Integrand size = 31, antiderivative size = 122 \[ \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=-\frac {3 (A-B) x}{2 a}+\frac {(4 A-3 B) \sin (c+d x)}{a d}-\frac {3 (A-B) \cos (c+d x) \sin (c+d x)}{2 a d}-\frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}-\frac {(4 A-3 B) \sin ^3(c+d x)}{3 a d} \] Output:
-3/2*(A-B)*x/a+(4*A-3*B)*sin(d*x+c)/a/d-3/2*(A-B)*cos(d*x+c)*sin(d*x+c)/a/ d-(A-B)*cos(d*x+c)^2*sin(d*x+c)/d/(a+a*sec(d*x+c))-1/3*(4*A-3*B)*sin(d*x+c )^3/a/d
Leaf count is larger than twice the leaf count of optimal. \(249\) vs. \(2(122)=244\).
Time = 1.78 (sec) , antiderivative size = 249, normalized size of antiderivative = 2.04 \[ \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (-36 (A-B) d x \cos \left (\frac {d x}{2}\right )-36 (A-B) d x \cos \left (c+\frac {d x}{2}\right )+69 A \sin \left (\frac {d x}{2}\right )-60 B \sin \left (\frac {d x}{2}\right )+21 A \sin \left (c+\frac {d x}{2}\right )-12 B \sin \left (c+\frac {d x}{2}\right )+18 A \sin \left (c+\frac {3 d x}{2}\right )-9 B \sin \left (c+\frac {3 d x}{2}\right )+18 A \sin \left (2 c+\frac {3 d x}{2}\right )-9 B \sin \left (2 c+\frac {3 d x}{2}\right )-2 A \sin \left (2 c+\frac {5 d x}{2}\right )+3 B \sin \left (2 c+\frac {5 d x}{2}\right )-2 A \sin \left (3 c+\frac {5 d x}{2}\right )+3 B \sin \left (3 c+\frac {5 d x}{2}\right )+A \sin \left (3 c+\frac {7 d x}{2}\right )+A \sin \left (4 c+\frac {7 d x}{2}\right )\right )}{24 a d (1+\cos (c+d x))} \] Input:
Integrate[(Cos[c + d*x]^3*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x]),x]
Output:
(Cos[(c + d*x)/2]*Sec[c/2]*(-36*(A - B)*d*x*Cos[(d*x)/2] - 36*(A - B)*d*x* Cos[c + (d*x)/2] + 69*A*Sin[(d*x)/2] - 60*B*Sin[(d*x)/2] + 21*A*Sin[c + (d *x)/2] - 12*B*Sin[c + (d*x)/2] + 18*A*Sin[c + (3*d*x)/2] - 9*B*Sin[c + (3* d*x)/2] + 18*A*Sin[2*c + (3*d*x)/2] - 9*B*Sin[2*c + (3*d*x)/2] - 2*A*Sin[2 *c + (5*d*x)/2] + 3*B*Sin[2*c + (5*d*x)/2] - 2*A*Sin[3*c + (5*d*x)/2] + 3* B*Sin[3*c + (5*d*x)/2] + A*Sin[3*c + (7*d*x)/2] + A*Sin[4*c + (7*d*x)/2])) /(24*a*d*(1 + Cos[c + d*x]))
Time = 0.57 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.89, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {3042, 4508, 3042, 4274, 3042, 3113, 2009, 3115, 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.
