Integrand size = 12, antiderivative size = 81 \[ \int \frac {1}{(-5-3 \sin (c+d x))^3} \, dx=-\frac {59 x}{2048}-\frac {59 \arctan \left (\frac {\cos (c+d x)}{3+\sin (c+d x)}\right )}{1024 d}-\frac {3 \cos (c+d x)}{32 d (5+3 \sin (c+d x))^2}-\frac {45 \cos (c+d x)}{512 d (5+3 \sin (c+d x))} \] Output:
-59/2048*x-59/1024*arctan(cos(d*x+c)/(3+sin(d*x+c)))/d-3/32*cos(d*x+c)/d/( 5+3*sin(d*x+c))^2-45/512*cos(d*x+c)/d/(5+3*sin(d*x+c))
Time = 0.66 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.41 \[ \int \frac {1}{(-5-3 \sin (c+d x))^3} \, dx=\frac {-59 \arctan \left (\frac {2 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}\right )+\frac {3 (-182 \cos (c+d x)+3 (59-9 \cos (2 (c+d x))+60 \sin (c+d x)-15 \sin (2 (c+d x))))}{(5+3 \sin (c+d x))^2}}{1024 d} \] Input:
Integrate[(-5 - 3*Sin[c + d*x])^(-3),x]
Output:
(-59*ArcTan[(2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])] + (3*(-182*Cos[c + d*x] + 3*(59 - 9*Cos[2*(c + d*x)] + 60*Sin[c + d*x] - 15*Sin[2*(c + d*x)])))/(5 + 3*Sin[c + d*x])^2)/(1024*d)
Time = 0.36 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.12, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {3042, 3143, 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.
| \(\displaystyle \int \frac {1}{(-3 \sin (c+d x)-5)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(-3 \sin (c+d x)-5)^3}dx\) |
| \(\Big \downarrow \) 3143 |
| \(\displaystyle -\frac {1}{32} \int \frac {10-3 \sin (c+d x)}{(3 \sin (c+d x)+5)^2}dx-\frac {3 \cos (c+d x)}{32 d (3 \sin (c+d x)+5)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {1}{32} \int \frac {10-3 \sin (c+d x)}{(3 \sin (c+d x)+5)^2}dx-\frac {3 \cos (c+d x)}{32 d (3 \sin (c+d x)+5)^2}\) |
| \(\Big \downarrow \) 3233 |
| \(\displaystyle \frac {1}{32} \left (\frac {1}{16} \int -\frac {59}{3 \sin (c+d x)+5}dx-\frac {45 \cos (c+d x)}{16 d (3 \sin (c+d x)+5)}\right )-\frac {3 \cos (c+d x)}{32 d (3 \sin (c+d x)+5)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{32} \left (-\frac {59}{16} \int \frac {1}{3 \sin (c+d x)+5}dx-\frac {45 \cos (c+d x)}{16 d (3 \sin (c+d x)+5)}\right )-\frac {3 \cos (c+d x)}{32 d (3 \sin (c+d x)+5)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{32} \left (-\frac {59}{16} \int \frac {1}{3 \sin (c+d x)+5}dx-\frac {45 \cos (c+d x)}{16 d (3 \sin (c+d x)+5)}\right )-\frac {3 \cos (c+d x)}{32 d (3 \sin (c+d x)+5)^2}\) |
| \(\Big \downarrow \) 3136 |
| \(\displaystyle \frac {1}{32} \left (-\frac {59}{16} \left (\frac {\arctan \left (\frac {\cos (c+d x)}{\sin (c+d x)+3}\right )}{2 d}+\frac {x}{4}\right )-\frac {45 \cos (c+d x)}{16 d (3 \sin (c+d x)+5)}\right )-\frac {3 \cos (c+d x)}{32 d (3 \sin (c+d x)+5)^2}\) |
Input:
Int[(-5 - 3*Sin[c + d*x])^(-3),x]
Output:
(-3*Cos[c + d*x])/(32*d*(5 + 3*Sin[c + d*x])^2) + ((-59*(x/4 + ArcTan[Cos[ c + d*x]/(3 + Sin[c + d*x])]/(2*d)))/16 - (45*Cos[c + d*x])/(16*d*(5 + 3*S in[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 = 0.33 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.12
| method | result | size |
