Integrand size = 41, antiderivative size = 204 \[ \int (a+b \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {1}{2} b \left (2 A b^2+6 a b B+6 a^2 C+b^2 C\right ) x+\frac {a \left (6 A b^2+6 a b B+a^2 (A+2 C)\right ) \text {arctanh}(\sin (c+d x))}{2 d}-\frac {b \left (9 a A b+4 a^2 B-2 b^2 B-6 a b C\right ) \sin (c+d x)}{2 d}-\frac {b^2 (4 A b+2 a B-b C) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {(3 A b+2 a B) (a+b \cos (c+d x))^2 \tan (c+d x)}{2 d}+\frac {A (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d} \] Output:
1/2*b*(2*A*b^2+6*B*a*b+6*C*a^2+C*b^2)*x+1/2*a*(6*A*b^2+6*B*a*b+a^2*(A+2*C) )*arctanh(sin(d*x+c))/d-1/2*b*(9*A*a*b+4*B*a^2-2*B*b^2-6*C*a*b)*sin(d*x+c) /d-1/2*b^2*(4*A*b+2*B*a-C*b)*cos(d*x+c)*sin(d*x+c)/d+1/2*(3*A*b+2*B*a)*(a+ b*cos(d*x+c))^2*tan(d*x+c)/d+1/2*A*(a+b*cos(d*x+c))^3*sec(d*x+c)*tan(d*x+c )/d
Time = 6.70 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.56 \[ \int (a+b \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {2 b \left (2 A b^2+6 a b B+6 a^2 C+b^2 C\right ) (c+d x)-2 a \left (6 A b^2+6 a b B+a^2 (A+2 C)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 a \left (6 A b^2+6 a b B+a^2 (A+2 C)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {a^3 A}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 a^2 (3 A b+a B) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}-\frac {a^3 A}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 a^2 (3 A b+a B) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}+4 b^2 (b B+3 a C) \sin (c+d x)+b^3 C \sin (2 (c+d x))}{4 d} \] Input:
Integrate[(a + b*Cos[c + d*x])^3*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*S ec[c + d*x]^3,x]
Output:
(2*b*(2*A*b^2 + 6*a*b*B + 6*a^2*C + b^2*C)*(c + d*x) - 2*a*(6*A*b^2 + 6*a* b*B + a^2*(A + 2*C))*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 2*a*(6*A*b ^2 + 6*a*b*B + a^2*(A + 2*C))*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + ( a^3*A)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (4*a^2*(3*A*b + a*B)*Sin[ (c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) - (a^3*A)/(Cos[(c + d* x)/2] + Sin[(c + d*x)/2])^2 + (4*a^2*(3*A*b + a*B)*Sin[(c + d*x)/2])/(Cos[ (c + d*x)/2] + Sin[(c + d*x)/2]) + 4*b^2*(b*B + 3*a*C)*Sin[c + d*x] + b^3* C*Sin[2*(c + d*x)])/(4*d)
Time = 1.48 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.96, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.317, Rules used = {3042, 3526, 3042, 3526, 3042, 3512, 27, 3042, 3502, 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.
| \(\displaystyle \int \sec ^3(c+d x) (a+b \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
| \(\Big \downarrow \) 3526 |
| \(\displaystyle \frac {1}{2} \int (a+b \cos (c+d x))^2 \left (-2 b (A-C) \cos ^2(c+d x)+(2 b B+a (A+2 C)) \cos (c+d x)+3 A b+2 a B\right ) \sec ^2(c+d x)dx+\frac {A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^3}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (-2 b (A-C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+(2 b B+a (A+2 C)) \sin \left (c+d x+\frac {\pi }{2}\right )+3 A b+2 a B\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^3}{2 d}\) |
| \(\Big \downarrow \) 3526 |
| \(\displaystyle \frac {1}{2} \left (\int (a+b \cos (c+d x)) \left ((A+2 C) a^2+6 b B a+6 A b^2-2 b (4 A b-C b+2 a B) \cos ^2(c+d x)+b (2 b B-a (A-4 C)) \cos (c+d x)\right ) \sec (c+d x)dx+\frac {(2 a B+3 A b) \tan (c+d x) (a+b \cos (c+d x))^2}{d}\right )+\frac {A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^3}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left ((A+2 C) a^2+6 b B a+6 A b^2-2 b (4 A b-C b+2 a B) \sin \left (c+d x+\frac {\pi }{2}\right )^2+b (2 b B-a (A-4 C)) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {(2 a B+3 A b) \tan (c+d x) (a+b \cos (c+d x))^2}{d}\right )+\frac {A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^3}{2 d}\) |
