Integrand size = 40, antiderivative size = 124 \[ \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=b^2 (b B+3 a C) x+\frac {a \left (a^2 B+6 b^2 B+6 a b C\right ) \text {arctanh}(\sin (c+d x))}{2 d}-\frac {b^2 (a B-2 b C) \sin (c+d x)}{2 d}+\frac {a^2 (2 b B+a C) \tan (c+d x)}{d}+\frac {a B (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d} \] Output:
b^2*(B*b+3*C*a)*x+1/2*a*(B*a^2+6*B*b^2+6*C*a*b)*arctanh(sin(d*x+c))/d-1/2* b^2*(B*a-2*C*b)*sin(d*x+c)/d+a^2*(2*B*b+C*a)*tan(d*x+c)/d+1/2*a*B*(a+b*cos (d*x+c))^2*sec(d*x+c)*tan(d*x+c)/d
Leaf count is larger than twice the leaf count of optimal. \(277\) vs. \(2(124)=248\).
Time = 4.91 (sec) , antiderivative size = 277, normalized size of antiderivative = 2.23 \[ \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {4 b^2 (b B+3 a C) (c+d x)-2 a \left (a^2 B+6 b^2 B+6 a b C\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 a \left (a^2 B+6 b^2 B+6 a b C\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {a^3 B}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 a^2 (3 b B+a C) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}-\frac {a^3 B}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 a^2 (3 b B+a C) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}+4 b^3 C \sin (c+d x)}{4 d} \] Input:
Integrate[(a + b*Cos[c + d*x])^3*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]
Output:
(4*b^2*(b*B + 3*a*C)*(c + d*x) - 2*a*(a^2*B + 6*b^2*B + 6*a*b*C)*Log[Cos[( c + d*x)/2] - Sin[(c + d*x)/2]] + 2*a*(a^2*B + 6*b^2*B + 6*a*b*C)*Log[Cos[ (c + d*x)/2] + Sin[(c + d*x)/2]] + (a^3*B)/(Cos[(c + d*x)/2] - Sin[(c + d* x)/2])^2 + (4*a^2*(3*b*B + a*C)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[ (c + d*x)/2]) - (a^3*B)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (4*a^2*( 3*b*B + a*C)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 4*b ^3*C*Sin[c + d*x])/(4*d)
Time = 1.07 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.02, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.325, Rules used = {3042, 3508, 3042, 3468, 3042, 3510, 25, 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.
| \(\displaystyle \int \sec ^4(c+d x) (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3 \left (B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
| \(\Big \downarrow \) 3508 |
| \(\displaystyle \int \sec ^3(c+d x) (a+b \cos (c+d x))^3 (B+C \cos (c+d x))dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3 \left (B+C \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
| \(\Big \downarrow \) 3468 |
| \(\displaystyle \frac {1}{2} \int (a+b \cos (c+d x)) \left (-b (a B-2 b C) \cos ^2(c+d x)+\left (B a^2+4 b C a+2 b^2 B\right ) \cos (c+d x)+2 a (2 b B+a C)\right ) \sec ^2(c+d x)dx+\frac {a B \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^2}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (-b (a B-2 b C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+\left (B a^2+4 b C a+2 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+2 a (2 b B+a C)\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a B \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^2}{2 d}\) |
| \(\Big \downarrow \) 3510 |
| \(\displaystyle \frac {1}{2} \left (\frac {2 a^2 (a C+2 b B) \tan (c+d x)}{d}-\int -\left (\left (-\left ((a B-2 b C) \cos ^2(c+d x) b^2\right )+2 (b B+3 a C) \cos (c+d x) b^2+a \left (B a^2+6 b C a+6 b^2 B\right )\right ) \sec (c+d x)\right )dx\right )+\frac {a B \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^2}{2 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{2} \left (\int \left (-\left ((a B-2 b C) \cos ^2(c+d x) b^2\right )+2 (b B+3 a C) \cos (c+d x) b^2+a \left (B a^2+6 b C a+6 b^2 B\right )\right ) \sec (c+d x)dx+\frac {2 a^2 (a C+2 b B) \tan (c+d