Integrand size = 31, antiderivative size = 221 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {1}{8} \left (9 a^2 A b+4 A b^3+3 a^3 B+12 a b^2 B\right ) x+\frac {\left (4 a^3 A+14 a A b^2+15 a^2 b B+5 b^3 B\right ) \sin (c+d x)}{5 d}+\frac {\left (9 a^2 A b+4 A b^3+3 a^3 B+12 a b^2 B\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 (7 A b+5 a B) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {a A \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}-\frac {a \left (4 a^2 A+12 A b^2+15 a b B\right ) \sin ^3(c+d x)}{15 d} \]
1/8*(9*A*a^2*b+4*A*b^3+3*B*a^3+12*B*a*b^2)*x+1/5*(4*A*a^3+14*A*a*b^2+15*B* a^2*b+5*B*b^3)*sin(d*x+c)/d+1/8*(9*A*a^2*b+4*A*b^3+3*B*a^3+12*B*a*b^2)*cos (d*x+c)*sin(d*x+c)/d+1/20*a^2*(7*A*b+5*B*a)*cos(d*x+c)^3*sin(d*x+c)/d+1/5* a*A*cos(d*x+c)^4*(a+b*sec(d*x+c))^2*sin(d*x+c)/d-1/15*a*(4*A*a^2+12*A*b^2+ 15*B*a*b)*sin(d*x+c)^3/d
Time = 2.54 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.80 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {60 \left (9 a^2 A b+4 A b^3+3 a^3 B+12 a b^2 B\right ) (c+d x)+60 \left (5 a^3 A+18 a A b^2+18 a^2 b B+8 b^3 B\right ) \sin (c+d x)+120 \left (3 a^2 A b+A b^3+a^3 B+3 a b^2 B\right ) \sin (2 (c+d x))+10 a \left (5 a^2 A+12 A b^2+12 a b B\right ) \sin (3 (c+d x))+15 a^2 (3 A b+a B) \sin (4 (c+d x))+6 a^3 A \sin (5 (c+d x))}{480 d} \]
(60*(9*a^2*A*b + 4*A*b^3 + 3*a^3*B + 12*a*b^2*B)*(c + d*x) + 60*(5*a^3*A + 18*a*A*b^2 + 18*a^2*b*B + 8*b^3*B)*Sin[c + d*x] + 120*(3*a^2*A*b + A*b^3 + a^3*B + 3*a*b^2*B)*Sin[2*(c + d*x)] + 10*a*(5*a^2*A + 12*A*b^2 + 12*a*b* B)*Sin[3*(c + d*x)] + 15*a^2*(3*A*b + a*B)*Sin[4*(c + d*x)] + 6*a^3*A*Sin[ 5*(c + d*x)])/(480*d)
Time = 1.27 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.93, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.484, Rules used = {3042, 4513, 25, 3042, 4562, 25, 3042, 4535, 3042, 3115, 24, 4532, 3042, 3492, 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.
