Integrand size = 40, antiderivative size = 143 \[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {(7 B-4 C) x}{2 a^2}-\frac {2 (8 B-5 C) \sin (c+d x)}{3 a^2 d}+\frac {(7 B-4 C) \cos (c+d x) \sin (c+d x)}{2 a^2 d}-\frac {(8 B-5 C) \cos (c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(B-C) \cos (c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2} \] Output:
1/2*(7*B-4*C)*x/a^2-2/3*(8*B-5*C)*sin(d*x+c)/a^2/d+1/2*(7*B-4*C)*cos(d*x+c )*sin(d*x+c)/a^2/d-1/3*(8*B-5*C)*cos(d*x+c)*sin(d*x+c)/a^2/d/(1+sec(d*x+c) )-1/3*(B-C)*cos(d*x+c)*sin(d*x+c)/d/(a+a*sec(d*x+c))^2
Leaf count is larger than twice the leaf count of optimal. \(315\) vs. \(2(143)=286\).
Time = 1.09 (sec) , antiderivative size = 315, normalized size of antiderivative = 2.20 \[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (36 (7 B-4 C) d x \cos \left (\frac {d x}{2}\right )+36 (7 B-4 C) d x \cos \left (c+\frac {d x}{2}\right )+84 B d x \cos \left (c+\frac {3 d x}{2}\right )-48 C d x \cos \left (c+\frac {3 d x}{2}\right )+84 B d x \cos \left (2 c+\frac {3 d x}{2}\right )-48 C d x \cos \left (2 c+\frac {3 d x}{2}\right )-381 B \sin \left (\frac {d x}{2}\right )+264 C \sin \left (\frac {d x}{2}\right )+147 B \sin \left (c+\frac {d x}{2}\right )-120 C \sin \left (c+\frac {d x}{2}\right )-239 B \sin \left (c+\frac {3 d x}{2}\right )+164 C \sin \left (c+\frac {3 d x}{2}\right )-63 B \sin \left (2 c+\frac {3 d x}{2}\right )+36 C \sin \left (2 c+\frac {3 d x}{2}\right )-15 B \sin \left (2 c+\frac {5 d x}{2}\right )+12 C \sin \left (2 c+\frac {5 d x}{2}\right )-15 B \sin \left (3 c+\frac {5 d x}{2}\right )+12 C \sin \left (3 c+\frac {5 d x}{2}\right )+3 B \sin \left (3 c+\frac {7 d x}{2}\right )+3 B \sin \left (4 c+\frac {7 d x}{2}\right )\right )}{48 a^2 d (1+\cos (c+d x))^2} \] Input:
Integrate[(Cos[c + d*x]^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[ c + d*x])^2,x]
Output:
(Cos[(c + d*x)/2]*Sec[c/2]*(36*(7*B - 4*C)*d*x*Cos[(d*x)/2] + 36*(7*B - 4* C)*d*x*Cos[c + (d*x)/2] + 84*B*d*x*Cos[c + (3*d*x)/2] - 48*C*d*x*Cos[c + ( 3*d*x)/2] + 84*B*d*x*Cos[2*c + (3*d*x)/2] - 48*C*d*x*Cos[2*c + (3*d*x)/2] - 381*B*Sin[(d*x)/2] + 264*C*Sin[(d*x)/2] + 147*B*Sin[c + (d*x)/2] - 120*C *Sin[c + (d*x)/2] - 239*B*Sin[c + (3*d*x)/2] + 164*C*Sin[c + (3*d*x)/2] - 63*B*Sin[2*c + (3*d*x)/2] + 36*C*Sin[2*c + (3*d*x)/2] - 15*B*Sin[2*c + (5* d*x)/2] + 12*C*Sin[2*c + (5*d*x)/2] - 15*B*Sin[3*c + (5*d*x)/2] + 12*C*Sin [3*c + (5*d*x)/2] + 3*B*Sin[3*c + (7*d*x)/2] + 3*B*Sin[4*c + (7*d*x)/2]))/ (48*a^2*d*(1 + Cos[c + d*x])^2)
Time = 0.97 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.99, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3042, 4560, 3042, 4508, 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.