x)}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^2}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\int \frac {-\left ((a B-2 b C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 b^2\right )+2 (b B+3 a C) \sin \left (c+d x+\frac {\pi }{2}\right ) b^2+a \left (B a^2+6 b C a+6 b^2 B\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 a^2 (a C+2 b B) \tan (c+d x)}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^2}{2 d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {1}{2} \left (\int \left (2 (b B+3 a C) \cos (c+d x) b^2+a \left (B a^2+6 b C a+6 b^2 B\right )\right ) \sec (c+d x)dx+\frac {2 a^2 (a C+2 b B) \tan (c+d x)}{d}-\frac {b^2 (a B-2 b C) \sin (c+d x)}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^2}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\int \frac {2 (b B+3 a C) \sin \left (c+d x+\frac {\pi }{2}\right ) b^2+a \left (B a^2+6 b C a+6 b^2 B\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 a^2 (a C+2 b B) \tan (c+d x)}{d}-\frac {b^2 (a B-2 b C) \sin (c+d x)}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^2}{2 d}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {1}{2} \left (a \left (a^2 B+6 a b C+6 b^2 B\right ) \int \sec (c+d x)dx+\frac {2 a^2 (a C+2 b B) \tan (c+d x)}{d}-\frac {b^2 (a B-2 b C) \sin (c+d x)}{d}+2 b^2 x (3 a C+b B)\right )+\frac {a B \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^2}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (a \left (a^2 B+6 a b C+6 b^2 B\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {2 a^2 (a C+2 b B) \tan (c+d x)}{d}-\frac {b^2 (a B-2 b C) \sin (c+d x)}{d}+2 b^2 x (3 a C+b B)\right )+\frac {a B \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^2}{2 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{2} \left (\frac {a \left (a^2 B+6 a b C+6 b^2 B\right ) \text {arctanh}(\sin (c+d x))}{d}+\frac {2 a^2 (a C+2 b B) \tan (c+d x)}{d}-\frac {b^2 (a B-2 b C) \sin (c+d x)}{d}+2 b^2 x (3 a C+b B)\right )+\frac {a B \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^2}{2 d}\) |
Input:
Int[(a + b*Cos[c + d*x])^3*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x ]^4,x]
Output:
(a*B*(a + b*Cos[c + d*x])^2*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (2*b^2*(b*B + 3*a*C)*x + (a*(a^2*B + 6*b^2*B + 6*a*b*C)*ArcTanh[Sin[c + d*x]])/d - (b ^2*(a*B - 2*b*C)*Sin[c + d*x])/d + (2*a^2*(2*b*B + a*C)*Tan[c + d*x])/d)/2
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_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[(-(b*c - a*d))*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((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 - 2)*(c + d*Sin[e + f*x])^(n + 1)*Simp[b*(b*c - a*d)*(B*c - A*d)*(m - 1) + a*d*(a*A*c + b*B*c - (A*b + a *B)*d)*(n + 1) + (b*(b*d*(B*c - A*d) + a*(A*c*d + B*(c^2 - 2*d^2)))*(n + 1) - a*(b*c - a*d)*(B*c - A*d)*(n + 2))*Sin[e + f*x] + b*(d*(A*b*c + a*B*c - a*A*d)*(m + n + 1) - b*B*(c^2*m + d^2*(n + 1)))*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] && GtQ[m, 1] && 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[(-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_)])^(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[((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_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 1.83 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.07
| method | result | size |
| parts | \(\frac {\left (B \,b^{3}+3 C a \,b^{2}\right ) \left (d x +c \right )}{d}+\frac {\left (3 B a \,b^{2}+3 a^{2} b C \right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {\left (3 B \,a^{2} b +a^{3} C \right ) \tan \left (d x +c \right )}{d}+\frac {B \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {\sin \left (d x +c \right ) C \,b^{3}}{d}\) | \(133\) |
