Integrand size = 31, antiderivative size = 98 \[ \int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=\frac {(3 A-2 B) x}{2 a}-\frac {2 (A-B) \sin (c+d x)}{a d}+\frac {(3 A-2 B) \cos (c+d x) \sin (c+d x)}{2 a d}-\frac {(A-B) \cos (c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))} \] Output:
1/2*(3*A-2*B)*x/a-2*(A-B)*sin(d*x+c)/a/d+1/2*(3*A-2*B)*cos(d*x+c)*sin(d*x+ c)/a/d-(A-B)*cos(d*x+c)*sin(d*x+c)/d/(a+a*sec(d*x+c))
Leaf count is larger than twice the leaf count of optimal. \(197\) vs. \(2(98)=196\).
Time = 1.46 (sec) , antiderivative size = 197, normalized size of antiderivative = 2.01 \[ \int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (4 (3 A-2 B) d x \cos \left (\frac {d x}{2}\right )+4 (3 A-2 B) d x \cos \left (c+\frac {d x}{2}\right )-20 A \sin \left (\frac {d x}{2}\right )+20 B \sin \left (\frac {d x}{2}\right )-4 A \sin \left (c+\frac {d x}{2}\right )+4 B \sin \left (c+\frac {d x}{2}\right )-3 A \sin \left (c+\frac {3 d x}{2}\right )+4 B \sin \left (c+\frac {3 d x}{2}\right )-3 A \sin \left (2 c+\frac {3 d x}{2}\right )+4 B \sin \left (2 c+\frac {3 d x}{2}\right )+A \sin \left (2 c+\frac {5 d x}{2}\right )+A \sin \left (3 c+\frac {5 d x}{2}\right )\right )}{8 a d (1+\cos (c+d x))} \] Input:
Integrate[(Cos[c + d*x]^2*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x]),x]
Output:
(Cos[(c + d*x)/2]*Sec[c/2]*(4*(3*A - 2*B)*d*x*Cos[(d*x)/2] + 4*(3*A - 2*B) *d*x*Cos[c + (d*x)/2] - 20*A*Sin[(d*x)/2] + 20*B*Sin[(d*x)/2] - 4*A*Sin[c + (d*x)/2] + 4*B*Sin[c + (d*x)/2] - 3*A*Sin[c + (3*d*x)/2] + 4*B*Sin[c + ( 3*d*x)/2] - 3*A*Sin[2*c + (3*d*x)/2] + 4*B*Sin[2*c + (3*d*x)/2] + A*Sin[2* c + (5*d*x)/2] + A*Sin[3*c + (5*d*x)/2]))/(8*a*d*(1 + Cos[c + d*x]))
Time = 0.52 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.93, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {3042, 4508, 3042, 4274, 3042, 3115, 24, 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.
| \(\displaystyle \int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{a \sec (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )}dx\) |
| \(\Big \downarrow \) 4508 |
| \(\displaystyle \frac {\int \cos ^2(c+d x) (a (3 A-2 B)-2 a (A-B) \sec (c+d x))dx}{a^2}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (3 A-2 B)-2 a (A-B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx}{a^2}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {a (3 A-2 B) \int \cos ^2(c+d x)dx-2 a (A-B) \int \cos (c+d x)dx}{a^2}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a (3 A-2 B) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-2 a (A-B) \int \sin \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {a (3 A-2 B) \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-2 a (A-B) \int \sin \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {a (3 A-2 B) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-2 a (A-B) \int \sin \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3117 |
| \(\displaystyle \frac {a (3 A-2 B) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-\frac {2 a (A-B) \sin (c+d x)}{d}}{a^2}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{d (a \sec (c+d x)+a)}\) |
Input:
Int[(Cos[c + d*x]^2*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x]),x]
Output:
-(((A - B)*Cos[c + d*x]*Sin[c + d*x])/(d*(a + a*Sec[c + d*x]))) + ((-2*a*( A - B)*Sin[c + d*x])/d + a*(3*A - 2*B)*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/ (2*d)))/a^2
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[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_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-(A*b - a*B))*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(b*f*(2*m + 1))), x] - Simp[1/(a^2*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Cs c[e + f*x])^n*Simp[b*B*n - a*A*(2*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[ e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B , 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && !GtQ[n, 0]
Time = 0.31 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.62
| method | result | size |
| parallelrisch | \(\frac {\left (A \cos \left (2 d x +2 c \right )+\left (-2 A +4 B \right ) \cos \left (d x +c \right )-7 A +8 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+6 \left (A -\frac {2 B}{3}\right ) d x}{4 d a}\) | \(61\) |
| derivativedivides | \(\frac {-\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +\frac {2 \left (-\frac {3 A}{2}+B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+2 \left (-\frac {A}{2}+B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\left (3 A -2 B \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}\) | \(100\) |
| default | \(\frac {-\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +\frac {2 \left (-\frac {3 A}{2}+B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+2 \left (-\frac {A}{2}+B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\left (3 A -2 B \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}\) | \(100\) |
