Integrand size = 12, antiderivative size = 81 \[ \int \frac {1}{(5+3 \cos (c+d x))^3} \, dx=\frac {59 x}{2048}-\frac {59 \arctan \left (\frac {\sin (c+d x)}{3+\cos (c+d x)}\right )}{1024 d}-\frac {3 \sin (c+d x)}{32 d (5+3 \cos (c+d x))^2}-\frac {45 \sin (c+d x)}{512 d (5+3 \cos (c+d x))} \] Output:
59/2048*x-59/1024*arctan(sin(d*x+c)/(3+cos(d*x+c)))/d-3/32*sin(d*x+c)/d/(5 +3*cos(d*x+c))^2-45/512*sin(d*x+c)/d/(5+3*cos(d*x+c))
Time = 0.28 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.69 \[ \int \frac {1}{(5+3 \cos (c+d x))^3} \, dx=-\frac {59 \arctan \left (2 \cot \left (\frac {1}{2} (c+d x)\right )\right )+\frac {3 (182 \sin (c+d x)+45 \sin (2 (c+d x)))}{(5+3 \cos (c+d x))^2}}{1024 d} \] Input:
Integrate[(5 + 3*Cos[c + d*x])^(-3),x]
Output:
-1/1024*(59*ArcTan[2*Cot[(c + d*x)/2]] + (3*(182*Sin[c + d*x] + 45*Sin[2*( c + d*x)]))/(5 + 3*Cos[c + d*x])^2)/d
Time = 0.38 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.12, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 3143, 25, 3042, 3233, 27, 3042, 3136}
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.
| \(\displaystyle \int \frac {1}{(3 \cos (c+d x)+5)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\left (3 \sin \left (c+d x+\frac {\pi }{2}\right )+5\right )^3}dx\) |
| \(\Big \downarrow \) 3143 |
| \(\displaystyle -\frac {1}{32} \int -\frac {10-3 \cos (c+d x)}{(3 \cos (c+d x)+5)^2}dx-\frac {3 \sin (c+d x)}{32 d (3 \cos (c+d x)+5)^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{32} \int \frac {10-3 \cos (c+d x)}{(3 \cos (c+d x)+5)^2}dx-\frac {3 \sin (c+d x)}{32 d (3 \cos (c+d x)+5)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{32} \int \frac {10-3 \sin \left (c+d x+\frac {\pi }{2}\right )}{\left (3 \sin \left (c+d x+\frac {\pi }{2}\right )+5\right )^2}dx-\frac {3 \sin (c+d x)}{32 d (3 \cos (c+d x)+5)^2}\) |
| \(\Big \downarrow \) 3233 |
| \(\displaystyle \frac {1}{32} \left (-\frac {1}{16} \int -\frac {59}{3 \cos (c+d x)+5}dx-\frac {45 \sin (c+d x)}{16 d (3 \cos (c+d x)+5)}\right )-\frac {3 \sin (c+d x)}{32 d (3 \cos (c+d x)+5)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{32} \left (\frac {59}{16} \int \frac {1}{3 \cos (c+d x)+5}dx-\frac {45 \sin (c+d x)}{16 d (3 \cos (c+d x)+5)}\right )-\frac {3 \sin (c+d x)}{32 d (3 \cos (c+d x)+5)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{32} \left (\frac {59}{16} \int \frac {1}{3 \sin \left (c+d x+\frac {\pi }{2}\right )+5}dx-\frac {45 \sin (c+d x)}{16 d (3 \cos (c+d x)+5)}\right )-\frac {3 \sin (c+d x)}{32 d (3 \cos (c+d x)+5)^2}\) |
| \(\Big \downarrow \) 3136 |
| \(\displaystyle \frac {1}{32} \left (\frac {59}{16} \left (\frac {x}{4}-\frac {\arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)+3}\right )}{2 d}\right )-\frac {45 \sin (c+d x)}{16 d (3 \cos (c+d x)+5)}\right )-\frac {3 \sin (c+d x)}{32 d (3 \cos (c+d x)+5)^2}\) |
Input:
Int[(5 + 3*Cos[c + d*x])^(-3),x]
Output:
(-3*Sin[c + d*x])/(32*d*(5 + 3*Cos[c + d*x])^2) + ((59*(x/4 - ArcTan[Sin[c + d*x]/(3 + Cos[c + d*x])]/(2*d)))/16 - (45*Sin[c + d*x])/(16*d*(5 + 3*Co s[c + d*x])))/32
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{q = Rt[ a^2 - b^2, 2]}, Simp[x/q, x] + Simp[(2/(d*q))*ArcTan[b*(Cos[c + d*x]/(a + q + b*Sin[c + d*x]))], x]] /; FreeQ[{a, b, c, d}, x] && GtQ[a^2 - b^2, 0] && PosQ[a]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos [c + d*x]*((a + b*Sin[c + d*x])^(n + 1)/(d*(n + 1)*(a^2 - b^2))), x] + Simp [1/((n + 1)*(a^2 - b^2)) Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(b*c - a*d))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*( m + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]
Time = 0.92 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.77
| method | result | size |
| derivativedivides | \(\frac {\frac {-\frac {69 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{128}-\frac {51 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{32}}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+4\right )^{2}}+\frac {59 \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}\right )}{1024}}{d}\) | \(62\) |