| \(\Big \downarrow \) 3512 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int 2 \left (-b \left (4 B a^2+9 A b a-6 b C a-2 b^2 B\right ) \cos ^2(c+d x)+b \left (6 C a^2+6 b B a+2 A b^2+b^2 C\right ) \cos (c+d x)+a \left ((A+2 C) a^2+6 b B a+6 A b^2\right )\right ) \sec (c+d x)dx-\frac {b^2 \sin (c+d x) \cos (c+d x) (2 a B+4 A b-b C)}{d}+\frac {(2 a B+3 A b) \tan (c+d x) (a+b \cos (c+d x))^2}{d}\right )+\frac {A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^3}{2 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \left (\int \left (-b \left (4 B a^2+9 A b a-6 b C a-2 b^2 B\right ) \cos ^2(c+d x)+b \left (6 C a^2+6 b B a+2 A b^2+b^2 C\right ) \cos (c+d x)+a \left ((A+2 C) a^2+6 b B a+6 A b^2\right )\right ) \sec (c+d x)dx-\frac {b^2 \sin (c+d x) \cos (c+d x) (2 a B+4 A b-b C)}{d}+\frac {(2 a B+3 A b) \tan (c+d x) (a+b \cos (c+d x))^2}{d}\right )+\frac {A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^3}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\int \frac {-b \left (4 B a^2+9 A b a-6 b C a-2 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+b \left (6 C a^2+6 b B a+2 A b^2+b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+a \left ((A+2 C) a^2+6 b B a+6 A b^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {b^2 \sin (c+d x) \cos (c+d x) (2 a B+4 A b-b C)}{d}+\frac {(2 a B+3 A b) \tan (c+d x) (a+b \cos (c+d x))^2}{d}\right )+\frac {A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^3}{2 d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {1}{2} \left (\int \left (a \left ((A+2 C) a^2+6 b B a+6 A b^2\right )+b \left (6 C a^2+6 b B a+2 A b^2+b^2 C\right ) \cos (c+d x)\right ) \sec (c+d x)dx-\frac {b \sin (c+d x) \left (4 a^2 B+9 a A b-6 a b C-2 b^2 B\right )}{d}-\frac {b^2 \sin (c+d x) \cos (c+d x) (2 a B+4 A b-b C)}{d}+\frac {(2 a B+3 A b) \tan (c+d x) (a+b \cos (c+d x))^2}{d}\right )+\frac {A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^3}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\int \frac {a \left ((A+2 C) a^2+6 b B a+6 A b^2\right )+b \left (6 C a^2+6 b B a+2 A b^2+b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {b \sin (c+d x) \left (4 a^2 B+9 a A b-6 a b C-2 b^2 B\right )}{d}-\frac {b^2 \sin (c+d x) \cos (c+d x) (2 a B+4 A b-b C)}{d}+\frac {(2 a B+3 A b) \tan (c+d x) (a+b \cos (c+d x))^2}{d}\right )+\frac {A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^3}{2 d}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {1}{2} \left (a \left (a^2 (A+2 C)+6 a b B+6 A b^2\right ) \int \sec (c+d x)dx-\frac {b \sin (c+d x) \left (4 a^2 B+9 a A b-6 a b C-2 b^2 B\right )}{d}+b x \left (6 a^2 C+6 a b B+2 A b^2+b^2 C\right )-\frac {b^2 \sin (c+d x) \cos (c+d x) (2 a B+4 A b-b C)}{d}+\frac {(2 a B+3 A b) \tan (c+d x) (a+b \cos (c+d x))^2}{d}\right )+\frac {A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^3}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (a \left (a^2 (A+2 C)+6 a b B+6 A b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {b \sin (c+d x) \left (4 a^2 B+9 a A b-6 a b C-2 b^2 B\right )}{d}+b x \left (6 a^2 C+6 a b B+2 A b^2+b^2 C\right )-\frac {b^2 \sin (c+d x) \cos (c+d x) (2 a B+4 A b-b C)}{d}+\frac {(2 a B+3 A b) \tan (c+d x) (a+b \cos (c+d x))^2}{d}\right )+\frac {A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^3}{2 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{2} \left (\frac {a \left (a^2 (A+2 C)+6 a b B+6 A b^2\right ) \text {arctanh}(\sin (c+d x))}{d}-\frac {b \sin (c+d x) \left (4 a^2 B+9 a A b-6 a b C-2 b^2 B\right )}{d}+b x \left (6 a^2 C+6 a b B+2 A b^2+b^2 C\right )-\frac {b^2 \sin (c+d x) \cos (c+d x) (2 a B+4 A b-b C)}{d}+\frac {(2 a B+3 A b) \tan (c+d x) (a+b \cos (c+d x))^2}{d}\right )+\frac {A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^3}{2 d}\) |
