Integrand size = 31, antiderivative size = 114 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\frac {C x}{a^3}-\frac {(A+C) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(3 A-7 C) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {(6 A-29 C) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )} \] Output:
C*x/a^3-1/5*(A+C)*cos(d*x+c)^2*sin(d*x+c)/d/(a+a*cos(d*x+c))^3-1/15*(3*A-7 *C)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^2+1/15*(6*A-29*C)*sin(d*x+c)/d/(a^3+a^ 3*cos(d*x+c))
Time = 1.94 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.99 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\frac {\sec \left (\frac {c}{2}\right ) \sec ^5\left (\frac {1}{2} (c+d x)\right ) \left (150 C d x \cos \left (\frac {d x}{2}\right )+150 C d x \cos \left (c+\frac {d x}{2}\right )+75 C d x \cos \left (c+\frac {3 d x}{2}\right )+75 C d x \cos \left (2 c+\frac {3 d x}{2}\right )+15 C d x \cos \left (2 c+\frac {5 d x}{2}\right )+15 C d x \cos \left (3 c+\frac {5 d x}{2}\right )+30 A \sin \left (\frac {d x}{2}\right )-370 C \sin \left (\frac {d x}{2}\right )-30 A \sin \left (c+\frac {d x}{2}\right )+270 C \sin \left (c+\frac {d x}{2}\right )+30 A \sin \left (c+\frac {3 d x}{2}\right )-230 C \sin \left (c+\frac {3 d x}{2}\right )+90 C \sin \left (2 c+\frac {3 d x}{2}\right )+6 A \sin \left (2 c+\frac {5 d x}{2}\right )-64 C \sin \left (2 c+\frac {5 d x}{2}\right )\right )}{480 a^3 d} \] Input:
Integrate[(Cos[c + d*x]*(A + C*Cos[c + d*x]^2))/(a + a*Cos[c + d*x])^3,x]
Output:
(Sec[c/2]*Sec[(c + d*x)/2]^5*(150*C*d*x*Cos[(d*x)/2] + 150*C*d*x*Cos[c + ( d*x)/2] + 75*C*d*x*Cos[c + (3*d*x)/2] + 75*C*d*x*Cos[2*c + (3*d*x)/2] + 15 *C*d*x*Cos[2*c + (5*d*x)/2] + 15*C*d*x*Cos[3*c + (5*d*x)/2] + 30*A*Sin[(d* x)/2] - 370*C*Sin[(d*x)/2] - 30*A*Sin[c + (d*x)/2] + 270*C*Sin[c + (d*x)/2 ] + 30*A*Sin[c + (3*d*x)/2] - 230*C*Sin[c + (3*d*x)/2] + 90*C*Sin[2*c + (3 *d*x)/2] + 6*A*Sin[2*c + (5*d*x)/2] - 64*C*Sin[2*c + (5*d*x)/2]))/(480*a^3 *d)
Time = 0.78 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.08, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {3042, 3521, 3042, 3447, 3042, 3498, 25, 3042, 3214, 3042, 3127}
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.
