Integrand size = 21, antiderivative size = 213 \[ \int \cos ^6(c+d x) (a+b \sec (c+d x))^4 \, dx=\frac {1}{16} \left (5 a^4+36 a^2 b^2+8 b^4\right ) x+\frac {4 a b \left (4 a^2+5 b^2\right ) \sin (c+d x)}{5 d}+\frac {\left (5 a^4+36 a^2 b^2+8 b^4\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {a^2 \left (5 a^2+32 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {7 a^3 b \cos ^4(c+d x) \sin (c+d x)}{15 d}+\frac {a^2 \cos ^5(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{6 d}-\frac {4 a b \left (4 a^2+5 b^2\right ) \sin ^3(c+d x)}{15 d} \] Output:
1/16*(5*a^4+36*a^2*b^2+8*b^4)*x+4/5*a*b*(4*a^2+5*b^2)*sin(d*x+c)/d+1/16*(5 *a^4+36*a^2*b^2+8*b^4)*cos(d*x+c)*sin(d*x+c)/d+1/24*a^2*(5*a^2+32*b^2)*cos (d*x+c)^3*sin(d*x+c)/d+7/15*a^3*b*cos(d*x+c)^4*sin(d*x+c)/d+1/6*a^2*cos(d* x+c)^5*(a+b*sec(d*x+c))^2*sin(d*x+c)/d-4/15*a*b*(4*a^2+5*b^2)*sin(d*x+c)^3 /d
Time = 0.71 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.73 \[ \int \cos ^6(c+d x) (a+b \sec (c+d x))^4 \, dx=\frac {60 \left (5 a^4+36 a^2 b^2+8 b^4\right ) (c+d x)+480 a b \left (5 a^2+6 b^2\right ) \sin (c+d x)+15 \left (15 a^4+96 a^2 b^2+16 b^4\right ) \sin (2 (c+d x))+80 a b \left (5 a^2+4 b^2\right ) \sin (3 (c+d x))+45 a^2 \left (a^2+4 b^2\right ) \sin (4 (c+d x))+48 a^3 b \sin (5 (c+d x))+5 a^4 \sin (6 (c+d x))}{960 d} \] Input:
Integrate[Cos[c + d*x]^6*(a + b*Sec[c + d*x])^4,x]
Output:
(60*(5*a^4 + 36*a^2*b^2 + 8*b^4)*(c + d*x) + 480*a*b*(5*a^2 + 6*b^2)*Sin[c + d*x] + 15*(15*a^4 + 96*a^2*b^2 + 16*b^4)*Sin[2*(c + d*x)] + 80*a*b*(5*a ^2 + 4*b^2)*Sin[3*(c + d*x)] + 45*a^2*(a^2 + 4*b^2)*Sin[4*(c + d*x)] + 48* a^3*b*Sin[5*(c + d*x)] + 5*a^4*Sin[6*(c + d*x)])/(960*d)
Time = 1.14 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.91, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 4328, 3042, 4562, 25, 3042, 4535, 3042, 3113, 2009, 4533, 3042, 3115, 24}
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.
