\(\int \frac {1}{(-5+3 \sin (c+d x))^3} \, dx\) [28]

Optimal result

Integrand size = 12, antiderivative size = 83 \[ \int \frac {1}{(-5+3 \sin (c+d x))^3} \, dx=-\frac {59 x}{2048}+\frac {59 \arctan \left (\frac {\cos (c+d x)}{3-\sin (c+d x)}\right )}{1024 d}+\frac {3 \cos (c+d x)}{32 d (5-3 \sin (c+d x))^2}+\frac {45 \cos (c+d x)}{512 d (5-3 \sin (c+d x))} \]

[Out]

-59/2048*x+59/1024*arctan(cos(d*x+c)/(3-sin(d*x+c)))/d+3/32*cos(d*x+c)/d/(5-3*sin(d*x+c))^2+45/512*cos(d*x+c)/
d/(5-3*sin(d*x+c))

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2743, 2833, 12, 2737} \[ \int \frac {1}{(-5+3 \sin (c+d x))^3} \, dx=\frac {59 \arctan \left (\frac {\cos (c+d x)}{3-\sin (c+d x)}\right )}{1024 d}+\frac {45 \cos (c+d x)}{512 d (5-3 \sin (c+d x))}+\frac {3 \cos (c+d x)}{32 d (5-3 \sin (c+d x))^2}-\frac {59 x}{2048} \]

[In]

Int[(-5 + 3*Sin[c + d*x])^(-3),x]

[Out]

(-59*x)/2048 + (59*ArcTan[Cos[c + d*x]/(3 - Sin[c + d*x])])/(1024*d) + (3*Cos[c + d*x])/(32*d*(5 - 3*Sin[c + d
*x])^2) + (45*Cos[c + d*x])/(512*d*(5 - 3*Sin[c + d*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2737

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{q = Rt[a^2 - b^2, 2]}, Simp[-x/q, x] - Sim
p[(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] && NegQ[a]

Rule 2743

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] + Dist[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] && Integ
erQ[2*n]

Rule 2833

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] + Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {3 \cos (c+d x)}{32 d (5-3 \sin (c+d x))^2}-\frac {1}{32} \int \frac {10+3 \sin (c+d x)}{(-5+3 \sin (c+d x))^2} \, dx \\ & = \frac {3 \cos (c+d x)}{32 d (5-3 \sin (c+d x))^2}+\frac {45 \cos (c+d x)}{512 d (5-3 \sin (c+d x))}+\frac {1}{512} \int \frac {59}{-5+3 \sin (c+d x)} \, dx \\ & = \frac {3 \cos (c+d x)}{32 d (5-3 \sin (c+d x))^2}+\frac {45 \cos (c+d x)}{512 d (5-3 \sin (c+d x))}+\frac {59}{512} \int \frac {1}{-5+3 \sin (c+d x)} \, dx \\ & = -\frac {59 x}{2048}+\frac {59 \arctan \left (\frac {\cos (c+d x)}{3-\sin (c+d x)}\right )}{1024 d}+\frac {3 \cos (c+d x)}{32 d (5-3 \sin (c+d x))^2}+\frac {45 \cos (c+d x)}{512 d (5-3 \sin (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.36 \[ \int \frac {1}{(-5+3 \sin (c+d x))^3} \, dx=\frac {59 \arctan \left (\frac {2 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}\right )+\frac {546 \cos (c+d x)+9 (-59+9 \cos (2 (c+d x))+60 \sin (c+d x)-15 \sin (2 (c+d x)))}{(5-3 \sin (c+d x))^2}}{1024 d} \]

[In]

Integrate[(-5 + 3*Sin[c + d*x])^(-3),x]

[Out]

(59*ArcTan[(2*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]))/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])] + (546*Cos[c + d*x
] + 9*(-59 + 9*Cos[2*(c + d*x)] + 60*Sin[c + d*x] - 15*Sin[2*(c + d*x)]))/(5 - 3*Sin[c + d*x])^2)/(1024*d)