\(\displaystyle \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^3 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
\(\Big \downarrow \) 4513 |
\(\displaystyle \frac {a A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d}-\frac {1}{5} \int -\cos ^4(c+d x) (a+b \sec (c+d x)) \left (b (2 a A+5 b B) \sec ^2(c+d x)+\left (4 A a^2+10 b B a+5 A b^2\right ) \sec (c+d x)+a (7 A b+5 a B)\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{5} \int \cos ^4(c+d x) (a+b \sec (c+d x)) \left (b (2 a A+5 b B) \sec ^2(c+d x)+\left (4 A a^2+10 b B a+5 A b^2\right ) \sec (c+d x)+a (7 A b+5 a B)\right )dx+\frac {a A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (b (2 a A+5 b B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+\left (4 A a^2+10 b B a+5 A b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+a (7 A b+5 a B)\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {a A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d}\) |
\(\Big \downarrow \) 4562 |
\(\displaystyle \frac {1}{5} \left (\frac {a^2 (5 a B+7 A b) \sin (c+d x) \cos ^3(c+d x)}{4 d}-\frac {1}{4} \int -\cos ^3(c+d x) \left (4 b^2 (2 a A+5 b B) \sec ^2(c+d x)+5 \left (3 B a^3+9 A b a^2+12 b^2 B a+4 A b^3\right ) \sec (c+d x)+4 a \left (4 A a^2+15 b B a+12 A b^2\right )\right )dx\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \int \cos ^3(c+d x) \left (4 b^2 (2 a A+5 b B) \sec ^2(c+d x)+5 \left (3 B a^3+9 A b a^2+12 b^2 B a+4 A b^3\right ) \sec (c+d x)+4 a \left (4 A a^2+15 b B a+12 A b^2\right )\right )dx+\frac {a^2 (5 a B+7 A b) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \int \frac {4 b^2 (2 a A+5 b B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+5 \left (3 B a^3+9 A b a^2+12 b^2 B a+4 A b^3\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+4 a \left (4 A a^2+15 b B a+12 A b^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {a^2 (5 a B+7 A b) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d}\) |
\(\Big \downarrow \) 4535 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\int \cos ^3(c+d x) \left (4 b^2 (2 a A+5 b B) \sec ^2(c+d x)+4 a \left (4 A a^2+15 b B a+12 A b^2\right )\right )dx+5 \left (3 a^3 B+9 a^2 A b+12 a b^2 B+4 A b^3\right ) \int \cos ^2(c+d x)dx\right )+\frac {a^2 (5 a B+7 A b) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\int \frac {4 b^2 (2 a A+5 b B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+4 a \left (4 A a^2+15 b B a+12 A b^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx+5 \left (3 a^3 B+9 a^2 A b+12 a b^2 B+4 A b^3\right ) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx\right )+\frac {a^2 (5 a B+7 A b) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\int \frac {4 b^2 (2 a A+5 b B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+4 a \left (4 A a^2+15 b B a+12 A b^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx+5 \left (3 a^3 B+9 a^2 A b+12 a b^2 B+4 A b^3\right ) \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )\right )+\frac {a^2 (5 a B+7 A b) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\int \frac {4 b^2 (2 a A+5 b B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+4 a \left (4 A a^2+15 b B a+12 A b^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx+5 \left (3 a^3 B+9 a^2 A b+12 a b^2 B+4 A b^3\right ) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )+\frac {a^2 (5 a B+7 A b) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d}\) |
\(\Big \downarrow \) 4532 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\int \cos (c+d x) \left (4 (2 a A+5 b B) b^2+4 a \left (4 A a^2+15 b B a+12 A b^2\right ) \cos ^2(c+d x)\right )dx+5 \left (3 a^3 B+9 a^2 A b+12 a b^2 B+4 A b^3\right ) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )+\frac {a^2 (5 a B+7 A b) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (4 (2 a A+5 b B) b^2+4 a \left (4 A a^2+15 b B a+12 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx+5 \left (3 a^3 B+9 a^2 A b+12 a b^2 B+4 A b^3\right ) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )+\frac {a^2 (5 a B+7 A b) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d}\) |