| norman | \(\frac {\frac {\left (3 A -2 B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a}+\frac {\left (3 A -2 B \right ) x}{2 a}-\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{a d}-\frac {\left (2 A -3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}+\frac {\left (3 A -2 B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2 a}-\frac {\left (5 A -4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}\) | \(152\) |
| risch | \(\frac {3 A x}{2 a}-\frac {x B}{a}+\frac {i A \,{\mathrm e}^{i \left (d x +c \right )}}{2 a d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} B}{2 a d}-\frac {i A \,{\mathrm e}^{-i \left (d x +c \right )}}{2 a d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B}{2 a d}-\frac {2 i A}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}+\frac {2 i B}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}+\frac {A \sin \left (2 d x +2 c \right )}{4 a d}\) | \(156\) |
Input:
int(cos(d*x+c)^2*(A+B*sec(d*x+c))/(a+a*sec(d*x+c)),x,method=_RETURNVERBOSE )
Output:
1/4*((A*cos(2*d*x+2*c)+(-2*A+4*B)*cos(d*x+c)-7*A+8*B)*tan(1/2*d*x+1/2*c)+6 *(A-2/3*B)*d*x)/d/a
Time = 0.09 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.83 \[ \int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=\frac {{\left (3 \, A - 2 \, B\right )} d x \cos \left (d x + c\right ) + {\left (3 \, A - 2 \, B\right )} d x + {\left (A \cos \left (d x + c\right )^{2} - {\left (A - 2 \, B\right )} \cos \left (d x + c\right ) - 4 \, A + 4 \, B\right )} \sin \left (d x + c\right )}{2 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \] Input:
integrate(cos(d*x+c)^2*(A+B*sec(d*x+c))/(a+a*sec(d*x+c)),x, algorithm="fri cas")
Output:
1/2*((3*A - 2*B)*d*x*cos(d*x + c) + (3*A - 2*B)*d*x + (A*cos(d*x + c)^2 - (A - 2*B)*cos(d*x + c) - 4*A + 4*B)*sin(d*x + c))/(a*d*cos(d*x + c) + a*d)
\[ \int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=\frac {\int \frac {A \cos ^{2}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate(cos(d*x+c)**2*(A+B*sec(d*x+c))/(a+a*sec(d*x+c)),x)
Output:
(Integral(A*cos(c + d*x)**2/(sec(c + d*x) + 1), x) + Integral(B*cos(c + d* x)**2*sec(c + d*x)/(sec(c + d*x) + 1), x))/a
Leaf count of result is larger than twice the leaf count of optimal. 225 vs. \(2 (94) = 188\).
Time = 0.11 (sec) , antiderivative size = 225, normalized size of antiderivative = 2.30 \[ \int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=-\frac {A {\left (\frac {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a + \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac {3 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + B {\left (\frac {2 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {2 \, \sin \left (d x + c\right )}{{\left (a + \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{d} \] Input:
integrate(cos(d*x+c)^2*(A+B*sec(d*x+c))/(a+a*sec(d*x+c)),x, algorithm="max ima")
Output:
-(A*((sin(d*x + c)/(cos(d*x + c) + 1) + 3*sin(d*x + c)^3/(cos(d*x + c) + 1 )^3)/(a + 2*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a*sin(d*x + c)^4/(cos( d*x + c) + 1)^4) - 3*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a + sin(d*x + c)/(a*(cos(d*x + c) + 1))) + B*(2*arctan(sin(d*x + c)/(cos(d*x + c) + 1)) /a - 2*sin(d*x + c)/((a + a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) - sin(d*x + c)/(a*(cos(d*x + c) + 1))))/d
Time = 0.15 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.26 \[ \int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=\frac {\frac {{\left (d x + c\right )} {\left (3 \, A - 2 \, B\right )}}{a} - \frac {2 \, {\left (A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a} - \frac {2 \, {\left (3 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a}}{2 \, d} \] Input:
integrate(cos(d*x+c)^2*(A+B*sec(d*x+c))/(a+a*sec(d*x+c)),x, algorithm="gia c")
Output:
1/2*((d*x + c)*(3*A - 2*B)/a - 2*(A*tan(1/2*d*x + 1/2*c) - B*tan(1/2*d*x + 1/2*c))/a - 2*(3*A*tan(1/2*d*x + 1/2*c)^3 - 2*B*tan(1/2*d*x + 1/2*c)^3 + A*tan(1/2*d*x + 1/2*c) - 2*B*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^ 2 + 1)^2*a))/d
Time = 12.36 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.09 \[ \int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=\frac {x\,\left (3\,A-2\,B\right )}{2\,a}-\frac {\left (3\,A-2\,B\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (A-2\,B\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A-B\right )}{a\,d} \] Input:
int((cos(c + d*x)^2*(A + B/cos(c + d*x)))/(a + a/cos(c + d*x)),x)
Output:
(x*(3*A - 2*B))/(2*a) - (tan(c/2 + (d*x)/2)^3*(3*A - 2*B) + tan(c/2 + (d*x )/2)*(A - 2*B))/(d*(a + 2*a*tan(c/2 + (d*x)/2)^2 + a*tan(c/2 + (d*x)/2)^4) ) - (tan(c/2 + (d*x)/2)*(A - B))/(a*d)
Time = 0.18 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.03 \[ \int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a +2 \cos \left (d x +c \right ) a -2 \cos \left (d x +c \right ) b -2 \sin \left (d x +c \right )^{2} a +2 \sin \left (d x +c \right )^{2} b +3 \sin \left (d x +c \right ) a d x -2 \sin \left (d x +c \right ) b d x -2 a +2 b}{2 \sin \left (d x +c \right ) a d} \] Input:
int(cos(d*x+c)^2*(A+B*sec(d*x+c))/(a+a*sec(d*x+c)),x)
Output:
(cos(c + d*x)*sin(c + d*x)**2*a + 2*cos(c + d*x)*a - 2*cos(c + d*x)*b - 2* sin(c + d*x)**2*a + 2*sin(c + d*x)**2*b + 3*sin(c + d*x)*a*d*x - 2*sin(c + d*x)*b*d*x - 2*a + 2*b)/(2*sin(c + d*x)*a*d)