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.10

method	result	size
derivativedivides	\(\frac {-\frac {50 \left (\frac {963 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64000}-\frac {11739 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{320000}+\frac {2313 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{64000}-\frac {273}{12800}\right )}{{\left (5 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+5\right )}^{2}}-\frac {59 \arctan \left (\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4}-\frac {3}{4}\right )}{1024}}{d}\)	\(91\)
default	\(\frac {-\frac {50 \left (\frac {963 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64000}-\frac {11739 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{320000}+\frac {2313 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{64000}-\frac {273}{12800}\right )}{{\left (5 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+5\right )}^{2}}-\frac {59 \arctan \left (\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4}-\frac {3}{4}\right )}{1024}}{d}\)	\(91\)
risch	\(\frac {-\frac {885 i {\mathrm e}^{2 i \left (d x +c \right )}}{256}+\frac {177 \,{\mathrm e}^{3 i \left (d x +c \right )}}{256}-\frac {723 \,{\mathrm e}^{i \left (d x +c \right )}}{256}+\frac {135 i}{256}}{\left (3 \,{\mathrm e}^{2 i \left (d x +c \right )}-3-10 i {\mathrm e}^{i \left (d x +c \right )}\right )^{2} d}+\frac {59 i \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i}{3}\right )}{2048 d}-\frac {59 i \ln \left ({\mathrm e}^{i \left (d x +c \right )}-3 i\right )}{2048 d}\)	\(109\)
parallelrisch	\(\frac {-1062+59 i \left (-59+9 \cos \left (2 d x +2 c \right )+60 \sin \left (d x +c \right )\right ) \ln \left (5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3-4 i\right )+59 i \left (59-9 \cos \left (2 d x +2 c \right )-60 \sin \left (d x +c \right )\right ) \ln \left (5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3+4 i\right )-2184 \cos \left (d x +c \right )+162 \cos \left (2 d x +2 c \right )+1080 \sin \left (d x +c \right )+540 \sin \left (2 d x +2 c \right )}{2048 d \left (-59+9 \cos \left (2 d x +2 c \right )+60 \sin \left (d x +c \right )\right )}\)	\(149\)

[In]

int(1/(-5+3*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(-50*(963/64000*tan(1/2*d*x+1/2*c)^3-11739/320000*tan(1/2*d*x+1/2*c)^2+2313/64000*tan(1/2*d*x+1/2*c)-273/1
2800)/(5*tan(1/2*d*x+1/2*c)^2-6*tan(1/2*d*x+1/2*c)+5)^2-59/1024*arctan(5/4*tan(1/2*d*x+1/2*c)-3/4))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.13 \[ \int \frac {1}{(-5+3 \sin (c+d x))^3} \, dx=-\frac {59 \, {\left (9 \, \cos \left (d x + c\right )^{2} + 30 \, \sin \left (d x + c\right ) - 34\right )} \arctan \left (\frac {5 \, \sin \left (d x + c\right ) - 3}{4 \, \cos \left (d x + c\right )}\right ) - 540 \, \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 1092 \, \cos \left (d x + c\right )}{2048 \, {\left (9 \, d \cos \left (d x + c\right )^{2} + 30 \, d \sin \left (d x + c\right ) - 34 \, d\right )}} \]

[In]

integrate(1/(-5+3*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/2048*(59*(9*cos(d*x + c)^2 + 30*sin(d*x + c) - 34)*arctan(1/4*(5*sin(d*x + c) - 3)/cos(d*x + c)) - 540*cos(
d*x + c)*sin(d*x + c) + 1092*cos(d*x + c))/(9*d*cos(d*x + c)^2 + 30*d*sin(d*x + c) - 34*d)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.95 (sec) , antiderivative size = 915, normalized size of antiderivative = 11.02 \[ \int \frac {1}{(-5+3 \sin (c+d x))^3} \, dx=\text {Too large to display} \]

[In]

integrate(1/(-5+3*sin(d*x+c))**3,x)

[Out]

Piecewise((x/(-5 + 3*sin(2*atan(3/5 - 4*I/5)))**3, Eq(c, -d*x + 2*atan(3/5 - 4*I/5))), (x/(-5 + 3*sin(2*atan(3
/5 + 4*I/5)))**3, Eq(c, -d*x + 2*atan(3/5 + 4*I/5))), (x/(3*sin(c) - 5)**3, Eq(d, 0)), (-36875*(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)**4/(640000*d*tan(c/2 + d*x/2)**4 - 15
36000*d*tan(c/2 + d*x/2)**3 + 2201600*d*tan(c/2 + d*x/2)**2 - 1536000*d*tan(c/2 + d*x/2) + 640000*d) + 88500*(
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)**3/(640000*d*tan(c/2 +
d*x/2)**4 - 1536000*d*tan(c/2 + d*x/2)**3 + 2201600*d*tan(c/2 + d*x/2)**2 - 1536000*d*tan(c/2 + d*x/2) + 64000
0*d) - 126850*(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/(6400
00*d*tan(c/2 + d*x/2)**4 - 1536000*d*tan(c/2 + d*x/2)**3 + 2201600*d*tan(c/2 + d*x/2)**2 - 1536000*d*tan(c/2 +
d*x/2) + 640000*d) + 88500*(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)/(640000*d*tan(c/2 + d*x/2)**4 - 1536000*d*tan(c/2 + d*x/2)**3 + 2201600*d*tan(c/2 + d*x/2)**2 - 1536000*
d*tan(c/2 + d*x/2) + 640000*d) - 36875*(atan(5*tan(c/2 + d*x/2)/4 - 3/4) + pi*floor((c/2 + d*x/2 - pi/2)/pi))/
(640000*d*tan(c/2 + d*x/2)**4 - 1536000*d*tan(c/2 + d*x/2)**3 + 2201600*d*tan(c/2 + d*x/2)**2 - 1536000*d*tan(
c/2 + d*x/2) + 640000*d) - 19260*tan(c/2 + d*x/2)**3/(640000*d*tan(c/2 + d*x/2)**4 - 1536000*d*tan(c/2 + d*x/2
)**3 + 2201600*d*tan(c/2 + d*x/2)**2 - 1536000*d*tan(c/2 + d*x/2) + 640000*d) + 46956*tan(c/2 + d*x/2)**2/(640
000*d*tan(c/2 + d*x/2)**4 - 1536000*d*tan(c/2 + d*x/2)**3 + 2201600*d*tan(c/2 + d*x/2)**2 - 1536000*d*tan(c/2
+ d*x/2) + 640000*d) - 46260*tan(c/2 + d*x/2)/(640000*d*tan(c/2 + d*x/2)**4 - 1536000*d*tan(c/2 + d*x/2)**3 +
2201600*d*tan(c/2 + d*x/2)**2 - 1536000*d*tan(c/2 + d*x/2) + 640000*d) + 27300/(640000*d*tan(c/2 + d*x/2)**4 -
1536000*d*tan(c/2 + d*x/2)**3 + 2201600*d*tan(c/2 + d*x/2)**2 - 1536000*d*tan(c/2 + d*x/2) + 640000*d), True)
)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 173 vs. \(2 (74) = 148\).

