Integrand size = 21, antiderivative size = 147 \[ \int \frac {\cos ^2(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {13 x}{2 a^3}-\frac {152 \sin (c+d x)}{15 a^3 d}+\frac {13 \cos (c+d x) \sin (c+d x)}{2 a^3 d}-\frac {\cos (c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {11 \cos (c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {76 \cos (c+d x) \sin (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )} \] Output:
13/2*x/a^3-152/15*sin(d*x+c)/a^3/d+13/2*cos(d*x+c)*sin(d*x+c)/a^3/d-1/5*co s(d*x+c)*sin(d*x+c)/d/(a+a*sec(d*x+c))^3-11/15*cos(d*x+c)*sin(d*x+c)/a/d/( a+a*sec(d*x+c))^2-76/15*cos(d*x+c)*sin(d*x+c)/d/(a^3+a^3*sec(d*x+c))
Time = 1.72 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.23 \[ \int \frac {\cos ^2(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {2 \cos \left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (-3 \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+46 \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )-508 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+15 \cos ^5\left (\frac {1}{2} (c+d x)\right ) (26 d x-12 \sin (c+d x)+\sin (2 (c+d x)))-3 \cos \left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {c}{2}\right )+46 \cos ^3\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {c}{2}\right )\right )}{15 a^3 d (1+\sec (c+d x))^3} \] Input:
Integrate[Cos[c + d*x]^2/(a + a*Sec[c + d*x])^3,x]
Output:
(2*Cos[(c + d*x)/2]*Sec[c + d*x]^3*(-3*Sec[c/2]*Sin[(d*x)/2] + 46*Cos[(c + d*x)/2]^2*Sec[c/2]*Sin[(d*x)/2] - 508*Cos[(c + d*x)/2]^4*Sec[c/2]*Sin[(d* x)/2] + 15*Cos[(c + d*x)/2]^5*(26*d*x - 12*Sin[c + d*x] + Sin[2*(c + d*x)] ) - 3*Cos[(c + d*x)/2]*Tan[c/2] + 46*Cos[(c + d*x)/2]^3*Tan[c/2]))/(15*a^3 *d*(1 + Sec[c + d*x])^3)
Time = 0.96 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.10, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.619, Rules used = {3042, 4304, 25, 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.
| \(\displaystyle \int \frac {\cos ^2(c+d x)}{(a \sec (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 4304 |
| \(\displaystyle -\frac {\int -\frac {\cos ^2(c+d x) (7 a-4 a \sec (c+d x))}{(\sec (c+d x) a+a)^2}dx}{5 a^2}-\frac {\sin (c+d x) \cos (c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {\cos ^2(c+d x) (7 a-4 a \sec (c+d x))}{(\sec (c+d x) a+a)^2}dx}{5 a^2}-\frac {\sin (c+d x) \cos (c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {7 a-4 a \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {\sin (c+d x) \cos (c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4508 |
| \(\displaystyle \frac {\frac {\int \frac {\cos ^2(c+d x) \left (43 a^2-33 a^2 \sec (c+d x)\right )}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {11 a \sin (c+d x) \cos (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {\sin (c+d x) \cos (c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {43 a^2-33 a^2 \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}-\frac {11 a \sin (c+d x) \cos (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {\sin (c+d x) \cos (c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4508 |
| \(\displaystyle \frac {\frac {\frac {\int \cos ^2(c+d x) \left (195 a^3-152 a^3 \sec (c+d x)\right )dx}{a^2}-\frac {76 a^2 \sin (c+d x) \cos (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {11 a \sin (c+d x) \cos (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {\sin (c+d x) \cos (c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {195 a^3-152 a^3 \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx}{a^2}-\frac {76 a^2 \sin (c+d x) \cos (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {11 a \sin (c+d x) \cos (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {\sin (c+d x) \cos (c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {\frac {\frac {195 a^3 \int \cos ^2(c+d x)dx-152 a^3 \int \cos (c+d x)dx}{a^2}-\frac {76 a^2 \sin (c+d x) \cos (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {11 a \sin (c+d x) \cos (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {\sin (c+d x) \cos (c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {195 a^3 \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-152 a^3 \int \sin \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {76 a^2 \sin (c+d x) \cos (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {11 a \sin (c+d x) \cos (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {\sin (c+d x) \cos (c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {\frac {\frac {195 a^3 \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-152 