Integrand size = 31, antiderivative size = 162 \[ \int (a+a \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {1}{2} a^4 (8 A+13 B) x+\frac {a^4 (13 A+8 B) \text {arctanh}(\sin (c+d x))}{2 d}-\frac {5 a^4 (A-B) \sin (c+d x)}{2 d}-\frac {(6 A+B) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{2 d}+\frac {(5 A+2 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \tan (c+d x)}{2 d}+\frac {a A (a+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d} \] Output:
1/2*a^4*(8*A+13*B)*x+1/2*a^4*(13*A+8*B)*arctanh(sin(d*x+c))/d-5/2*a^4*(A-B )*sin(d*x+c)/d-1/2*(6*A+B)*(a^4+a^4*cos(d*x+c))*sin(d*x+c)/d+1/2*(5*A+2*B) *(a^2+a^2*cos(d*x+c))^2*tan(d*x+c)/d+1/2*a*A*(a+a*cos(d*x+c))^3*sec(d*x+c) *tan(d*x+c)/d
Leaf count is larger than twice the leaf count of optimal. \(343\) vs. \(2(162)=324\).
Time = 10.67 (sec) , antiderivative size = 343, normalized size of antiderivative = 2.12 \[ \int (a+a \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {1}{64} a^4 (1+\cos (c+d x))^4 \sec ^8\left (\frac {1}{2} (c+d x)\right ) \left (2 (8 A+13 B) x-\frac {2 (13 A+8 B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {2 (13 A+8 B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {4 (A+4 B) \cos (d x) \sin (c)}{d}+\frac {B \cos (2 d x) \sin (2 c)}{d}+\frac {4 (A+4 B) \cos (c) \sin (d x)}{d}+\frac {B \cos (2 c) \sin (2 d x)}{d}+\frac {A}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 (4 A+B) \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {A}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 (4 A+B) \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right ) \] Input:
Integrate[(a + a*Cos[c + d*x])^4*(A + B*Cos[c + d*x])*Sec[c + d*x]^3,x]
Output:
(a^4*(1 + Cos[c + d*x])^4*Sec[(c + d*x)/2]^8*(2*(8*A + 13*B)*x - (2*(13*A + 8*B)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/d + (2*(13*A + 8*B)*Log[C os[(c + d*x)/2] + Sin[(c + d*x)/2]])/d + (4*(A + 4*B)*Cos[d*x]*Sin[c])/d + (B*Cos[2*d*x]*Sin[2*c])/d + (4*(A + 4*B)*Cos[c]*Sin[d*x])/d + (B*Cos[2*c] *Sin[2*d*x])/d + A/(d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + (4*(4*A + B)*Sin[(d*x)/2])/(d*(Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d* x)/2])) - A/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (4*(4*A + B)*Sin [(d*x)/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) ))/64
Time = 1.27 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.95, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.484, Rules used = {3042, 3454, 3042, 3454, 3042, 3455, 27, 3042, 3447, 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.