| \(\displaystyle \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 4560 |
| \(\displaystyle \int \frac {\cos ^2(c+d x) (B+C \sec (c+d x))}{(a \sec (c+d x)+a)^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {B+C \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 4508 |
| \(\displaystyle \frac {\int \frac {\cos ^2(c+d x) (a (5 B-2 C)-3 a (B-C) \sec (c+d x))}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {(B-C) \sin (c+d x) \cos (c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (5 B-2 C)-3 a (B-C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}-\frac {(B-C) \sin (c+d x) \cos (c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4508 |
| \(\displaystyle \frac {\frac {\int \cos ^2(c+d x) \left (3 a^2 (7 B-4 C)-2 a^2 (8 B-5 C) \sec (c+d x)\right )dx}{a^2}-\frac {(8 B-5 C) \sin (c+d x) \cos (c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {(B-C) \sin (c+d x) \cos (c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {3 a^2 (7 B-4 C)-2 a^2 (8 B-5 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx}{a^2}-\frac {(8 B-5 C) \sin (c+d x) \cos (c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {(B-C) \sin (c+d x) \cos (c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {\frac {3 a^2 (7 B-4 C) \int \cos ^2(c+d x)dx-2 a^2 (8 B-5 C) \int \cos (c+d x)dx}{a^2}-\frac {(8 B-5 C) \sin (c+d x) \cos (c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {(B-C) \sin (c+d x) \cos (c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 a^2 (7 B-4 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-2 a^2 (8 B-5 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {(8 B-5 C) \sin (c+d x) \cos (c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {(B-C) \sin (c+d x) \cos (c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {\frac {3 a^2 (7 B-4 C) \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-2 a^2 (8 B-5 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {(8 B-5 C) \sin (c+d x) \cos (c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {(B-C) \sin (c+d x) \cos (c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\frac {3 a^2 (7 B-4 C) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-2 a^2 (8 B-5 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {(8 B-5 C) \sin (c+d x) \cos (c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {(B-C) \sin (c+d x) \cos (c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3117 |
| \(\displaystyle \frac {\frac {3 a^2 (7 B-4 C) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-\frac {2 a^2 (8 B-5 C) \sin (c+d x)}{d}}{a^2}-\frac {(8 B-5 C) \sin (c+d x) \cos (c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {(B-C) \sin (c+d x) \cos (c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
Input:
Int[(Cos[c + d*x]^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d* x])^2,x]
Output:
-1/3*((B - C)*Cos[c + d*x]*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^2) + (-(( (8*B - 5*C)*Cos[c + d*x]*Sin[c + d*x])/(d*(1 + Sec[c + d*x]))) + ((-2*a^2* (8*B - 5*C)*Sin[c + d*x])/d + 3*a^2*(7*B - 4*C)*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)))/a^2)/(3*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]
Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_. )*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.) *(x_)]*(d_.))^(n_.), x_Symbol] :> Simp[1/b^2 Int[(a + b*Csc[e + f*x])^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Time = 0.40 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.62
| method | result | size |
| parallelrisch | \(\frac {-163 \left (\frac {12 \left (B -C \right ) \cos \left (2 d x +2 c \right )}{163}-\frac {3 B \cos \left (3 d x +3 c \right )}{163}+\left (B -\frac {112 C}{163}\right ) \cos \left (d x +c \right )+\frac {140 B}{163}-\frac {92 C}{163}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+168 d \left (B -\frac {4 C}{7}\right ) x}{48 a^{2} d}\) | \(88\) |
| derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +\frac {4 \left (-\frac {5 B}{2}+C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+4 \left (-\frac {3 B}{2}+C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+2 \left (7 B -4 C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\) | \(131\) |
| default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +\frac {4 \left (-\frac {5 B}{2}+C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+4 \left (-\frac {3 B}{2}+C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+2 \left (7 B -4 C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\) | \(131\) |
| risch | \(\frac {7 B x}{2 a^{2}}-\frac {2 x C}{a^{2}}-\frac {i B \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 a^{2} d}+\frac {i B \,{\mathrm e}^{i \left (d x +c \right )}}{a^{2} d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} C}{2 a^{2} d}-\frac {i B \,{\mathrm e}^{-i \left (d x +c \right )}}{a^{2} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} C}{2 a^{2} d}+\frac {i B \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 a^{2} d}-\frac {2 i \left (12 B \,{\mathrm e}^{2 i \left (d x +c \right )}-9 C \,{\mathrm e}^{2 i \left (d x +c \right )}+21 B \,{\mathrm e}^{i \left (d x +c \right )}-15 C \,{\mathrm e}^{i \left (d x +c \right )}+11 B -8 C \right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}\) | \(207\) |
| norman | \(\frac {\frac {\left (7 B -4 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{a}-\frac {\left (7 B -4 C \right ) x}{2 a}+\frac {\left (B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{6 a d}-\frac {\left (7 B -4 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a}+\frac {\left (7 B -4 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{2 a}-\frac {\left (10 B -7 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 a d}-\frac {\left (12 B -7 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{a d}+\frac {\left (13 B -9 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}-\frac {\left (19 B -13 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{6 a d}+\frac {\left (71 B -41 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) a}\) | \(272\) |
Input:
int(cos(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x,method =_RETURNVERBOSE)
Output:
1/48*(-163*(12/163*(B-C)*cos(2*d*x+2*c)-3/163*B*cos(3*d*x+3*c)+(B-112/163* C)*cos(d*x+c)+140/163*B-92/163*C)*tan(1/2*d*x+1/2*c)*sec(1/2*d*x+1/2*c)^2+ 168*d*(B-4/7*C)*x)/a^2/d
Time = 0.08 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.97 \[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {3 \, {\left (7 \, B - 4 \, C\right )} d x \cos \left (d x + c\right )^{2} + 6 \, {\left (7 \, B - 4 \, C\right )} d x \cos \left (d x + c\right ) + 3 \, {\left (7 \, B - 4 \, C\right )} d x + {\left (3 \, B \cos \left (d x + c\right )^{3} - 6 \, {\left (B - C\right )} \cos \left (d x + c\right )^{2} - {\left (43 \, B - 28 \, C\right )} \cos \left (d x + c\right ) - 32 \, B + 20 \, C\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \] Input:
integrate(cos(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x, algorithm="fricas")
Output:
1/6*(3*(7*B - 4*C)*d*x*cos(d*x + c)^2 + 6*(7*B - 4*C)*d*x*cos(d*x + c) + 3 *(7*B - 4*C)*d*x + (3*B*cos(d*x + c)^3 - 6*(B - C)*cos(d*x + c)^2 - (43*B - 28*C)*cos(d*x + c) - 32*B + 20*C)*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)
\[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {B \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \cos ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \] Input:
integrate(cos(d*x+c)**3*(B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**2 ,x)
Output:
(Integral(B*cos(c + d*x)**3*sec(c + d*x)/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x) + Integral(C*cos(c + d*x)**3*sec(c + d*x)**2/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x))/a**2
Leaf count of result is larger than twice the leaf count of optimal. 283 vs. \(2 (133) = 266\).