Input:
Int[(a + b*Cos[c + d*x])^3*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^3,x]
Output:
(A*(a + b*Cos[c + d*x])^3*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (b*(2*A*b^2 + 6*a*b*B + 6*a^2*C + b^2*C)*x + (a*(6*A*b^2 + 6*a*b*B + a^2*(A + 2*C))*Arc Tanh[Sin[c + d*x]])/d - (b*(9*a*A*b + 4*a^2*B - 2*b^2*B - 6*a*b*C)*Sin[c + d*x])/d - (b^2*(4*A*b + 2*a*B - b*C)*Cos[c + d*x]*Sin[c + d*x])/d + ((3*A *b + 2*a*B)*(a + b*Cos[c + d*x])^2*Tan[c + d*x])/d)/2
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[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
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[(-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a + b*Si n[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3)) Int[(a + b*Si n[e + f*x])^m*Simp[a*C*d + A*b*c*(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e + f*x]^2 , x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, 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 = 2.22 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.90
| method | result | size |
| parts | \(\frac {A \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {\left (3 A \,a^{2} b +B \,a^{3}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (B \,b^{3}+3 C a \,b^{2}\right ) \sin \left (d x +c \right )}{d}+\frac {\left (A \,b^{3}+3 B a \,b^{2}+3 a^{2} b C \right ) \left (d x +c \right )}{d}+\frac {\left (3 a A \,b^{2}+3 B \,a^{2} b +a^{3} C \right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {C \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(184\) |
| derivativedivides | \(\frac {A \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \,a^{3} \tan \left (d x +c \right )+a^{3} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 A \,a^{2} b \tan \left (d x +c \right )+3 B \,a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 a^{2} b C \left (d x +c \right )+3 a A \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 B a \,b^{2} \left (d x +c \right )+3 C \sin \left (d x +c \right ) a \,b^{2}+A \,b^{3} \left (d x +c \right )+B \sin \left (d x +c \right ) b^{3}+C \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(212\) |
| default | \(\frac {A \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \,a^{3} \tan \left (d x +c \right )+a^{3} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 A \,a^{2} b \tan \left (d x +c \right )+3 B \,a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 a^{2} b C \left (d x +c \right )+3 a A \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 B a \,b^{2} \left (d x +c \right )+3 C \sin \left (d x +c \right ) a \,b^{2}+A \,b^{3} \left (d x +c \right )+B \sin \left (d x +c \right ) b^{3}+C \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(212\) |
| parallelrisch | \(\frac {-4 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (6 A \,b^{2}+6 B a b +a^{2} \left (A +2 C \right )\right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+4 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (6 A \,b^{2}+6 B a b +a^{2} \left (A +2 C \right )\right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+8 b \left (3 a^{2} C +3 B a b +\left (A +\frac {C}{2}\right ) b^{2}\right ) x d \cos \left (2 d x +2 c \right )+2 \left (12 A \,a^{2} b +4 B \,a^{3}+C \,b^{3}\right ) \sin \left (2 d x +2 c \right )+4 \left (B \,b^{3}+3 C a \,b^{2}\right ) \sin \left (3 d x +3 c \right )+C \sin \left (4 d x +4 c \right ) b^{3}+4 \left (2 A \,a^{3}+B \,b^{3}+3 C a \,b^{2}\right ) \sin \left (d x +c \right )+8 b \left (3 a^{2} C +3 B a b +\left (A +\frac {C}{2}\right ) b^{2}\right ) x d}{8 d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(270\) |