| \(\displaystyle \int \sec ^3(c+d x) (a \cos (c+d x)+a)^4 (A+B \cos (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^4 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {1}{2} \int (\cos (c+d x) a+a)^3 (a (5 A+2 B)-2 a (A-B) \cos (c+d x)) \sec ^2(c+d x)dx+\frac {a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3 \left (a (5 A+2 B)-2 a (A-B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {1}{2} \left (\int (\cos (c+d x) a+a)^2 \left (a^2 (13 A+8 B)-2 a^2 (6 A+B) \cos (c+d x)\right ) \sec (c+d x)dx+\frac {(5 A+2 B) \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}\right )+\frac {a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (a^2 (13 A+8 B)-2 a^2 (6 A+B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {(5 A+2 B) \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}\right )+\frac {a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int 2 (\cos (c+d x) a+a) \left (a^3 (13 A+8 B)-5 a^3 (A-B) \cos (c+d x)\right ) \sec (c+d x)dx-\frac {(6 A+B) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d}+\frac {(5 A+2 B) \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}\right )+\frac {a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \left (\int (\cos (c+d x) a+a) \left (a^3 (13 A+8 B)-5 a^3 (A-B) \cos (c+d x)\right ) \sec (c+d x)dx-\frac {(6 A+B) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d}+\frac {(5 A+2 B) \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}\right )+\frac {a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (a^3 (13 A+8 B)-5 a^3 (A-B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {(6 A+B) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d}+\frac {(5 A+2 B) \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}\right )+\frac {a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {1}{2} \left (\int \left (-5 (A-B) \cos ^2(c+d x) a^4+(13 A+8 B) a^4+\left (a^4 (13 A+8 B)-5 a^4 (A-B)\right ) \cos (c+d x)\right ) \sec (c+d x)dx-\frac {(6 A+B) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d}+\frac {(5 A+2 B) \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}\right )+\frac {a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\int \frac {-5 (A-B) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^4+(13 A+8 B) a^4+\left (a^4 (13 A+8 B)-5 a^4 (A-B)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {(6 A+B) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d}+\frac {(5 A+2 B) \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}\right )+\frac {a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {1}{2} \left (\int \left ((13 A+8 B) a^4+(8 A+13 B) \cos (c+d x) a^4\right ) \sec (c+d x)dx-\frac {5 a^4 (A-B) \sin (c+d x)}{d}-\frac {(6 A+B) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d}+\frac {(5 A+2 B) \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}\right )+\frac {a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\int \frac {(13 A+8 B) a^4+(8 A+13 B) \sin \left (c+d x+\frac {\pi }{2}\right ) a^4}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {5 a^4 (A-B) \sin (c+d x)}{d}-\frac {(6 A+B) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d}+\frac {(5 A+2 B) \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}\right )+\frac {a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {1}{2} \left (a^4 (13 A+8 B) \int \sec (c+d x)dx-\frac {5 a^4 (A-B) \sin (c+d x)}{d}-\frac {(6 A+B) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d}+a^4 x (8 A+13 B)+\frac {(5 A+2 B) \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}\right )+\frac {a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (a^4 (13 A+8 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {5 a^4 (A-B) \sin (c+d x)}{d}-\frac {(6 A+B) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d}+a^4 x (8 A+13 B)+\frac {(5 A+2 B) \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}\right )+\frac {a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{2} \left (\frac {a^4 (13 A+8 B) \text {arctanh}(\sin (c+d x))}{d}-\frac {5 a^4 (A-B) \sin (c+d x)}{d}-\frac {(6 A+B) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d}+a^4 x (8 A+13 B)+\frac {(5 A+2 B) \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d}\right )+\frac {a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}\) |
Input:
Int[(a + a*Cos[c + d*x])^4*(A + B*Cos[c + d*x])*Sec[c + d*x]^3,x]
Output:
(a*A*(a + a*Cos[c + d*x])^3*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (a^4*(8*A + 13*B)*x + (a^4*(13*A + 8*B)*ArcTanh[Sin[c + d*x]])/d - (5*a^4*(A - B)*Sin [c + d*x])/d - ((6*A + B)*(a^4 + a^4*Cos[c + d*x])*Sin[c + d*x])/d + ((5*A + 2*B)*(a^2 + a^2*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_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, 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] :> Sim p[(-b^2)*(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)*(b*c + a*d))), x] - Simp[b/(d*(n + 1)*(b*c + a*d)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp [a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n + 1) - B *(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f , A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0 ])
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(-b)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1 ) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && !LtQ[n, -1 ] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 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[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 27.70 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.04
| method | result | size |
| parallelrisch | \(\frac {\left (-13 \left (\cos \left (2 d x +2 c \right )+1\right ) \left (A +\frac {8 B}{13}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+13 \left (\cos \left (2 d x +2 c \right )+1\right ) \left (A +\frac {8 B}{13}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+8 x \left (A +\frac {13 B}{8}\right ) d \cos \left (2 d x +2 c \right )+\left (8 A +\frac {5 B}{2}\right ) \sin \left (2 d x +2 c \right )+\left (A +4 B \right ) \sin \left (3 d x +3 c \right )+\frac {\sin \left (4 d x +4 c \right ) B}{4}+\left (3 A +4 B \right ) \sin \left (d x +c \right )+8 x \left (A +\frac {13 B}{8}\right ) d \right ) a^{4}}{2 d \left (\cos \left (2 d x +2 c \right )+1\right )}\) | \(168\) |
| parts | \(\frac {a^{4} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {\left (a^{4} A +4 B \,a^{4}\right ) \sin \left (d x +c \right )}{d}+\frac {\left (4 a^{4} A +B \,a^{4}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (4 a^{4} A +6 B \,a^{4}\right ) \left (d x +c \right )}{d}+\frac {\left (6 a^{4} A +4 B \,a^{4}\right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {B \,a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(168\) |
| derivativedivides | \(\frac {a^{4} A \sin \left (d x +c \right )+B \,a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 a^{4} A \left (d x +c \right )+4 B \,a^{4} \sin \left (d x +c \right )+6 a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 B \,a^{4} \left (d x +c \right )+4 a^{4} A \tan \left (d x +c \right )+4 B \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{4} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \,a^{4} \tan \left (d x +c \right )}{d}\) | \(177\) |
| default | \(\frac {a^{4} A \sin \left (d x +c \right )+B \,a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 a^{4} A \left (d x +c \right )+4 B \,a^{4} \sin \left (d x +c \right )+6 a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 B \,a^{4} \left (d x +c \right )+4 a^{4} A \tan \left (d x +c \right )+4 B \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{4} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \,a^{4} \tan \left (d x +c \right )}{d}\) | \(177\) |
| risch | \(4 a^{4} x A +\frac {13 a^{4} B x}{2}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )} B \,a^{4}}{8 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} a^{4} A}{2 d}-\frac {2 i {\mathrm e}^{i \left (d x +c \right )} B \,a^{4}}{d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} a^{4} A}{2 d}+\frac {2 i {\mathrm e}^{-i \left (d x +c \right )} B \,a^{4}}{d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} B \,a^{4}}{8 d}-\frac {i a^{4} \left (A \,{\mathrm e}^{3 i \left (d x +c \right )}-8 A \,{\mathrm e}^{2 i \left (d x +c \right )}-2 B \,{\mathrm e}^{2 i \left (d x +c \right )}-A \,{\mathrm e}^{i \left (d x +c \right )}-8 A -2 B \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {13 a^{4} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {4 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d}-\frac {13 a^{4} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {4 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}\) | \(294\) |
