Integrand size = 38, antiderivative size = 116 \[ \int \frac {\cos (c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\frac {C x}{a^3}+\frac {(B-C) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(2 B-7 C) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {(4 B-29 C) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )} \] Output:
C*x/a^3+1/5*(B-C)*cos(d*x+c)^2*sin(d*x+c)/d/(a+a*cos(d*x+c))^3-1/15*(2*B-7 *C)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^2+1/15*(4*B-29*C)*sin(d*x+c)/d/(a^3+a^ 3*cos(d*x+c))
Leaf count is larger than twice the leaf count of optimal. \(241\) vs. \(2(116)=232\).
Time = 1.56 (sec) , antiderivative size = 241, normalized size of antiderivative = 2.08 \[ \int \frac {\cos (c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\frac {\sec \left (\frac {c}{2}\right ) \sec ^5\left (\frac {1}{2} (c+d x)\right ) \left (150 C d x \cos \left (\frac {d x}{2}\right )+150 C d x \cos \left (c+\frac {d x}{2}\right )+75 C d x \cos \left (c+\frac {3 d x}{2}\right )+75 C d x \cos \left (2 c+\frac {3 d x}{2}\right )+15 C d x \cos \left (2 c+\frac {5 d x}{2}\right )+15 C d x \cos \left (3 c+\frac {5 d x}{2}\right )+80 B \sin \left (\frac {d x}{2}\right )-370 C \sin \left (\frac {d x}{2}\right )-60 B \sin \left (c+\frac {d x}{2}\right )+270 C \sin \left (c+\frac {d x}{2}\right )+40 B \sin \left (c+\frac {3 d x}{2}\right )-230 C \sin \left (c+\frac {3 d x}{2}\right )-30 B \sin \left (2 c+\frac {3 d x}{2}\right )+90 C \sin \left (2 c+\frac {3 d x}{2}\right )+14 B \sin \left (2 c+\frac {5 d x}{2}\right )-64 C \sin \left (2 c+\frac {5 d x}{2}\right )\right )}{480 a^3 d} \] Input:
Integrate[(Cos[c + d*x]*(B*Cos[c + d*x] + 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] + 80*B*Sin[(d* x)/2] - 370*C*Sin[(d*x)/2] - 60*B*Sin[c + (d*x)/2] + 270*C*Sin[c + (d*x)/2 ] + 40*B*Sin[c + (3*d*x)/2] - 230*C*Sin[c + (3*d*x)/2] - 30*B*Sin[2*c + (3 *d*x)/2] + 90*C*Sin[2*c + (3*d*x)/2] + 14*B*Sin[2*c + (5*d*x)/2] - 64*C*Si n[2*c + (5*d*x)/2]))/(480*a^3*d)
Time = 0.91 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.08, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.342, Rules used = {3042, 3508, 3042, 3456, 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.