| \(\displaystyle \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 3521 |
| \(\displaystyle \frac {\int \frac {\cos (c+d x) (a (3 A-2 C)+5 a C \cos (c+d x))}{(\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A+C) \sin (c+d x) \cos ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a (3 A-2 C)+5 a C \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {(A+C) \sin (c+d x) \cos ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {\int \frac {5 a C \cos ^2(c+d x)+a (3 A-2 C) \cos (c+d x)}{(\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A+C) \sin (c+d x) \cos ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {5 a C \sin \left (c+d x+\frac {\pi }{2}\right )^2+a (3 A-2 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {(A+C) \sin (c+d x) \cos ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3498 |
| \(\displaystyle \frac {-\frac {\int -\frac {2 (3 A-7 C) a^2+15 C \cos (c+d x) a^2}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {a (3 A-7 C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \sin (c+d x) \cos ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {2 (3 A-7 C) a^2+15 C \cos (c+d x) a^2}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {a (3 A-7 C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \sin (c+d x) \cos ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {2 (3 A-7 C) a^2+15 C \sin \left (c+d x+\frac {\pi }{2}\right ) a^2}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {a (3 A-7 C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \sin (c+d x) \cos ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {\frac {a^2 (6 A-29 C) \int \frac {1}{\cos (c+d x) a+a}dx+15 a C x}{3 a^2}-\frac {a (3 A-7 C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \sin (c+d x) \cos ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a^2 (6 A-29 C) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+15 a C x}{3 a^2}-\frac {a (3 A-7 C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \sin (c+d x) \cos ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3127 |
| \(\displaystyle \frac {\frac {\frac {a^2 (6 A-29 C) \sin (c+d x)}{d (a \cos (c+d x)+a)}+15 a C x}{3 a^2}-\frac {a (3 A-7 C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \sin (c+d x) \cos ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
Input:
Int[(Cos[c + d*x]*(A + C*Cos[c + d*x]^2))/(a + a*Cos[c + d*x])^3,x]
Output:
-1/5*((A + C)*Cos[c + d*x]^2*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^3) + (- 1/3*(a*(3*A - 7*C)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^2) + (15*a*C*x + (a^2*(6*A - 29*C)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])))/(3*a^2))/(5*a^2)
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b ^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*b - a* B + b*C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(a*f*(2*m + 1))), x] + Simp[1 /(a^2*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*A*(m + 1) + m*(b *B - a*C) + b*C*(2*m + 1)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && EqQ[a^2 - b^2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[a*(A + C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(f*(b*c - a*d)*(2*m + 1))), x] + Simp[1/(b*(b*c - a*d)*(2*m + 1)) I nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(m + n + 2) + C*(b* c*(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)]
Time = 0.21 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.57
| method | result | size |
| parallelrisch | \(\frac {\left (-3 A -3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+20 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C +\left (15 A -105 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+60 d x C}{60 a^{3} d}\) | \(65\) |
| derivativedivides | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}+\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-7 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+8 C \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\) | \(88\) |
| default | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}+\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-7 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+8 C \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\) | \(88\) |
| risch | \(\frac {C x}{a^{3}}-\frac {2 i \left (45 C \,{\mathrm e}^{4 i \left (d x +c \right )}-15 A \,{\mathrm e}^{3 i \left (d x +c \right )}+135 C \,{\mathrm e}^{3 i \left (d x +c \right )}-15 A \,{\mathrm e}^{2 i \left (d x +c \right )}+185 C \,{\mathrm e}^{2 i \left (d x +c \right )}-15 A \,{\mathrm e}^{i \left (d x +c \right )}+115 C \,{\mathrm e}^{i \left (d x +c \right )}-3 A +32 C \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\) | \(121\) |
| norman | \(\frac {\frac {C x}{a}+\frac {C x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{a}+\frac {3 C x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a}+\frac {3 C x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{a}+\frac {\left (A -9 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{10 a d}+\frac {\left (A -7 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}-\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{20 a d}+\frac {\left (7 A -43 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{10 a d}+\frac {\left (9 A -59 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 a d}-\frac {\left (9 A -11 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{60 a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} a^{2}}\) | \(226\) |
