Integrand size = 12, antiderivative size = 88 \[ \int \frac {1}{(3-5 \cos (c+d x))^2} \, dx=-\frac {3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-2 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{64 d}+\frac {3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+2 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{64 d}-\frac {5 \sin (c+d x)}{16 d (3-5 \cos (c+d x))} \]
-3/64*ln(cos(1/2*d*x+1/2*c)-2*sin(1/2*d*x+1/2*c))/d+3/64*ln(cos(1/2*d*x+1/ 2*c)+2*sin(1/2*d*x+1/2*c))/d-5/16*sin(d*x+c)/d/(3-5*cos(d*x+c))
Time = 0.20 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.58 \[ \int \frac {1}{(3-5 \cos (c+d x))^2} \, dx=\frac {9 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-2 \sin \left (\frac {1}{2} (c+d x)\right )\right )-15 \cos (c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-2 \sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+2 \sin \left (\frac {1}{2} (c+d x)\right )\right )\right )-9 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+2 \sin \left (\frac {1}{2} (c+d x)\right )\right )+20 \sin (c+d x)}{64 d (-3+5 \cos (c+d x))} \]
(9*Log[Cos[(c + d*x)/2] - 2*Sin[(c + d*x)/2]] - 15*Cos[c + d*x]*(Log[Cos[( c + d*x)/2] - 2*Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + 2*Sin[(c + d*x) /2]]) - 9*Log[Cos[(c + d*x)/2] + 2*Sin[(c + d*x)/2]] + 20*Sin[c + d*x])/(6 4*d*(-3 + 5*Cos[c + d*x]))
Time = 0.26 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.52, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 3143, 27, 3042, 3138, 220}
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.
\(\displaystyle \int \frac {1}{(3-5 \cos (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (3-5 \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 3143 |
\(\displaystyle \frac {1}{16} \int -\frac {3}{3-5 \cos (c+d x)}dx-\frac {5 \sin (c+d x)}{16 d (3-5 \cos (c+d x))}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {3}{16} \int \frac {1}{3-5 \cos (c+d x)}dx-\frac {5 \sin (c+d x)}{16 d (3-5 \cos (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {3}{16} \int \frac {1}{3-5 \sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {5 \sin (c+d x)}{16 d (3-5 \cos (c+d x))}\) |
\(\Big \downarrow \) 3138 |
\(\displaystyle -\frac {3 \int \frac {1}{8 \tan ^2\left (\frac {1}{2} (c+d x)\right )-2}d\tan \left (\frac {1}{2} (c+d x)\right )}{8 d}-\frac {5 \sin (c+d x)}{16 d (3-5 \cos (c+d x))}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle \frac {3 \text {arctanh}\left (2 \tan \left (\frac {1}{2} (c+d x)\right )\right )}{32 d}-\frac {5 \sin (c+d x)}{16 d (3-5 \cos (c+d x))}\) |
3.1.39.3.1 Defintions of rubi rules used
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_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
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 = 0.64 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.81
method | result | size |
norman | \(-\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 d \left (4 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}-\frac {3 \ln \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{64 d}+\frac {3 \ln \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{64 d}\) | \(71\) |
derivativedivides | \(\frac {-\frac {5}{64 \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {3 \ln \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{64}-\frac {5}{64 \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {3 \ln \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{64}}{d}\) | \(72\) |
default | \(\frac {-\frac {5}{64 \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {3 \ln \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{64}-\frac {5}{64 \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {3 \ln \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{64}}{d}\) | \(72\) |
risch | \(-\frac {i \left (3 \,{\mathrm e}^{i \left (d x +c \right )}-5\right )}{8 d \left (5 \,{\mathrm e}^{2 i \left (d x +c \right )}-6 \,{\mathrm e}^{i \left (d x +c \right )}+5\right )}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {3}{5}+\frac {4 i}{5}\right )}{64 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {3}{5}-\frac {4 i}{5}\right )}{64 d}\) | \(85\) |
parallelrisch | \(\frac {-15 \ln \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (d x +c \right )+15 \ln \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (d x +c \right )+20 \sin \left (d x +c \right )+9 \ln \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-9 \ln \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{64 d \left (-3+5 \cos \left (d x +c \right )\right )}\) | \(103\) |
-5/16/d*tan(1/2*d*x+1/2*c)/(4*tan(1/2*d*x+1/2*c)^2-1)-3/64/d*ln(2*tan(1/2* d*x+1/2*c)-1)+3/64/d*ln(2*tan(1/2*d*x+1/2*c)+1)
Time = 0.26 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(3-5 \cos (c+d x))^2} \, dx=\frac {3 \, {\left (5 \, \cos \left (d x + c\right ) - 3\right )} \log \left (-\frac {3}{2} \, \cos \left (d x + c\right ) + 2 \, \sin \left (d x + c\right ) + \frac {5}{2}\right ) - 3 \, {\left (5 \, \cos \left (d x + c\right ) - 3\right )} \log \left (-\frac {3}{2} \, \cos \left (d x + c\right ) - 2 \, \sin \left (d x + c\right ) + \frac {5}{2}\right ) + 40 \, \sin \left (d x + c\right )}{128 \, {\left (5 \, d \cos \left (d x + c\right ) - 3 \, d\right )}} \]
1/128*(3*(5*cos(d*x + c) - 3)*log(-3/2*cos(d*x + c) + 2*sin(d*x + c) + 5/2 ) - 3*(5*cos(d*x + c) - 3)*log(-3/2*cos(d*x + c) - 2*sin(d*x + c) + 5/2) + 40*sin(d*x + c))/(5*d*cos(d*x + c) - 3*d)
Leaf count of result is larger than twice the leaf count of optimal. 240 vs. \(2 (78) = 156\).