| norman | \(\frac {\left (4 a^{4} A +\frac {13}{2} B \,a^{4}\right ) x +\left (-20 a^{4} A -\frac {65}{2} B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (-20 a^{4} A -\frac {65}{2} B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (4 a^{4} A +\frac {13}{2} B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (4 a^{4} A +\frac {13}{2} B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (4 a^{4} A +\frac {13}{2} B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}+\left (12 a^{4} A +\frac {39}{2} B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (12 a^{4} A +\frac {39}{2} B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\frac {a^{4} \left (53 A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {5 a^{4} \left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{d}+\frac {11 a^{4} \left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {4 a^{4} \left (3 A -8 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {6 a^{4} \left (7 A +4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}-\frac {3 a^{4} \left (9 A +5 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}-\frac {2 a^{4} \left (11 A -4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}-\frac {a^{4} \left (13 A +8 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a^{4} \left (13 A +8 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(458\) |
Input:
int((a+a*cos(d*x+c))^4*(A+B*cos(d*x+c))*sec(d*x+c)^3,x,method=_RETURNVERBO SE)
Output:
1/2*(-13*(cos(2*d*x+2*c)+1)*(A+8/13*B)*ln(tan(1/2*d*x+1/2*c)-1)+13*(cos(2* d*x+2*c)+1)*(A+8/13*B)*ln(tan(1/2*d*x+1/2*c)+1)+8*x*(A+13/8*B)*d*cos(2*d*x +2*c)+(8*A+5/2*B)*sin(2*d*x+2*c)+(A+4*B)*sin(3*d*x+3*c)+1/4*sin(4*d*x+4*c) *B+(3*A+4*B)*sin(d*x+c)+8*x*(A+13/8*B)*d)*a^4/d/(cos(2*d*x+2*c)+1)
Time = 0.09 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.96 \[ \int (a+a \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {2 \, {\left (8 \, A + 13 \, B\right )} a^{4} d x \cos \left (d x + c\right )^{2} + {\left (13 \, A + 8 \, B\right )} a^{4} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (13 \, A + 8 \, B\right )} a^{4} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (B a^{4} \cos \left (d x + c\right )^{3} + 2 \, {\left (A + 4 \, B\right )} a^{4} \cos \left (d x + c\right )^{2} + 2 \, {\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right ) + A a^{4}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate((a+a*cos(d*x+c))^4*(A+B*cos(d*x+c))*sec(d*x+c)^3,x, algorithm="f ricas")
Output:
1/4*(2*(8*A + 13*B)*a^4*d*x*cos(d*x + c)^2 + (13*A + 8*B)*a^4*cos(d*x + c) ^2*log(sin(d*x + c) + 1) - (13*A + 8*B)*a^4*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(B*a^4*cos(d*x + c)^3 + 2*(A + 4*B)*a^4*cos(d*x + c)^2 + 2*(4* A + B)*a^4*cos(d*x + c) + A*a^4)*sin(d*x + c))/(d*cos(d*x + c)^2)
Timed out. \[ \int (a+a \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+a*cos(d*x+c))**4*(A+B*cos(d*x+c))*sec(d*x+c)**3,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.23 \[ \int (a+a \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {16 \, {\left (d x + c\right )} A a^{4} + {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} + 24 \, {\left (d x + c\right )} B a^{4} - A a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, A a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 8 \, B a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, A a^{4} \sin \left (d x + c\right ) + 16 \, B a^{4} \sin \left (d x + c\right ) + 16 \, A a^{4} \tan \left (d x + c\right ) + 4 \, B a^{4} \tan \left (d x + c\right )}{4 \, d} \] Input:
integrate((a+a*cos(d*x+c))^4*(A+B*cos(d*x+c))*sec(d*x+c)^3,x, algorithm="m axima")
Output:
1/4*(16*(d*x + c)*A*a^4 + (2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^4 + 24*(d*x + c)*B*a^4 - A*a^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c ) + 1) + log(sin(d*x + c) - 1)) + 12*A*a^4*(log(sin(d*x + c) + 1) - log(si n(d*x + c) - 1)) + 8*B*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 4*A*a^4*sin(d*x + c) + 16*B*a^4*sin(d*x + c) + 16*A*a^4*tan(d*x + c) + 4*B*a^4*tan(d*x + c))/d
Time = 0.38 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.42 \[ \int (a+a \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {{\left (8 \, A a^{4} + 13 \, B a^{4}\right )} {\left (d x + c\right )} + {\left (13 \, A a^{4} + 8 \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (13 \, A a^{4} + 8 \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (5 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 5 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 7 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 7 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 11 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 11 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1\right )}^{2}}}{2 \, d} \] Input:
integrate((a+a*cos(d*x+c))^4*(A+B*cos(d*x+c))*sec(d*x+c)^3,x, algorithm="g iac")
Output:
1/2*((8*A*a^4 + 13*B*a^4)*(d*x + c) + (13*A*a^4 + 8*B*a^4)*log(abs(tan(1/2 *d*x + 1/2*c) + 1)) - (13*A*a^4 + 8*B*a^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(5*A*a^4*tan(1/2*d*x + 1/2*c)^7 - 5*B*a^4*tan(1/2*d*x + 1/2*c)^7 + 7*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 7*B*a^4*tan(1/2*d*x + 1/2*c)^5 - 9*A*a^4 *tan(1/2*d*x + 1/2*c)^3 + 9*B*a^4*tan(1/2*d*x + 1/2*c)^3 - 11*A*a^4*tan(1/ 2*d*x + 1/2*c) - 11*B*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^4 - 1)^2)/d
Time = 42.74 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.50 \[ \int (a+a \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {A\,a^4\,\sin \left (c+d\,x\right )}{d}+\frac {4\,B\,a^4\,\sin \left (c+d\,x\right )}{d}+\frac {8\,A\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {13\,A\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {13\,B\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {8\,B\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {4\,A\,a^4\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {A\,a^4\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {B\,a^4\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {B\,a^4\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d} \] Input:
int(((A + B*cos(c + d*x))*(a + a*cos(c + d*x))^4)/cos(c + d*x)^3,x)
Output:
(A*a^4*sin(c + d*x))/d + (4*B*a^4*sin(c + d*x))/d + (8*A*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (13*A*a^4*atanh(sin(c/2 + (d*x)/2)/cos (c/2 + (d*x)/2)))/d + (13*B*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2) ))/d + (8*B*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (4*A*a^4 *sin(c + d*x))/(d*cos(c + d*x)) + (A*a^4*sin(c + d*x))/(2*d*cos(c + d*x)^2 ) + (B*a^4*sin(c + d*x))/(d*cos(c + d*x)) + (B*a^4*cos(c + d*x)*sin(c + d* x))/(2*d)
Time = 0.18 (sec) , antiderivative size = 428, normalized size of antiderivative = 2.64 \[ \int (a+a \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {a^{4} \left (\cos \left (d x +c \right )^{2} \sin \left (d x +c \right )^{3} b -\cos \left (d x +c \right )^{2} \sin \left (d x +c \right ) b -13 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a -8 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b +13 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a +8 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b +13 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a +8 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b -13 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a -8 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b +2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a +8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b +8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a d x +13 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b d x -3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a -8 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b -8 \cos \left (d x +c \right ) a d x -13 \cos \left (d x +c \right ) b d x +8 \sin \left (d x +c \right )^{3} a +2 \sin \left (d x +c \right )^{3} b -8 \sin \left (d x +c \right ) a -2 \sin \left (d x +c \right ) b \right )}{2 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int((a+a*cos(d*x+c))^4*(A+B*cos(d*x+c))*sec(d*x+c)^3,x)
Output:
(a**4*(cos(c + d*x)**2*sin(c + d*x)**3*b - cos(c + d*x)**2*sin(c + d*x)*b - 13*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a - 8*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b + 13*cos(c + d*x)*log(tan ((c + d*x)/2) - 1)*a + 8*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*b + 13*cos (c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a + 8*cos(c + d*x)*log (tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b - 13*cos(c + d*x)*log(tan((c + d* x)/2) + 1)*a - 8*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*b + 2*cos(c + d*x) *sin(c + d*x)**3*a + 8*cos(c + d*x)*sin(c + d*x)**3*b + 8*cos(c + d*x)*sin (c + d*x)**2*a*d*x + 13*cos(c + d*x)*sin(c + d*x)**2*b*d*x - 3*cos(c + d*x )*sin(c + d*x)*a - 8*cos(c + d*x)*sin(c + d*x)*b - 8*cos(c + d*x)*a*d*x - 13*cos(c + d*x)*b*d*x + 8*sin(c + d*x)**3*a + 2*sin(c + d*x)**3*b - 8*sin( c + d*x)*a - 2*sin(c + d*x)*b))/(2*cos(c + d*x)*d*(sin(c + d*x)**2 - 1))