\(\Big \downarrow \) 3492 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 \left (3 a^3 B+9 a^2 A b+12 a b^2 B+4 A b^3\right ) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-\frac {\int \left (4 \left (4 A a^3+15 b B a^2+14 A b^2 a+5 b^3 B\right )-4 a \left (4 A a^2+15 b B a+12 A b^2\right ) \sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}\right )+\frac {a^2 (5 a B+7 A b) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (\frac {a^2 (5 a B+7 A b) \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {1}{4} \left (5 \left (3 a^3 B+9 a^2 A b+12 a b^2 B+4 A b^3\right ) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-\frac {\frac {4}{3} a \left (4 a^2 A+15 a b B+12 A b^2\right ) \sin ^3(c+d x)-4 \left (4 a^3 A+15 a^2 b B+14 a A b^2+5 b^3 B\right ) \sin (c+d x)}{d}\right )\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d}\) |
(a*A*Cos[c + d*x]^4*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(5*d) + ((a^2*(7* A*b + 5*a*B)*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + (5*(9*a^2*A*b + 4*A*b^3 + 3*a^3*B + 12*a*b^2*B)*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)) - (-4*(4 *a^3*A + 14*a*A*b^2 + 15*a^2*b*B + 5*b^3*B)*Sin[c + d*x] + (4*a*(4*a^2*A + 12*A*b^2 + 15*a*b*B)*Sin[c + d*x]^3)/3)/d)/4)/5
3.4.1.3.1 Defintions of rubi rules used
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[sin[(e_.) + (f_.)*(x_)]^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[-f^(-1) Subst[Int[(1 - x^2)^((m - 1)/2)*(A + C - C*x^2 ), x], x, Cos[e + f*x]], x] /; FreeQ[{e, f, A, C}, x] && IGtQ[(m + 1)/2, 0]
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[1/(d*n) Int[(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^(n + 1)*Simp[ a*(a*B*n - A*b*(m - n - 1)) + (2*a*b*B*n + A*(b^2*n + a^2*(1 + n)))*Csc[e + f*x] + b*(b*B*n + a*A*(m + n))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] & & LeQ[n, -1]
Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Int[(C + A*Sin[e + f*x]^2)/Sin[e + f*x]^(m + 2), x] /; FreeQ[ {e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && ILtQ[(m + 1)/2, 0]
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 = 3.70 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.81
method | result | size |
parallelrisch | \(\frac {120 \left (3 A \,a^{2} b +A \,b^{3}+B \,a^{3}+3 B a \,b^{2}\right ) \sin \left (2 d x +2 c \right )+10 \left (5 a^{3} A +12 A a \,b^{2}+12 B \,a^{2} b \right ) \sin \left (3 d x +3 c \right )+15 \left (3 A \,a^{2} b +B \,a^{3}\right ) \sin \left (4 d x +4 c \right )+6 a^{3} A \sin \left (5 d x +5 c \right )+60 \left (5 a^{3} A +18 A a \,b^{2}+18 B \,a^{2} b +8 B \,b^{3}\right ) \sin \left (d x +c \right )+540 \left (A \,a^{2} b +\frac {4}{9} A \,b^{3}+\frac {1}{3} B \,a^{3}+\frac {4}{3} B a \,b^{2}\right ) d x}{480 d}\) | \(179\) |
derivativedivides | \(\frac {\frac {a^{3} A \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+3 A \,a^{2} b \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+B \,a^{3} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+A a \,b^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+B \,a^{2} b \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+A \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 B a \,b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \sin \left (d x +c \right ) b^{3}}{d}\) | \(227\) |
