Integrand size = 36, antiderivative size = 52 \[ \int (a+b \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {1}{2} (2 a B+b C) x+\frac {(b B+a C) \sin (c+d x)}{d}+\frac {b C \cos (c+d x) \sin (c+d x)}{2 d} \] Output:
1/2*(2*B*a+C*b)*x+(B*b+C*a)*sin(d*x+c)/d+1/2*b*C*cos(d*x+c)*sin(d*x+c)/d
Time = 0.08 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.98 \[ \int (a+b \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {2 b c C+4 a B d x+2 b C d x+4 (b B+a C) \sin (c+d x)+b C \sin (2 (c+d x))}{4 d} \] Input:
Integrate[(a + b*Cos[c + d*x])*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x],x]
Output:
(2*b*c*C + 4*a*B*d*x + 2*b*C*d*x + 4*(b*B + a*C)*Sin[c + d*x] + b*C*Sin[2* (c + d*x)])/(4*d)
Time = 0.30 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3042, 3508, 3042, 3213}
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.
\(\displaystyle \int \sec (c+d x) (a+b \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 3508 |
\(\displaystyle \int (a+b \cos (c+d x)) (B+C \cos (c+d x))dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (B+C \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
\(\Big \downarrow \) 3213 |
\(\displaystyle \frac {(a C+b B) \sin (c+d x)}{d}+\frac {1}{2} x (2 a B+b C)+\frac {b C \sin (c+d x) \cos (c+d x)}{2 d}\) |
Input:
Int[(a + b*Cos[c + d*x])*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x], x]
Output:
((2*a*B + b*C)*x)/2 + ((b*B + a*C)*Sin[c + d*x])/d + (b*C*Cos[c + d*x]*Sin [c + d*x])/(2*d)
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.) *(x_)]), x_Symbol] :> Simp[(2*a*c + b*d)*(x/2), x] + (-Simp[(b*c + a*d)*(Co s[e + f*x]/f), x] - Simp[b*d*Cos[e + f*x]*(Sin[e + f*x]/(2*f)), x]) /; Free Q[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_ .) + (f_.)*(x_)]^2), x_Symbol] :> Simp[1/b^2 Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Time = 0.29 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.98
method | result | size |
risch | \(x B a +\frac {b C x}{2}+\frac {\sin \left (d x +c \right ) B b}{d}+\frac {a C \sin \left (d x +c \right )}{d}+\frac {\sin \left (2 d x +2 c \right ) C b}{4 d}\) | \(51\) |
parallelrisch | \(\frac {4 B a d x +2 C b d x +4 B \sin \left (d x +c \right ) b +C \sin \left (2 d x +2 c \right ) b +4 C \sin \left (d x +c \right ) a}{4 d}\) | \(51\) |
derivativedivides | \(\frac {C b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \sin \left (d x +c \right ) b +C \sin \left (d x +c \right ) a +B a \left (d x +c \right )}{d}\) | \(57\) |
default | \(\frac {C b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \sin \left (d x +c \right ) b +C \sin \left (d x +c \right ) a +B a \left (d x +c \right )}{d}\) | \(57\) |
parts | \(\frac {\left (B b +C a \right ) \sin \left (d x +c \right )}{d}+\frac {B a \left (d x +c \right )}{d}+\frac {C b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(58\) |
norman | \(\frac {\left (B a +\frac {C b}{2}\right ) x +\left (B a +\frac {C b}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (3 B a +\frac {3 C b}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (3 B a +\frac {3 C b}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\frac {\left (2 B b +2 C a -C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}+\frac {\left (2 B b +2 C a +C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {4 \left (B b +C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}\) | \(169\) |
Input:
int((a+b*cos(d*x+c))*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c),x,method=_RE TURNVERBOSE)
Output:
x*B*a+1/2*b*C*x+sin(d*x+c)/d*B*b+a*C*sin(d*x+c)/d+1/4/d*sin(2*d*x+2*c)*C*b
Time = 0.08 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.81 \[ \int (a+b \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {{\left (2 \, B a + C b\right )} d x + {\left (C b \cos \left (d x + c\right ) + 2 \, C a + 2 \, B b\right )} \sin \left (d x + c\right )}{2 \, d} \] Input:
integrate((a+b*cos(d*x+c))*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c),x, alg orithm="fricas")
Output:
1/2*((2*B*a + C*b)*d*x + (C*b*cos(d*x + c) + 2*C*a + 2*B*b)*sin(d*x + c))/ d
\[ \int (a+b \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\int \left (B + C \cos {\left (c + d x \right )}\right ) \left (a + b \cos {\left (c + d x \right )}\right ) \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx \] Input:
integrate((a+b*cos(d*x+c))*(B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c),x)
Output:
Integral((B + C*cos(c + d*x))*(a + b*cos(c + d*x))*cos(c + d*x)*sec(c + d* x), x)
Time = 0.04 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.06 \[ \int (a+b \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {4 \, {\left (d x + c\right )} B a + {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C b + 4 \, C a \sin \left (d x + c\right ) + 4 \, B b \sin \left (d x + c\right )}{4 \, d} \] Input:
integrate((a+b*cos(d*x+c))*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c),x, alg orithm="maxima")
Output:
1/4*(4*(d*x + c)*B*a + (2*d*x + 2*c + sin(2*d*x + 2*c))*C*b + 4*C*a*sin(d* x + c) + 4*B*b*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 121 vs. \(2 (48) = 96\).
Time = 0.12 (sec) , antiderivative size = 121, normalized size of antiderivative = 2.33 \[ \int (a+b \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {{\left (2 \, B a + C b\right )} {\left (d x + c\right )} + \frac {2 \, {\left (2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \] Input:
integrate((a+b*cos(d*x+c))*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c),x, alg orithm="giac")
Output:
1/2*((2*B*a + C*b)*(d*x + c) + 2*(2*C*a*tan(1/2*d*x + 1/2*c)^3 + 2*B*b*tan (1/2*d*x + 1/2*c)^3 - C*b*tan(1/2*d*x + 1/2*c)^3 + 2*C*a*tan(1/2*d*x + 1/2 *c) + 2*B*b*tan(1/2*d*x + 1/2*c) + C*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^2)/d
Time = 0.55 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.96 \[ \int (a+b \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=B\,a\,x+\frac {C\,b\,x}{2}+\frac {B\,b\,\sin \left (c+d\,x\right )}{d}+\frac {C\,a\,\sin \left (c+d\,x\right )}{d}+\frac {C\,b\,\sin \left (2\,c+2\,d\,x\right )}{4\,d} \] Input:
int(((B*cos(c + d*x) + C*cos(c + d*x)^2)*(a + b*cos(c + d*x)))/cos(c + d*x ),x)
Output:
B*a*x + (C*b*x)/2 + (B*b*sin(c + d*x))/d + (C*a*sin(c + d*x))/d + (C*b*sin (2*c + 2*d*x))/(4*d)
Time = 0.16 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.02 \[ \int (a+b \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right ) b c +2 \sin \left (d x +c \right ) a c +2 \sin \left (d x +c \right ) b^{2}+2 a b d x +b c d x}{2 d} \] Input:
int((a+b*cos(d*x+c))*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c),x)
Output:
(cos(c + d*x)*sin(c + d*x)*b*c + 2*sin(c + d*x)*a*c + 2*sin(c + d*x)*b**2 + 2*a*b*d*x + b*c*d*x)/(2*d)