Integrand size = 29, antiderivative size = 47 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {1}{2} a (A+2 B) x+\frac {a (A+B) \sin (c+d x)}{d}+\frac {a A \cos (c+d x) \sin (c+d x)}{2 d} \] Output:
1/2*a*(A+2*B)*x+a*(A+B)*sin(d*x+c)/d+1/2*a*A*cos(d*x+c)*sin(d*x+c)/d
Time = 0.13 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.94 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {a (2 A c+2 A d x+4 B d x+4 (A+B) \sin (c+d x)+A \sin (2 (c+d x)))}{4 d} \] Input:
Integrate[Cos[c + d*x]^2*(a + a*Sec[c + d*x])*(A + B*Sec[c + d*x]),x]
Output:
(a*(2*A*c + 2*A*d*x + 4*B*d*x + 4*(A + B)*Sin[c + d*x] + A*Sin[2*(c + d*x) ]))/(4*d)
Time = 0.39 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {3042, 4484, 25, 3042, 4274, 24, 3042, 3117}
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.
\(\displaystyle \int \cos ^2(c+d x) (a \sec (c+d x)+a) (A+B \sec (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right ) \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
\(\Big \downarrow \) 4484 |
\(\displaystyle \frac {a A \sin (c+d x) \cos (c+d x)}{2 d}-\frac {1}{2} \int -\cos (c+d x) (2 a (A+B)+a (A+2 B) \sec (c+d x))dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \int \cos (c+d x) (2 a (A+B)+a (A+2 B) \sec (c+d x))dx+\frac {a A \sin (c+d x) \cos (c+d x)}{2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \int \frac {2 a (A+B)+a (A+2 B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a A \sin (c+d x) \cos (c+d x)}{2 d}\) |
\(\Big \downarrow \) 4274 |
\(\displaystyle \frac {1}{2} (2 a (A+B) \int \cos (c+d x)dx+a (A+2 B) \int 1dx)+\frac {a A \sin (c+d x) \cos (c+d x)}{2 d}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {1}{2} (2 a (A+B) \int \cos (c+d x)dx+a x (A+2 B))+\frac {a A \sin (c+d x) \cos (c+d x)}{2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (2 a (A+B) \int \sin \left (c+d x+\frac {\pi }{2}\right )dx+a x (A+2 B)\right )+\frac {a A \sin (c+d x) \cos (c+d x)}{2 d}\) |
\(\Big \downarrow \) 3117 |
\(\displaystyle \frac {1}{2} \left (\frac {2 a (A+B) \sin (c+d x)}{d}+a x (A+2 B)\right )+\frac {a A \sin (c+d x) \cos (c+d x)}{2 d}\) |
Input:
Int[Cos[c + d*x]^2*(a + a*Sec[c + d*x])*(A + B*Sec[c + d*x]),x]
Output:
(a*A*Cos[c + d*x]*Sin[c + d*x])/(2*d) + (a*(A + 2*B)*x + (2*a*(A + B)*Sin[ c + d*x])/d)/2
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_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))*(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[n*(B*a + A*b) + (B*b*n + A*a*(n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && LeQ[n, -1]
Time = 0.21 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.89
method | result | size |
parallelrisch | \(\frac {\left (\frac {A \sin \left (2 d x +2 c \right )}{2}+\left (2 A +2 B \right ) \sin \left (d x +c \right )+\left (A +2 B \right ) x d \right ) a}{2 d}\) | \(42\) |
risch | \(\frac {a A x}{2}+a x B +\frac {a A \sin \left (d x +c \right )}{d}+\frac {\sin \left (d x +c \right ) B a}{d}+\frac {a A \sin \left (2 d x +2 c \right )}{4 d}\) | \(51\) |
derivativedivides | \(\frac {a A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a A \sin \left (d x +c \right )+B a \sin \left (d x +c \right )+B a \left (d x +c \right )}{d}\) | \(57\) |
default | \(\frac {a A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a A \sin \left (d x +c \right )+B a \sin \left (d x +c \right )+B a \left (d x +c \right )}{d}\) | \(57\) |
