Integrand size = 33, antiderivative size = 124 \[ \int \frac {\cos ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=-\frac {(3 A+2 C) x}{2 a}+\frac {(4 A+3 C) \sin (c+d x)}{a d}-\frac {(3 A+2 C) \cos (c+d x) \sin (c+d x)}{2 a d}-\frac {(A+C) \cos ^2(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}-\frac {(4 A+3 C) \sin ^3(c+d x)}{3 a d} \] Output:
-1/2*(3*A+2*C)*x/a+(4*A+3*C)*sin(d*x+c)/a/d-1/2*(3*A+2*C)*cos(d*x+c)*sin(d *x+c)/a/d-(A+C)*cos(d*x+c)^2*sin(d*x+c)/d/(a+a*sec(d*x+c))-1/3*(4*A+3*C)*s in(d*x+c)^3/a/d
Time = 1.47 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.81 \[ \int \frac {\cos ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (-12 (3 A+2 C) d x \cos \left (\frac {d x}{2}\right )-12 (3 A+2 C) d x \cos \left (c+\frac {d x}{2}\right )+69 A \sin \left (\frac {d x}{2}\right )+60 C \sin \left (\frac {d x}{2}\right )+21 A \sin \left (c+\frac {d x}{2}\right )+12 C \sin \left (c+\frac {d x}{2}\right )+18 A \sin \left (c+\frac {3 d x}{2}\right )+12 C \sin \left (c+\frac {3 d x}{2}\right )+18 A \sin \left (2 c+\frac {3 d x}{2}\right )+12 C \sin \left (2 c+\frac {3 d x}{2}\right )-2 A \sin \left (2 c+\frac {5 d x}{2}\right )-2 A \sin \left (3 c+\frac {5 d x}{2}\right )+A \sin \left (3 c+\frac {7 d x}{2}\right )+A \sin \left (4 c+\frac {7 d x}{2}\right )\right )}{24 a d (1+\cos (c+d x))} \] Input:
Integrate[(Cos[c + d*x]^3*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x]),x]
Output:
(Cos[(c + d*x)/2]*Sec[c/2]*(-12*(3*A + 2*C)*d*x*Cos[(d*x)/2] - 12*(3*A + 2 *C)*d*x*Cos[c + (d*x)/2] + 69*A*Sin[(d*x)/2] + 60*C*Sin[(d*x)/2] + 21*A*Si n[c + (d*x)/2] + 12*C*Sin[c + (d*x)/2] + 18*A*Sin[c + (3*d*x)/2] + 12*C*Si n[c + (3*d*x)/2] + 18*A*Sin[2*c + (3*d*x)/2] + 12*C*Sin[2*c + (3*d*x)/2] - 2*A*Sin[2*c + (5*d*x)/2] - 2*A*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 = 0.59 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.88, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.303, Rules used = {3042, 4573, 25, 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.
| \(\displaystyle \int \frac {\cos ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a \sec (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )}dx\) |
| \(\Big \downarrow \) 4573 |
| \(\displaystyle -\frac {\int -\cos ^3(c+d x) (a (4 A+3 C)-a (3 A+2 C) \sec (c+d x))dx}{a^2}-\frac {(A+C) \sin (c+d x) \cos ^2(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \cos ^3(c+d x) (a (4 A+3 C)-a (3 A+2 C) \sec (c+d x))dx}{a^2}-\frac {(A+C) \sin (c+d x) \cos ^2(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (4 A+3 C)-a (3 A+2 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx}{a^2}-\frac {(A+C) \sin (c+d x) \cos ^2(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {a (4 A+3 C) \int \cos ^3(c+d x)dx-a (3 A+2 C) \int \cos ^2(c+d x)dx}{a^2}-\frac {(A+C) \sin (c+d x) \cos ^2(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a (4 A+3 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^3dx-a (3 A+2 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx}{a^2}-\frac {(A+C) \sin (c+d x) \cos ^2(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3113 |
| \(\displaystyle \frac {-\frac {a (4 A+3 C) \int \left (1-\sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}-a (3 A+2 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx}{a^2}-\frac {(A+C) \sin (c+d x) \cos ^2(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-a (3 A+2 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {a (4 A+3 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}}{a^2}-\frac {(A+C) \sin (c+d x) \cos ^2(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {-a (3 A+2 C) \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {a (4 A+3 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}}{a^2}-\frac {(A+C) \sin (c+d x) \cos ^2(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {-\frac {a (4 A+3 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}-a (3 A+2 C) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )}{a^2}-\frac {(A+C) \sin (c+d x) \cos ^2(c+d x)}{d (a \sec (c+d x)+a)}\) |
Input:
Int[(Cos[c + d*x]^3*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x]),x]
Output:
-(((A + C)*Cos[c + d*x]^2*Sin[c + d*x])/(d*(a + a*Sec[c + d*x]))) + (-(a*( 3*A + 2*C)*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d))) - (a*(4*A + 3*C)*(-S in[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[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. ))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-a) *(A + C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(a*f*(2*m + 1))), x] + Simp[1/(a*b*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*C sc[e + f*x])^n*Simp[b*C*n + A*b*(2*m + n + 1) - (a*(A*(m + n + 1) - C*(m - n)))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x] && EqQ[ a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Time = 0.36 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.59
| method | result | size |
| parallelrisch | \(\frac {\left (-A \cos \left (2 d x +2 c \right )+A \cos \left (3 d x +3 c \right )+\left (17 A +12 C \right ) \cos \left (d x +c \right )+31 A +24 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-18 d \left (A +\frac {2 C}{3}\right ) x}{12 d a}\) | \(73\) |
| derivativedivides | \(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -\frac {4 \left (\left (-\frac {5 A}{4}-\frac {C}{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-\frac {4 A}{3}-C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-\frac {3 A}{4}-\frac {C}{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}-\left (3 A +2 C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}\) | \(123\) |
| default | \(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -\frac {4 \left (\left (-\frac {5 A}{4}-\frac {C}{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-\frac {4 A}{3}-C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-\frac {3 A}{4}-\frac {C}{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}-\left (3 A +2 C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}\) | \(123\) |
| risch | \(-\frac {3 A x}{2 a}-\frac {x C}{a}-\frac {7 i A \,{\mathrm e}^{i \left (d x +c \right )}}{8 a d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} C}{2 a d}+\frac {7 i A \,{\mathrm e}^{-i \left (d x +c \right )}}{8 a d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} C}{2 a d}+\frac {2 i A}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}+\frac {2 i C}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}+\frac {A \sin \left (3 d x +3 c \right )}{12 a d}-\frac {A \sin \left (2 d x +2 c \right )}{4 a d}\) | \(174\) |
| norman | \(\frac {\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{a d}+\frac {\left (3 A +2 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a}+\frac {\left (7 A +4 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{a d}+\frac {\left (3 A +2 C \right ) x}{2 a}+\frac {\left (A +6 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 a d}-\frac {\left (3 A +2 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{a}-\frac {\left (3 A +2 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{2 a}-\frac {\left (4 A +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {\left (13 A +12 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}\) | \(237\) |
Input:
int(cos(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x,method=_RETURNVERBO SE)
Output:
1/12*((-A*cos(2*d*x+2*c)+A*cos(3*d*x+3*c)+(17*A+12*C)*cos(d*x+c)+31*A+24*C )*tan(1/2*d*x+1/2*c)-18*d*(A+2/3*C)*x)/d/a
Time = 0.08 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.78 \[ \int \frac {\cos ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=-\frac {3 \, {\left (3 \, A + 2 \, C\right )} d x \cos \left (d x + c\right ) + 3 \, {\left (3 \, A + 2 \, C\right )} d x - {\left (2 \, A \cos \left (d x + c\right )^{3} - A \cos \left (d x + c\right )^{2} + {\left (7 \, A + 6 \, C\right )} \cos \left (d x + c\right ) + 16 \, A + 12 \, C\right )} \sin \left (d x + c\right )}{6 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \] Input:
integrate(cos(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x, algorithm="f ricas")
Output:
-1/6*(3*(3*A + 2*C)*d*x*cos(d*x + c) + 3*(3*A + 2*C)*d*x - (2*A*cos(d*x + c)^3 - A*cos(d*x + c)^2 + (7*A + 6*C)*cos(d*x + c) + 16*A + 12*C)*sin(d*x + c))/(a*d*cos(d*x + c) + a*d)
\[ \int \frac {\cos ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {\int \frac {A \cos ^{3}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \cos ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate(cos(d*x+c)**3*(A+C*sec(d*x+c)**2)/(a+a*sec(d*x+c)),x)
Output:
(Integral(A*cos(c + d*x)**3/(sec(c + d*x) + 1), x) + Integral(C*cos(c + d* x)**3*sec(c + d*x)**2/(sec(c + d*x) + 1), x))/a
Leaf count of result is larger than twice the leaf count of optimal. 269 vs. \(2 (118) = 236\).