| \(\displaystyle \int \cos ^6(c+d x) (a+b \sec (c+d x))^4 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^4}{\csc \left (c+d x+\frac {\pi }{2}\right )^6}dx\) |
| \(\Big \downarrow \) 4328 |
| \(\displaystyle \frac {1}{6} \int \cos ^5(c+d x) (a+b \sec (c+d x)) \left (14 b a^2+\left (5 a^2+18 b^2\right ) \sec (c+d x) a+3 b \left (a^2+2 b^2\right ) \sec ^2(c+d x)\right )dx+\frac {a^2 \sin (c+d x) \cos ^5(c+d x) (a+b \sec (c+d x))^2}{6 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{6} \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (14 b a^2+\left (5 a^2+18 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right ) a+3 b \left (a^2+2 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^5}dx+\frac {a^2 \sin (c+d x) \cos ^5(c+d x) (a+b \sec (c+d x))^2}{6 d}\) |
| \(\Big \downarrow \) 4562 |
| \(\displaystyle \frac {1}{6} \left (\frac {14 a^3 b \sin (c+d x) \cos ^4(c+d x)}{5 d}-\frac {1}{5} \int -\cos ^4(c+d x) \left (5 \left (5 a^2+32 b^2\right ) a^2+24 b \left (4 a^2+5 b^2\right ) \sec (c+d x) a+15 b^2 \left (a^2+2 b^2\right ) \sec ^2(c+d x)\right )dx\right )+\frac {a^2 \sin (c+d x) \cos ^5(c+d x) (a+b \sec (c+d x))^2}{6 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{5} \int \cos ^4(c+d x) \left (5 \left (5 a^2+32 b^2\right ) a^2+24 b \left (4 a^2+5 b^2\right ) \sec (c+d x) a+15 b^2 \left (a^2+2 b^2\right ) \sec ^2(c+d x)\right )dx+\frac {14 a^3 b \sin (c+d x) \cos ^4(c+d x)}{5 d}\right )+\frac {a^2 \sin (c+d x) \cos ^5(c+d x) (a+b \sec (c+d x))^2}{6 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{5} \int \frac {5 \left (5 a^2+32 b^2\right ) a^2+24 b \left (4 a^2+5 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right ) a+15 b^2 \left (a^2+2 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {14 a^3 b \sin (c+d x) \cos ^4(c+d x)}{5 d}\right )+\frac {a^2 \sin (c+d x) \cos ^5(c+d x) (a+b \sec (c+d x))^2}{6 d}\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{5} \left (24 a b \left (4 a^2+5 b^2\right ) \int \cos ^3(c+d x)dx+\int \cos ^4(c+d x) \left (5 \left (5 a^2+32 b^2\right ) a^2+15 b^2 \left (a^2+2 b^2\right ) \sec ^2(c+d x)\right )dx\right )+\frac {14 a^3 b \sin (c+d x) \cos ^4(c+d x)}{5 d}\right )+\frac {a^2 \sin (c+d x) \cos ^5(c+d x) (a+b \sec (c+d x))^2}{6 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{5} \left (24 a b \left (4 a^2+5 b^2\right ) \int \sin \left (c+d x+\frac {\pi }{2}\right )^3dx+\int \frac {5 \left (5 a^2+32 b^2\right ) a^2+15 b^2 \left (a^2+2 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx\right )+\frac {14 a^3 b \sin (c+d x) \cos ^4(c+d x)}{5 d}\right )+\frac {a^2 \sin (c+d x) \cos ^5(c+d x) (a+b \sec (c+d x))^2}{6 d}\) |
| \(\Big \downarrow \) 3113 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{5} \left (\int \frac {5 \left (5 a^2+32 b^2\right ) a^2+15 b^2 \left (a^2+2 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx-\frac {24 a b \left (4 a^2+5 b^2\right ) \int \left (1-\sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}\right )+\frac {14 a^3 b \sin (c+d x) \cos ^4(c+d x)}{5 d}\right )+\frac {a^2 \sin (c+d x) \cos ^5(c+d x) (a+b \sec (c+d x))^2}{6 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{5} \left (\int \frac {5 \left (5 a^2+32 b^2\right ) a^2+15 b^2 \left (a^2+2 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx-\frac {24 a b \left (4 a^2+5 b^2\right ) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {14 a^3 b \sin (c+d x) \cos ^4(c+d x)}{5 d}\right )+\frac {a^2 \sin (c+d x) \cos ^5(c+d x) (a+b \sec (c+d x))^2}{6 d}\) |
