Integrand size = 12, antiderivative size = 115 \[ \int \frac {1}{(-3+5 \sin (c+d x))^3} \, dx=\frac {43 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-3 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{2048 d}-\frac {43 \log \left (3 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2048 d}-\frac {5 \cos (c+d x)}{32 d (3-5 \sin (c+d x))^2}+\frac {45 \cos (c+d x)}{512 d (3-5 \sin (c+d x))} \]
43/2048*ln(cos(1/2*d*x+1/2*c)-3*sin(1/2*d*x+1/2*c))/d-43/2048*ln(3*cos(1/2 *d*x+1/2*c)-sin(1/2*d*x+1/2*c))/d-5/32*cos(d*x+c)/d/(3-5*sin(d*x+c))^2+45/ 512*cos(d*x+c)/d/(3-5*sin(d*x+c))
Time = 0.37 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.59 \[ \int \frac {1}{(-3+5 \sin (c+d x))^3} \, dx=\frac {43 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-3 \sin \left (\frac {1}{2} (c+d x)\right )\right )-43 \log \left (3 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\frac {40}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-3 \sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+60 \left (\frac {3}{\cos \left (\frac {1}{2} (c+d x)\right )-3 \sin \left (\frac {1}{2} (c+d x)\right )}+\frac {1}{3 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}\right ) \sin \left (\frac {1}{2} (c+d x)\right )+\frac {40}{\left (-3 \cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}}{2048 d} \]
(43*Log[Cos[(c + d*x)/2] - 3*Sin[(c + d*x)/2]] - 43*Log[3*Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 40/(Cos[(c + d*x)/2] - 3*Sin[(c + d*x)/2])^2 + 60*( 3/(Cos[(c + d*x)/2] - 3*Sin[(c + d*x)/2]) + (3*Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^(-1))*Sin[(c + d*x)/2] + 40/(-3*Cos[(c + d*x)/2] + Sin[(c + d*x) /2])^2)/(2048*d)
Time = 0.39 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.89, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {3042, 3143, 3042, 3233, 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.
| \(\displaystyle \int \frac {1}{(5 \sin (c+d x)-3)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(5 \sin (c+d x)-3)^3}dx\) |
| \(\Big \downarrow \) 3143 |
| \(\displaystyle \frac {1}{32} \int \frac {5 \sin (c+d x)+6}{(3-5 \sin (c+d x))^2}dx-\frac {5 \cos (c+d x)}{32 d (3-5 \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{32} \int \frac {5 \sin (c+d x)+6}{(3-5 \sin (c+d x))^2}dx-\frac {5 \cos (c+d x)}{32 d (3-5 \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 3233 |
| \(\displaystyle \frac {1}{32} \left (\frac {1}{16} \int -\frac {43}{3-5 \sin (c+d x)}dx+\frac {45 \cos (c+d x)}{16 d (3-5 \sin (c+d x))}\right )-\frac {5 \cos (c+d x)}{32 d (3-5 \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{32} \left (\frac {45 \cos (c+d x)}{16 d (3-5 \sin (c+d x))}-\frac {43}{16} \int \frac {1}{3-5 \sin (c+d x)}dx\right )-\frac {5 \cos (c+d x)}{32 d (3-5 \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{32} \left (\frac {45 \cos (c+d x)}{16 d (3-5 \sin (c+d x))}-\frac {43}{16} \int \frac {1}{3-5 \sin (c+d x)}dx\right )-\frac {5 \cos (c+d x)}{32 d (3-5 \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 3139 |
| \(\displaystyle \frac {1}{32} \left (\frac {45 \cos (c+d x)}{16 d (3-5 \sin (c+d x))}-\frac {43 \int \frac {1}{3 \tan ^2\left (\frac {1}{2} (c+d x)\right )-10 \tan \left (\frac {1}{2} (c+d x)\right )+3}d\tan \left (\frac {1}{2} (c+d x)\right )}{8 d}\right )-\frac {5 \cos (c+d x)}{32 d (3-5 \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 1081 |
| \(\displaystyle \frac {1}{32} \left (\frac {45 \cos (c+d x)}{16 d (3-5 \sin (c+d x))}-\frac {129 \int \left (\frac {1}{8 \left (1-3 \tan \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {1}{24 \left (3-\tan \left (\frac {1}{2} (c+d x)\right )\right )}\right )d\tan \left (\frac {1}{2} (c+d x)\right )}{8 d}\right )-\frac {5 \cos (c+d x)}{32 d (3-5 \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{32} \left (\frac {45 \cos (c+d x)}{16 d (3-5 \sin (c+d x))}-\frac {129 \left (\frac {1}{24} \log \left (3-\tan \left (\frac {1}{2} (c+d x)\right )\right )-\frac {1}{24} \log \left (1-3 \tan \left (\frac {1}{2} (c+d x)\right )\right )\right )}{8 d}\right )-\frac {5 \cos (c+d x)}{32 d (3-5 \sin (c+d x))^2}\) |
((-129*(-1/24*Log[1 - 3*Tan[(c + d*x)/2]] + Log[3 - Tan[(c + d*x)/2]]/24)) /(8*d) + (45*Cos[c + d*x])/(16*d*(3 - 5*Sin[c + d*x])))/32 - (5*Cos[c + d* x])/(32*d*(3 - 5*Sin[c + d*x])^2)