| \(\displaystyle \int \frac {\cos (c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \left (B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 3508 |
| \(\displaystyle \int \frac {\cos ^2(c+d x) (B+C \cos (c+d x))}{(a \cos (c+d x)+a)^3}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (B+C \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 3456 |
| \(\displaystyle \frac {\int \frac {\cos (c+d x) (2 a (B-C)+5 a C \cos (c+d x))}{(\cos (c+d x) a+a)^2}dx}{5 a^2}+\frac {(B-C) \sin (c+d x) \cos ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \left (2 a (B-C)+5 a C \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}+\frac {(B-C) \sin (c+d x) \cos ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {\int \frac {5 a C \cos ^2(c+d x)+2 a (B-C) \cos (c+d x)}{(\cos (c+d x) a+a)^2}dx}{5 a^2}+\frac {(B-C) \sin (c+d x) \cos ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {5 a C \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 a (B-C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}+\frac {(B-C) \sin (c+d x) \cos ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3498 |
| \(\displaystyle \frac {-\frac {\int -\frac {2 (2 B-7 C) a^2+15 C \cos (c+d x) a^2}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {a (2 B-7 C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}+\frac {(B-C) \sin (c+d x) \cos ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {2 (2 B-7 C) a^2+15 C \cos (c+d x) a^2}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {a (2 B-7 C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}+\frac {(B-C) \sin (c+d x) \cos ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {2 (2 B-7 C) a^2+15 C \sin \left (c+d x+\frac {\pi }{2}\right ) a^2}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {a (2 B-7 C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}+\frac {(B-C) \sin (c+d x) \cos ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {\frac {a^2 (4 B-29 C) \int \frac {1}{\cos (c+d x) a+a}dx+15 a C x}{3 a^2}-\frac {a (2 B-7 C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}+\frac {(B-C) \sin (c+d x) \cos ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a^2 (4 B-29 C) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+15 a C x}{3 a^2}-\frac {a (2 B-7 C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}+\frac {(B-C) \sin (c+d x) \cos ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3127 |
| \(\displaystyle \frac {\frac {\frac {a^2 (4 B-29 C) \sin (c+d x)}{d (a \cos (c+d x)+a)}+15 a C x}{3 a^2}-\frac {a (2 B-7 C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}+\frac {(B-C) \sin (c+d x) \cos ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
Input:
Int[(Cos[c + d*x]*(B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(a + a*Cos[c + d*x] )^3,x]
Output:
((B - C)*Cos[c + d*x]^2*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) + (-1/3 *(a*(2*B - 7*C)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^2) + (15*a*C*x + (a^ 2*(4*B - 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_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/( a*f*(2*m + 1))), x] - Simp[1/(a*b*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x], x] /; Fre eQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] & & NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (In tegerQ[2*n] || EqQ[c, 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_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_ .) + (f_.)*(x_)]^2), x_Symbol] :> Simp[1/b^2 Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Time = 0.26 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.59
| method | result | size |
| parallelrisch | \(\frac {3 \left (B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+10 \left (-B +2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+15 \left (B -7 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+60 d x C}{60 a^{3} d}\) | \(69\) |
| derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -7 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+8 C \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\) | \(102\) |
| default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -7 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+8 C \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\) | \(102\) |
| risch | \(\frac {C x}{a^{3}}+\frac {2 i \left (15 B \,{\mathrm e}^{4 i \left (d x +c \right )}-45 C \,{\mathrm e}^{4 i \left (d x +c \right )}+30 B \,{\mathrm e}^{3 i \left (d x +c \right )}-135 C \,{\mathrm e}^{3 i \left (d x +c \right )}+40 B \,{\mathrm e}^{2 i \left (d x +c \right )}-185 C \,{\mathrm e}^{2 i \left (d x +c \right )}+20 B \,{\mathrm e}^{i \left (d x +c \right )}-115 C \,{\mathrm e}^{i \left (d x +c \right )}+7 B -32 C \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\) | \(133\) |
| norman | \(\frac {\frac {C x}{a}+\frac {C x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{a}+\frac {3 C x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a}+\frac {3 C x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{a}-\frac {\left (B -11 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{60 a d}+\frac {\left (B -7 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {\left (B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{20 a d}-\frac {\left (B +9 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{10 a d}+\frac {\left (3 B -43 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{10 a d}+\frac {\left (7 B -59 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} a^{2}}\) | \(226\) |