a^3 \int \sin \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {76 a^2 \sin (c+d x) \cos (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {11 a \sin (c+d x) \cos (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {\sin (c+d x) \cos (c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\frac {\frac {195 a^3 \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-152 a^3 \int \sin \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {76 a^2 \sin (c+d x) \cos (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {11 a \sin (c+d x) \cos (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {\sin (c+d x) \cos (c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3117 |
| \(\displaystyle \frac {\frac {\frac {195 a^3 \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-\frac {152 a^3 \sin (c+d x)}{d}}{a^2}-\frac {76 a^2 \sin (c+d x) \cos (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {11 a \sin (c+d x) \cos (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {\sin (c+d x) \cos (c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
Input:
Int[Cos[c + d*x]^2/(a + a*Sec[c + d*x])^3,x]
Output:
-1/5*(Cos[c + d*x]*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^3) + ((-11*a*Cos[ c + d*x]*Sin[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2) + ((-76*a^2*Cos[c + d* x]*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])) + ((-152*a^3*Sin[c + d*x])/d + 1 95*a^3*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)))/a^2)/(3*a^2))/(5*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_), x_Symbol] :> Simp[(-Cot[e + f*x])*(a + b*Csc[e + f*x])^m*((d*Csc [e + f*x])^n/(f*(2*m + 1))), x] + Simp[1/(a^2*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*(a*(2*m + n + 1) - b*(m + n + 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && LtQ [m, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m])
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 = 0.58 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.52
| method | result | size |
| parallelrisch | \(\frac {-\frac {1001 \left (\cos \left (d x +c \right )+\frac {928 \cos \left (2 d x +2 c \right )}{3003}+\frac {15 \cos \left (3 d x +3 c \right )}{1001}-\frac {5 \cos \left (4 d x +4 c \right )}{2002}+\frac {331}{462}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{160}+\frac {13 d x}{2}}{d \,a^{3}}\) | \(77\) |
| derivativedivides | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}+\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {-28 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-20 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+52 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\) | \(101\) |
| default | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}+\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {-28 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-20 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+52 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\) | \(101\) |
| risch | \(\frac {13 x}{2 a^{3}}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{8 d \,a^{3}}+\frac {3 i {\mathrm e}^{i \left (d x +c \right )}}{2 d \,a^{3}}-\frac {3 i {\mathrm e}^{-i \left (d x +c \right )}}{2 d \,a^{3}}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d \,a^{3}}-\frac {2 i \left (150 \,{\mathrm e}^{4 i \left (d x +c \right )}+525 \,{\mathrm e}^{3 i \left (d x +c \right )}+745 \,{\mathrm e}^{2 i \left (d x +c \right )}+485 \,{\mathrm e}^{i \left (d x +c \right )}+127\right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\) | \(148\) |
| norman | \(\frac {\frac {13 x}{2 a}-\frac {51 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}-\frac {131 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 a d}-\frac {97 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 a d}+\frac {17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{30 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{20 a d}+\frac {13 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a}+\frac {13 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2 a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} a^{2}}\) | \(154\) |
Input:
int(cos(d*x+c)^2/(a+a*sec(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
13/160*(-77*(cos(d*x+c)+928/3003*cos(2*d*x+2*c)+15/1001*cos(3*d*x+3*c)-5/2 002*cos(4*d*x+4*c)+331/462)*tan(1/2*d*x+1/2*c)*sec(1/2*d*x+1/2*c)^4+80*d*x )/d/a^3