| default | \(\frac {\frac {-\frac {69 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{128}-\frac {51 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{32}}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+4\right )^{2}}+\frac {59 \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}\right )}{1024}}{d}\) | \(62\) |
| risch | \(-\frac {3 i \left (59 \,{\mathrm e}^{3 i \left (d x +c \right )}+295 \,{\mathrm e}^{2 i \left (d x +c \right )}+241 \,{\mathrm e}^{i \left (d x +c \right )}+45\right )}{256 d \left (3 \,{\mathrm e}^{2 i \left (d x +c \right )}+10 \,{\mathrm e}^{i \left (d x +c \right )}+3\right )^{2}}-\frac {59 i \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {1}{3}\right )}{2048 d}+\frac {59 i \ln \left ({\mathrm e}^{i \left (d x +c \right )}+3\right )}{2048 d}\) | \(105\) |
| parallelrisch | \(\frac {59 i \left (-9 \cos \left (2 d x +2 c \right )-59-60 \cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 i\right )+59 i \left (9 \cos \left (2 d x +2 c \right )+59+60 \cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 i\right )-2184 \sin \left (d x +c \right )-540 \sin \left (2 d x +2 c \right )}{2048 d \left (9 \cos \left (2 d x +2 c \right )+59+60 \cos \left (d x +c \right )\right )}\) | \(123\) |
Input:
int(1/(5+3*cos(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/d*(1/4*(-69/128*tan(1/2*d*x+1/2*c)^3-51/32*tan(1/2*d*x+1/2*c))/(tan(1/2* d*x+1/2*c)^2+4)^2+59/1024*arctan(1/2*tan(1/2*d*x+1/2*c)))
Time = 0.08 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.11 \[ \int \frac {1}{(5+3 \cos (c+d x))^3} \, dx=-\frac {59 \, {\left (9 \, \cos \left (d x + c\right )^{2} + 30 \, \cos \left (d x + c\right ) + 25\right )} \arctan \left (\frac {5 \, \cos \left (d x + c\right ) + 3}{4 \, \sin \left (d x + c\right )}\right ) + 12 \, {\left (45 \, \cos \left (d x + c\right ) + 91\right )} \sin \left (d x + c\right )}{2048 \, {\left (9 \, d \cos \left (d x + c\right )^{2} + 30 \, d \cos \left (d x + c\right ) + 25 \, d\right )}} \] Input:
integrate(1/(5+3*cos(d*x+c))^3,x, algorithm="fricas")
Output:
-1/2048*(59*(9*cos(d*x + c)^2 + 30*cos(d*x + c) + 25)*arctan(1/4*(5*cos(d* x + c) + 3)/sin(d*x + c)) + 12*(45*cos(d*x + c) + 91)*sin(d*x + c))/(9*d*c os(d*x + c)^2 + 30*d*cos(d*x + c) + 25*d)
Result contains complex when optimal does not.
Time = 1.45 (sec) , antiderivative size = 359, normalized size of antiderivative = 4.43 \[ \int \frac {1}{(5+3 \cos (c+d x))^3} \, dx=\begin {cases} \frac {x}{\left (5 + 3 \cosh {\left (2 \operatorname {atanh}{\left (2 \right )} \right )}\right )^{3}} & \text {for}\: c = - d x - 2 i \operatorname {atanh}{\left (2 \right )} \vee c = - d x + 2 i \operatorname {atanh}{\left (2 \right )} \\\frac {x}{\left (3 \cos {\left (c \right )} + 5\right )^{3}} & \text {for}\: d = 0 \\\frac {59 \left (\operatorname {atan}{\left (\frac {\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2} \right )} + \pi \left \lfloor {\frac {\frac {c}{2} + \frac {d x}{2} - \frac {\pi }{2}}{\pi }}\right \rfloor \right ) \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{1024 d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 8192 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 16384 d} + \frac {472 \left (\operatorname {atan}{\left (\frac {\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2} \right )} + \pi \left \lfloor {\frac {\frac {c}{2} + \frac {d x}{2} - \frac {\pi }{2}}{\pi }}\right \rfloor \right ) \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{1024 d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 8192 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 16384 d} + \frac {944 \left (\operatorname {atan}{\left (\frac {\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2} \right )} + \pi \left \lfloor {\frac {\frac {c}{2} + \frac {d x}{2} - \frac {\pi }{2}}{\pi }}\right \rfloor \right )}{1024 d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 8192 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 16384 d} - \frac {138 \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{1024 d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 8192 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 16384 d} - \frac {408 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{1024 d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 8192 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 16384 d} & \text {otherwise} \end {cases} \] Input:
integrate(1/(5+3*cos(d*x+c))**3,x)
Output:
Piecewise((x/(5 + 3*cosh(2*atanh(2)))**3, Eq(c, -d*x - 2*I*atanh(2)) | Eq( c, -d*x + 2*I*atanh(2))), (x/(3*cos(c) + 5)**3, Eq(d, 0)), (59*(atan(tan(c /2 + d*x/2)/2) + pi*floor((c/2 + d*x/2 - pi/2)/pi))*tan(c/2 + d*x/2)**4/(1 024*d*tan(c/2 + d*x/2)**4 + 8192*d*tan(c/2 + d*x/2)**2 + 16384*d) + 472*(a tan(tan(c/2 + d*x/2)/2) + pi*floor((c/2 + d*x/2 - pi/2)/pi))*tan(c/2 + d*x /2)**2/(1024*d*tan(c/2 + d*x/2)**4 + 8192*d*tan(c/2 + d*x/2)**2 + 16384*d) + 944*(atan(tan(c/2 + d*x/2)/2) + pi*floor((c/2 + d*x/2 - pi/2)/pi))/(102 4*d*tan(c/2 + d*x/2)**4 + 8192*d*tan(c/2 + d*x/2)**2 + 16384*d) - 138*tan( c/2 + d*x/2)**3/(1024*d*tan(c/2 + d*x/2)**4 + 8192*d*tan(c/2 + d*x/2)**2 + 16384*d) - 408*tan(c/2 + d*x/2)/(1024*d*tan(c/2 + d*x/2)**4 + 8192*d*tan( c/2 + d*x/2)**2 + 16384*d), True))
Time = 0.11 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.37 \[ \int \frac {1}{(5+3 \cos (c+d x))^3} \, dx=-\frac {\frac {6 \, {\left (\frac {68 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {23 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{\frac {8 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {\sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 16} - 59 \, \arctan \left (\frac {\sin \left (d x + c\right )}{2 \, {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}{1024 \, d} \] Input:
integrate(1/(5+3*cos(d*x+c))^3,x, algorithm="maxima")
Output:
-1/1024*(6*(68*sin(d*x + c)/(cos(d*x + c) + 1) + 23*sin(d*x + c)^3/(cos(d* x + c) + 1)^3)/(8*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + sin(d*x + c)^4/(co s(d*x + c) + 1)^4 + 16) - 59*arctan(1/2*sin(d*x + c)/(cos(d*x + c) + 1)))/ d
Time = 0.15 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.93 \[ \int \frac {1}{(5+3 \cos (c+d x))^3} \, dx=\frac {59 \, d x + 59 \, c - \frac {12 \, {\left (23 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 68 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 4\right )}^{2}} - 118 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 3}\right )}{2048 \, d} \] Input:
integrate(1/(5+3*cos(d*x+c))^3,x, algorithm="giac")
Output:
1/2048*(59*d*x + 59*c - 12*(23*tan(1/2*d*x + 1/2*c)^3 + 68*tan(1/2*d*x + 1 /2*c))/(tan(1/2*d*x + 1/2*c)^2 + 4)^2 - 118*arctan(sin(d*x + c)/(cos(d*x + c) + 3)))/d
Time = 36.60 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.19 \[ \int \frac {1}{(5+3 \cos (c+d x))^3} \, dx=\frac {59\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}\right )}{1024\,d}-\frac {59\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{1024\,d}-\frac {\frac {69\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{512}+\frac {51\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{128}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+16\right )} \] Input:
int(1/(3*cos(c + d*x) + 5)^3,x)
Output:
(59*atan(tan(c/2 + (d*x)/2)/2))/(1024*d) - (59*(atan(tan(c/2 + (d*x)/2)) - (d*x)/2))/(1024*d) - ((51*tan(c/2 + (d*x)/2))/128 + (69*tan(c/2 + (d*x)/2 )^3)/512)/(d*(8*tan(c/2 + (d*x)/2)^2 + tan(c/2 + (d*x)/2)^4 + 16))
Time = 0.16 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.31 \[ \int \frac {1}{(5+3 \cos (c+d x))^3} \, dx=\frac {1770 \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}\right ) \cos \left (d x +c \right )-531 \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}\right ) \sin \left (d x +c \right )^{2}+2006 \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}\right )-270 \cos \left (d x +c \right ) \sin \left (d x +c \right )-546 \sin \left (d x +c \right )}{1024 d \left (30 \cos \left (d x +c \right )-9 \sin \left (d x +c \right )^{2}+34\right )} \] Input:
int(1/(5+3*cos(d*x+c))^3,x)
Output:
(1770*atan(tan((c + d*x)/2)/2)*cos(c + d*x) - 531*atan(tan((c + d*x)/2)/2) *sin(c + d*x)**2 + 2006*atan(tan((c + d*x)/2)/2) - 270*cos(c + d*x)*sin(c + d*x) - 546*sin(c + d*x))/(1024*d*(30*cos(c + d*x) - 9*sin(c + d*x)**2 + 34))