Input:
int(cos(d*x+c)*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x,method=_ RETURNVERBOSE)
Output:
1/60*(3*(B-C)*tan(1/2*d*x+1/2*c)^5+10*(-B+2*C)*tan(1/2*d*x+1/2*c)^3+15*(B- 7*C)*tan(1/2*d*x+1/2*c)+60*d*x*C)/a^3/d
Time = 0.08 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.18 \[ \int \frac {\cos (c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\frac {15 \, C d x \cos \left (d x + c\right )^{3} + 45 \, C d x \cos \left (d x + c\right )^{2} + 45 \, C d x \cos \left (d x + c\right ) + 15 \, C d x + {\left ({\left (7 \, B - 32 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, B - 17 \, C\right )} \cos \left (d x + c\right ) + 2 \, B - 22 \, C\right )} \sin \left (d x + c\right )}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \] Input:
integrate(cos(d*x+c)*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x, a lgorithm="fricas")
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 + ((7*B - 32*C)*cos(d*x + c)^2 + 3*(2*B - 17*C)*cos(d*x + c) + 2*B - 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 = 1.91 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.30 \[ \int \frac {\cos (c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\begin {cases} \frac {B \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{20 a^{3} d} - \frac {B \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{3} d} + \frac {B \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{4 a^{3} d} + \frac {C x}{a^{3}} - \frac {C \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{20 a^{3} d} + \frac {C \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a^{3} d} - \frac {7 C \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{4 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \left (B \cos {\left (c \right )} + C \cos ^{2}{\left (c \right )}\right ) \cos {\left (c \right )}}{\left (a \cos {\left (c \right )} + a\right )^{3}} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)*(B*cos(d*x+c)+C*cos(d*x+c)**2)/(a+a*cos(d*x+c))**3,x)
Output:
Piecewise((B*tan(c/2 + d*x/2)**5/(20*a**3*d) - B*tan(c/2 + d*x/2)**3/(6*a* *3*d) + B*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*(B*cos(c) + C*cos(c)**2)*cos(c)/(a*cos(c) + a)**3, Tr ue))
Time = 0.13 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.38 \[ \int \frac {\cos (c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=-\frac {C {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {120 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} - \frac {B {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \] Input:
integrate(cos(d*x+c)*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x, a lgorithm="maxima")
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) - B*(15*sin(d*x + c)/(cos(d*x + c) + 1 ) - 10*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)/d
Time = 0.15 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.03 \[ \int \frac {\cos (c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\frac {\frac {60 \, {\left (d x + c\right )} C}{a^{3}} + \frac {3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 10 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 20 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 105 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \] Input:
integrate(cos(d*x+c)*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x, a lgorithm="giac")
Output:
1/60*(60*(d*x + c)*C/a^3 + (3*B*a^12*tan(1/2*d*x + 1/2*c)^5 - 3*C*a^12*tan (1/2*d*x + 1/2*c)^5 - 10*B*a^12*tan(1/2*d*x + 1/2*c)^3 + 20*C*a^12*tan(1/2 *d*x + 1/2*c)^3 + 15*B*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 = 0.32 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.16 \[ \int \frac {\cos (c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\frac {C\,x}{a^3}-\frac {{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {B\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6}-\frac {C\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}\right )-{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {B\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}-\frac {7\,C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}\right )-\frac {B\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{20}+\frac {C\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{20}}{a^3\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5} \] Input:
int((cos(c + d*x)*(B*cos(c + d*x) + 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)^2*((B*sin(c/2 + (d*x)/2)^3)/6 - (C*sin(c/2 + (d*x)/2)^3)/3) - cos(c/2 + (d*x)/2)^4*((B*sin(c/2 + (d*x)/2))/4 - (7*C* sin(c/2 + (d*x)/2))/4) - (B*sin(c/2 + (d*x)/2)^5)/20 + (C*sin(c/2 + (d*x)/ 2)^5)/20)/(a^3*d*cos(c/2 + (d*x)/2)^5)
Time = 0.16 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.81 \[ \int \frac {\cos (c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} b -3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} c -10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b +20 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} c +15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b -105 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) c +60 c d x}{60 a^{3} d} \] Input:
int(cos(d*x+c)*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x)
Output:
(3*tan((c + d*x)/2)**5*b - 3*tan((c + d*x)/2)**5*c - 10*tan((c + d*x)/2)** 3*b + 20*tan((c + d*x)/2)**3*c + 15*tan((c + d*x)/2)*b - 105*tan((c + d*x) /2)*c + 60*c*d*x)/(60*a**3*d)