Integrand size = 12, antiderivative size = 90 \[ \int \frac {1}{(3-5 \sin (c+d x))^2} \, dx=\frac {3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-3 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{64 d}-\frac {3 \log \left (3 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{64 d}+\frac {5 \cos (c+d x)}{16 d (3-5 \sin (c+d x))} \]
3/64*ln(cos(1/2*d*x+1/2*c)-3*sin(1/2*d*x+1/2*c))/d-3/64*ln(3*cos(1/2*d*x+1 /2*c)-sin(1/2*d*x+1/2*c))/d+5/16*cos(d*x+c)/d/(3-5*sin(d*x+c))
Time = 0.21 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.44 \[ \int \frac {1}{(3-5 \sin (c+d x))^2} \, dx=\frac {9 \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-3 \sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (3 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+20 \left (\frac {3}{\cos \left (\frac {1}{2} (c+d x)\right )-3 \sin \left (\frac {1}{2} (c+d x)\right )}+\frac {1}{3 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}\right ) \sin \left (\frac {1}{2} (c+d x)\right )}{192 d} \]
(9*(Log[Cos[(c + d*x)/2] - 3*Sin[(c + d*x)/2]] - Log[3*Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]) + 20*(3/(Cos[(c + d*x)/2] - 3*Sin[(c + d*x)/2]) + (3*Co s[(c + d*x)/2] - Sin[(c + d*x)/2])^(-1))*Sin[(c + d*x)/2])/(192*d)
Time = 0.28 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.80, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {3042, 3143, 27, 3042, 3139, 1081, 2009}
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.
| \(\displaystyle \int \frac {1}{(3-5 \sin (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(3-5 \sin (c+d x))^2}dx\) |
| \(\Big \downarrow \) 3143 |
| \(\displaystyle \frac {1}{16} \int -\frac {3}{3-5 \sin (c+d x)}dx+\frac {5 \cos (c+d x)}{16 d (3-5 \sin (c+d x))}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {5 \cos (c+d x)}{16 d (3-5 \sin (c+d x))}-\frac {3}{16} \int \frac {1}{3-5 \sin (c+d x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5 \cos (c+d x)}{16 d (3-5 \sin (c+d x))}-\frac {3}{16} \int \frac {1}{3-5 \sin (c+d x)}dx\) |
| \(\Big \downarrow \) 3139 |
| \(\displaystyle \frac {5 \cos (c+d x)}{16 d (3-5 \sin (c+d x))}-\frac {3 \int \frac {1}{3 \tan ^2\left (\frac {1}{2} (c+d x)\right )-10 \tan \left (\frac {1}{2} (c+d x)\right )+3}d\tan \left (\frac {1}{2} (c+d x)\right )}{8 d}\) |
| \(\Big \downarrow \) 1081 |
| \(\displaystyle \frac {5 \cos (c+d x)}{16 d (3-5 \sin (c+d x))}-\frac {9 \int \left (\frac {1}{8 \left (1-3 \tan \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {1}{24 \left (3-\tan \left (\frac {1}{2} (c+d x)\right )\right )}\right )d\tan \left (\frac {1}{2} (c+d x)\right )}{8 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {5 \cos (c+d x)}{16 d (3-5 \sin (c+d x))}-\frac {9 \left (\frac {1}{24} \log \left (3-\tan \left (\frac {1}{2} (c+d x)\right )\right )-\frac {1}{24} \log \left (1-3 \tan \left (\frac {1}{2} (c+d x)\right )\right )\right )}{8 d}\) |
