Integrand size = 21, antiderivative size = 100 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^3 \, dx=\frac {1}{2} b \left (3 a^2+2 b^2\right ) x+\frac {a \left (2 a^2+9 b^2\right ) \sin (c+d x)}{3 d}+\frac {7 a^2 b \cos (c+d x) \sin (c+d x)}{6 d}+\frac {a^2 \cos ^2(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{3 d} \]
1/2*b*(3*a^2+2*b^2)*x+1/3*a*(2*a^2+9*b^2)*sin(d*x+c)/d+7/6*a^2*b*cos(d*x+c )*sin(d*x+c)/d+1/3*a^2*cos(d*x+c)^2*(a+b*sec(d*x+c))*sin(d*x+c)/d
Time = 0.23 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.80 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^3 \, dx=\frac {18 a^2 b c+12 b^3 c+18 a^2 b d x+12 b^3 d x+9 a \left (a^2+4 b^2\right ) \sin (c+d x)+9 a^2 b \sin (2 (c+d x))+a^3 \sin (3 (c+d x))}{12 d} \]
(18*a^2*b*c + 12*b^3*c + 18*a^2*b*d*x + 12*b^3*d*x + 9*a*(a^2 + 4*b^2)*Sin [c + d*x] + 9*a^2*b*Sin[2*(c + d*x)] + a^3*Sin[3*(c + d*x)])/(12*d)
Time = 0.59 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 4328, 3042, 4535, 3042, 3117, 4533, 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.
\(\displaystyle \int \cos ^3(c+d x) (a+b \sec (c+d x))^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^3}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
\(\Big \downarrow \) 4328 |
\(\displaystyle \frac {1}{3} \int \cos ^2(c+d x) \left (7 b a^2+\left (2 a^2+9 b^2\right ) \sec (c+d x) a+b \left (a^2+3 b^2\right ) \sec ^2(c+d x)\right )dx+\frac {a^2 \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} \int \frac {7 b a^2+\left (2 a^2+9 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right ) a+b \left (a^2+3 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a^2 \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))}{3 d}\) |
\(\Big \downarrow \) 4535 |
\(\displaystyle \frac {1}{3} \left (a \left (2 a^2+9 b^2\right ) \int \cos (c+d x)dx+\int \cos ^2(c+d x) \left (7 b a^2+b \left (a^2+3 b^2\right ) \sec ^2(c+d x)\right )dx\right )+\frac {a^2 \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} \left (a \left (2 a^2+9 b^2\right ) \int \sin \left (c+d x+\frac {\pi }{2}\right )dx+\int \frac {7 b a^2+b \left (a^2+3 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx\right )+\frac {a^2 \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))}{3 d}\) |
\(\Big \downarrow \) 3117 |
\(\displaystyle \frac {1}{3} \left (\int \frac {7 b a^2+b \left (a^2+3 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a \left (2 a^2+9 b^2\right ) \sin (c+d x)}{d}\right )+\frac {a^2 \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))}{3 d}\) |
\(\Big \downarrow \) 4533 |
\(\displaystyle \frac {1}{3} \left (\frac {3}{2} b \left (3 a^2+2 b^2\right ) \int 1dx+\frac {a \left (2 a^2+9 b^2\right ) \sin (c+d x)}{d}+\frac {7 a^2 b \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {a^2 \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))}{3 d}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {1}{3} \left (\frac {a \left (2 a^2+9 b^2\right ) \sin (c+d x)}{d}+\frac {3}{2} b x \left (3 a^2+2 b^2\right )+\frac {7 a^2 b \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {a^2 \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))}{3 d}\) |
(a^2*Cos[c + d*x]^2*(a + b*Sec[c + d*x])*Sin[c + d*x])/(3*d) + ((3*b*(3*a^ 2 + 2*b^2)*x)/2 + (a*(2*a^2 + 9*b^2)*Sin[c + d*x])/d + (7*a^2*b*Cos[c + d* x]*Sin[c + d*x])/(2*d))/3
3.5.72.3.1 Defintions of rubi rules used
Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]
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]
Time = 0.67 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.67
method | result | size |
parallelrisch | \(\frac {9 a^{2} b \sin \left (2 d x +2 c \right )+a^{3} \sin \left (3 d x +3 c \right )+9 \left (a^{3}+4 a \,b^{2}\right ) \sin \left (d x +c \right )+18 d \left (a^{2}+\frac {2 b^{2}}{3}\right ) x b}{12 d}\) | \(67\) |