| risch | \(x A \,b^{3}+3 x B a \,b^{2}+3 x \,a^{2} b C +\frac {b^{3} C x}{2}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} C a \,b^{2}}{2 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} C a \,b^{2}}{2 d}-\frac {i a^{2} \left (A a \,{\mathrm e}^{3 i \left (d x +c \right )}-6 A b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 B a \,{\mathrm e}^{2 i \left (d x +c \right )}-a A \,{\mathrm e}^{i \left (d x +c \right )}-6 A b -2 B a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {i {\mathrm e}^{i \left (d x +c \right )} B \,b^{3}}{2 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B \,b^{3}}{2 d}+\frac {i C \,b^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {i C \,b^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,b^{2}}{d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B b}{d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}-\frac {A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,b^{2}}{d}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B b}{d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}\) | \(408\) |
| norman | \(\frac {\left (A \,b^{3}+3 B a \,b^{2}+3 a^{2} b C +\frac {1}{2} C \,b^{3}\right ) x +\left (-5 A \,b^{3}-15 B a \,b^{2}-15 a^{2} b C -\frac {5}{2} C \,b^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (-5 A \,b^{3}-15 B a \,b^{2}-15 a^{2} b C -\frac {5}{2} C \,b^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (A \,b^{3}+3 B a \,b^{2}+3 a^{2} b C +\frac {1}{2} C \,b^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (A \,b^{3}+3 B a \,b^{2}+3 a^{2} b C +\frac {1}{2} C \,b^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (A \,b^{3}+3 B a \,b^{2}+3 a^{2} b C +\frac {1}{2} C \,b^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}+\left (3 A \,b^{3}+9 B a \,b^{2}+9 a^{2} b C +\frac {3}{2} C \,b^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (3 A \,b^{3}+9 B a \,b^{2}+9 a^{2} b C +\frac {3}{2} C \,b^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\frac {\left (A \,a^{3}-6 A \,a^{2} b -2 B \,a^{3}+2 B \,b^{3}+6 C a \,b^{2}-C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{d}+\frac {\left (A \,a^{3}+6 A \,a^{2} b +2 B \,a^{3}+2 B \,b^{3}+6 C a \,b^{2}+C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (15 A \,a^{3}-30 A \,a^{2} b -10 B \,a^{3}-2 B \,b^{3}-6 C a \,b^{2}+3 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}+\frac {\left (15 A \,a^{3}+30 A \,a^{2} b +10 B \,a^{3}-2 B \,b^{3}-6 C a \,b^{2}-3 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}+\frac {4 \left (5 A \,a^{3}-2 B \,b^{3}-6 C a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {2 \left (3 A \,a^{3}-12 A \,a^{2} b -4 B \,a^{3}+2 B \,b^{3}+6 C a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d}+\frac {2 \left (3 A \,a^{3}+12 A \,a^{2} b +4 B \,a^{3}+2 B \,b^{3}+6 C a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}-\frac {a \left (A \,a^{2}+6 A \,b^{2}+6 B a b +2 a^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a \left (A \,a^{2}+6 A \,b^{2}+6 B a b +2 a^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(766\) |
Input:
int((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3,x,meth od=_RETURNVERBOSE)
Output:
A*a^3/d*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+(3*A*a^2 *b+B*a^3)/d*tan(d*x+c)+(B*b^3+3*C*a*b^2)/d*sin(d*x+c)+(A*b^3+3*B*a*b^2+3*C *a^2*b)/d*(d*x+c)+(3*A*a*b^2+3*B*a^2*b+C*a^3)/d*ln(sec(d*x+c)+tan(d*x+c))+ C*b^3/d*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)