(-9*(-1/24*Log[1 - 3*Tan[(c + d*x)/2]] + Log[3 - Tan[(c + d*x)/2]]/24))/(8 *d) + (5*Cos[c + d*x])/(16*d*(3 - 5*Sin[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_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[c Int[ExpandIntegrand[1/((b/2 - q/2 + c*x)*(b/2 + q/2 + c*x)), x], x], x]] /; FreeQ[{a, b, c}, x] && NiceSqrtQ[b^2 - 4*a*c]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + 2*b*e*x + a *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 = 0.26 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.76
| method | result | size |
| derivativedivides | \(\frac {-\frac {5}{16 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3\right )}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3\right )}{64}-\frac {5}{48 \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {3 \ln \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{64}}{d}\) | \(68\) |
| default | \(\frac {-\frac {5}{16 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3\right )}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3\right )}{64}-\frac {5}{48 \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {3 \ln \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{64}}{d}\) | \(68\) |
| risch | \(\frac {3 \,{\mathrm e}^{i \left (d x +c \right )}-5 i}{8 d \left (5 \,{\mathrm e}^{2 i \left (d x +c \right )}-5-6 i {\mathrm e}^{i \left (d x +c \right )}\right )}+\frac {3 \ln \left (-\frac {4}{5}-\frac {3 i}{5}+{\mathrm e}^{i \left (d x +c \right )}\right )}{64 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {4}{5}-\frac {3 i}{5}\right )}{64 d}\) | \(86\) |
| norman | \(\frac {\frac {5}{8 d}-\frac {25 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{24 d}}{3 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+3}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3\right )}{64 d}+\frac {3 \ln \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{64 d}\) | \(87\) |
| parallelrisch | \(\frac {-45 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3\right ) \sin \left (d x +c \right )+45 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{3}\right ) \sin \left (d x +c \right )+27 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3\right )-27 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{3}\right )+100 \sin \left (d x +c \right )-60 \cos \left (d x +c \right )-60}{192 d \left (-3+5 \sin \left (d x +c \right )\right )}\) | \(104\) |
1/d*(-5/16/(tan(1/2*d*x+1/2*c)-3)-3/64*ln(tan(1/2*d*x+1/2*c)-3)-5/48/(3*ta n(1/2*d*x+1/2*c)-1)+3/64*ln(3*tan(1/2*d*x+1/2*c)-1))
Time = 0.28 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.98 \[ \int \frac {1}{(3-5 \sin (c+d x))^2} \, dx=-\frac {3 \, {\left (5 \, \sin \left (d x + c\right ) - 3\right )} \log \left (4 \, \cos \left (d x + c\right ) - 3 \, \sin \left (d x + c\right ) + 5\right ) - 3 \, {\left (5 \, \sin \left (d x + c\right ) - 3\right )} \log \left (-4 \, \cos \left (d x + c\right ) - 3 \, \sin \left (d x + c\right ) + 5\right ) + 40 \, \cos \left (d x + c\right )}{128 \, {\left (5 \, d \sin \left (d x + c\right ) - 3 \, d\right )}} \]
-1/128*(3*(5*sin(d*x + c) - 3)*log(4*cos(d*x + c) - 3*sin(d*x + c) + 5) - 3*(5*sin(d*x + c) - 3)*log(-4*cos(d*x + c) - 3*sin(d*x + c) + 5) + 40*cos( d*x + c))/(5*d*sin(d*x + c) - 3*d)
Leaf count of result is larger than twice the leaf count of optimal. 462 vs. \(2 (78) = 156\).