| \(\Big \downarrow \) 4533 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{5} \left (\frac {15}{4} \left (5 a^4+36 a^2 b^2+8 b^4\right ) \int \cos ^2(c+d x)dx-\frac {24 a b \left (4 a^2+5 b^2\right ) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+\frac {5 a^2 \left (5 a^2+32 b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {14 a^3 b \sin (c+d x) \cos ^4(c+d x)}{5 d}\right )+\frac {a^2 \sin (c+d x) \cos ^5(c+d x) (a+b \sec (c+d x))^2}{6 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{5} \left (\frac {15}{4} \left (5 a^4+36 a^2 b^2+8 b^4\right ) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {24 a b \left (4 a^2+5 b^2\right ) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+\frac {5 a^2 \left (5 a^2+32 b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {14 a^3 b \sin (c+d x) \cos ^4(c+d x)}{5 d}\right )+\frac {a^2 \sin (c+d x) \cos ^5(c+d x) (a+b \sec (c+d x))^2}{6 d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{5} \left (\frac {15}{4} \left (5 a^4+36 a^2 b^2+8 b^4\right ) \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {24 a b \left (4 a^2+5 b^2\right ) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+\frac {5 a^2 \left (5 a^2+32 b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {14 a^3 b \sin (c+d x) \cos ^4(c+d x)}{5 d}\right )+\frac {a^2 \sin (c+d x) \cos ^5(c+d x) (a+b \sec (c+d x))^2}{6 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {a^2 \sin (c+d x) \cos ^5(c+d x) (a+b \sec (c+d x))^2}{6 d}+\frac {1}{6} \left (\frac {14 a^3 b \sin (c+d x) \cos ^4(c+d x)}{5 d}+\frac {1}{5} \left (-\frac {24 a b \left (4 a^2+5 b^2\right ) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+\frac {5 a^2 \left (5 a^2+32 b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {15}{4} \left (5 a^4+36 a^2 b^2+8 b^4\right ) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )\right )\) |
Input:
Int[Cos[c + d*x]^6*(a + b*Sec[c + d*x])^4,x]
Output:
(a^2*Cos[c + d*x]^5*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(6*d) + ((14*a^3* b*Cos[c + d*x]^4*Sin[c + d*x])/(5*d) + ((5*a^2*(5*a^2 + 32*b^2)*Cos[c + d* x]^3*Sin[c + d*x])/(4*d) + (15*(5*a^4 + 36*a^2*b^2 + 8*b^4)*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)))/4 - (24*a*b*(4*a^2 + 5*b^2)*(-Sin[c + d*x] + Sin[c + d*x]^3/3))/d)/5)/6
Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp and[(1 - x^2)^((n - 1)/2), x], x], x, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_), x_Symbol] :> Simp[a^2*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)* ((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/(d*n) Int[(a + b*Csc[e + f*x])^(m - 3)*(d*Csc[e + f*x])^(n + 1)*Simp[a^2*b*(m - 2*n - 2) - a*(3*b^2*n + a^2* (n + 1))*Csc[e + f*x] - b*(b^2*n + a^2*(m + n - 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 2] && ((Int egerQ[m] && LtQ[n, -1]) || (IntegersQ[m + 1/2, 2*n] && LeQ[n, -1]))