Time = 0.12 (sec) , antiderivative size = 269, normalized size of antiderivative = 2.17 \[ \int \frac {\cos ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {A {\left (\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {16 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a + \frac {3 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} - \frac {9 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac {3 \, \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - 3 \, C {\left (\frac {2 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {2 \, \sin \left (d x + c\right )}{{\left (a + \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{3 \, d} \] Input:
integrate(cos(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x, algorithm="m axima")
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*C*(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 = 0.31 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.23 \[ \int \frac {\cos ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=-\frac {\frac {3 \, {\left (d x + c\right )} {\left (3 \, A + 2 \, C\right )}}{a} - \frac {6 \, {\left (A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a} - \frac {2 \, {\left (15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 16 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} a}}{6 \, d} \] Input:
integrate(cos(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x, algorithm="g iac")
Output:
-1/6*(3*(d*x + c)*(3*A + 2*C)/a - 6*(A*tan(1/2*d*x + 1/2*c) + C*tan(1/2*d* x + 1/2*c))/a - 2*(15*A*tan(1/2*d*x + 1/2*c)^5 + 6*C*tan(1/2*d*x + 1/2*c)^ 5 + 16*A*tan(1/2*d*x + 1/2*c)^3 + 12*C*tan(1/2*d*x + 1/2*c)^3 + 9*A*tan(1/ 2*d*x + 1/2*c) + 6*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^3 *a))/d
Time = 12.72 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.92 \[ \int \frac {\cos ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {7\,A\,\sin \left (c+d\,x\right )}{4\,a\,d}-\frac {C\,x}{a}-\frac {3\,A\,x}{2\,a}+\frac {C\,\sin \left (c+d\,x\right )}{a\,d}-\frac {A\,\sin \left (2\,c+2\,d\,x\right )}{4\,a\,d}+\frac {A\,\sin \left (3\,c+3\,d\,x\right )}{12\,a\,d}+\frac {A\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a\,d}+\frac {C\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a\,d} \] Input:
int((cos(c + d*x)^3*(A + C/cos(c + d*x)^2))/(a + a/cos(c + d*x)),x)
Output:
(7*A*sin(c + d*x))/(4*a*d) - (C*x)/a - (3*A*x)/(2*a) + (C*sin(c + d*x))/(a *d) - (A*sin(2*c + 2*d*x))/(4*a*d) + (A*sin(3*c + 3*d*x))/(12*a*d) + (A*ta n(c/2 + (d*x)/2))/(a*d) + (C*tan(c/2 + (d*x)/2))/(a*d)
Time = 0.16 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.91 \[ \int \frac {\cos ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {-3 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a -6 \cos \left (d x +c \right ) a -6 \cos \left (d x +c \right ) c -2 \sin \left (d x +c \right )^{4} a +12 \sin \left (d x +c \right )^{2} a +6 \sin \left (d x +c \right )^{2} c -9 \sin \left (d x +c \right ) a d x -6 \sin \left (d x +c \right ) c d x +6 a +6 c}{6 \sin \left (d x +c \right ) a d} \] Input:
int(cos(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x)
Output:
( - 3*cos(c + d*x)*sin(c + d*x)**2*a - 6*cos(c + d*x)*a - 6*cos(c + d*x)*c - 2*sin(c + d*x)**4*a + 12*sin(c + d*x)**2*a + 6*sin(c + d*x)**2*c - 9*si n(c + d*x)*a*d*x - 6*sin(c + d*x)*c*d*x + 6*a + 6*c)/(6*sin(c + d*x)*a*d)