Time = 0.71 (sec) , antiderivative size = 240, normalized size of antiderivative = 2.73 \[ \int \frac {1}{(3-5 \cos (c+d x))^2} \, dx=\begin {cases} \frac {x}{\left (3 - 5 \cos {\left (2 \operatorname {atan}{\left (\frac {1}{2} \right )} \right )}\right )^{2}} & \text {for}\: c = - d x - 2 \operatorname {atan}{\left (\frac {1}{2} \right )} \vee c = - d x + 2 \operatorname {atan}{\left (\frac {1}{2} \right )} \\\frac {x}{\left (3 - 5 \cos {\left (c \right )}\right )^{2}} & \text {for}\: d = 0 \\- \frac {12 \log {\left (2 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - 1 \right )} \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{256 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 64 d} + \frac {3 \log {\left (2 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - 1 \right )}}{256 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 64 d} + \frac {12 \log {\left (2 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 1 \right )} \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{256 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 64 d} - \frac {3 \log {\left (2 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 1 \right )}}{256 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 64 d} - \frac {20 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{256 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 64 d} & \text {otherwise} \end {cases} \]
Piecewise((x/(3 - 5*cos(2*atan(1/2)))**2, Eq(c, -d*x - 2*atan(1/2)) | Eq(c , -d*x + 2*atan(1/2))), (x/(3 - 5*cos(c))**2, Eq(d, 0)), (-12*log(2*tan(c/ 2 + d*x/2) - 1)*tan(c/2 + d*x/2)**2/(256*d*tan(c/2 + d*x/2)**2 - 64*d) + 3 *log(2*tan(c/2 + d*x/2) - 1)/(256*d*tan(c/2 + d*x/2)**2 - 64*d) + 12*log(2 *tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2)**2/(256*d*tan(c/2 + d*x/2)**2 - 64 *d) - 3*log(2*tan(c/2 + d*x/2) + 1)/(256*d*tan(c/2 + d*x/2)**2 - 64*d) - 2 0*tan(c/2 + d*x/2)/(256*d*tan(c/2 + d*x/2)**2 - 64*d), True))
Time = 0.28 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.07 \[ \int \frac {1}{(3-5 \cos (c+d x))^2} \, dx=-\frac {\frac {20 \, \sin \left (d x + c\right )}{{\left (\frac {4 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} - 3 \, \log \left (\frac {2 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right ) + 3 \, \log \left (\frac {2 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{64 \, d} \]
-1/64*(20*sin(d*x + c)/((4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 1)*(cos(d *x + c) + 1)) - 3*log(2*sin(d*x + c)/(cos(d*x + c) + 1) + 1) + 3*log(2*sin (d*x + c)/(cos(d*x + c) + 1) - 1))/d
Time = 0.28 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.77 \[ \int \frac {1}{(3-5 \cos (c+d x))^2} \, dx=-\frac {\frac {20 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - 3 \, \log \left ({\left | 2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 3 \, \log \left ({\left | 2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{64 \, d} \]
-1/64*(20*tan(1/2*d*x + 1/2*c)/(4*tan(1/2*d*x + 1/2*c)^2 - 1) - 3*log(abs( 2*tan(1/2*d*x + 1/2*c) + 1)) + 3*log(abs(2*tan(1/2*d*x + 1/2*c) - 1)))/d
Time = 14.32 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.53 \[ \int \frac {1}{(3-5 \cos (c+d x))^2} \, dx=\frac {3\,\mathrm {atanh}\left (2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{32\,d}-\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64\,d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\frac {1}{4}\right )} \]