Time = 0.75 (sec) , antiderivative size = 462, normalized size of antiderivative = 5.13 \[ \int \frac {1}{(3-5 \sin (c+d x))^2} \, dx=\begin {cases} \frac {x}{\left (3 - 5 \sin {\left (2 \operatorname {atan}{\left (\frac {1}{3} \right )} \right )}\right )^{2}} & \text {for}\: c = - d x + 2 \operatorname {atan}{\left (\frac {1}{3} \right )} \\\frac {x}{\left (3 - 5 \sin {\left (2 \operatorname {atan}{\left (3 \right )} \right )}\right )^{2}} & \text {for}\: c = - d x + 2 \operatorname {atan}{\left (3 \right )} \\\frac {x}{\left (3 - 5 \sin {\left (c \right )}\right )^{2}} & \text {for}\: d = 0 \\- \frac {27 \log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - 3 \right )} \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{576 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 1920 d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 576 d} + \frac {90 \log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - 3 \right )} \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{576 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 1920 d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 576 d} - \frac {27 \log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - 3 \right )}}{576 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 1920 d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 576 d} + \frac {27 \log {\left (3 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - 1 \right )} \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{576 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 1920 d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 576 d} - \frac {90 \log {\left (3 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - 1 \right )} \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{576 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 1920 d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 576 d} + \frac {27 \log {\left (3 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - 1 \right )}}{576 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 1920 d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 576 d} - \frac {200 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{576 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 1920 d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 576 d} + \frac {120}{576 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 1920 d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 576 d} & \text {otherwise} \end {cases} \]
Piecewise((x/(3 - 5*sin(2*atan(1/3)))**2, Eq(c, -d*x + 2*atan(1/3))), (x/( 3 - 5*sin(2*atan(3)))**2, Eq(c, -d*x + 2*atan(3))), (x/(3 - 5*sin(c))**2, Eq(d, 0)), (-27*log(tan(c/2 + d*x/2) - 3)*tan(c/2 + d*x/2)**2/(576*d*tan(c /2 + d*x/2)**2 - 1920*d*tan(c/2 + d*x/2) + 576*d) + 90*log(tan(c/2 + d*x/2 ) - 3)*tan(c/2 + d*x/2)/(576*d*tan(c/2 + d*x/2)**2 - 1920*d*tan(c/2 + d*x/ 2) + 576*d) - 27*log(tan(c/2 + d*x/2) - 3)/(576*d*tan(c/2 + d*x/2)**2 - 19 20*d*tan(c/2 + d*x/2) + 576*d) + 27*log(3*tan(c/2 + d*x/2) - 1)*tan(c/2 + d*x/2)**2/(576*d*tan(c/2 + d*x/2)**2 - 1920*d*tan(c/2 + d*x/2) + 576*d) - 90*log(3*tan(c/2 + d*x/2) - 1)*tan(c/2 + d*x/2)/(576*d*tan(c/2 + d*x/2)**2 - 1920*d*tan(c/2 + d*x/2) + 576*d) + 27*log(3*tan(c/2 + d*x/2) - 1)/(576* d*tan(c/2 + d*x/2)**2 - 1920*d*tan(c/2 + d*x/2) + 576*d) - 200*tan(c/2 + d *x/2)/(576*d*tan(c/2 + d*x/2)**2 - 1920*d*tan(c/2 + d*x/2) + 576*d) + 120/ (576*d*tan(c/2 + d*x/2)**2 - 1920*d*tan(c/2 + d*x/2) + 576*d), True))
Time = 0.19 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.28 \[ \int \frac {1}{(3-5 \sin (c+d x))^2} \, dx=\frac {\frac {40 \, {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 3\right )}}{\frac {10 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {3 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 3} + 9 \, \log \left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right ) - 9 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 3\right )}{192 \, d} \]
1/192*(40*(5*sin(d*x + c)/(cos(d*x + c) + 1) - 3)/(10*sin(d*x + c)/(cos(d* x + c) + 1) - 3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 3) + 9*log(3*sin(d*x + c)/(cos(d*x + c) + 1) - 1) - 9*log(sin(d*x + c)/(cos(d*x + c) + 1) - 3) )/d
Time = 0.30 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.90 \[ \int \frac {1}{(3-5 \sin (c+d x))^2} \, dx=-\frac {\frac {40 \, {\left (5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3\right )}}{3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 10 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3} - 9 \, \log \left ({\left | 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 9 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \right |}\right )}{192 \, d} \]
-1/192*(40*(5*tan(1/2*d*x + 1/2*c) - 3)/(3*tan(1/2*d*x + 1/2*c)^2 - 10*tan (1/2*d*x + 1/2*c) + 3) - 9*log(abs(3*tan(1/2*d*x + 1/2*c) - 1)) + 9*log(ab s(tan(1/2*d*x + 1/2*c) - 3)))/d
Time = 6.11 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.71 \[ \int \frac {1}{(3-5 \sin (c+d x))^2} \, dx=\frac {3\,\mathrm {atanh}\left (\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}-\frac {5}{4}\right )}{32\,d}-\frac {\frac {25\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{72}-\frac {5}{24}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\frac {10\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+1\right )} \]