Integrand size = 41, antiderivative size = 225 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{8} a^4 (28 A+35 B+48 C) x+\frac {a^4 C \text {arctanh}(\sin (c+d x))}{d}+\frac {a^4 (28 A+35 B+40 C) \sin (c+d x)}{8 d}+\frac {a (4 A+5 B) \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{20 d}+\frac {A \cos ^4(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{5 d}+\frac {(28 A+35 B+20 C) \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{60 d}+\frac {(28 A+35 B+32 C) \cos (c+d x) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{24 d} \] Output:
1/8*a^4*(28*A+35*B+48*C)*x+a^4*C*arctanh(sin(d*x+c))/d+1/8*a^4*(28*A+35*B+ 40*C)*sin(d*x+c)/d+1/20*a*(4*A+5*B)*cos(d*x+c)^3*(a+a*sec(d*x+c))^3*sin(d* x+c)/d+1/5*A*cos(d*x+c)^4*(a+a*sec(d*x+c))^4*sin(d*x+c)/d+1/60*(28*A+35*B+ 20*C)*cos(d*x+c)^2*(a^2+a^2*sec(d*x+c))^2*sin(d*x+c)/d+1/24*(28*A+35*B+32* C)*cos(d*x+c)*(a^4+a^4*sec(d*x+c))*sin(d*x+c)/d
Time = 2.61 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.81 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^4 \left (1680 A d x+2100 B d x+2880 C d x-480 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+480 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+60 (49 A+56 B+54 C) \sin (c+d x)+120 (8 A+7 B+4 C) \sin (2 (c+d x))+290 A \sin (3 (c+d x))+160 B \sin (3 (c+d x))+40 C \sin (3 (c+d x))+60 A \sin (4 (c+d x))+15 B \sin (4 (c+d x))+6 A \sin (5 (c+d x))\right )}{480 d} \] Input:
Integrate[Cos[c + d*x]^5*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Se c[c + d*x]^2),x]
Output:
(a^4*(1680*A*d*x + 2100*B*d*x + 2880*C*d*x - 480*C*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 480*C*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 60*(4 9*A + 56*B + 54*C)*Sin[c + d*x] + 120*(8*A + 7*B + 4*C)*Sin[2*(c + d*x)] + 290*A*Sin[3*(c + d*x)] + 160*B*Sin[3*(c + d*x)] + 40*C*Sin[3*(c + d*x)] + 60*A*Sin[4*(c + d*x)] + 15*B*Sin[4*(c + d*x)] + 6*A*Sin[5*(c + d*x)]))/(4 80*d)
Time = 1.40 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.08, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.317, Rules used = {3042, 4574, 3042, 4505, 3042, 4505, 27, 3042, 4505, 27, 3042, 4484, 2009}
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.
| \(\displaystyle \int \cos ^5(c+d x) (a \sec (c+d x)+a)^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^4 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
| \(\Big \downarrow \) 4574 |
| \(\displaystyle \frac {\int \cos ^4(c+d x) (\sec (c+d x) a+a)^4 (a (4 A+5 B)+5 a C \sec (c+d x))dx}{5 a}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^4}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^4 \left (a (4 A+5 B)+5 a C \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx}{5 a}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^4}{5 d}\) |
| \(\Big \downarrow \) 4505 |
| \(\displaystyle \frac {\frac {1}{4} \int \cos ^3(c+d x) (\sec (c+d x) a+a)^3 \left ((28 A+35 B+20 C) a^2+20 C \sec (c+d x) a^2\right )dx+\frac {a^2 (4 A+5 B) \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{4 d}}{5 a}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^4}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{4} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3 \left ((28 A+35 B+20 C) a^2+20 C \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {a^2 (4 A+5 B) \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{4 d}}{5 a}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^4}{5 d}\) |
| \(\Big \downarrow \) 4505 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \int 5 \cos ^2(c+d x) (\sec (c+d x) a+a)^2 \left ((28 A+35 B+32 C) a^3+12 C \sec (c+d x) a^3\right )dx+\frac {a^3 (28 A+35 B+20 C) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{3 d}\right )+\frac {a^2 (4 A+5 B) \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{4 d}}{5 a}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^4}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {5}{3} \int \cos ^2(c+d x) (\sec (c+d x) a+a)^2 \left ((28 A+35 B+32 C) a^3+12 C \sec (c+d x) a^3\right )dx+\frac {a^3 (28 A+35 B+20 C) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{3 d}\right )+\frac {a^2 (4 A+5 B) \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{4 d}}{5 a}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^4}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {5}{3} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left ((28 A+35 B+32 C) a^3+12 C \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a^3 (28 A+35 B+20 C) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{3 d}\right )+\frac {a^2 (4 A+5 B) \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{4 d}}{5 a}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^4}{5 d}\) |
