Integrand size = 12, antiderivative size = 58 \[ \int \frac {1}{(5-3 \sin (c+d x))^2} \, dx=\frac {5 x}{64}-\frac {5 \arctan \left (\frac {\cos (c+d x)}{3-\sin (c+d x)}\right )}{32 d}-\frac {3 \cos (c+d x)}{16 d (5-3 \sin (c+d x))} \] Output:
5/64*x-5/32*arctan(cos(d*x+c)/(3-sin(d*x+c)))/d-3/16*cos(d*x+c)/d/(5-3*sin (d*x+c))
Time = 0.30 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.57 \[ \int \frac {1}{(5-3 \sin (c+d x))^2} \, dx=\frac {-25 \arctan \left (\frac {2 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}\right )+\frac {6 (-5+5 \cos (c+d x)+3 \sin (c+d x))}{-5+3 \sin (c+d x)}}{160 d} \] Input:
Integrate[(5 - 3*Sin[c + d*x])^(-2),x]
Output:
(-25*ArcTan[(2*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]))/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])] + (6*(-5 + 5*Cos[c + d*x] + 3*Sin[c + d*x]))/(-5 + 3*Si n[c + d*x]))/(160*d)
Time = 0.27 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {3042, 3143, 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.
\(\displaystyle \int \frac {1}{(5-3 \sin (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(5-3 \sin (c+d x))^2}dx\) |
\(\Big \downarrow \) 3143 |
\(\displaystyle -\frac {1}{16} \int -\frac {5}{5-3 \sin (c+d x)}dx-\frac {3 \cos (c+d x)}{16 d (5-3 \sin (c+d x))}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {5}{16} \int \frac {1}{5-3 \sin (c+d x)}dx-\frac {3 \cos (c+d x)}{16 d (5-3 \sin (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5}{16} \int \frac {1}{5-3 \sin (c+d x)}dx-\frac {3 \cos (c+d x)}{16 d (5-3 \sin (c+d x))}\) |
\(\Big \downarrow \) 3136 |
\(\displaystyle \frac {5}{16} \left (\frac {x}{4}-\frac {\arctan \left (\frac {\cos (c+d x)}{3-\sin (c+d x)}\right )}{2 d}\right )-\frac {3 \cos (c+d x)}{16 d (5-3 \sin (c+d x))}\) |
Input:
Int[(5 - 3*Sin[c + d*x])^(-2),x]
Output:
(5*(x/4 - ArcTan[Cos[c + d*x]/(3 - Sin[c + d*x])]/(2*d)))/16 - (3*Cos[c + d*x])/(16*d*(5 - 3*Sin[c + d*x]))
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]
Time = 0.24 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.09
method | result | size |
derivativedivides | \(\frac {\frac {\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{200}-\frac {3}{40}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{5}+1}+\frac {5 \arctan \left (\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4}-\frac {3}{4}\right )}{32}}{d}\) | \(63\) |
default | \(\frac {\frac {\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{200}-\frac {3}{40}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{5}+1}+\frac {5 \arctan \left (\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4}-\frac {3}{4}\right )}{32}}{d}\) | \(63\) |
risch | \(-\frac {5 \,{\mathrm e}^{i \left (d x +c \right )}-3 i}{8 d \left (3 \,{\mathrm e}^{2 i \left (d x +c \right )}-3-10 i {\mathrm e}^{i \left (d x +c \right )}\right )}+\frac {5 i \ln \left ({\mathrm e}^{i \left (d x +c \right )}-3 i\right )}{64 d}-\frac {5 i \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i}{3}\right )}{64 d}\) | \(86\) |
parallelrisch | \(\frac {\left (125 i-75 i \sin \left (d x +c \right )\right ) \ln \left (5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3-4 i\right )+\left (75 i \sin \left (d x +c \right )-125 i\right ) \ln \left (5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3+4 i\right )+60 \cos \left (d x +c \right )-36 \sin \left (d x +c \right )+60}{960 \sin \left (d x +c \right ) d -1600 d}\) | \(93\) |
Input:
int(1/(5-3*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(2*(9/400*tan(1/2*d*x+1/2*c)-3/80)/(tan(1/2*d*x+1/2*c)^2-6/5*tan(1/2*d *x+1/2*c)+1)+5/32*arctan(5/4*tan(1/2*d*x+1/2*c)-3/4))
Time = 0.12 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.02 \[ \int \frac {1}{(5-3 \sin (c+d x))^2} \, dx=\frac {5 \, {\left (3 \, \sin \left (d x + c\right ) - 5\right )} \arctan \left (\frac {5 \, \sin \left (d x + c\right ) - 3}{4 \, \cos \left (d x + c\right )}\right ) + 12 \, \cos \left (d x + c\right )}{64 \, {\left (3 \, d \sin \left (d x + c\right ) - 5 \, d\right )}} \] Input:
integrate(1/(5-3*sin(d*x+c))^2,x, algorithm="fricas")
Output:
1/64*(5*(3*sin(d*x + c) - 5)*arctan(1/4*(5*sin(d*x + c) - 3)/cos(d*x + c)) + 12*cos(d*x + c))/(3*d*sin(d*x + c) - 5*d)
Result contains complex when optimal does not.