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*m)), x] + Simp[(C*m + A*(m + 1))/(b^2*m) Int[(b*Csc[e + f*x])^(m + 2), x], x] /; Fr eeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* (B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b Int[(b*Cs c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) , x] /; FreeQ[{b, e, f, A, B, C, m}, x]
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 _)), x_Symbol] :> Simp[A*a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n)), x] + Si mp[1/(d*n) Int[(d*Csc[e + f*x])^(n + 1)*Simp[n*(B*a + A*b) + (n*(a*C + B* b) + A*a*(n + 1))*Csc[e + f*x] + b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[ {a, b, d, e, f, A, B, C}, x] && LtQ[n, -1]
Time = 2.53 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.72
| method | result | size |
| parallelrisch | \(\frac {\left (225 a^{4}+1440 a^{2} b^{2}+240 b^{4}\right ) \sin \left (2 d x +2 c \right )+\left (400 b \,a^{3}+320 a \,b^{3}\right ) \sin \left (3 d x +3 c \right )+\left (45 a^{4}+180 a^{2} b^{2}\right ) \sin \left (4 d x +4 c \right )+48 b \,a^{3} \sin \left (5 d x +5 c \right )+5 a^{4} \sin \left (6 d x +6 c \right )+\left (2400 b \,a^{3}+2880 a \,b^{3}\right ) \sin \left (d x +c \right )+300 x d \left (a^{4}+\frac {36}{5} a^{2} b^{2}+\frac {8}{5} b^{4}\right )}{960 d}\) | \(153\) |
| derivativedivides | \(\frac {a^{4} \left (\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+\frac {4 b \,a^{3} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+6 a^{2} b^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {4 a \,b^{3} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+b^{4} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(174\) |
| default | \(\frac {a^{4} \left (\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+\frac {4 b \,a^{3} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+6 a^{2} b^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {4 a \,b^{3} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+b^{4} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(174\) |
| risch | \(\frac {5 a^{4} x}{16}+\frac {9 a^{2} b^{2} x}{4}+\frac {x \,b^{4}}{2}+\frac {5 a^{3} b \sin \left (d x +c \right )}{2 d}+\frac {3 \sin \left (d x +c \right ) a \,b^{3}}{d}+\frac {a^{4} \sin \left (6 d x +6 c \right )}{192 d}+\frac {b \,a^{3} \sin \left (5 d x +5 c \right )}{20 d}+\frac {3 a^{4} \sin \left (4 d x +4 c \right )}{64 d}+\frac {3 \sin \left (4 d x +4 c \right ) a^{2} b^{2}}{16 d}+\frac {5 b \,a^{3} \sin \left (3 d x +3 c \right )}{12 d}+\frac {\sin \left (3 d x +3 c \right ) a \,b^{3}}{3 d}+\frac {15 \sin \left (2 d x +2 c \right ) a^{4}}{64 d}+\frac {3 \sin \left (2 d x +2 c \right ) a^{2} b^{2}}{2 d}+\frac {\sin \left (2 d x +2 c \right ) b^{4}}{4 d}\) | \(215\) |
Input:
int(cos(d*x+c)^6*(a+b*sec(d*x+c))^4,x,method=_RETURNVERBOSE)
Output:
1/960*((225*a^4+1440*a^2*b^2+240*b^4)*sin(2*d*x+2*c)+(400*a^3*b+320*a*b^3) *sin(3*d*x+3*c)+(45*a^4+180*a^2*b^2)*sin(4*d*x+4*c)+48*b*a^3*sin(5*d*x+5*c )+5*a^4*sin(6*d*x+6*c)+(2400*a^3*b+2880*a*b^3)*sin(d*x+c)+300*x*d*(a^4+36/ 5*a^2*b^2+8/5*b^4))/d