| \(\Big \downarrow \) 4505 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {5}{3} \left (\frac {1}{2} \int 3 \cos (c+d x) (\sec (c+d x) a+a) \left ((28 A+35 B+40 C) a^4+8 C \sec (c+d x) a^4\right )dx+\frac {(28 A+35 B+32 C) \sin (c+d x) \cos (c+d x) \left (a^5 \sec (c+d x)+a^5\right )}{2 d}\right )+\frac {a^3 (28 A+35 B+20 C) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{3 d}\right )+\frac {a^2 (4 A+5 B) \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{4 d}}{5 a}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^4}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {5}{3} \left (\frac {3}{2} \int \cos (c+d x) (\sec (c+d x) a+a) \left ((28 A+35 B+40 C) a^4+8 C \sec (c+d x) a^4\right )dx+\frac {(28 A+35 B+32 C) \sin (c+d x) \cos (c+d x) \left (a^5 \sec (c+d x)+a^5\right )}{2 d}\right )+\frac {a^3 (28 A+35 B+20 C) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{3 d}\right )+\frac {a^2 (4 A+5 B) \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{4 d}}{5 a}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^4}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {5}{3} \left (\frac {3}{2} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left ((28 A+35 B+40 C) a^4+8 C \csc \left (c+d x+\frac {\pi }{2}\right ) a^4\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {(28 A+35 B+32 C) \sin (c+d x) \cos (c+d x) \left (a^5 \sec (c+d x)+a^5\right )}{2 d}\right )+\frac {a^3 (28 A+35 B+20 C) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{3 d}\right )+\frac {a^2 (4 A+5 B) \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{4 d}}{5 a}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^4}{5 d}\) |
| \(\Big \downarrow \) 4484 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {5}{3} \left (\frac {3}{2} \left (\frac {a^5 (28 A+35 B+40 C) \sin (c+d x)}{d}-\int \left (-\left ((28 A+35 B+48 C) a^5\right )-8 C \sec (c+d x) a^5\right )dx\right )+\frac {(28 A+35 B+32 C) \sin (c+d x) \cos (c+d x) \left (a^5 \sec (c+d x)+a^5\right )}{2 d}\right )+\frac {a^3 (28 A+35 B+20 C) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{3 d}\right )+\frac {a^2 (4 A+5 B) \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{4 d}}{5 a}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^4}{5 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {a^2 (4 A+5 B) \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{4 d}+\frac {1}{4} \left (\frac {5}{3} \left (\frac {3}{2} \left (\frac {a^5 (28 A+35 B+40 C) \sin (c+d x)}{d}+a^5 x (28 A+35 B+48 C)+\frac {8 a^5 C \text {arctanh}(\sin (c+d x))}{d}\right )+\frac {(28 A+35 B+32 C) \sin (c+d x) \cos (c+d x) \left (a^5 \sec (c+d x)+a^5\right )}{2 d}\right )+\frac {a^3 (28 A+35 B+20 C) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{3 d}\right )}{5 a}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^4}{5 d}\) |
Input:
Int[Cos[c + d*x]^5*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
Output:
(A*Cos[c + d*x]^4*(a + a*Sec[c + d*x])^4*Sin[c + d*x])/(5*d) + ((a^2*(4*A + 5*B)*Cos[c + d*x]^3*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(4*d) + ((a^3*( 28*A + 35*B + 20*C)*Cos[c + d*x]^2*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(3 *d) + (5*(((28*A + 35*B + 32*C)*Cos[c + d*x]*(a^5 + a^5*Sec[c + d*x])*Sin[ c + d*x])/(2*d) + (3*(a^5*(28*A + 35*B + 48*C)*x + (8*a^5*C*ArcTanh[Sin[c + d*x]])/d + (a^5*(28*A + 35*B + 40*C)*Sin[c + d*x])/d))/2))/3)/4)/(5*a)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n)), x] + Simp[1/(d*n) Int[(d*Csc[e + f*x])^( n + 1)*Simp[n*(B*a + A*b) + (B*b*n + A*a*(n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && LeQ[n, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[a*A*Cot [e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*n)), x] - Sim p[b/(a*d*n) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Sim p[a*A*(m - n - 1) - b*B*n - (a*B*n + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0 ] && GtQ[m, 1/2] && LtQ[n, -1]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/(b*d*n) Int[(a + b*Csc[e + f*x])^m*(d*Csc[ e + f*x])^(n + 1)*Simp[a*A*m - b*B*n - b*(A*(m + n + 1) + C*n)*Csc[e + f*x] , x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])