| derivativedivides | \(\frac {C \tan \left (d x +c \right ) a^{3}+B \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 a^{2} b C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 B \tan \left (d x +c \right ) a^{2} b +3 C a \,b^{2} \left (d x +c \right )+3 B a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \sin \left (d x +c \right ) b^{3}+B \,b^{3} \left (d x +c \right )}{d}\) | \(141\) |
| default | \(\frac {C \tan \left (d x +c \right ) a^{3}+B \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 a^{2} b C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 B \tan \left (d x +c \right ) a^{2} b +3 C a \,b^{2} \left (d x +c \right )+3 B a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \sin \left (d x +c \right ) b^{3}+B \,b^{3} \left (d x +c \right )}{d}\) | \(141\) |
| parallelrisch | \(\frac {-a \left (1+\cos \left (2 d x +2 c \right )\right ) \left (B \,a^{2}+6 B \,b^{2}+6 a b C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+a \left (1+\cos \left (2 d x +2 c \right )\right ) \left (B \,a^{2}+6 B \,b^{2}+6 a b C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+2 b^{2} d x \left (B b +3 C a \right ) \cos \left (2 d x +2 c \right )+\left (6 B \,a^{2} b +2 a^{3} C \right ) \sin \left (2 d x +2 c \right )+C \sin \left (3 d x +3 c \right ) b^{3}+\left (2 B \,a^{3}+C \,b^{3}\right ) \sin \left (d x +c \right )+2 b^{2} d x \left (B b +3 C a \right )}{2 d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(201\) |
| risch | \(x B \,b^{3}+3 a \,b^{2} C x -\frac {i {\mathrm e}^{i \left (d x +c \right )} C \,b^{3}}{2 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} C \,b^{3}}{2 d}-\frac {i a^{2} \left (B a \,{\mathrm e}^{3 i \left (d x +c \right )}-6 B b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 C a \,{\mathrm e}^{2 i \left (d x +c \right )}-B a \,{\mathrm e}^{i \left (d x +c \right )}-6 B b -2 C a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {B \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B \,b^{2}}{d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b C}{d}-\frac {B \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B \,b^{2}}{d}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b C}{d}\) | \(272\) |
| norman | \(\frac {\left (-B \,b^{3}-3 C a \,b^{2}\right ) x +\left (-6 B \,b^{3}-18 C a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (-2 B \,b^{3}-6 C a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (-2 B \,b^{3}-6 C a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (B \,b^{3}+3 C a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{16}+\left (2 B \,b^{3}+6 C a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (2 B \,b^{3}+6 C a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}+\left (6 B \,b^{3}+18 C a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\frac {\left (B \,a^{3}-6 B \,a^{2} b -2 a^{3} C +2 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{d}+\frac {\left (5 B \,a^{3}-18 B \,a^{2} b -6 a^{3} C +2 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{d}+\frac {\left (5 B \,a^{3}+30 B \,a^{2} b +10 a^{3} C -6 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}+\frac {\left (9 B \,a^{3}-6 B \,a^{2} b -2 a^{3} C -6 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d}-\frac {\left (B \,a^{3}+6 B \,a^{2} b +2 a^{3} C +2 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {\left (5 B \,a^{3}-30 B \,a^{2} b -10 a^{3} C -6 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}-\frac {\left (5 B \,a^{3}+18 B \,a^{2} b +6 a^{3} C +2 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}-\frac {\left (9 B \,a^{3}+6 B \,a^{2} b +2 a^{3} C -6 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{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 )^{3}}-\frac {a \left (B \,a^{2}+6 B \,b^{2}+6 a b C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a \left (B \,a^{2}+6 B \,b^{2}+6 a b C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(636\) |
