + 1/2*c)^5 + 400*B*b*tan(1/2*d*x + 1/2*c)^5 - 160*A*a*tan(1/2*d*x + 1/2*c)^3 - 30*B* a*tan(1/2*d*x + 1/2*c)^3 - 320*C*a*tan(1/2*d*x + 1/2*c)^3 - 30*A*b*tan(1/2 *d*x + 1/2*c)^3 - 320*B*b*tan(1/2*d*x + 1/2*c)^3 - 120*C*b*tan(1/2*d*x + 1 /2*c)^3 + 120*A*a*tan(1/2*d*x + 1/2*c) + 75*B*a*tan(1/2*d*x + 1/2*c) + 120 *C*a*tan(1/2*d*x + 1/2*c) + 75*A*b*tan(1/2*d*x + 1/2*c) + 120*B*b*tan(1/2* d*x + 1/2*c) + 60*C*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5 )/d
Time = 2.81 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.83 \[ \int (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,A\,b}{8}+\frac {3\,B\,a}{8}+\frac {C\,b}{2}\right )}{\frac {3\,A\,b}{2}+\frac {3\,B\,a}{2}+2\,C\,b}\right )\,\left (\frac {3\,A\,b}{4}+\frac {3\,B\,a}{4}+C\,b\right )}{d}-\frac {\left (2\,A\,a-\frac {5\,A\,b}{4}-\frac {5\,B\,a}{4}+2\,B\,b+2\,C\,a-C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {A\,b}{2}-\frac {8\,A\,a}{3}+\frac {B\,a}{2}-\frac {16\,B\,b}{3}-\frac {16\,C\,a}{3}+2\,C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {116\,A\,a}{15}+\frac {20\,B\,b}{3}+\frac {20\,C\,a}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {8\,A\,a}{3}-\frac {A\,b}{2}-\frac {B\,a}{2}-\frac {16\,B\,b}{3}-\frac {16\,C\,a}{3}-2\,C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a+\frac {5\,A\,b}{4}+\frac {5\,B\,a}{4}+2\,B\,b+2\,C\,a+C\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \] Input:
int(((a + b*cos(c + d*x))*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + d*x)^6,x)
Output:
(atanh((4*tan(c/2 + (d*x)/2)*((3*A*b)/8 + (3*B*a)/8 + (C*b)/2))/((3*A*b)/2 + (3*B*a)/2 + 2*C*b))*((3*A*b)/4 + (3*B*a)/4 + C*b))/d - (tan(c/2 + (d*x) /2)^9*(2*A*a - (5*A*b)/4 - (5*B*a)/4 + 2*B*b + 2*C*a - C*b) - tan(c/2 + (d *x)/2)^3*((8*A*a)/3 + (A*b)/2 + (B*a)/2 + (16*B*b)/3 + (16*C*a)/3 + 2*C*b) - tan(c/2 + (d*x)/2)^7*((8*A*a)/3 - (A*b)/2 - (B*a)/2 + (16*B*b)/3 + (16* C*a)/3 - 2*C*b) + tan(c/2 + (d*x)/2)*(2*A*a + (5*A*b)/4 + (5*B*a)/4 + 2*B* b + 2*C*a + C*b) + tan(c/2 + (d*x)/2)^5*((116*A*a)/15 + (20*B*b)/3 + (20*C *a)/3))/(d*(5*tan(c/2 + (d*x)/2)^2 - 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 - 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 - 1))
Time = 0.18 (sec) , antiderivative size = 540, normalized size of antiderivative = 3.27 \[ \int (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx =\text {Too large to display} \] Input:
int((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x)
Output:
( - 45*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a*b - 30*cos (c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*b*c + 90*cos(c + d*x)* log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a*b + 60*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b*c - 45*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a*b - 30*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*b*c + 45*cos(c + d*x )*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a*b + 30*cos(c + d*x)*log(tan( (c + d*x)/2) + 1)*sin(c + d*x)**4*b*c - 90*cos(c + d*x)*log(tan((c + d*x)/ 2) + 1)*sin(c + d*x)**2*a*b - 60*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*si n(c + d*x)**2*b*c + 45*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a*b + 30*cos (c + d*x)*log(tan((c + d*x)/2) + 1)*b*c - 45*cos(c + d*x)*sin(c + d*x)**3* a*b - 30*cos(c + d*x)*sin(c + d*x)**3*b*c + 75*cos(c + d*x)*sin(c + d*x)*a *b + 30*cos(c + d*x)*sin(c + d*x)*b*c + 32*sin(c + d*x)**5*a**2 + 40*sin(c + d*x)**5*a*c + 40*sin(c + d*x)**5*b**2 - 80*sin(c + d*x)**3*a**2 - 100*s in(c + d*x)**3*a*c - 100*sin(c + d*x)**3*b**2 + 60*sin(c + d*x)*a**2 + 60* sin(c + d*x)*a*c + 60*sin(c + d*x)*b**2)/(60*cos(c + d*x)*d*(sin(c + d*x)* *4 - 2*sin(c + d*x)**2 + 1))