Time = 0.10 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.02 \[ \int (a+b \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {2 \, {\left (6 \, C a^{2} b + 6 \, B a b^{2} + {\left (2 \, A + C\right )} b^{3}\right )} d x \cos \left (d x + c\right )^{2} + {\left ({\left (A + 2 \, C\right )} a^{3} + 6 \, B a^{2} b + 6 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (A + 2 \, C\right )} a^{3} + 6 \, B a^{2} b + 6 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (C b^{3} \cos \left (d x + c\right )^{3} + A a^{3} + 2 \, {\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3, x, algorithm="fricas")
Output:
1/4*(2*(6*C*a^2*b + 6*B*a*b^2 + (2*A + C)*b^3)*d*x*cos(d*x + c)^2 + ((A + 2*C)*a^3 + 6*B*a^2*b + 6*A*a*b^2)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - ( (A + 2*C)*a^3 + 6*B*a^2*b + 6*A*a*b^2)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(C*b^3*cos(d*x + c)^3 + A*a^3 + 2*(3*C*a*b^2 + B*b^3)*cos(d*x + c)^ 2 + 2*(B*a^3 + 3*A*a^2*b)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2)
Timed out. \[ \int (a+b \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+b*cos(d*x+c))**3*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)* *3,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.19 \[ \int (a+b \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {12 \, {\left (d x + c\right )} C a^{2} b + 12 \, {\left (d x + c\right )} B a b^{2} + 4 \, {\left (d x + c\right )} A b^{3} + {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C b^{3} - A a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, C a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, B a^{2} b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, A a b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, C a b^{2} \sin \left (d x + c\right ) + 4 \, B b^{3} \sin \left (d x + c\right ) + 4 \, B a^{3} \tan \left (d x + c\right ) + 12 \, A a^{2} b \tan \left (d x + c\right )}{4 \, d} \] Input:
integrate((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3, x, algorithm="maxima")
Output:
1/4*(12*(d*x + c)*C*a^2*b + 12*(d*x + c)*B*a*b^2 + 4*(d*x + c)*A*b^3 + (2* d*x + 2*c + sin(2*d*x + 2*c))*C*b^3 - A*a^3*(2*sin(d*x + c)/(sin(d*x + c)^ 2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 2*C*a^3*(log(sin (d*x + c) + 1) - log(sin(d*x + c) - 1)) + 6*B*a^2*b*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 6*A*a*b^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 12*C*a*b^2*sin(d*x + c) + 4*B*b^3*sin(d*x + c) + 4*B*a^3*tan (d*x + c) + 12*A*a^2*b*tan(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 538 vs. \(2 (192) = 384\).
Time = 0.18 (sec) , antiderivative size = 538, normalized size of antiderivative = 2.64 \[ \int (a+b \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx =\text {Too large to display} \] Input:
integrate((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3, x, algorithm="giac")
Output:
1/2*((6*C*a^2*b + 6*B*a*b^2 + 2*A*b^3 + C*b^3)*(d*x + c) + (A*a^3 + 2*C*a^ 3 + 6*B*a^2*b + 6*A*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (A*a^3 + 2 *C*a^3 + 6*B*a^2*b + 6*A*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(A* a^3*tan(1/2*d*x + 1/2*c)^7 - 2*B*a^3*tan(1/2*d*x + 1/2*c)^7 - 6*A*a^2*b*ta n(1/2*d*x + 1/2*c)^7 + 6*C*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 2*B*b^3*tan(1/2* d*x + 1/2*c)^7 - C*b^3*tan(1/2*d*x + 1/2*c)^7 + 3*A*a^3*tan(1/2*d*x + 1/2* c)^5 - 2*B*a^3*tan(1/2*d*x + 1/2*c)^5 - 6*A*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 6*C*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 2*B*b^3*tan(1/2*d*x + 1/2*c)^5 + 3*C*b ^3*tan(1/2*d*x + 1/2*c)^5 + 3*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 2*B*a^3*tan(1 /2*d*x + 1/2*c)^3 + 6*A*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 6*C*a*b^2*tan(1/2*d *x + 1/2*c)^3 - 2*B*b^3*tan(1/2*d*x + 1/2*c)^3 - 3*C*b^3*tan(1/2*d*x + 1/2 *c)^3 + A*a^3*tan(1/2*d*x + 1/2*c) + 2*B*a^3*tan(1/2*d*x + 1/2*c) + 6*A*a^ 2*b*tan(1/2*d*x + 1/2*c) + 6*C*a*b^2*tan(1/2*d*x + 1/2*c) + 2*B*b^3*tan(1/ 2*d*x + 1/2*c) + C*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^4 - 1)^ 2)/d