Input:
int((a+b*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x,method =_RETURNVERBOSE)
Output:
(B*b^3+3*C*a*b^2)/d*(d*x+c)+(3*B*a*b^2+3*C*a^2*b)/d*ln(sec(d*x+c)+tan(d*x+ c))+(3*B*a^2*b+C*a^3)/d*tan(d*x+c)+B*a^3/d*(1/2*sec(d*x+c)*tan(d*x+c)+1/2* ln(sec(d*x+c)+tan(d*x+c)))+sin(d*x+c)/d*C*b^3
Time = 0.09 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.35 \[ \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {4 \, {\left (3 \, C a b^{2} + B b^{3}\right )} d x \cos \left (d x + c\right )^{2} + {\left (B a^{3} + 6 \, C a^{2} b + 6 \, B a b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (B a^{3} + 6 \, C a^{2} b + 6 \, B a b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, C b^{3} \cos \left (d x + c\right )^{2} + B a^{3} + 2 \, {\left (C a^{3} + 3 \, B a^{2} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate((a+b*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="fricas")
Output:
1/4*(4*(3*C*a*b^2 + B*b^3)*d*x*cos(d*x + c)^2 + (B*a^3 + 6*C*a^2*b + 6*B*a *b^2)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (B*a^3 + 6*C*a^2*b + 6*B*a*b^ 2)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(2*C*b^3*cos(d*x + c)^2 + B*a ^3 + 2*(C*a^3 + 3*B*a^2*b)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2)
Timed out. \[ \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+b*cos(d*x+c))**3*(B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**4 ,x)
Output:
Timed out
Time = 0.05 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.36 \[ \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {12 \, {\left (d x + c\right )} C a b^{2} + 4 \, {\left (d x + c\right )} B b^{3} - B a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, C a^{2} b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, B a b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, C b^{3} \sin \left (d x + c\right ) + 4 \, C a^{3} \tan \left (d x + c\right ) + 12 \, B a^{2} b \tan \left (d x + c\right )}{4 \, d} \] Input:
integrate((a+b*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="maxima")
Output:
1/4*(12*(d*x + c)*C*a*b^2 + 4*(d*x + c)*B*b^3 - B*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)) + 6*C*a^ 2*b*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 6*B*a*b^2*(log(sin(d *x + c) + 1) - log(sin(d*x + c) - 1)) + 4*C*b^3*sin(d*x + c) + 4*C*a^3*tan (d*x + c) + 12*B*a^2*b*tan(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 239 vs. \(2 (118) = 236\).
Time = 0.17 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.93 \[ \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {\frac {4 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + 2 \, {\left (3 \, C a b^{2} + B b^{3}\right )} {\left (d x + c\right )} + {\left (B a^{3} + 6 \, C a^{2} b + 6 \, B a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (B a^{3} + 6 \, C a^{2} b + 6 \, B a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \] Input:
integrate((a+b*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="giac")
Output:
1/2*(4*C*b^3*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 + 1) + 2*(3*C*a* b^2 + B*b^3)*(d*x + c) + (B*a^3 + 6*C*a^2*b + 6*B*a*b^2)*log(abs(tan(1/2*d *x + 1/2*c) + 1)) - (B*a^3 + 6*C*a^2*b + 6*B*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(B*a^3*tan(1/2*d*x + 1/2*c)^3 - 2*C*a^3*tan(1/2*d*x + 1/2 *c)^3 - 6*B*a^2*b*tan(1/2*d*x + 1/2*c)^3 + B*a^3*tan(1/2*d*x + 1/2*c) + 2* C*a^3*tan(1/2*d*x + 1/2*c) + 6*B*a^2*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2)/d