Time = 1.50 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.62
| method | result | size |
| parallelrisch | \(\frac {29 \left (-\frac {48 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{29}+\frac {48 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{29}+\frac {12 \left (8 A +7 B +4 C \right ) \sin \left (2 d x +2 c \right )}{29}+\left (A +\frac {16 B}{29}+\frac {4 C}{29}\right ) \sin \left (3 d x +3 c \right )+\frac {3 \left (2 A +\frac {B}{2}\right ) \sin \left (4 d x +4 c \right )}{29}+\frac {3 A \sin \left (5 d x +5 c \right )}{145}+\frac {6 \left (49 A +56 B +54 C \right ) \sin \left (d x +c \right )}{29}+\frac {168 \left (A +\frac {5 B}{4}+\frac {12 C}{7}\right ) x d}{29}\right ) a^{4}}{48 d}\) | \(139\) |
| derivativedivides | \(\frac {a^{4} A \sin \left (d x +c \right )+B \,a^{4} \left (d x +c \right )+a^{4} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a^{4} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 B \,a^{4} \sin \left (d x +c \right )+4 a^{4} C \left (d x +c \right )+2 a^{4} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+6 B \,a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+6 a^{4} C \sin \left (d x +c \right )+4 a^{4} A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {4 B \,a^{4} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+4 a^{4} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {a^{4} A \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+B \,a^{4} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {a^{4} C \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}}{d}\) | \(340\) |
| default | \(\frac {a^{4} A \sin \left (d x +c \right )+B \,a^{4} \left (d x +c \right )+a^{4} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a^{4} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 B \,a^{4} \sin \left (d x +c \right )+4 a^{4} C \left (d x +c \right )+2 a^{4} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+6 B \,a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+6 a^{4} C \sin \left (d x +c \right )+4 a^{4} A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {4 B \,a^{4} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+4 a^{4} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {a^{4} A \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+B \,a^{4} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {a^{4} C \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}}{d}\) | \(340\) |
| risch | \(\frac {7 a^{4} A x}{2}+\frac {35 a^{4} x B}{8}+6 a^{4} x C +\frac {49 i a^{4} A \,{\mathrm e}^{-i \left (d x +c \right )}}{16 d}+\frac {7 i {\mathrm e}^{-i \left (d x +c \right )} B \,a^{4}}{2 d}+\frac {27 i {\mathrm e}^{-i \left (d x +c \right )} a^{4} C}{8 d}-\frac {27 i {\mathrm e}^{i \left (d x +c \right )} a^{4} C}{8 d}-\frac {7 i {\mathrm e}^{i \left (d x +c \right )} B \,a^{4}}{2 d}-\frac {49 i a^{4} A \,{\mathrm e}^{i \left (d x +c \right )}}{16 d}+\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}-\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}+\frac {a^{4} A \sin \left (5 d x +5 c \right )}{80 d}+\frac {a^{4} A \sin \left (4 d x +4 c \right )}{8 d}+\frac {\sin \left (4 d x +4 c \right ) B \,a^{4}}{32 d}+\frac {29 a^{4} A \sin \left (3 d x +3 c \right )}{48 d}+\frac {\sin \left (3 d x +3 c \right ) B \,a^{4}}{3 d}+\frac {\sin \left (3 d x +3 c \right ) a^{4} C}{12 d}+\frac {2 \sin \left (2 d x +2 c \right ) a^{4} A}{d}+\frac {7 \sin \left (2 d x +2 c \right ) B \,a^{4}}{4 d}+\frac {\sin \left (2 d x +2 c \right ) a^{4} C}{d}\) | \(341\) |
Input:
int(cos(d*x+c)^5*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x,meth od=_RETURNVERBOSE)
Output:
29/48*(-48/29*C*ln(tan(1/2*d*x+1/2*c)-1)+48/29*C*ln(tan(1/2*d*x+1/2*c)+1)+ 12/29*(8*A+7*B+4*C)*sin(2*d*x+2*c)+(A+16/29*B+4/29*C)*sin(3*d*x+3*c)+3/29* (2*A+1/2*B)*sin(4*d*x+4*c)+3/145*A*sin(5*d*x+5*c)+6/29*(49*A+56*B+54*C)*si n(d*x+c)+168/29*(A+5/4*B+12/7*C)*x*d)*a^4/d