Time = 0.92 (sec) , antiderivative size = 384, normalized size of antiderivative = 6.62 \[ \int \frac {1}{(5-3 \sin (c+d x))^2} \, dx =\text {Too large to display} \] Input:
integrate(1/(5-3*sin(d*x+c))**2,x)
Output:
Piecewise((x/(5 - 3*sin(2*atan(3/5 - 4*I/5)))**2, Eq(c, -d*x + 2*atan(3/5 - 4*I/5))), (x/(5 - 3*sin(2*atan(3/5 + 4*I/5)))**2, Eq(c, -d*x + 2*atan(3/ 5 + 4*I/5))), (x/(5 - 3*sin(c))**2, Eq(d, 0)), (125*(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/(800* d*tan(c/2 + d*x/2)**2 - 960*d*tan(c/2 + d*x/2) + 800*d) - 150*(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 )/(800*d*tan(c/2 + d*x/2)**2 - 960*d*tan(c/2 + d*x/2) + 800*d) + 125*(atan (5*tan(c/2 + d*x/2)/4 - 3/4) + pi*floor((c/2 + d*x/2 - pi/2)/pi))/(800*d*t an(c/2 + d*x/2)**2 - 960*d*tan(c/2 + d*x/2) + 800*d) + 36*tan(c/2 + d*x/2) /(800*d*tan(c/2 + d*x/2)**2 - 960*d*tan(c/2 + d*x/2) + 800*d) - 60/(800*d* tan(c/2 + d*x/2)**2 - 960*d*tan(c/2 + d*x/2) + 800*d), True))
Time = 0.11 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.60 \[ \int \frac {1}{(5-3 \sin (c+d x))^2} \, dx=-\frac {\frac {12 \, {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 5\right )}}{\frac {6 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {5 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 5} - 25 \, \arctan \left (\frac {5 \, \sin \left (d x + c\right )}{4 \, {\left (\cos \left (d x + c\right ) + 1\right )}} - \frac {3}{4}\right )}{160 \, d} \] Input:
integrate(1/(5-3*sin(d*x+c))^2,x, algorithm="maxima")
Output:
-1/160*(12*(3*sin(d*x + c)/(cos(d*x + c) + 1) - 5)/(6*sin(d*x + c)/(cos(d* x + c) + 1) - 5*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 5) - 25*arctan(5/4*s in(d*x + c)/(cos(d*x + c) + 1) - 3/4))/d
Time = 0.12 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.66 \[ \int \frac {1}{(5-3 \sin (c+d x))^2} \, dx=\frac {25 \, d x + 25 \, c + \frac {24 \, {\left (3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5\right )}}{5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5} + 50 \, \arctan \left (\frac {3 \, \cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 3}{\cos \left (d x + c\right ) + 3 \, \sin \left (d x + c\right ) - 9}\right )}{320 \, d} \] Input:
integrate(1/(5-3*sin(d*x+c))^2,x, algorithm="giac")
Output:
1/320*(25*d*x + 25*c + 24*(3*tan(1/2*d*x + 1/2*c) - 5)/(5*tan(1/2*d*x + 1/ 2*c)^2 - 6*tan(1/2*d*x + 1/2*c) + 5) + 50*arctan((3*cos(d*x + c) - sin(d*x + c) + 3)/(cos(d*x + c) + 3*sin(d*x + c) - 9)))/d
Time = 26.01 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.43 \[ \int \frac {1}{(5-3 \sin (c+d x))^2} \, dx=\frac {5\,\mathrm {atan}\left (\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}-\frac {3}{4}\right )}{32\,d}-\frac {5\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{32\,d}+\frac {\frac {9\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{200}-\frac {3}{40}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\frac {6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{5}+1\right )} \] Input:
int(1/(3*sin(c + d*x) - 5)^2,x)
Output:
(5*atan((5*tan(c/2 + (d*x)/2))/4 - 3/4))/(32*d) - (5*(atan(tan(c/2 + (d*x) /2)) - (d*x)/2))/(32*d) + ((9*tan(c/2 + (d*x)/2))/200 - 3/40)/(d*(tan(c/2 + (d*x)/2)^2 - (6*tan(c/2 + (d*x)/2))/5 + 1))
Time = 0.16 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.10 \[ \int \frac {1}{(5-3 \sin (c+d x))^2} \, dx=\frac {15 \mathit {atan} \left (\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4}-\frac {3}{4}\right ) \sin \left (d x +c \right )-25 \mathit {atan} \left (\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4}-\frac {3}{4}\right )+6 \cos \left (d x +c \right )}{32 d \left (3 \sin \left (d x +c \right )-5\right )} \] Input:
int(1/(5-3*sin(d*x+c))^2,x)
Output:
(15*atan((5*tan((c + d*x)/2) - 3)/4)*sin(c + d*x) - 25*atan((5*tan((c + d* x)/2) - 3)/4) + 6*cos(c + d*x))/(32*d*(3*sin(c + d*x) - 5))