default | \(\frac {\frac {a^{3} A \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+3 A \,a^{2} b \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+B \,a^{3} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+A a \,b^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+B \,a^{2} b \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+A \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 B a \,b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \sin \left (d x +c \right ) b^{3}}{d}\) | \(227\) |
risch | \(\frac {9 A \,a^{2} b x}{8}+\frac {x A \,b^{3}}{2}+\frac {3 a^{3} x B}{8}+\frac {3 x B a \,b^{2}}{2}+\frac {5 a^{3} A \sin \left (d x +c \right )}{8 d}+\frac {9 \sin \left (d x +c \right ) A a \,b^{2}}{4 d}+\frac {9 \sin \left (d x +c \right ) B \,a^{2} b}{4 d}+\frac {\sin \left (d x +c \right ) B \,b^{3}}{d}+\frac {a^{3} A \sin \left (5 d x +5 c \right )}{80 d}+\frac {3 \sin \left (4 d x +4 c \right ) A \,a^{2} b}{32 d}+\frac {\sin \left (4 d x +4 c \right ) B \,a^{3}}{32 d}+\frac {5 a^{3} A \sin \left (3 d x +3 c \right )}{48 d}+\frac {\sin \left (3 d x +3 c \right ) A a \,b^{2}}{4 d}+\frac {\sin \left (3 d x +3 c \right ) B \,a^{2} b}{4 d}+\frac {3 \sin \left (2 d x +2 c \right ) A \,a^{2} b}{4 d}+\frac {\sin \left (2 d x +2 c \right ) A \,b^{3}}{4 d}+\frac {\sin \left (2 d x +2 c \right ) B \,a^{3}}{4 d}+\frac {3 \sin \left (2 d x +2 c \right ) B a \,b^{2}}{4 d}\) | \(278\) |
norman | \(\frac {\left (-\frac {9}{8} A \,a^{2} b -\frac {1}{2} A \,b^{3}-\frac {3}{8} B \,a^{3}-\frac {3}{2} B a \,b^{2}\right ) x +\left (-\frac {27}{4} A \,a^{2} b -3 A \,b^{3}-\frac {9}{4} B \,a^{3}-9 B a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (-\frac {9}{4} A \,a^{2} b -A \,b^{3}-\frac {3}{4} B \,a^{3}-3 B a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (-\frac {9}{4} A \,a^{2} b -A \,b^{3}-\frac {3}{4} B \,a^{3}-3 B a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (\frac {9}{4} A \,a^{2} b +A \,b^{3}+\frac {3}{4} B \,a^{3}+3 B a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (\frac {9}{4} A \,a^{2} b +A \,b^{3}+\frac {3}{4} B \,a^{3}+3 B a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}+\left (\frac {9}{8} A \,a^{2} b +\frac {1}{2} A \,b^{3}+\frac {3}{8} B \,a^{3}+\frac {3}{2} B a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{16}+\left (\frac {27}{4} A \,a^{2} b +3 A \,b^{3}+\frac {9}{4} B \,a^{3}+9 B a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\frac {\left (8 a^{3} A -15 A \,a^{2} b +24 A a \,b^{2}-4 A \,b^{3}-5 B \,a^{3}+24 B \,a^{2} b -12 B a \,b^{2}+8 B \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{4 d}-\frac {\left (8 a^{3} A +15 A \,a^{2} b +24 A a \,b^{2}+4 A \,b^{3}+5 B \,a^{3}+24 B \,a^{2} b +12 B a \,b^{2}+8 B \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {\left (40 a^{3} A -117 A \,a^{2} b +24 A a \,b^{2}-12 A \,b^{3}-39 B \,a^{3}+24 B \,a^{2} b -36 B a \,b^{2}-24 B \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{12 d}+\frac {\left (40 a^{3} A +117 A \,a^{2} b +24 A a \,b^{2}+12 A \,b^{3}+39 B \,a^{3}+24 B \,a^{2} b +36 B a \,b^{2}-24 B \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}+\frac {\left (344 a^{3} A -405 A \,a^{2} b -600 A a \,b^{2}+180 A \,b^{3}-135 B \,a^{3}-600 B \,a^{2} b +540 B a \,b^{2}-360 B \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{60 d}-\frac {\left (344 a^{3} A +405 A \,a^{2} b -600 A a \,b^{2}-180 A \,b^{3}+135 B \,a^{3}-600 B \,a^{2} b -540 B a \,b^{2}-360 B \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{60 d}-\frac {\left (872 a^{3} A -45 A \,a^{2} b +120 A a \,b^{2}+180 A \,b^{3}-15 B \,a^{3}+120 B \,a^{2} b +540 B a \,b^{2}+360 B \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{60 d}+\frac {\left (872 a^{3} A +45 A \,a^{2} b +120 A a \,b^{2}-180 A \,b^{3}+15 B \,a^{3}+120 B \,a^{2} b -540 B a \,b^{2}+360 B \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{60 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5} \left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}\) | \(890\) |