Time = 0.10 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.70 \[ \int \cos ^6(c+d x) (a+b \sec (c+d x))^4 \, dx=\frac {15 \, {\left (5 \, a^{4} + 36 \, a^{2} b^{2} + 8 \, b^{4}\right )} d x + {\left (40 \, a^{4} \cos \left (d x + c\right )^{5} + 192 \, a^{3} b \cos \left (d x + c\right )^{4} + 512 \, a^{3} b + 640 \, a b^{3} + 10 \, {\left (5 \, a^{4} + 36 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{3} + 64 \, {\left (4 \, a^{3} b + 5 \, a b^{3}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (5 \, a^{4} + 36 \, a^{2} b^{2} + 8 \, b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \] Input:
integrate(cos(d*x+c)^6*(a+b*sec(d*x+c))^4,x, algorithm="fricas")
Output:
1/240*(15*(5*a^4 + 36*a^2*b^2 + 8*b^4)*d*x + (40*a^4*cos(d*x + c)^5 + 192* a^3*b*cos(d*x + c)^4 + 512*a^3*b + 640*a*b^3 + 10*(5*a^4 + 36*a^2*b^2)*cos (d*x + c)^3 + 64*(4*a^3*b + 5*a*b^3)*cos(d*x + c)^2 + 15*(5*a^4 + 36*a^2*b ^2 + 8*b^4)*cos(d*x + c))*sin(d*x + c))/d
Timed out. \[ \int \cos ^6(c+d x) (a+b \sec (c+d x))^4 \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**6*(a+b*sec(d*x+c))**4,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.80 \[ \int \cos ^6(c+d x) (a+b \sec (c+d x))^4 \, dx=-\frac {5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{4} - 256 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a^{3} b - 180 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} b^{2} + 1280 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a b^{3} - 240 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} b^{4}}{960 \, d} \] Input:
integrate(cos(d*x+c)^6*(a+b*sec(d*x+c))^4,x, algorithm="maxima")
Output:
-1/960*(5*(4*sin(2*d*x + 2*c)^3 - 60*d*x - 60*c - 9*sin(4*d*x + 4*c) - 48* sin(2*d*x + 2*c))*a^4 - 256*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin (d*x + c))*a^3*b - 180*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2 *c))*a^2*b^2 + 1280*(sin(d*x + c)^3 - 3*sin(d*x + c))*a*b^3 - 240*(2*d*x + 2*c + sin(2*d*x + 2*c))*b^4)/d
Leaf count of result is larger than twice the leaf count of optimal. 550 vs. \(2 (199) = 398\).
Time = 0.17 (sec) , antiderivative size = 550, normalized size of antiderivative = 2.58 \[ \int \cos ^6(c+d x) (a+b \sec (c+d x))^4 \, dx =\text {Too large to display} \] Input:
integrate(cos(d*x+c)^6*(a+b*sec(d*x+c))^4,x, algorithm="giac")
Output:
1/240*(15*(5*a^4 + 36*a^2*b^2 + 8*b^4)*(d*x + c) - 2*(165*a^4*tan(1/2*d*x + 1/2*c)^11 - 960*a^3*b*tan(1/2*d*x + 1/2*c)^11 + 900*a^2*b^2*tan(1/2*d*x + 1/2*c)^11 - 960*a*b^3*tan(1/2*d*x + 1/2*c)^11 + 120*b^4*tan(1/2*d*x + 1/ 2*c)^11 - 25*a^4*tan(1/2*d*x + 1/2*c)^9 - 2240*a^3*b*tan(1/2*d*x + 1/2*c)^ 9 + 1260*a^2*b^2*tan(1/2*d*x + 1/2*c)^9 - 3520*a*b^3*tan(1/2*d*x + 1/2*c)^ 9 + 360*b^4*tan(1/2*d*x + 1/2*c)^9 + 450*a^4*tan(1/2*d*x + 1/2*c)^7 - 4992 *a^3*b*tan(1/2*d*x + 1/2*c)^7 + 360*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 - 5760* a*b^3*tan(1/2*d*x + 1/2*c)^7 + 240*b^4*tan(1/2*d*x + 1/2*c)^7 - 450*a^4*ta n(1/2*d*x + 1/2*c)^5 - 4992*a^3*b*tan(1/2*d*x + 1/2*c)^5 - 360*a^2*b^2*tan (1/2*d*x + 1/2*c)^5 - 5760*a*b^3*tan(1/2*d*x + 1/2*c)^5 - 240*b^4*tan(1/2* d*x + 1/2*c)^5 + 25*a^4*tan(1/2*d*x + 1/2*c)^3 - 2240*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 1260*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 3520*a*b^3*tan(1/2*d*x + 1/2*c)^3 - 360*b^4*tan(1/2*d*x + 1/2*c)^3 - 165*a^4*tan(1/2*d*x + 1/2*c) - 960*a^3*b*tan(1/2*d*x + 1/2*c) - 900*a^2*b^2*tan(1/2*d*x + 1/2*c) - 960*a *b^3*tan(1/2*d*x + 1/2*c) - 120*b^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1 /2*c)^2 + 1)^6)/d
Time = 10.68 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.00 \[ \int \cos ^6(c+d x) (a+b \sec (c+d x))^4 \, dx=\frac {5\,a^4\,x}{16}+\frac {b^4\,x}{2}+\frac {9\,a^2\,b^2\,x}{4}+\frac {15\,a^4\,\sin \left (2\,c+2\,d\,x\right )}{64\,d}+\frac {3\,a^4\,\sin \left (4\,c+4\,d\,x\right )}{64\,d}+\frac {a^4\,\sin \left (6\,c+6\,d\,x\right )}{192\,d}+\frac {b^4\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {a\,b^3\,\sin \left (3\,c+3\,d\,x\right )}{3\,d}+\frac {5\,a^3\,b\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {a^3\,b\,\sin \left (5\,c+5\,d\,x\right )}{20\,d}+\frac {3\,a^2\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d}+\frac {3\,a^2\,b^2\,\sin \left (4\,c+4\,d\,x\right )}{16\,d}+\frac {3\,a\,b^3\,\sin \left (c+d\,x\right )}{d}+\frac {5\,a^3\,b\,\sin \left (c+d\,x\right )}{2\,d} \] Input:
int(cos(c + d*x)^6*(a + b/cos(c + d*x))^4,x)
Output:
(5*a^4*x)/16 + (b^4*x)/2 + (9*a^2*b^2*x)/4 + (15*a^4*sin(2*c + 2*d*x))/(64 *d) + (3*a^4*sin(4*c + 4*d*x))/(64*d) + (a^4*sin(6*c + 6*d*x))/(192*d) + ( b^4*sin(2*c + 2*d*x))/(4*d) + (a*b^3*sin(3*c + 3*d*x))/(3*d) + (5*a^3*b*si n(3*c + 3*d*x))/(12*d) + (a^3*b*sin(5*c + 5*d*x))/(20*d) + (3*a^2*b^2*sin( 2*c + 2*d*x))/(2*d) + (3*a^2*b^2*sin(4*c + 4*d*x))/(16*d) + (3*a*b^3*sin(c + d*x))/d + (5*a^3*b*sin(c + d*x))/(2*d)
Time = 0.17 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.99 \[ \int \cos ^6(c+d x) (a+b \sec (c+d x))^4 \, dx=\frac {40 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5} a^{4}-130 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a^{4}-360 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a^{2} b^{2}+165 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{4}+900 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b^{2}+120 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{4}+192 \sin \left (d x +c \right )^{5} a^{3} b -640 \sin \left (d x +c \right )^{3} a^{3} b -320 \sin \left (d x +c \right )^{3} a \,b^{3}+960 \sin \left (d x +c \right ) a^{3} b +960 \sin \left (d x +c \right ) a \,b^{3}+75 a^{4} d x +540 a^{2} b^{2} d x +120 b^{4} d x}{240 d} \] Input:
int(cos(d*x+c)^6*(a+b*sec(d*x+c))^4,x)
Output:
(40*cos(c + d*x)*sin(c + d*x)**5*a**4 - 130*cos(c + d*x)*sin(c + d*x)**3*a **4 - 360*cos(c + d*x)*sin(c + d*x)**3*a**2*b**2 + 165*cos(c + d*x)*sin(c + d*x)*a**4 + 900*cos(c + d*x)*sin(c + d*x)*a**2*b**2 + 120*cos(c + d*x)*s in(c + d*x)*b**4 + 192*sin(c + d*x)**5*a**3*b - 640*sin(c + d*x)**3*a**3*b - 320*sin(c + d*x)**3*a*b**3 + 960*sin(c + d*x)*a**3*b + 960*sin(c + d*x) *a*b**3 + 75*a**4*d*x + 540*a**2*b**2*d*x + 120*b**4*d*x)/(240*d)