| derivativedivides | \(\frac {-\frac {50 \left (\frac {963 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{64000}+\frac {11739 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{320000}+\frac {2313 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{64000}+\frac {273}{12800}\right )}{\left (5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+5\right )^{2}}-\frac {59 \arctan \left (\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4}+\frac {3}{4}\right )}{1024}}{d}\) | \(91\) |
| default | \(\frac {-\frac {50 \left (\frac {963 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{64000}+\frac {11739 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{320000}+\frac {2313 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{64000}+\frac {273}{12800}\right )}{\left (5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+5\right )^{2}}-\frac {59 \arctan \left (\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4}+\frac {3}{4}\right )}{1024}}{d}\) | \(91\) |
| risch | \(-\frac {3 \left (295 i {\mathrm e}^{2 i \left (d x +c \right )}+59 \,{\mathrm e}^{3 i \left (d x +c \right )}-241 \,{\mathrm e}^{i \left (d x +c \right )}-45 i\right )}{256 \left (3 \,{\mathrm e}^{2 i \left (d x +c \right )}-3+10 i {\mathrm e}^{i \left (d x +c \right )}\right )^{2} d}-\frac {59 i \ln \left ({\mathrm e}^{i \left (d x +c \right )}+3 i\right )}{2048 d}+\frac {59 i \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i}{3}\right )}{2048 d}\) | \(109\) |
| parallelrisch | \(\frac {1062+59 i \left (-59+9 \cos \left (2 d x +2 c \right )-60 \sin \left (d x +c \right )\right ) \ln \left (5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+3-4 i\right )+59 i \left (59-9 \cos \left (2 d x +2 c \right )+60 \sin \left (d x +c \right )\right ) \ln \left (5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+3+4 i\right )+2184 \cos \left (d x +c \right )-162 \cos \left (2 d x +2 c \right )+1080 \sin \left (d x +c \right )+540 \sin \left (2 d x +2 c \right )}{2048 d \left (-59+9 \cos \left (2 d x +2 c \right )-60 \sin \left (d x +c \right )\right )}\) | \(149\) |
Input:
int(1/(-5-3*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/d*(-50*(963/64000*tan(1/2*d*x+1/2*c)^3+11739/320000*tan(1/2*d*x+1/2*c)^2 +2313/64000*tan(1/2*d*x+1/2*c)+273/12800)/(5*tan(1/2*d*x+1/2*c)^2+6*tan(1/ 2*d*x+1/2*c)+5)^2-59/1024*arctan(5/4*tan(1/2*d*x+1/2*c)+3/4))
Time = 0.11 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.16 \[ \int \frac {1}{(-5-3 \sin (c+d x))^3} \, dx=-\frac {59 \, {\left (9 \, \cos \left (d x + c\right )^{2} - 30 \, \sin \left (d x + c\right ) - 34\right )} \arctan \left (\frac {5 \, \sin \left (d x + c\right ) + 3}{4 \, \cos \left (d x + c\right )}\right ) - 540 \, \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 1092 \, \cos \left (d x + c\right )}{2048 \, {\left (9 \, d \cos \left (d x + c\right )^{2} - 30 \, d \sin \left (d x + c\right ) - 34 \, d\right )}} \] Input:
integrate(1/(-5-3*sin(d*x+c))^3,x, algorithm="fricas")
Output:
-1/2048*(59*(9*cos(d*x + c)^2 - 30*sin(d*x + c) - 34)*arctan(1/4*(5*sin(d* x + c) + 3)/cos(d*x + c)) - 540*cos(d*x + c)*sin(d*x + c) - 1092*cos(d*x + c))/(9*d*cos(d*x + c)^2 - 30*d*sin(d*x + c) - 34*d)
Result contains complex when optimal does not.