1/480*(120*(3*A*a^2*b+A*b^3+B*a^3+3*B*a*b^2)*sin(2*d*x+2*c)+10*(5*A*a^3+12 *A*a*b^2+12*B*a^2*b)*sin(3*d*x+3*c)+15*(3*A*a^2*b+B*a^3)*sin(4*d*x+4*c)+6* a^3*A*sin(5*d*x+5*c)+60*(5*A*a^3+18*A*a*b^2+18*B*a^2*b+8*B*b^3)*sin(d*x+c) +540*(A*a^2*b+4/9*A*b^3+1/3*B*a^3+4/3*B*a*b^2)*d*x)/d
Time = 0.30 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.79 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {15 \, {\left (3 \, B a^{3} + 9 \, A a^{2} b + 12 \, B a b^{2} + 4 \, A b^{3}\right )} d x + {\left (24 \, A a^{3} \cos \left (d x + c\right )^{4} + 64 \, A a^{3} + 240 \, B a^{2} b + 240 \, A a b^{2} + 120 \, B b^{3} + 30 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left (4 \, A a^{3} + 15 \, B a^{2} b + 15 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (3 \, B a^{3} + 9 \, A a^{2} b + 12 \, B a b^{2} + 4 \, A b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \]
1/120*(15*(3*B*a^3 + 9*A*a^2*b + 12*B*a*b^2 + 4*A*b^3)*d*x + (24*A*a^3*cos (d*x + c)^4 + 64*A*a^3 + 240*B*a^2*b + 240*A*a*b^2 + 120*B*b^3 + 30*(B*a^3 + 3*A*a^2*b)*cos(d*x + c)^3 + 8*(4*A*a^3 + 15*B*a^2*b + 15*A*a*b^2)*cos(d *x + c)^2 + 15*(3*B*a^3 + 9*A*a^2*b + 12*B*a*b^2 + 4*A*b^3)*cos(d*x + c))* sin(d*x + c))/d
Timed out. \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\text {Timed out} \]
Time = 0.25 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.98 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A a^{3} + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} + 45 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} b - 480 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{2} b - 480 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a b^{2} + 360 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a b^{2} + 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A b^{3} + 480 \, B b^{3} \sin \left (d x + c\right )}{480 \, d} \]
1/480*(32*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*A*a^3 + 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B*a^3 + 45*(12 *d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a^2*b - 480*(sin(d* x + c)^3 - 3*sin(d*x + c))*B*a^2*b - 480*(sin(d*x + c)^3 - 3*sin(d*x + c)) *A*a*b^2 + 360*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a*b^2 + 120*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*b^3 + 480*B*b^3*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 672 vs. \(2 (209) = 418\).
Time = 0.35 (sec) , antiderivative size = 672, normalized size of antiderivative = 3.04 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {15 \, {\left (3 \, B a^{3} + 9 \, A a^{2} b + 12 \, B a b^{2} + 4 \, A b^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (120 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 225 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 360 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 360 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 180 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 60 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 120 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 160 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 30 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 90 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 960 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 960 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 360 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 120 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 480 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 464 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1200 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1200 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 