Time = 0.12 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.98 \[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=-\frac {B {\left (\frac {6 \, {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} + \frac {2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {42 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )} - C {\left (\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {24 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {12 \, \sin \left (d x + c\right )}{{\left (a^{2} + \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{6 \, d} \] Input:
integrate(cos(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x, algorithm="maxima")
Output:
-1/6*(B*(6*(3*sin(d*x + c)/(cos(d*x + c) + 1) + 5*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^2 + 2*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a^2*sin(d* x + c)^4/(cos(d*x + c) + 1)^4) + (21*sin(d*x + c)/(cos(d*x + c) + 1) - sin (d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 42*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2) - C*((15*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(c os(d*x + c) + 1)^3)/a^2 - 24*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 + 12*sin(d*x + c)/((a^2 + a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1))))/d
Time = 0.28 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.15 \[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {\frac {3 \, {\left (d x + c\right )} {\left (7 \, B - 4 \, C\right )}}{a^{2}} - \frac {6 \, {\left (5 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{2}} + \frac {B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 21 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \] Input:
integrate(cos(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x, algorithm="giac")
Output:
1/6*(3*(d*x + c)*(7*B - 4*C)/a^2 - 6*(5*B*tan(1/2*d*x + 1/2*c)^3 - 2*C*tan (1/2*d*x + 1/2*c)^3 + 3*B*tan(1/2*d*x + 1/2*c) - 2*C*tan(1/2*d*x + 1/2*c)) /((tan(1/2*d*x + 1/2*c)^2 + 1)^2*a^2) + (B*a^4*tan(1/2*d*x + 1/2*c)^3 - C* a^4*tan(1/2*d*x + 1/2*c)^3 - 21*B*a^4*tan(1/2*d*x + 1/2*c) + 15*C*a^4*tan( 1/2*d*x + 1/2*c))/a^6)/d
Time = 12.09 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.08 \[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {x\,\left (7\,B-4\,C\right )}{2\,a^2}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (B-C\right )}{2\,a^2}+\frac {4\,B-2\,C}{2\,a^2}\right )}{d}-\frac {\left (5\,B-2\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (3\,B-2\,C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (B-C\right )}{6\,a^2\,d} \] Input:
int((cos(c + d*x)^3*(B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + a/cos(c + d* x))^2,x)
Output:
(x*(7*B - 4*C))/(2*a^2) - (tan(c/2 + (d*x)/2)*((3*(B - C))/(2*a^2) + (4*B - 2*C)/(2*a^2)))/d - (tan(c/2 + (d*x)/2)^3*(5*B - 2*C) + tan(c/2 + (d*x)/2 )*(3*B - 2*C))/(d*(2*a^2*tan(c/2 + (d*x)/2)^2 + a^2*tan(c/2 + (d*x)/2)^4 + a^2)) + (tan(c/2 + (d*x)/2)^3*(B - C))/(6*a^2*d)
Time = 0.16 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.22 \[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {-9 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b +6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} c +21 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b d x -12 \cos \left (d x +c \right ) \sin \left (d x +c \right ) c d x -2 \cos \left (d x +c \right ) b +2 \cos \left (d x +c \right ) c -3 \sin \left (d x +c \right )^{4} b -31 \sin \left (d x +c \right )^{2} b +22 \sin \left (d x +c \right )^{2} c +21 \sin \left (d x +c \right ) b d x -12 \sin \left (d x +c \right ) c d x +2 b -2 c}{6 \sin \left (d x +c \right ) a^{2} d \left (\cos \left (d x +c \right )+1\right )} \] Input:
int(cos(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x)
Output:
( - 9*cos(c + d*x)*sin(c + d*x)**2*b + 6*cos(c + d*x)*sin(c + d*x)**2*c + 21*cos(c + d*x)*sin(c + d*x)*b*d*x - 12*cos(c + d*x)*sin(c + d*x)*c*d*x - 2*cos(c + d*x)*b + 2*cos(c + d*x)*c - 3*sin(c + d*x)**4*b - 31*sin(c + d*x )**2*b + 22*sin(c + d*x)**2*c + 21*sin(c + d*x)*b*d*x - 12*sin(c + d*x)*c* d*x + 2*b - 2*c)/(6*sin(c + d*x)*a**2*d*(cos(c + d*x) + 1))