Integrand size = 12, antiderivative size = 58 \[ \int \frac {1}{(5-3 \cos (c+d x))^2} \, dx=\frac {5 x}{64}+\frac {5 \arctan \left (\frac {\sin (c+d x)}{3-\cos (c+d x)}\right )}{32 d}+\frac {3 \sin (c+d x)}{16 d (5-3 \cos (c+d x))} \] Output:
5/64*x+5/32*arctan(sin(d*x+c)/(3-cos(d*x+c)))/d+3/16*sin(d*x+c)/d/(5-3*cos (d*x+c))
Time = 0.22 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.74 \[ \int \frac {1}{(5-3 \cos (c+d x))^2} \, dx=\frac {5 \arctan \left (2 \tan \left (\frac {1}{2} (c+d x)\right )\right )-\frac {6 \sin (c+d x)}{-5+3 \cos (c+d x)}}{32 d} \] Input:
Integrate[(5 - 3*Cos[c + d*x])^(-2),x]
Output:
(5*ArcTan[2*Tan[(c + d*x)/2]] - (6*Sin[c + d*x])/(-5 + 3*Cos[c + d*x]))/(3 2*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.
\(\displaystyle \int \frac {1}{(5-3 \cos (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (5-3 \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 3143 |
\(\displaystyle \frac {3 \sin (c+d x)}{16 d (5-3 \cos (c+d x))}-\frac {1}{16} \int -\frac {5}{5-3 \cos (c+d x)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {5}{16} \int \frac {1}{5-3 \cos (c+d x)}dx+\frac {3 \sin (c+d x)}{16 d (5-3 \cos (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5}{16} \int \frac {1}{5-3 \sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {3 \sin (c+d x)}{16 d (5-3 \cos (c+d x))}\) |
\(\Big \downarrow \) 3136 |
\(\displaystyle \frac {5}{16} \left (\frac {\arctan \left (\frac {\sin (c+d x)}{3-\cos (c+d x)}\right )}{2 d}+\frac {x}{4}\right )+\frac {3 \sin (c+d x)}{16 d (5-3 \cos (c+d x))}\) |
Input:
Int[(5 - 3*Cos[c + d*x])^(-2),x]
Output:
(5*(x/4 + ArcTan[Sin[c + d*x]/(3 - Cos[c + d*x])]/(2*d)))/16 + (3*Sin[c + d*x])/(16*d*(5 - 3*Cos[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 = 0.96 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.79
method | result | size |
derivativedivides | \(\frac {\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{64 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\frac {1}{4}\right )}+\frac {5 \arctan \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32}}{d}\) | \(46\) |
default | \(\frac {\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{64 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\frac {1}{4}\right )}+\frac {5 \arctan \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32}}{d}\) | \(46\) |
parallelrisch | \(\frac {\left (-15 i \cos \left (d x +c \right )+25 i\right ) \ln \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )+\left (15 i \cos \left (d x +c \right )-25 i\right ) \ln \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )-12 \sin \left (d x +c \right )}{192 d \cos \left (d x +c \right )-320 d}\) | \(82\) |
risch | \(\frac {i \left (5 \,{\mathrm e}^{i \left (d x +c \right )}-3\right )}{8 d \left (3 \,{\mathrm e}^{2 i \left (d x +c \right )}-10 \,{\mathrm e}^{i \left (d x +c \right )}+3\right )}-\frac {5 i \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {1}{3}\right )}{64 d}+\frac {5 i \ln \left ({\mathrm e}^{i \left (d x +c \right )}-3\right )}{64 d}\) | \(83\) |
Input:
int(1/(5-3*cos(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(3/64*tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2+1/4)+5/32*arctan(2*tan( 1/2*d*x+1/2*c)))
Time = 0.09 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.02 \[ \int \frac {1}{(5-3 \cos (c+d x))^2} \, dx=-\frac {5 \, {\left (3 \, \cos \left (d x + c\right ) - 5\right )} \arctan \left (\frac {5 \, \cos \left (d x + c\right ) - 3}{4 \, \sin \left (d x + c\right )}\right ) + 12 \, \sin \left (d x + c\right )}{64 \, {\left (3 \, d \cos \left (d x + c\right ) - 5 \, d\right )}} \] Input:
integrate(1/(5-3*cos(d*x+c))^2,x, algorithm="fricas")
Output:
-1/64*(5*(3*cos(d*x + c) - 5)*arctan(1/4*(5*cos(d*x + c) - 3)/sin(d*x + c) ) + 12*sin(d*x + c))/(3*d*cos(d*x + c) - 5*d)
Result contains complex when optimal does not.
