Integrand size = 31, antiderivative size = 77 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{2} b (A+2 C) x+\frac {a (2 A+3 C) \sin (c+d x)}{3 d}+\frac {A b \cos (c+d x) \sin (c+d x)}{2 d}+\frac {a A \cos ^2(c+d x) \sin (c+d x)}{3 d} \]
1/2*b*(A+2*C)*x+1/3*a*(2*A+3*C)*sin(d*x+c)/d+1/2*A*b*cos(d*x+c)*sin(d*x+c) /d+1/3*a*A*cos(d*x+c)^2*sin(d*x+c)/d
Time = 0.09 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.83 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {6 A b c+6 A b d x+12 b C d x+3 a (3 A+4 C) \sin (c+d x)+3 A b \sin (2 (c+d x))+a A \sin (3 (c+d x))}{12 d} \]
(6*A*b*c + 6*A*b*d*x + 12*b*C*d*x + 3*a*(3*A + 4*C)*Sin[c + d*x] + 3*A*b*S in[2*(c + d*x)] + a*A*Sin[3*(c + d*x)])/(12*d)
Time = 0.51 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {3042, 4563, 25, 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.
\(\displaystyle \int \cos ^3(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
\(\Big \downarrow \) 4563 |
\(\displaystyle \frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d}-\frac {1}{3} \int -\cos ^2(c+d x) \left (3 b C \sec ^2(c+d x)+a (2 A+3 C) \sec (c+d x)+3 A b\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{3} \int \cos ^2(c+d x) \left (3 b C \sec ^2(c+d x)+a (2 A+3 C) \sec (c+d x)+3 A b\right )dx+\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} \int \frac {3 b C \csc \left (c+d x+\frac {\pi }{2}\right )^2+a (2 A+3 C) \csc \left (c+d x+\frac {\pi }{2}\right )+3 A b}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d}\) |
\(\Big \downarrow \) 4535 |
\(\displaystyle \frac {1}{3} \left (a (2 A+3 C) \int \cos (c+d x)dx+\int \cos ^2(c+d x) \left (3 b C \sec ^2(c+d x)+3 A b\right )dx\right )+\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} \left (a (2 A+3 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )dx+\int \frac {3 b C \csc \left (c+d x+\frac {\pi }{2}\right )^2+3 A b}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx\right )+\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d}\) |
\(\Big \downarrow \) 3117 |
\(\displaystyle \frac {1}{3} \left (\int \frac {3 b C \csc \left (c+d x+\frac {\pi }{2}\right )^2+3 A b}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a (2 A+3 C) \sin (c+d x)}{d}\right )+\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d}\) |
\(\Big \downarrow \) 4533 |
\(\displaystyle \frac {1}{3} \left (\frac {3}{2} b (A+2 C) \int 1dx+\frac {a (2 A+3 C) \sin (c+d x)}{d}+\frac {3 A b \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {1}{3} \left (\frac {a (2 A+3 C) \sin (c+d x)}{d}+\frac {3 A b \sin (c+d x) \cos (c+d x)}{2 d}+\frac {3}{2} b x (A+2 C)\right )+\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d}\) |
(a*A*Cos[c + d*x]^2*Sin[c + d*x])/(3*d) + ((3*b*(A + 2*C)*x)/2 + (a*(2*A + 3*C)*Sin[c + d*x])/d + (3*A*b*Cos[c + d*x]*Sin[c + d*x])/(2*d))/3
3.7.43.3.1 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_)]*(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_)]^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] + Simp[1/(d*n) Int[(d*Csc[e + f*x] )^(n + 1)*Simp[A*b*n + a*(C*n + 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, C}, x] && LtQ[n, -1]
Time = 0.35 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.73
method | result | size |
parallelrisch | \(\frac {3 A b \sin \left (2 d x +2 c \right )+a A \sin \left (3 d x +3 c \right )+9 \left (A +\frac {4 C}{3}\right ) a \sin \left (d x +c \right )+6 b \left (A +2 C \right ) x d}{12 d}\) | \(56\) |
derivativedivides | \(\frac {\frac {a A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+A b \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C a \sin \left (d x +c \right )+C b \left (d x +c \right )}{d}\) | \(68\) |
default | \(\frac {\frac {a A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+A b \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C a \sin \left (d x +c \right )+C b \left (d x +c \right )}{d}\) | \(68\) |
risch | \(\frac {A b x}{2}+b x C +\frac {3 a A \sin \left (d x +c \right )}{4 d}+\frac {\sin \left (d x +c \right ) C a}{d}+\frac {a A \sin \left (3 d x +3 c \right )}{12 d}+\frac {A b \sin \left (2 d x +2 c \right )}{4 d}\) | \(68\) |
norman | \(\frac {\left (\frac {1}{2} A b +C b \right ) x +\left (\frac {1}{2} A b +C b \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (\frac {1}{2} A b +C b \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (\frac {1}{2} A b +C b \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (-A b -2 C b \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-A b -2 C b \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\frac {\left (2 a A -A b +2 C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}+\frac {\left (2 a A +A b +2 C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 A \left (4 a -3 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}-\frac {2 A \left (4 a +3 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {4 a \left (A -3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}\) | \(273\) |
Time = 0.25 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.73 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (A + 2 \, C\right )} b d x + {\left (2 \, A a \cos \left (d x + c\right )^{2} + 3 \, A b \cos \left (d x + c\right ) + 2 \, {\left (2 \, A + 3 \, C\right )} a\right )} \sin \left (d x + c\right )}{6 \, d} \]
1/6*(3*(A + 2*C)*b*d*x + (2*A*a*cos(d*x + c)^2 + 3*A*b*cos(d*x + c) + 2*(2 *A + 3*C)*a)*sin(d*x + c))/d
\[ \int \cos ^3(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right ) \cos ^{3}{\left (c + d x \right )}\, dx \]
Time = 0.22 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.87 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=-\frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A b - 12 \, {\left (d x + c\right )} C b - 12 \, C a \sin \left (d x + c\right )}{12 \, d} \]
-1/12*(4*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a - 3*(2*d*x + 2*c + sin(2*d* x + 2*c))*A*b - 12*(d*x + c)*C*b - 12*C*a*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 153 vs. \(2 (69) = 138\).
Time = 0.30 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.99 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (A b + 2 \, C b\right )} {\left (d x + c\right )} + \frac {2 \, {\left (6 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]
1/6*(3*(A*b + 2*C*b)*(d*x + c) + 2*(6*A*a*tan(1/2*d*x + 1/2*c)^5 + 6*C*a*t an(1/2*d*x + 1/2*c)^5 - 3*A*b*tan(1/2*d*x + 1/2*c)^5 + 4*A*a*tan(1/2*d*x + 1/2*c)^3 + 12*C*a*tan(1/2*d*x + 1/2*c)^3 + 6*A*a*tan(1/2*d*x + 1/2*c) + 6 *C*a*tan(1/2*d*x + 1/2*c) + 3*A*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2 *c)^2 + 1)^3)/d
Time = 15.48 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.87 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {A\,b\,x}{2}+C\,b\,x+\frac {3\,A\,a\,\sin \left (c+d\,x\right )}{4\,d}+\frac {C\,a\,\sin \left (c+d\,x\right )}{d}+\frac {A\,a\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {A\,b\,\sin \left (2\,c+2\,d\,x\right )}{4\,d} \]