Time = 0.10 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.92 \[ \int \frac {\cos ^2(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {195 \, d x \cos \left (d x + c\right )^{3} + 585 \, d x \cos \left (d x + c\right )^{2} + 585 \, d x \cos \left (d x + c\right ) + 195 \, d x + {\left (15 \, \cos \left (d x + c\right )^{4} - 45 \, \cos \left (d x + c\right )^{3} - 479 \, \cos \left (d x + c\right )^{2} - 717 \, \cos \left (d x + c\right ) - 304\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \] Input:
integrate(cos(d*x+c)^2/(a+a*sec(d*x+c))^3,x, algorithm="fricas")
Output:
1/30*(195*d*x*cos(d*x + c)^3 + 585*d*x*cos(d*x + c)^2 + 585*d*x*cos(d*x + c) + 195*d*x + (15*cos(d*x + c)^4 - 45*cos(d*x + c)^3 - 479*cos(d*x + c)^2 - 717*cos(d*x + c) - 304)*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*c os(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)
\[ \int \frac {\cos ^2(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {\cos ^{2}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \] Input:
integrate(cos(d*x+c)**2/(a+a*sec(d*x+c))**3,x)
Output:
Integral(cos(c + d*x)**2/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x)/a**3
Time = 0.11 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.25 \[ \int \frac {\cos ^2(c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {\frac {60 \, {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{3} + \frac {2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {465 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {40 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {780 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{60 \, d} \] Input:
integrate(cos(d*x+c)^2/(a+a*sec(d*x+c))^3,x, algorithm="maxima")
Output:
-1/60*(60*(5*sin(d*x + c)/(cos(d*x + c) + 1) + 7*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^3 + 2*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (465*sin(d*x + c)/(cos(d*x + c) + 1) - 40* sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^ 5)/a^3 - 780*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3)/d
Time = 0.16 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.77 \[ \int \frac {\cos ^2(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\frac {390 \, {\left (d x + c\right )}}{a^{3}} - \frac {60 \, {\left (7 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{3}} - \frac {3 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 465 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \] Input:
integrate(cos(d*x+c)^2/(a+a*sec(d*x+c))^3,x, algorithm="giac")
Output:
1/60*(390*(d*x + c)/a^3 - 60*(7*tan(1/2*d*x + 1/2*c)^3 + 5*tan(1/2*d*x + 1 /2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*a^3) - (3*a^12*tan(1/2*d*x + 1/2*c) ^5 - 40*a^12*tan(1/2*d*x + 1/2*c)^3 + 465*a^12*tan(1/2*d*x + 1/2*c))/a^15) /d
Time = 9.79 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.93 \[ \int \frac {\cos ^2(c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-46\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+508\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+420\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-120\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-390\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (c+d\,x\right )}{60\,a^3\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5} \] Input:
int(cos(c + d*x)^2/(a + a/cos(c + d*x))^3,x)
Output:
-(3*sin(c/2 + (d*x)/2) - 46*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2) + 508* cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2) + 420*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2) - 120*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2) - 390*cos(c/2 + (d*x)/2)^5*(c + d*x))/(60*a^3*d*cos(c/2 + (d*x)/2)^5)
Time = 0.17 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.91 \[ \int \frac {\cos ^2(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}+34 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}-388 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+390 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} d x -1310 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+780 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} d x -765 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+390 d x}{60 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )} \] Input:
int(cos(d*x+c)^2/(a+a*sec(d*x+c))^3,x)
Output:
( - 3*tan((c + d*x)/2)**9 + 34*tan((c + d*x)/2)**7 - 388*tan((c + d*x)/2)* *5 + 390*tan((c + d*x)/2)**4*d*x - 1310*tan((c + d*x)/2)**3 + 780*tan((c + d*x)/2)**2*d*x - 765*tan((c + d*x)/2) + 390*d*x)/(60*a**3*d*(tan((c + d*x )/2)**4 + 2*tan((c + d*x)/2)**2 + 1))