3.1.44.3.1 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]
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 = 0.27 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.87
| method | result | size |
| derivativedivides | \(\frac {\frac {25}{128 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3\right )^{2}}+\frac {15}{512 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3\right )}-\frac {43 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3\right )}{2048}-\frac {25}{1152 \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {155}{4608 \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {43 \ln \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2048}}{d}\) | \(100\) |
| default | \(\frac {\frac {25}{128 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3\right )^{2}}+\frac {15}{512 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3\right )}-\frac {43 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3\right )}{2048}-\frac {25}{1152 \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {155}{4608 \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {43 \ln \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2048}}{d}\) | \(100\) |
| risch | \(\frac {-387 i {\mathrm e}^{2 i \left (d x +c \right )}+215 \,{\mathrm e}^{3 i \left (d x +c \right )}-325 \,{\mathrm e}^{i \left (d x +c \right )}+225 i}{256 \left (5 \,{\mathrm e}^{2 i \left (d x +c \right )}-5-6 i {\mathrm e}^{i \left (d x +c \right )}\right )^{2} d}-\frac {43 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {4}{5}-\frac {3 i}{5}\right )}{2048 d}+\frac {43 \ln \left (-\frac {4}{5}-\frac {3 i}{5}+{\mathrm e}^{i \left (d x +c \right )}\right )}{2048 d}\) | \(109\) |
| norman | \(\frac {\frac {55}{256 d}+\frac {3245 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2304 d}+\frac {125 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{768 d}-\frac {1225 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{768 d}}{{\left (3 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+3\right )}^{2}}-\frac {43 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3\right )}{2048 d}+\frac {43 \ln \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2048 d}\) | \(119\) |
| parallelrisch | \(\frac {\left (9675 \cos \left (2 d x +2 c \right )+23220 \sin \left (d x +c \right )-16641\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{3}\right )+\left (-9675 \cos \left (2 d x +2 c \right )-23220 \sin \left (d x +c \right )+16641\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3\right )-3960 \cos \left (d x +c \right )+5500 \cos \left (2 d x +2 c \right )+13200 \sin \left (d x +c \right )+8100 \sin \left (2 d x +2 c \right )-9460}{18432 d \left (-43+25 \cos \left (2 d x +2 c \right )+60 \sin \left (d x +c \right )\right )}\) | \(137\) |
1/d*(25/128/(tan(1/2*d*x+1/2*c)-3)^2+15/512/(tan(1/2*d*x+1/2*c)-3)-43/2048 *ln(tan(1/2*d*x+1/2*c)-3)-25/1152/(3*tan(1/2*d*x+1/2*c)-1)^2-155/4608/(3*t an(1/2*d*x+1/2*c)-1)+43/2048*ln(3*tan(1/2*d*x+1/2*c)-1))
Time = 0.27 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.16 \[ \int \frac {1}{(-3+5 \sin (c+d x))^3} \, dx=-\frac {43 \, {\left (25 \, \cos \left (d x + c\right )^{2} + 30 \, \sin \left (d x + c\right ) - 34\right )} \log \left (4 \, \cos \left (d x + c\right ) - 3 \, \sin \left (d x + c\right ) + 5\right ) - 43 \, {\left (25 \, \cos \left (d x + c\right )^{2} + 30 \, \sin \left (d x + c\right ) - 34\right )} \log \left (-4 \, \cos \left (d x + c\right ) - 3 \, \sin \left (d x + c\right ) + 5\right ) - 1800 \, \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 440 \, \cos \left (d x + c\right )}{4096 \, {\left (25 \, d \cos \left (d x + c\right )^{2} + 30 \, d \sin \left (d x + c\right ) - 34 \, d\right )}} \]
-1/4096*(43*(25*cos(d*x + c)^2 + 30*sin(d*x + c) - 34)*log(4*cos(d*x + c) - 3*sin(d*x + c) + 5) - 43*(25*cos(d*x + c)^2 + 30*sin(d*x + c) - 34)*log( -4*cos(d*x + c) - 3*sin(d*x + c) + 5) - 1800*cos(d*x + c)*sin(d*x + c) + 4 40*cos(d*x + c))/(25*d*cos(d*x + c)^2 + 30*d*sin(d*x + c) - 34*d)
Leaf count of result is larger than twice the leaf count of optimal. 1224 vs. \(2 (102) = 204\).