Input:
int(cos(d*x+c)*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x,method=_RETURNVERBO SE)
Output:
1/60*((-3*A-3*C)*tan(1/2*d*x+1/2*c)^5+20*tan(1/2*d*x+1/2*c)^3*C+(15*A-105* C)*tan(1/2*d*x+1/2*c)+60*d*x*C)/a^3/d
Time = 0.09 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.20 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\frac {15 \, C d x \cos \left (d x + c\right )^{3} + 45 \, C d x \cos \left (d x + c\right )^{2} + 45 \, C d x \cos \left (d x + c\right ) + 15 \, C d x + {\left ({\left (3 \, A - 32 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, A - 17 \, C\right )} \cos \left (d x + c\right ) + 3 \, A - 22 \, C\right )} \sin \left (d x + c\right )}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \] Input:
integrate(cos(d*x+c)*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x, algorithm="f ricas")
Output:
1/15*(15*C*d*x*cos(d*x + c)^3 + 45*C*d*x*cos(d*x + c)^2 + 45*C*d*x*cos(d*x + c) + 15*C*d*x + ((3*A - 32*C)*cos(d*x + c)^2 + 3*(3*A - 17*C)*cos(d*x + c) + 3*A - 22*C)*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)
Time = 1.38 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.12 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\begin {cases} - \frac {A \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{20 a^{3} d} + \frac {A \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{4 a^{3} d} + \frac {C x}{a^{3}} - \frac {C \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{20 a^{3} d} + \frac {C \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a^{3} d} - \frac {7 C \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{4 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \left (A + C \cos ^{2}{\left (c \right )}\right ) \cos {\left (c \right )}}{\left (a \cos {\left (c \right )} + a\right )^{3}} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)*(A+C*cos(d*x+c)**2)/(a+a*cos(d*x+c))**3,x)
Output:
Piecewise((-A*tan(c/2 + d*x/2)**5/(20*a**3*d) + A*tan(c/2 + d*x/2)/(4*a**3 *d) + C*x/a**3 - C*tan(c/2 + d*x/2)**5/(20*a**3*d) + C*tan(c/2 + d*x/2)**3 /(3*a**3*d) - 7*C*tan(c/2 + d*x/2)/(4*a**3*d), Ne(d, 0)), (x*(A + C*cos(c) **2)*cos(c)/(a*cos(c) + a)**3, True))
Time = 0.15 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.23 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=-\frac {C {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {120 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} - \frac {3 \, A {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \] Input:
integrate(cos(d*x+c)*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x, algorithm="m axima")
Output:
-1/60*(C*((105*sin(d*x + c)/(cos(d*x + c) + 1) - 20*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 - 120*arctan(si n(d*x + c)/(cos(d*x + c) + 1))/a^3) - 3*A*(5*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3)/d
Time = 0.33 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.91 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\frac {\frac {60 \, {\left (d x + c\right )} C}{a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 20 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 105 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \] Input:
integrate(cos(d*x+c)*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x, algorithm="g iac")
Output:
1/60*(60*(d*x + c)*C/a^3 - (3*A*a^12*tan(1/2*d*x + 1/2*c)^5 + 3*C*a^12*tan (1/2*d*x + 1/2*c)^5 - 20*C*a^12*tan(1/2*d*x + 1/2*c)^3 - 15*A*a^12*tan(1/2 *d*x + 1/2*c) + 105*C*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d
Time = 0.23 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.02 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\frac {C\,x}{a^3}+\frac {{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}-\frac {7\,C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}\right )-\frac {A\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{20}-\frac {C\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{20}+\frac {C\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}}{a^3\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5} \] Input:
int((cos(c + d*x)*(A + C*cos(c + d*x)^2))/(a + a*cos(c + d*x))^3,x)
Output:
(C*x)/a^3 + (cos(c/2 + (d*x)/2)^4*((A*sin(c/2 + (d*x)/2))/4 - (7*C*sin(c/2 + (d*x)/2))/4) - (A*sin(c/2 + (d*x)/2)^5)/20 - (C*sin(c/2 + (d*x)/2)^5)/2 0 + (C*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^3)/3)/(a^3*d*cos(c/2 + (d*x )/2)^5)
Time = 0.16 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.70 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\frac {-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} a -3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} c +20 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} c +15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -105 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) c +60 c d x}{60 a^{3} d} \] Input:
int(cos(d*x+c)*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x)
Output:
( - 3*tan((c + d*x)/2)**5*a - 3*tan((c + d*x)/2)**5*c + 20*tan((c + d*x)/2 )**3*c + 15*tan((c + d*x)/2)*a - 105*tan((c + d*x)/2)*c + 60*c*d*x)/(60*a* *3*d)