Time = 0.09 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.68 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (28 \, A + 35 \, B + 48 \, C\right )} a^{4} d x + 60 \, C a^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 60 \, C a^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (24 \, A a^{4} \cos \left (d x + c\right )^{4} + 30 \, {\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right )^{3} + 8 \, {\left (34 \, A + 20 \, B + 5 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 15 \, {\left (28 \, A + 27 \, B + 16 \, C\right )} a^{4} \cos \left (d x + c\right ) + 8 \, {\left (83 \, A + 100 \, B + 100 \, C\right )} a^{4}\right )} \sin \left (d x + c\right )}{120 \, d} \] Input:
integrate(cos(d*x+c)^5*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2), x, algorithm="fricas")
Output:
1/120*(15*(28*A + 35*B + 48*C)*a^4*d*x + 60*C*a^4*log(sin(d*x + c) + 1) - 60*C*a^4*log(-sin(d*x + c) + 1) + (24*A*a^4*cos(d*x + c)^4 + 30*(4*A + B)* a^4*cos(d*x + c)^3 + 8*(34*A + 20*B + 5*C)*a^4*cos(d*x + c)^2 + 15*(28*A + 27*B + 16*C)*a^4*cos(d*x + c) + 8*(83*A + 100*B + 100*C)*a^4)*sin(d*x + c ))/d
Timed out. \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**5*(a+a*sec(d*x+c))**4*(A+B*sec(d*x+c)+C*sec(d*x+c)** 2),x)
Output:
Timed out
Time = 0.05 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.48 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A a^{4} - 960 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{4} + 60 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} + 480 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} - 640 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{4} + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} + 720 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} + 480 \, {\left (d x + c\right )} B a^{4} - 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{4} + 480 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{4} + 1920 \, {\left (d x + c\right )} C a^{4} + 240 \, C a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 480 \, A a^{4} \sin \left (d x + c\right ) + 1920 \, B a^{4} \sin \left (d x + c\right ) + 2880 \, C a^{4} \sin \left (d x + c\right )}{480 \, d} \] Input:
integrate(cos(d*x+c)^5*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2), x, algorithm="maxima")
Output:
1/480*(32*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*A*a^4 - 960*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^4 + 60*(12*d*x + 12*c + sin(4*d *x + 4*c) + 8*sin(2*d*x + 2*c))*A*a^4 + 480*(2*d*x + 2*c + sin(2*d*x + 2*c ))*A*a^4 - 640*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^4 + 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B*a^4 + 720*(2*d*x + 2*c + sin(2 *d*x + 2*c))*B*a^4 + 480*(d*x + c)*B*a^4 - 160*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^4 + 480*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^4 + 1920*(d*x + c) *C*a^4 + 240*C*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 480*A *a^4*sin(d*x + c) + 1920*B*a^4*sin(d*x + c) + 2880*C*a^4*sin(d*x + c))/d
Time = 0.34 (sec) , antiderivative size = 337, normalized size of antiderivative = 1.50 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {120 \, C a^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 120 \, C a^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 15 \, {\left (28 \, A a^{4} + 35 \, B a^{4} + 48 \, C a^{4}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (420 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 525 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 600 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 1960 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 2450 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 2720 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 3584 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4480 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4720 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3160 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3950 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3680 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1500 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1395 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1080 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{120 \, d} \] Input:
integrate(cos(d*x+c)^5*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2), x, algorithm="giac")
Output:
1/120*(120*C*a^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 120*C*a^4*log(abs(ta n(1/2*d*x + 1/2*c) - 1)) + 15*(28*A*a^4 + 35*B*a^4 + 48*C*a^4)*(d*x + c) + 2*(420*A*a^4*tan(1/2*d*x + 1/2*c)^9 + 525*B*a^4*tan(1/2*d*x + 1/2*c)^9 + 600*C*a^4*tan(1/2*d*x + 1/2*c)^9 + 1960*A*a^4*tan(1/2*d*x + 1/2*c)^7 + 245 0*B*a^4*tan(1/2*d*x + 1/2*c)^7 + 2720*C*a^4*tan(1/2*d*x + 1/2*c)^7 + 3584* A*a^4*tan(1/2*d*x + 1/2*c)^5 + 4480*B*a^4*tan(1/2*d*x + 1/2*c)^5 + 4720*C* a^4*tan(1/2*d*x + 1/2*c)^5 + 3160*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 3950*B*a^ 4*tan(1/2*d*x + 1/2*c)^3 + 3680*C*a^4*tan(1/2*d*x + 1/2*c)^3 + 1500*A*a^4* tan(1/2*d*x + 1/2*c) + 1395*B*a^4*tan(1/2*d*x + 1/2*c) + 1080*C*a^4*tan(1/ 2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d
Time = 13.25 (sec) , antiderivative size = 1151, normalized size of antiderivative = 5.12 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \] Input:
int(cos(c + d*x)^5*(a + a/cos(c + d*x))^4*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)
Output:
(tan(c/2 + (d*x)/2)^9*(7*A*a^4 + (35*B*a^4)/4 + 10*C*a^4) + tan(c/2 + (d*x )/2)^7*((98*A*a^4)/3 + (245*B*a^4)/6 + (136*C*a^4)/3) + tan(c/2 + (d*x)/2) ^3*((158*A*a^4)/3 + (395*B*a^4)/6 + (184*C*a^4)/3) + tan(c/2 + (d*x)/2)^5* ((896*A*a^4)/15 + (224*B*a^4)/3 + (236*C*a^4)/3) + tan(c/2 + (d*x)/2)*(25* A*a^4 + (93*B*a^4)/4 + 18*C*a^4))/(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)) - (C*a^4*atan((C*a^4*(tan(c/2 + (d*x)/2)*(392*A^2*a^8 + (1225*B^2*a^8)/2 + 1184*C^2*a^8 + 980*A*B*a^8 + 1344*A*C*a^8 + 1680*B*C* a^8) + C*a^4*(112*A*a^4 + 140*B*a^4 + 224*C*a^4))*1i + C*a^4*(tan(c/2 + (d *x)/2)*(392*A^2*a^8 + (1225*B^2*a^8)/2 + 1184*C^2*a^8 + 980*A*B*a^8 + 1344 *A*C*a^8 + 1680*B*C*a^8) - C*a^4*(112*A*a^4 + 140*B*a^4 + 224*C*a^4))*1i)/ (1920*C^3*a^12 + 2464*A*C^2*a^12 + 784*A^2*C*a^12 + 3080*B*C^2*a^12 + 1225 *B^2*C*a^12 + C*a^4*(tan(c/2 + (d*x)/2)*(392*A^2*a^8 + (1225*B^2*a^8)/2 + 1184*C^2*a^8 + 980*A*B*a^8 + 1344*A*C*a^8 + 1680*B*C*a^8) + C*a^4*(112*A*a ^4 + 140*B*a^4 + 224*C*a^4)) - C*a^4*(tan(c/2 + (d*x)/2)*(392*A^2*a^8 + (1 225*B^2*a^8)/2 + 1184*C^2*a^8 + 980*A*B*a^8 + 1344*A*C*a^8 + 1680*B*C*a^8) - C*a^4*(112*A*a^4 + 140*B*a^4 + 224*C*a^4)) + 1960*A*B*C*a^12))*2i)/d - (a^4*atan(((a^4*(tan(c/2 + (d*x)/2)*(392*A^2*a^8 + (1225*B^2*a^8)/2 + 1184 *C^2*a^8 + 980*A*B*a^8 + 1344*A*C*a^8 + 1680*B*C*a^8) - (a^4*(28*A + 35*B + 48*C)*(112*A*a^4 + 140*B*a^4 + 224*C*a^4)*1i)/8)*(28*A + 35*B + 48*C)...
Time = 0.16 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.96 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^{4} \left (-120 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a -30 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b +540 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a +435 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b +240 \cos \left (d x +c \right ) \sin \left (d x +c \right ) c -120 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) c +120 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) c +24 \sin \left (d x +c \right )^{5} a -320 \sin \left (d x +c \right )^{3} a -160 \sin \left (d x +c \right )^{3} b -40 \sin \left (d x +c \right )^{3} c +960 \sin \left (d x +c \right ) a +960 \sin \left (d x +c \right ) b +840 \sin \left (d x +c \right ) c +420 a c +420 a d x +525 b c +525 b d x +720 c^{2}+720 c d x \right )}{120 d} \] Input:
int(cos(d*x+c)^5*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
Output:
(a**4*( - 120*cos(c + d*x)*sin(c + d*x)**3*a - 30*cos(c + d*x)*sin(c + d*x )**3*b + 540*cos(c + d*x)*sin(c + d*x)*a + 435*cos(c + d*x)*sin(c + d*x)*b + 240*cos(c + d*x)*sin(c + d*x)*c - 120*log(tan((c + d*x)/2) - 1)*c + 120 *log(tan((c + d*x)/2) + 1)*c + 24*sin(c + d*x)**5*a - 320*sin(c + d*x)**3* a - 160*sin(c + d*x)**3*b - 40*sin(c + d*x)**3*c + 960*sin(c + d*x)*a + 96 0*sin(c + d*x)*b + 840*sin(c + d*x)*c + 420*a*c + 420*a*d*x + 525*b*c + 52 5*b*d*x + 720*c**2 + 720*c*d*x))/(120*d)