Time = 2.52 (sec) , antiderivative size = 3879, normalized size of antiderivative = 19.01 \[ \int (a+b \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\text {Too large to display} \] Input:
int(((a + b*cos(c + d*x))^3*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + d*x)^3,x)
Output:
(tan(c/2 + (d*x)/2)^7*(A*a^3 - 2*B*a^3 + 2*B*b^3 - C*b^3 - 6*A*a^2*b + 6*C *a*b^2) + tan(c/2 + (d*x)/2)^3*(3*A*a^3 + 2*B*a^3 - 2*B*b^3 - 3*C*b^3 + 6* A*a^2*b - 6*C*a*b^2) - tan(c/2 + (d*x)/2)^5*(2*B*a^3 - 3*A*a^3 + 2*B*b^3 - 3*C*b^3 + 6*A*a^2*b + 6*C*a*b^2) + tan(c/2 + (d*x)/2)*(A*a^3 + 2*B*a^3 + 2*B*b^3 + C*b^3 + 6*A*a^2*b + 6*C*a*b^2))/(d*(tan(c/2 + (d*x)/2)^8 - 2*tan (c/2 + (d*x)/2)^4 + 1)) - (atan(((((A*a^3)/2 + C*a^3 + 3*A*a*b^2 + 3*B*a^2 *b)*(16*A*a^3 + 32*A*b^3 + 32*C*a^3 + 16*C*b^3 + 96*A*a*b^2 + 96*B*a*b^2 + 96*B*a^2*b + 96*C*a^2*b) + tan(c/2 + (d*x)/2)*(8*A^2*a^6 + 32*A^2*b^6 + 3 2*C^2*a^6 + 8*C^2*b^6 + 288*A^2*a^2*b^4 + 96*A^2*a^4*b^2 + 288*B^2*a^2*b^4 + 288*B^2*a^4*b^2 + 96*C^2*a^2*b^4 + 288*C^2*a^4*b^2 + 32*A*C*a^6 + 32*A* C*b^6 + 192*A*B*a*b^5 + 96*A*B*a^5*b + 96*B*C*a*b^5 + 192*B*C*a^5*b + 576* A*B*a^3*b^3 + 192*A*C*a^2*b^4 + 192*A*C*a^4*b^2 + 576*B*C*a^3*b^3))*((A*a^ 3)/2 + C*a^3 + 3*A*a*b^2 + 3*B*a^2*b)*1i - (((A*a^3)/2 + C*a^3 + 3*A*a*b^2 + 3*B*a^2*b)*(16*A*a^3 + 32*A*b^3 + 32*C*a^3 + 16*C*b^3 + 96*A*a*b^2 + 96 *B*a*b^2 + 96*B*a^2*b + 96*C*a^2*b) - tan(c/2 + (d*x)/2)*(8*A^2*a^6 + 32*A ^2*b^6 + 32*C^2*a^6 + 8*C^2*b^6 + 288*A^2*a^2*b^4 + 96*A^2*a^4*b^2 + 288*B ^2*a^2*b^4 + 288*B^2*a^4*b^2 + 96*C^2*a^2*b^4 + 288*C^2*a^4*b^2 + 32*A*C*a ^6 + 32*A*C*b^6 + 192*A*B*a*b^5 + 96*A*B*a^5*b + 96*B*C*a*b^5 + 192*B*C*a^ 5*b + 576*A*B*a^3*b^3 + 192*A*C*a^2*b^4 + 192*A*C*a^4*b^2 + 576*B*C*a^3*b^ 3))*((A*a^3)/2 + C*a^3 + 3*A*a*b^2 + 3*B*a^2*b)*1i)/((((A*a^3)/2 + C*a^...
Time = 0.19 (sec) , antiderivative size = 633, normalized size of antiderivative = 3.10 \[ \int (a+b \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx =\text {Too large to display} \] Input:
int((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3,x)
Output:
(cos(c + d*x)**2*sin(c + d*x)**3*b**3*c - cos(c + d*x)**2*sin(c + d*x)*b** 3*c - cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**4 - 2*cos( c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**3*c - 12*cos(c + d*x )*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2*b**2 + cos(c + d*x)*log(t an((c + d*x)/2) - 1)*a**4 + 2*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**3* c + 12*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**2*b**2 + cos(c + d*x)*log (tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**4 + 2*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**3*c + 12*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2*b**2 - cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a **4 - 2*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a**3*c - 12*cos(c + d*x)*lo g(tan((c + d*x)/2) + 1)*a**2*b**2 + 6*cos(c + d*x)*sin(c + d*x)**3*a*b**2* c + 2*cos(c + d*x)*sin(c + d*x)**3*b**4 + 6*cos(c + d*x)*sin(c + d*x)**2*a **2*b*c*d*x + 8*cos(c + d*x)*sin(c + d*x)**2*a*b**3*d*x + cos(c + d*x)*sin (c + d*x)**2*b**3*c*d*x - cos(c + d*x)*sin(c + d*x)*a**4 - 6*cos(c + d*x)* sin(c + d*x)*a*b**2*c - 2*cos(c + d*x)*sin(c + d*x)*b**4 - 6*cos(c + d*x)* a**2*b*c*d*x - 8*cos(c + d*x)*a*b**3*d*x - cos(c + d*x)*b**3*c*d*x + 8*sin (c + d*x)**3*a**3*b - 8*sin(c + d*x)*a**3*b)/(2*cos(c + d*x)*d*(sin(c + d* x)**2 - 1))