Time = 1.96 (sec) , antiderivative size = 921, normalized size of antiderivative = 11.37 \[ \int \frac {1}{(-5-3 \sin (c+d x))^3} \, dx=\text {Too large to display} \] Input:
integrate(1/(-5-3*sin(d*x+c))**3,x)
Output:
Piecewise((x/(-5 + 3*sin(2*atan(3/5 - 4*I/5)))**3, Eq(c, -d*x - 2*atan(3/5 - 4*I/5))), (x/(-5 + 3*sin(2*atan(3/5 + 4*I/5)))**3, Eq(c, -d*x - 2*atan( 3/5 + 4*I/5))), (x/(-3*sin(c) - 5)**3, Eq(d, 0)), (-36875*(atan(5*tan(c/2 + d*x/2)/4 + 3/4) + pi*floor((c/2 + d*x/2 - pi/2)/pi))*tan(c/2 + d*x/2)**4 /(640000*d*tan(c/2 + d*x/2)**4 + 1536000*d*tan(c/2 + d*x/2)**3 + 2201600*d *tan(c/2 + d*x/2)**2 + 1536000*d*tan(c/2 + d*x/2) + 640000*d) - 88500*(ata n(5*tan(c/2 + d*x/2)/4 + 3/4) + pi*floor((c/2 + d*x/2 - pi/2)/pi))*tan(c/2 + d*x/2)**3/(640000*d*tan(c/2 + d*x/2)**4 + 1536000*d*tan(c/2 + d*x/2)**3 + 2201600*d*tan(c/2 + d*x/2)**2 + 1536000*d*tan(c/2 + d*x/2) + 640000*d) - 126850*(atan(5*tan(c/2 + d*x/2)/4 + 3/4) + pi*floor((c/2 + d*x/2 - pi/2) /pi))*tan(c/2 + d*x/2)**2/(640000*d*tan(c/2 + d*x/2)**4 + 1536000*d*tan(c/ 2 + d*x/2)**3 + 2201600*d*tan(c/2 + d*x/2)**2 + 1536000*d*tan(c/2 + d*x/2) + 640000*d) - 88500*(atan(5*tan(c/2 + d*x/2)/4 + 3/4) + pi*floor((c/2 + d *x/2 - pi/2)/pi))*tan(c/2 + d*x/2)/(640000*d*tan(c/2 + d*x/2)**4 + 1536000 *d*tan(c/2 + d*x/2)**3 + 2201600*d*tan(c/2 + d*x/2)**2 + 1536000*d*tan(c/2 + d*x/2) + 640000*d) - 36875*(atan(5*tan(c/2 + d*x/2)/4 + 3/4) + pi*floor ((c/2 + d*x/2 - pi/2)/pi))/(640000*d*tan(c/2 + d*x/2)**4 + 1536000*d*tan(c /2 + d*x/2)**3 + 2201600*d*tan(c/2 + d*x/2)**2 + 1536000*d*tan(c/2 + d*x/2 ) + 640000*d) - 19260*tan(c/2 + d*x/2)**3/(640000*d*tan(c/2 + d*x/2)**4 + 1536000*d*tan(c/2 + d*x/2)**3 + 2201600*d*tan(c/2 + d*x/2)**2 + 1536000...
Leaf count of result is larger than twice the leaf count of optimal. 173 vs. \(2 (73) = 146\).