Time = 1.41 (sec) , antiderivative size = 1224, normalized size of antiderivative = 10.64 \[ \int \frac {1}{(-3+5 \sin (c+d x))^3} \, dx=\text {Too large to display} \]
Piecewise((x/(-3 + 5*sin(2*atan(1/3)))**3, Eq(c, -d*x + 2*atan(1/3))), (x/ (-3 + 5*sin(2*atan(3)))**3, Eq(c, -d*x + 2*atan(3))), (x/(5*sin(c) - 3)**3 , Eq(d, 0)), (-3483*log(tan(c/2 + d*x/2) - 3)*tan(c/2 + d*x/2)**4/(165888* d*tan(c/2 + d*x/2)**4 - 1105920*d*tan(c/2 + d*x/2)**3 + 2174976*d*tan(c/2 + d*x/2)**2 - 1105920*d*tan(c/2 + d*x/2) + 165888*d) + 23220*log(tan(c/2 + d*x/2) - 3)*tan(c/2 + d*x/2)**3/(165888*d*tan(c/2 + d*x/2)**4 - 1105920*d *tan(c/2 + d*x/2)**3 + 2174976*d*tan(c/2 + d*x/2)**2 - 1105920*d*tan(c/2 + d*x/2) + 165888*d) - 45666*log(tan(c/2 + d*x/2) - 3)*tan(c/2 + d*x/2)**2/ (165888*d*tan(c/2 + d*x/2)**4 - 1105920*d*tan(c/2 + d*x/2)**3 + 2174976*d* tan(c/2 + d*x/2)**2 - 1105920*d*tan(c/2 + d*x/2) + 165888*d) + 23220*log(t an(c/2 + d*x/2) - 3)*tan(c/2 + d*x/2)/(165888*d*tan(c/2 + d*x/2)**4 - 1105 920*d*tan(c/2 + d*x/2)**3 + 2174976*d*tan(c/2 + d*x/2)**2 - 1105920*d*tan( c/2 + d*x/2) + 165888*d) - 3483*log(tan(c/2 + d*x/2) - 3)/(165888*d*tan(c/ 2 + d*x/2)**4 - 1105920*d*tan(c/2 + d*x/2)**3 + 2174976*d*tan(c/2 + d*x/2) **2 - 1105920*d*tan(c/2 + d*x/2) + 165888*d) + 3483*log(3*tan(c/2 + d*x/2) - 1)*tan(c/2 + d*x/2)**4/(165888*d*tan(c/2 + d*x/2)**4 - 1105920*d*tan(c/ 2 + d*x/2)**3 + 2174976*d*tan(c/2 + d*x/2)**2 - 1105920*d*tan(c/2 + d*x/2) + 165888*d) - 23220*log(3*tan(c/2 + d*x/2) - 1)*tan(c/2 + d*x/2)**3/(1658 88*d*tan(c/2 + d*x/2)**4 - 1105920*d*tan(c/2 + d*x/2)**3 + 2174976*d*tan(c /2 + d*x/2)**2 - 1105920*d*tan(c/2 + d*x/2) + 165888*d) + 45666*log(3*t...
Time = 0.19 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.70 \[ \int \frac {1}{(-3+5 \sin (c+d x))^3} \, dx=\frac {\frac {40 \, {\left (\frac {735 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {649 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {75 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - 99\right )}}{\frac {60 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {118 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {60 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {9 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - 9} + 387 \, \log \left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right ) - 387 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 3\right )}{18432 \, d} \]
1/18432*(40*(735*sin(d*x + c)/(cos(d*x + c) + 1) - 649*sin(d*x + c)^2/(cos (d*x + c) + 1)^2 - 75*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 99)/(60*sin(d* x + c)/(cos(d*x + c) + 1) - 118*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 60*s in(d*x + c)^3/(cos(d*x + c) + 1)^3 - 9*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 9) + 387*log(3*sin(d*x + c)/(cos(d*x + c) + 1) - 1) - 387*log(sin(d*x + c)/(cos(d*x + c) + 1) - 3))/d
Time = 0.32 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.93 \[ \int \frac {1}{(-3+5 \sin (c+d x))^3} \, dx=\frac {\frac {40 \, {\left (75 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 649 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 735 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 99\right )}}{{\left (3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 10 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3\right )}^{2}} + 387 \, \log \left ({\left | 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - 387 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \right |}\right )}{18432 \, d} \]
1/18432*(40*(75*tan(1/2*d*x + 1/2*c)^3 + 649*tan(1/2*d*x + 1/2*c)^2 - 735* tan(1/2*d*x + 1/2*c) + 99)/(3*tan(1/2*d*x + 1/2*c)^2 - 10*tan(1/2*d*x + 1/ 2*c) + 3)^2 + 387*log(abs(3*tan(1/2*d*x + 1/2*c) - 1)) - 387*log(abs(tan(1 /2*d*x + 1/2*c) - 3)))/d
Time = 7.33 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(-3+5 \sin (c+d x))^3} \, dx=\frac {43\,\mathrm {atanh}\left (\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}-\frac {5}{4}\right )}{1024\,d}+\frac {\frac {125\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6912}+\frac {3245\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{20736}-\frac {1225\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{6912}+\frac {55}{2304}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-\frac {20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {118\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{9}-\frac {20\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+1\right )} \]