Time = 1.24 (sec) , antiderivative size = 249, normalized size of antiderivative = 2.01 \[ \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {\frac {C\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {C\,b^3\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {B\,a^3\,\sin \left (c+d\,x\right )}{2}+\frac {C\,b^3\,\sin \left (c+d\,x\right )}{4}+\frac {3\,B\,a^2\,b\,\sin \left (2\,c+2\,d\,x\right )}{2}}{d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {1}{2}\right )}-\frac {2\,\left (\frac {B\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,1{}\mathrm {i}}{2}-B\,b^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+B\,a\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,3{}\mathrm {i}-3\,C\,a\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+C\,a^2\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,3{}\mathrm {i}\right )}{d} \] Input:
int(((B*cos(c + d*x) + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^3)/cos(c + d *x)^4,x)
Output:
((C*a^3*sin(2*c + 2*d*x))/2 + (C*b^3*sin(3*c + 3*d*x))/4 + (B*a^3*sin(c + d*x))/2 + (C*b^3*sin(c + d*x))/4 + (3*B*a^2*b*sin(2*c + 2*d*x))/2)/(d*(cos (2*c + 2*d*x)/2 + 1/2)) - (2*((B*a^3*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*1i)/2 - B*b^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + B* a*b^2*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*3i - 3*C*a*b^2*atan (sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + C*a^2*b*atan((sin(c/2 + (d*x)/2) *1i)/cos(c/2 + (d*x)/2))*3i))/d
Time = 0.17 (sec) , antiderivative size = 454, normalized size of antiderivative = 3.66 \[ \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {-2 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{3} c -6 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b^{2}-6 a \,b^{2} c d x +2 \sin \left (d x +c \right )^{3} b^{3} c -\sin \left (d x +c \right ) a^{3} b -2 \sin \left (d x +c \right ) b^{3} c -2 b^{4} c -6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a \,b^{3}+6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2} b c +6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a \,b^{3}-6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2} b c +6 \sin \left (d x +c \right )^{2} a \,b^{2} c^{2}+2 \sin \left (d x +c \right )^{2} b^{4} d x -6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{2} b c +6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{2} b c -2 b^{4} d x -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{3} b +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{3} b +6 \sin \left (d x +c \right )^{2} a \,b^{2} c d x +6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a \,b^{3}-6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a \,b^{3}+2 \sin \left (d x +c \right )^{2} b^{4} c +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{3} b -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{3} b -6 a \,b^{2} c^{2}}{2 d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int((a+b*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x)
Output:
( - 2*cos(c + d*x)*sin(c + d*x)*a**3*c - 6*cos(c + d*x)*sin(c + d*x)*a**2* b**2 - log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**3*b - 6*log(tan((c + d *x)/2) - 1)*sin(c + d*x)**2*a**2*b*c - 6*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a*b**3 + log(tan((c + d*x)/2) - 1)*a**3*b + 6*log(tan((c + d*x)/2 ) - 1)*a**2*b*c + 6*log(tan((c + d*x)/2) - 1)*a*b**3 + log(tan((c + d*x)/2 ) + 1)*sin(c + d*x)**2*a**3*b + 6*log(tan((c + d*x)/2) + 1)*sin(c + d*x)** 2*a**2*b*c + 6*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a*b**3 - log(tan( (c + d*x)/2) + 1)*a**3*b - 6*log(tan((c + d*x)/2) + 1)*a**2*b*c - 6*log(ta n((c + d*x)/2) + 1)*a*b**3 + 2*sin(c + d*x)**3*b**3*c + 6*sin(c + d*x)**2* a*b**2*c**2 + 6*sin(c + d*x)**2*a*b**2*c*d*x + 2*sin(c + d*x)**2*b**4*c + 2*sin(c + d*x)**2*b**4*d*x - sin(c + d*x)*a**3*b - 2*sin(c + d*x)*b**3*c - 6*a*b**2*c**2 - 6*a*b**2*c*d*x - 2*b**4*c - 2*b**4*d*x)/(2*d*(sin(c + d*x )**2 - 1))