Time = 0.69 (sec) , antiderivative size = 192, normalized size of antiderivative = 3.31 \[ \int \frac {1}{(5-3 \cos (c+d x))^2} \, dx=\begin {cases} \frac {x}{\left (5 - 3 \cosh {\left (2 \operatorname {atanh}{\left (\frac {1}{2} \right )} \right )}\right )^{2}} & \text {for}\: c = - d x - 2 i \operatorname {atanh}{\left (\frac {1}{2} \right )} \vee c = - d x + 2 i \operatorname {atanh}{\left (\frac {1}{2} \right )} \\\frac {x}{\left (5 - 3 \cos {\left (c \right )}\right )^{2}} & \text {for}\: d = 0 \\\frac {20 \left (\operatorname {atan}{\left (2 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} \right )} + \pi \left \lfloor {\frac {\frac {c}{2} + \frac {d x}{2} - \frac {\pi }{2}}{\pi }}\right \rfloor \right ) \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{128 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 32 d} + \frac {5 \left (\operatorname {atan}{\left (2 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} \right )} + \pi \left \lfloor {\frac {\frac {c}{2} + \frac {d x}{2} - \frac {\pi }{2}}{\pi }}\right \rfloor \right )}{128 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 32 d} + \frac {6 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{128 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 32 d} & \text {otherwise} \end {cases} \] Input:
integrate(1/(5-3*cos(d*x+c))**2,x)
Output:
Piecewise((x/(5 - 3*cosh(2*atanh(1/2)))**2, Eq(c, -d*x - 2*I*atanh(1/2)) | Eq(c, -d*x + 2*I*atanh(1/2))), (x/(5 - 3*cos(c))**2, Eq(d, 0)), (20*(atan (2*tan(c/2 + d*x/2)) + pi*floor((c/2 + d*x/2 - pi/2)/pi))*tan(c/2 + d*x/2) **2/(128*d*tan(c/2 + d*x/2)**2 + 32*d) + 5*(atan(2*tan(c/2 + d*x/2)) + pi* floor((c/2 + d*x/2 - pi/2)/pi))/(128*d*tan(c/2 + d*x/2)**2 + 32*d) + 6*tan (c/2 + d*x/2)/(128*d*tan(c/2 + d*x/2)**2 + 32*d), True))
Time = 0.11 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.19 \[ \int \frac {1}{(5-3 \cos (c+d x))^2} \, dx=\frac {\frac {6 \, \sin \left (d x + c\right )}{{\left (\frac {4 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + 5 \, \arctan \left (\frac {2 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{32 \, d} \] Input:
integrate(1/(5-3*cos(d*x+c))^2,x, algorithm="maxima")
Output:
1/32*(6*sin(d*x + c)/((4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)*(cos(d*x + c) + 1)) + 5*arctan(2*sin(d*x + c)/(cos(d*x + c) + 1)))/d
Time = 0.13 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.05 \[ \int \frac {1}{(5-3 \cos (c+d x))^2} \, dx=\frac {5 \, d x + 5 \, c + \frac {12 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} - 10 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) - 3}\right )}{64 \, d} \] Input:
integrate(1/(5-3*cos(d*x+c))^2,x, algorithm="giac")
Output:
1/64*(5*d*x + 5*c + 12*tan(1/2*d*x + 1/2*c)/(4*tan(1/2*d*x + 1/2*c)^2 + 1) - 10*arctan(sin(d*x + c)/(cos(d*x + c) - 3)))/d
Time = 42.56 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.16 \[ \int \frac {1}{(5-3 \cos (c+d x))^2} \, dx=\frac {5\,\mathrm {atan}\left (2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{32\,d}-\frac {5\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{32\,d}+\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64\,d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {1}{4}\right )} \] Input:
int(1/(3*cos(c + d*x) - 5)^2,x)
Output:
(5*atan(2*tan(c/2 + (d*x)/2)))/(32*d) - (5*(atan(tan(c/2 + (d*x)/2)) - (d* x)/2))/(32*d) + (3*tan(c/2 + (d*x)/2))/(64*d*(tan(c/2 + (d*x)/2)^2 + 1/4))
Time = 0.17 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.03 \[ \int \frac {1}{(5-3 \cos (c+d x))^2} \, dx=\frac {15 \mathit {atan} \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (d x +c \right )-25 \mathit {atan} \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6 \sin \left (d x +c \right )}{32 d \left (3 \cos \left (d x +c \right )-5\right )} \] Input:
int(1/(5-3*cos(d*x+c))^2,x)
Output:
(15*atan(2*tan((c + d*x)/2))*cos(c + d*x) - 25*atan(2*tan((c + d*x)/2)) - 6*sin(c + d*x))/(32*d*(3*cos(c + d*x) - 5))