derivativedivides | \(\frac {\frac {a^{3} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+3 a^{2} b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a \,b^{2} \sin \left (d x +c \right )+b^{3} \left (d x +c \right )}{d}\) | \(76\) |
default | \(\frac {\frac {a^{3} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+3 a^{2} b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a \,b^{2} \sin \left (d x +c \right )+b^{3} \left (d x +c \right )}{d}\) | \(76\) |
risch | \(\frac {3 a^{2} b x}{2}+b^{3} x +\frac {3 a^{3} \sin \left (d x +c \right )}{4 d}+\frac {3 \sin \left (d x +c \right ) a \,b^{2}}{d}+\frac {a^{3} \sin \left (3 d x +3 c \right )}{12 d}+\frac {3 a^{2} b \sin \left (2 d x +2 c \right )}{4 d}\) | \(78\) |
norman | \(\frac {\left (\frac {3}{2} a^{2} b +b^{3}\right ) x +\left (\frac {3}{2} a^{2} b +b^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (\frac {3}{2} a^{2} b +b^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (\frac {3}{2} a^{2} b +b^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (-3 a^{2} b -2 b^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-3 a^{2} b -2 b^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\frac {a \left (2 a^{2}-3 a b +6 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}+\frac {a \left (2 a^{2}+3 a b +6 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {4 a \left (a^{2}-9 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d}-\frac {2 a^{2} \left (4 a -9 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}-\frac {2 a^{2} \left (4 a +9 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}\) | \(302\) |
1/12*(9*a^2*b*sin(2*d*x+2*c)+a^3*sin(3*d*x+3*c)+9*(a^3+4*a*b^2)*sin(d*x+c) +18*d*(a^2+2/3*b^2)*x*b)/d
Time = 0.25 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.66 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^3 \, dx=\frac {3 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} d x + {\left (2 \, a^{3} \cos \left (d x + c\right )^{2} + 9 \, a^{2} b \cos \left (d x + c\right ) + 4 \, a^{3} + 18 \, a b^{2}\right )} \sin \left (d x + c\right )}{6 \, d} \]
1/6*(3*(3*a^2*b + 2*b^3)*d*x + (2*a^3*cos(d*x + c)^2 + 9*a^2*b*cos(d*x + c ) + 4*a^3 + 18*a*b^2)*sin(d*x + c))/d
\[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^3 \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right )^{3} \cos ^{3}{\left (c + d x \right )}\, dx \]
Time = 0.20 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.73 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^3 \, dx=-\frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{3} - 9 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} b - 12 \, {\left (d x + c\right )} b^{3} - 36 \, a b^{2} \sin \left (d x + c\right )}{12 \, d} \]
-1/12*(4*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^3 - 9*(2*d*x + 2*c + sin(2*d* x + 2*c))*a^2*b - 12*(d*x + c)*b^3 - 36*a*b^2*sin(d*x + c))/d
Time = 0.33 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.70 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^3 \, dx=\frac {3 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (6 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]
1/6*(3*(3*a^2*b + 2*b^3)*(d*x + c) + 2*(6*a^3*tan(1/2*d*x + 1/2*c)^5 - 9*a ^2*b*tan(1/2*d*x + 1/2*c)^5 + 18*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 4*a^3*tan( 1/2*d*x + 1/2*c)^3 + 36*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*a^3*tan(1/2*d*x + 1/2*c) + 9*a^2*b*tan(1/2*d*x + 1/2*c) + 18*a*b^2*tan(1/2*d*x + 1/2*c))/(t an(1/2*d*x + 1/2*c)^2 + 1)^3)/d
Time = 13.54 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.77 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^3 \, dx=b^3\,x+\frac {3\,a^3\,\sin \left (c+d\,x\right )}{4\,d}+\frac {a^3\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {3\,a^2\,b\,x}{2}+\frac {3\,a^2\,b\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {3\,a\,b^2\,\sin \left (c+d\,x\right )}{d} \]