Time = 0.11 (sec) , antiderivative size = 173, normalized size of antiderivative = 2.14 \[ \int \frac {1}{(-5-3 \sin (c+d x))^3} \, dx=-\frac {\frac {12 \, {\left (\frac {3855 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3913 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {1605 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + 2275\right )}}{\frac {60 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {86 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {60 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {25 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 25} + 1475 \, \arctan \left (\frac {5 \, \sin \left (d x + c\right )}{4 \, {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {3}{4}\right )}{25600 \, d} \] Input:
integrate(1/(-5-3*sin(d*x+c))^3,x, algorithm="maxima")
Output:
-1/25600*(12*(3855*sin(d*x + c)/(cos(d*x + c) + 1) + 3913*sin(d*x + c)^2/( cos(d*x + c) + 1)^2 + 1605*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 2275)/(60 *sin(d*x + c)/(cos(d*x + c) + 1) + 86*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 60*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 25*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 25) + 1475*arctan(5/4*sin(d*x + c)/(cos(d*x + c) + 1) + 3/4))/d
Time = 0.13 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.49 \[ \int \frac {1}{(-5-3 \sin (c+d x))^3} \, dx=-\frac {1475 \, d x + 1475 \, c + \frac {24 \, {\left (1605 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3913 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3855 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2275\right )}}{{\left (5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5\right )}^{2}} + 2950 \, \arctan \left (-\frac {3 \, \cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 3}{\cos \left (d x + c\right ) - 3 \, \sin \left (d x + c\right ) - 9}\right )}{51200 \, d} \] Input:
integrate(1/(-5-3*sin(d*x+c))^3,x, algorithm="giac")
Output:
-1/51200*(1475*d*x + 1475*c + 24*(1605*tan(1/2*d*x + 1/2*c)^3 + 3913*tan(1 /2*d*x + 1/2*c)^2 + 3855*tan(1/2*d*x + 1/2*c) + 2275)/(5*tan(1/2*d*x + 1/2 *c)^2 + 6*tan(1/2*d*x + 1/2*c) + 5)^2 + 2950*arctan(-(3*cos(d*x + c) + sin (d*x + c) + 3)/(cos(d*x + c) - 3*sin(d*x + c) - 9)))/d
Time = 26.05 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.38 \[ \int \frac {1}{(-5-3 \sin (c+d x))^3} \, dx=\frac {59\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{1024\,d}-\frac {59\,\mathrm {atan}\left (\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+\frac {3}{4}\right )}{1024\,d}-\frac {\frac {963\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{1280}+\frac {11739\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{6400}+\frac {2313\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{1280}+\frac {273}{256}}{d\,{\left (5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+5\right )}^2} \] Input:
int(-1/(3*sin(c + d*x) + 5)^3,x)
Output:
(59*(atan(tan(c/2 + (d*x)/2)) - (d*x)/2))/(1024*d) - (59*atan((5*tan(c/2 + (d*x)/2))/4 + 3/4))/(1024*d) - ((2313*tan(c/2 + (d*x)/2))/1280 + (11739*t an(c/2 + (d*x)/2)^2)/6400 + (963*tan(c/2 + (d*x)/2)^3)/1280 + 273/256)/(d* (6*tan(c/2 + (d*x)/2) + 5*tan(c/2 + (d*x)/2)^2 + 5)^2)
Time = 0.18 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.62 \[ \int \frac {1}{(-5-3 \sin (c+d x))^3} \, dx=\frac {-531 \mathit {atan} \left (\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4}+\frac {3}{4}\right ) \sin \left (d x +c \right )^{2}-1770 \mathit {atan} \left (\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4}+\frac {3}{4}\right ) \sin \left (d x +c \right )-1475 \mathit {atan} \left (\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4}+\frac {3}{4}\right )-270 \cos \left (d x +c \right ) \sin \left (d x +c \right )-546 \cos \left (d x +c \right )-81 \sin \left (d x +c \right )^{2}-270 \sin \left (d x +c \right )-225}{1024 d \left (9 \sin \left (d x +c \right )^{2}+30 \sin \left (d x +c \right )+25\right )} \] Input:
int(1/(-5-3*sin(d*x+c))^3,x)
Output:
( - 531*atan((5*tan((c + d*x)/2) + 3)/4)*sin(c + d*x)**2 - 1770*atan((5*ta n((c + d*x)/2) + 3)/4)*sin(c + d*x) - 1475*atan((5*tan((c + d*x)/2) + 3)/4 ) - 270*cos(c + d*x)*sin(c + d*x) - 546*cos(c + d*x) - 81*sin(c + d*x)**2 - 270*sin(c + d*x) - 225)/(1024*d*(9*sin(c + d*x)**2 + 30*sin(c + d*x) + 2 5))