norman | \(\frac {\frac {a \left (A +2 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {a \left (A +2 B \right ) x}{2}+\frac {2 a A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}-\frac {a \left (A +2 B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}+\frac {a \left (A +2 B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2}+\frac {a \left (A +2 B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2}-\frac {a \left (3 A +2 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}\) | \(163\) |
Input:
int(cos(d*x+c)^2*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x,method=_RETURNVERBOSE )
Output:
1/2*(1/2*A*sin(2*d*x+2*c)+(2*A+2*B)*sin(d*x+c)+(A+2*B)*x*d)*a/d
Time = 0.08 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.81 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {{\left (A + 2 \, B\right )} a d x + {\left (A a \cos \left (d x + c\right ) + 2 \, {\left (A + B\right )} a\right )} \sin \left (d x + c\right )}{2 \, d} \] Input:
integrate(cos(d*x+c)^2*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="fri cas")
Output:
1/2*((A + 2*B)*a*d*x + (A*a*cos(d*x + c) + 2*(A + B)*a)*sin(d*x + c))/d
\[ \int \cos ^2(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=a \left (\int A \cos ^{2}{\left (c + d x \right )}\, dx + \int A \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int B \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int B \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate(cos(d*x+c)**2*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x)
Output:
a*(Integral(A*cos(c + d*x)**2, x) + Integral(A*cos(c + d*x)**2*sec(c + d*x ), x) + Integral(B*cos(c + d*x)**2*sec(c + d*x), x) + Integral(B*cos(c + d *x)**2*sec(c + d*x)**2, x))
Time = 0.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.17 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a + 4 \, {\left (d x + c\right )} B a + 4 \, A a \sin \left (d x + c\right ) + 4 \, B a \sin \left (d x + c\right )}{4 \, d} \] Input:
integrate(cos(d*x+c)^2*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="max ima")
Output:
1/4*((2*d*x + 2*c + sin(2*d*x + 2*c))*A*a + 4*(d*x + c)*B*a + 4*A*a*sin(d* x + c) + 4*B*a*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 93 vs. \(2 (43) = 86\).
Time = 0.13 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.98 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {{\left (A a + 2 \, B a\right )} {\left (d x + c\right )} + \frac {2 \, {\left (A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \] Input:
integrate(cos(d*x+c)^2*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="gia c")
Output:
1/2*((A*a + 2*B*a)*(d*x + c) + 2*(A*a*tan(1/2*d*x + 1/2*c)^3 + 2*B*a*tan(1 /2*d*x + 1/2*c)^3 + 3*A*a*tan(1/2*d*x + 1/2*c) + 2*B*a*tan(1/2*d*x + 1/2*c ))/(tan(1/2*d*x + 1/2*c)^2 + 1)^2)/d
Time = 11.68 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.06 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {A\,a\,x}{2}+B\,a\,x+\frac {A\,a\,\sin \left (c+d\,x\right )}{d}+\frac {B\,a\,\sin \left (c+d\,x\right )}{d}+\frac {A\,a\,\sin \left (2\,c+2\,d\,x\right )}{4\,d} \] Input:
int(cos(c + d*x)^2*(A + B/cos(c + d*x))*(a + a/cos(c + d*x)),x)
Output:
(A*a*x)/2 + B*a*x + (A*a*sin(c + d*x))/d + (B*a*sin(c + d*x))/d + (A*a*sin (2*c + 2*d*x))/(4*d)
Time = 0.15 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.02 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {a \left (\cos \left (d x +c \right ) \sin \left (d x +c \right ) a +2 \sin \left (d x +c \right ) a +2 \sin \left (d x +c \right ) b +a d x +2 b d x \right )}{2 d} \] Input:
int(cos(d*x+c)^2*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x)
Output:
(a*(cos(c + d*x)*sin(c + d*x)*a + 2*sin(c + d*x)*a + 2*sin(c + d*x)*b + a* d*x + 2*b*d*x))/(2*d)