| \(\displaystyle \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{a \sec (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )}dx\) |
| \(\Big \downarrow \) 4508 |
| \(\displaystyle \frac {\int \cos ^3(c+d x) (a (4 A-3 B)-3 a (A-B) \sec (c+d x))dx}{a^2}-\frac {(A-B) \sin (c+d x) \cos ^2(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (4 A-3 B)-3 a (A-B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx}{a^2}-\frac {(A-B) \sin (c+d x) \cos ^2(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {a (4 A-3 B) \int \cos ^3(c+d x)dx-3 a (A-B) \int \cos ^2(c+d x)dx}{a^2}-\frac {(A-B) \sin (c+d x) \cos ^2(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a (4 A-3 B) \int \sin \left (c+d x+\frac {\pi }{2}\right )^3dx-3 a (A-B) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx}{a^2}-\frac {(A-B) \sin (c+d x) \cos ^2(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3113 |
| \(\displaystyle \frac {-\frac {a (4 A-3 B) \int \left (1-\sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}-3 a (A-B) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx}{a^2}-\frac {(A-B) \sin (c+d x) \cos ^2(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-3 a (A-B) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {a (4 A-3 B) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}}{a^2}-\frac {(A-B) \sin (c+d x) \cos ^2(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {-3 a (A-B) \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {a (4 A-3 B) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}}{a^2}-\frac {(A-B) \sin (c+d x) \cos ^2(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {-\frac {a (4 A-3 B) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}-3 a (A-B) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )}{a^2}-\frac {(A-B) \sin (c+d x) \cos ^2(c+d x)}{d (a \sec (c+d x)+a)}\) |
Input:
Int[(Cos[c + d*x]^3*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x]),x]
Output:
-(((A - B)*Cos[c + d*x]^2*Sin[c + d*x])/(d*(a + a*Sec[c + d*x]))) + (-3*a* (A - B)*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)) - (a*(4*A - 3*B)*(-Sin[c + d*x] + Sin[c + d*x]^3/3))/d)/a^2
Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp and[(1 - x^2)^((n - 1)/2), x], x], x, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]
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[(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 = 0.40 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.64
| method | result | size |
| parallelrisch | \(\frac {\left (\left (-A +3 B \right ) \cos \left (2 d x +2 c \right )+A \cos \left (3 d x +3 c \right )+\left (17 A -6 B \right ) \cos \left (d x +c \right )+31 A -21 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-18 d x \left (A -B \right )}{12 d a}\) | \(78\) |
| derivativedivides | \(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -\frac {2 \left (\left (\frac {3 B}{2}-\frac {5 A}{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-\frac {8 A}{3}+2 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (\frac {B}{2}-\frac {3 A}{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}-3 \left (A -B \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}\) | \(122\) |
| default | \(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -\frac {2 \left (\left (\frac {3 B}{2}-\frac {5 A}{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-\frac {8 A}{3}+2 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (\frac {B}{2}-\frac {3 A}{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}-3 \left (A -B \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}\) | \(122\) |
| risch | \(-\frac {3 A x}{2 a}+\frac {3 x B}{2 a}-\frac {7 i A \,{\mathrm e}^{i \left (d x +c \right )}}{8 a d}+\frac {i {\mathrm e}^{i \left (d x +c \right )} B}{2 a d}+\frac {7 i A \,{\mathrm e}^{-i \left (d x +c \right )}}{8 a d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} B}{2 a d}+\frac {2 i A}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}-\frac {2 i B}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}+\frac {A \sin \left (3 d x +3 c \right )}{12 a d}-\frac {A \sin \left (2 d x +2 c \right )}{4 a d}+\frac {\sin \left (2 d x +2 c \right ) B}{4 a d}\) | \(192\) |
| norman | \(\frac {\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{a d}-\frac {3 \left (A -B \right ) x}{2 a}-\frac {9 \left (A -B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 a}-\frac {9 \left (A -B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2 a}-\frac {3 \left (A -B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2 a}+\frac {2 \left (2 A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}+\frac {2 \left (4 A -3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{a d}+\frac {\left (25 A -21 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}\) | \(194\) |
Input:
int(cos(d*x+c)^3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c)),x,method=_RETURNVERBOSE )
Output:
1/12*(((-A+3*B)*cos(2*d*x+2*c)+A*cos(3*d*x+3*c)+(17*A-6*B)*cos(d*x+c)+31*A -21*B)*tan(1/2*d*x+1/2*c)-18*d*x*(A-B))/d/a
Time = 0.08 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.80 \[ \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=-\frac {9 \, {\left (A - B\right )} d x \cos \left (d x + c\right ) + 9 \, {\left (A - B\right )} d x - {\left (2 \, A \cos \left (d x + c\right )^{3} - {\left (A - 3 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (7 \, A - 3 \, B\right )} \cos \left (d x + c\right ) + 16 \, A - 12 \, B\right )} \sin \left (d x + c\right )}{6 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \] Input:
integrate(cos(d*x+c)^3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c)),x, algorithm="fri cas")
Output:
-1/6*(9*(A - B)*d*x*cos(d*x + c) + 9*(A - B)*d*x - (2*A*cos(d*x + c)^3 - ( A - 3*B)*cos(d*x + c)^2 + (7*A - 3*B)*cos(d*x + c) + 16*A - 12*B)*sin(d*x + c))/(a*d*cos(d*x + c) + a*d)
\[ \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=\frac {\int \frac {A \cos ^{3}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate(cos(d*x+c)**3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c)),x)
Output:
(Integral(A*cos(c + d*x)**3/(sec(c + d*x) + 1), x) + Integral(B*cos(c + d* x)**3*sec(c + d*x)/(sec(c + d*x) + 1), x))/a
Leaf count of result is larger than twice the leaf count of optimal. 310 vs. \(2 (116) = 232\).