720 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 160 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 90 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 960 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 960 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 360 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 480 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 225 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 360 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 360 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 180 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, B b^{3} \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} \]
1/120*(15*(3*B*a^3 + 9*A*a^2*b + 12*B*a*b^2 + 4*A*b^3)*(d*x + c) + 2*(120* A*a^3*tan(1/2*d*x + 1/2*c)^9 - 75*B*a^3*tan(1/2*d*x + 1/2*c)^9 - 225*A*a^2 *b*tan(1/2*d*x + 1/2*c)^9 + 360*B*a^2*b*tan(1/2*d*x + 1/2*c)^9 + 360*A*a*b ^2*tan(1/2*d*x + 1/2*c)^9 - 180*B*a*b^2*tan(1/2*d*x + 1/2*c)^9 - 60*A*b^3* tan(1/2*d*x + 1/2*c)^9 + 120*B*b^3*tan(1/2*d*x + 1/2*c)^9 + 160*A*a^3*tan( 1/2*d*x + 1/2*c)^7 - 30*B*a^3*tan(1/2*d*x + 1/2*c)^7 - 90*A*a^2*b*tan(1/2* d*x + 1/2*c)^7 + 960*B*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 960*A*a*b^2*tan(1/2* d*x + 1/2*c)^7 - 360*B*a*b^2*tan(1/2*d*x + 1/2*c)^7 - 120*A*b^3*tan(1/2*d* x + 1/2*c)^7 + 480*B*b^3*tan(1/2*d*x + 1/2*c)^7 + 464*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 1200*B*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 1200*A*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 720*B*b^3*tan(1/2*d*x + 1/2*c)^5 + 160*A*a^3*tan(1/2*d*x + 1/ 2*c)^3 + 30*B*a^3*tan(1/2*d*x + 1/2*c)^3 + 90*A*a^2*b*tan(1/2*d*x + 1/2*c) ^3 + 960*B*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 960*A*a*b^2*tan(1/2*d*x + 1/2*c) ^3 + 360*B*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 120*A*b^3*tan(1/2*d*x + 1/2*c)^3 + 480*B*b^3*tan(1/2*d*x + 1/2*c)^3 + 120*A*a^3*tan(1/2*d*x + 1/2*c) + 75* B*a^3*tan(1/2*d*x + 1/2*c) + 225*A*a^2*b*tan(1/2*d*x + 1/2*c) + 360*B*a^2* b*tan(1/2*d*x + 1/2*c) + 360*A*a*b^2*tan(1/2*d*x + 1/2*c) + 180*B*a*b^2*ta n(1/2*d*x + 1/2*c) + 60*A*b^3*tan(1/2*d*x + 1/2*c) + 120*B*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d
Time = 14.93 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.25 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {A\,b^3\,x}{2}+\frac {3\,B\,a^3\,x}{8}+\frac {9\,A\,a^2\,b\,x}{8}+\frac {3\,B\,a\,b^2\,x}{2}+\frac {5\,A\,a^3\,\sin \left (c+d\,x\right )}{8\,d}+\frac {B\,b^3\,\sin \left (c+d\,x\right )}{d}+\frac {5\,A\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{48\,d}+\frac {A\,a^3\,\sin \left (5\,c+5\,d\,x\right )}{80\,d}+\frac {A\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a^3\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {3\,A\,a^2\,b\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {A\,a\,b^2\,\sin \left (3\,c+3\,d\,x\right )}{4\,d}+\frac {3\,A\,a^2\,b\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {3\,B\,a\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a^2\,b\,\sin \left (3\,c+3\,d\,x\right )}{4\,d}+\frac {9\,A\,a\,b^2\,\sin \left (c+d\,x\right )}{4\,d}+\frac {9\,B\,a^2\,b\,\sin \left (c+d\,x\right )}{4\,d} \]
(A*b^3*x)/2 + (3*B*a^3*x)/8 + (9*A*a^2*b*x)/8 + (3*B*a*b^2*x)/2 + (5*A*a^3 *sin(c + d*x))/(8*d) + (B*b^3*sin(c + d*x))/d + (5*A*a^3*sin(3*c + 3*d*x)) /(48*d) + (A*a^3*sin(5*c + 5*d*x))/(80*d) + (A*b^3*sin(2*c + 2*d*x))/(4*d) + (B*a^3*sin(2*c + 2*d*x))/(4*d) + (B*a^3*sin(4*c + 4*d*x))/(32*d) + (3*A *a^2*b*sin(2*c + 2*d*x))/(4*d) + (A*a*b^2*sin(3*c + 3*d*x))/(4*d) + (3*A*a ^2*b*sin(4*c + 4*d*x))/(32*d) + (3*B*a*b^2*sin(2*c + 2*d*x))/(4*d) + (B*a^ 2*b*sin(3*c + 3*d*x))/(4*d) + (9*A*a*b^2*sin(c + d*x))/(4*d) + (9*B*a^2*b* sin(c + d*x))/(4*d)