Time = 0.27 (sec) , antiderivative size = 173, normalized size of antiderivative = 2.08 \[ \int \frac {1}{(-5+3 \sin (c+d x))^3} \, dx=\frac {\frac {12 \, {\left (\frac {3855 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {3913 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {1605 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - 2275\right )}}{\frac {60 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {86 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {60 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {25 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - 25} - 1475 \, \arctan \left (\frac {5 \, \sin \left (d x + c\right )}{4 \, {\left (\cos \left (d x + c\right ) + 1\right )}} - \frac {3}{4}\right )}{25600 \, d} \]

[In]

integrate(1/(-5+3*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/25600*(12*(3855*sin(d*x + c)/(cos(d*x + c) + 1) - 3913*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1605*sin(d*x +
c)^3/(cos(d*x + c) + 1)^3 - 2275)/(60*sin(d*x + c)/(cos(d*x + c) + 1) - 86*sin(d*x + c)^2/(cos(d*x + c) + 1)^2
+ 60*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 25*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 25) - 1475*arctan(5/4*sin
(d*x + c)/(cos(d*x + c) + 1) - 3/4))/d

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.47 \[ \int \frac {1}{(-5+3 \sin (c+d x))^3} \, dx=-\frac {1475 \, d x + 1475 \, c + \frac {24 \, {\left (1605 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3913 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3855 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2275\right )}}{{\left (5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5\right )}^{2}} + 2950 \, \arctan \left (\frac {3 \, \cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 3}{\cos \left (d x + c\right ) + 3 \, \sin \left (d x + c\right ) - 9}\right )}{51200 \, d} \]

[In]

integrate(1/(-5+3*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/51200*(1475*d*x + 1475*c + 24*(1605*tan(1/2*d*x + 1/2*c)^3 - 3913*tan(1/2*d*x + 1/2*c)^2 + 3855*tan(1/2*d*x
+ 1/2*c) - 2275)/(5*tan(1/2*d*x + 1/2*c)^2 - 6*tan(1/2*d*x + 1/2*c) + 5)^2 + 2950*arctan((3*cos(d*x + c) - si
n(d*x + c) + 3)/(cos(d*x + c) + 3*sin(d*x + c) - 9)))/d

Mupad [B] (verification not implemented)

Time = 6.18 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.35 \[ \int \frac {1}{(-5+3 \sin (c+d x))^3} \, dx=\frac {59\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{1024\,d}-\frac {59\,\mathrm {atan}\left (\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}-\frac {3}{4}\right )}{1024\,d}-\frac {\frac {963\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{1280}-\frac {11739\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{6400}+\frac {2313\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{1280}-\frac {273}{256}}{d\,{\left (5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+5\right )}^2} \]

[In]

int(1/(3*sin(c + d*x) - 5)^3,x)

[Out]

(59*(atan(tan(c/2 + (d*x)/2)) - (d*x)/2))/(1024*d) - (59*atan((5*tan(c/2 + (d*x)/2))/4 - 3/4))/(1024*d) - ((23
13*tan(c/2 + (d*x)/2))/1280 - (11739*tan(c/2 + (d*x)/2)^2)/6400 + (963*tan(c/2 + (d*x)/2)^3)/1280 - 273/256)/(
d*(5*tan(c/2 + (d*x)/2)^2 - 6*tan(c/2 + (d*x)/2) + 5)^2)