Time = 0.11 (sec) , antiderivative size = 310, normalized size of antiderivative = 2.54 \[ \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=\frac {A {\left (\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {16 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a + \frac {3 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} - \frac {9 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac {3 \, \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - 3 \, B {\left (\frac {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a + \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac {3 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{3 \, d} \] Input:
integrate(cos(d*x+c)^3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c)),x, algorithm="max ima")
Output:
1/3*(A*((9*sin(d*x + c)/(cos(d*x + c) + 1) + 16*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/(a + 3*a*sin(d*x + c)^ 2/(cos(d*x + c) + 1)^2 + 3*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + a*sin(d *x + c)^6/(cos(d*x + c) + 1)^6) - 9*arctan(sin(d*x + c)/(cos(d*x + c) + 1) )/a + 3*sin(d*x + c)/(a*(cos(d*x + c) + 1))) - 3*B*((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))) )/d
Time = 0.15 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.24 \[ \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=-\frac {\frac {9 \, {\left (d x + c\right )} {\left (A - B\right )}}{a} - \frac {6 \, {\left (A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a} - \frac {2 \, {\left (15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 16 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} a}}{6 \, d} \] Input:
integrate(cos(d*x+c)^3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c)),x, algorithm="gia c")
Output:
-1/6*(9*(d*x + c)*(A - B)/a - 6*(A*tan(1/2*d*x + 1/2*c) - B*tan(1/2*d*x + 1/2*c))/a - 2*(15*A*tan(1/2*d*x + 1/2*c)^5 - 9*B*tan(1/2*d*x + 1/2*c)^5 + 16*A*tan(1/2*d*x + 1/2*c)^3 - 12*B*tan(1/2*d*x + 1/2*c)^3 + 9*A*tan(1/2*d* x + 1/2*c) - 3*B*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*a)) /d
Time = 12.93 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.13 \[ \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=\frac {\left (5\,A-3\,B\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {16\,A}{3}-4\,B\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (3\,A-B\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )}-\frac {3\,x\,\left (A-B\right )}{2\,a}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A-B\right )}{a\,d} \] Input:
int((cos(c + d*x)^3*(A + B/cos(c + d*x)))/(a + a/cos(c + d*x)),x)
Output:
(tan(c/2 + (d*x)/2)^5*(5*A - 3*B) + tan(c/2 + (d*x)/2)^3*((16*A)/3 - 4*B) + tan(c/2 + (d*x)/2)*(3*A - B))/(d*(a + 3*a*tan(c/2 + (d*x)/2)^2 + 3*a*tan (c/2 + (d*x)/2)^4 + a*tan(c/2 + (d*x)/2)^6)) - (3*x*(A - B))/(2*a) + (tan( c/2 + (d*x)/2)*(A - B))/(a*d)
Time = 0.23 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.07 \[ \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=\frac {-3 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a +3 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b -6 \cos \left (d x +c \right ) a +6 \cos \left (d x +c \right ) b -2 \sin \left (d x +c \right )^{4} a +12 \sin \left (d x +c \right )^{2} a -6 \sin \left (d x +c \right )^{2} b -9 \sin \left (d x +c \right ) a d x +9 \sin \left (d x +c \right ) b d x +6 a -6 b}{6 \sin \left (d x +c \right ) a d} \] Input:
int(cos(d*x+c)^3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c)),x)
Output:
( - 3*cos(c + d*x)*sin(c + d*x)**2*a + 3*cos(c + d*x)*sin(c + d*x)**2*b - 6*cos(c + d*x)*a + 6*cos(c + d*x)*b - 2*sin(c + d*x)**4*a + 12*sin(c + d*x )**2*a - 6*sin(c + d*x)**2*b - 9*sin(c + d*x)*a*d*x + 9*sin(c + d*x)*b*d*x + 6*a - 6*b)/(6*sin(c + d*x)*a*d)