\(\int \frac {1}{(5+3 \sin (c+d x))^3} \, dx\) [20]

Optimal result

Integrand size = 12, antiderivative size = 81 \[ \int \frac {1}{(5+3 \sin (c+d x))^3} \, dx=\frac {59 x}{2048}+\frac {59 \arctan \left (\frac {\cos (c+d x)}{3+\sin (c+d x)}\right )}{1024 d}+\frac {3 \cos (c+d x)}{32 d (5+3 \sin (c+d x))^2}+\frac {45 \cos (c+d x)}{512 d (5+3 \sin (c+d x))} \]

[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 = 81, 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, 2736} \[ \int \frac {1}{(5+3 \sin (c+d x))^3} \, dx=\frac {59 \arctan \left (\frac {\cos (c+d x)}{\sin (c+d x)+3}\right )}{1024 d}+\frac {45 \cos (c+d x)}{512 d (3 \sin (c+d x)+5)}+\frac {3 \cos (c+d x)}{32 d (3 \sin (c+d x)+5)^2}+\frac {59 x}{2048} \]

[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 2736

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]

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 = 0.27 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.40 \[ \int \frac {1}{(5+3 \sin (c+d x))^3} \, dx=\frac {59 \arctan \left (\frac {2 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}\right )+\frac {546 \cos (c+d x)+9 (-59+9 \cos (2 (c+d x))-60 \sin (c+d x)+15 \sin (2 (c+d x)))}{(5+3 \sin (c+d x))^2}}{1024 d} \]

[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 = 0.30 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.12

method	result	size
derivativedivides	\(\frac {\frac {\frac {963 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{1280}+\frac {11739 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6400}+\frac {2313 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1280}+\frac {273}{256}}{{\left (5 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+5\right )}^{2}}+\frac {59 \arctan \left (\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4}+\frac {3}{4}\right )}{1024}}{d}\)	\(91\)
default	\(\frac {\frac {\frac {963 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{1280}+\frac {11739 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6400}+\frac {2313 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1280}+\frac {273}{256}}{{\left (5 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+5\right )}^{2}}+\frac {59 \arctan \left (\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4}+\frac {3}{4}\right )}{1024}}{d}\)	\(91\)
risch	\(\frac {\frac {885 i {\mathrm e}^{2 i \left (d x +c \right )}}{256}+\frac {177 \,{\mathrm e}^{3 i \left (d x +c \right )}}{256}-\frac {723 \,{\mathrm e}^{i \left (d x +c \right )}}{256}-\frac {135 i}{256}}{\left (3 \,{\mathrm e}^{2 i \left (d x +c \right )}-3+10 i {\mathrm e}^{i \left (d x +c \right )}\right )^{2} d}+\frac {59 i \ln \left ({\mathrm e}^{i \left (d x +c \right )}+3 i\right )}{2048 d}-\frac {59 i \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i}{3}\right )}{2048 d}\)	\(109\)
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)+273/12
800)/(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 = 0.28 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.16 \[ \int \frac {1}{(5+3 \sin (c+d x))^3} \, dx=\frac {59 \, {\left (9 \, \cos \left (d x + c\right )^{2} - 30 \, \sin \left (d x + c\right ) - 34\right )} \arctan \left (\frac {5 \, \sin \left (d x + c\right ) + 3}{4 \, \cos \left (d x + c\right )}\right ) - 540 \, \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 1092 \, \cos \left (d x + c\right )}{2048 \, {\left (9 \, d \cos \left (d x + c\right )^{2} - 30 \, d \sin \left (d x + c\right ) - 34 \, d\right )}} \]

[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 = 1.92 (sec) , antiderivative size = 918, normalized size of antiderivative = 11.33 \[ \int \frac {1}{(5+3 \sin (c+d x))^3} \, dx=\text {Too large to display} \]

[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 + 15360
00*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*(ata
n(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) + 640000*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/(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) + 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*t
an(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))/(64
0000*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/(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) + 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 + 220
1600*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 + 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), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 173 vs. \(2 (73) = 146\).

Time = 0.27 (sec) , antiderivative size = 173, normalized size of antiderivative = 2.14 \[ \int \frac {1}{(5+3 \sin (c+d x))^3} \, dx=\frac {\frac {12 \, {\left (\frac {3855 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3913 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {1605 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + 2275\right )}}{\frac {60 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {86 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {60 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {25 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 25} + 1475 \, \arctan \left (\frac {5 \, \sin \left (d x + c\right )}{4 \, {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {3}{4}\right )}{25600 \, d} \]

[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 = 0.32 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.49 \[ \int \frac {1}{(5+3 \sin (c+d x))^3} \, dx=\frac {1475 \, d x + 1475 \, c + \frac {24 \, {\left (1605 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3913 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3855 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2275\right )}}{{\left (5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5\right )}^{2}} + 2950 \, \arctan \left (-\frac {3 \, \cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 3}{\cos \left (d x + c\right ) - 3 \, \sin \left (d x + c\right ) - 9}\right )}{51200 \, d} \]

[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 = 6.16 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.37 \[ \int \frac {1}{(5+3 \sin (c+d x))^3} \, dx=\frac {59\,\mathrm {atan}\left (\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+\frac {3}{4}\right )}{1024\,d}-\frac {59\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{1024\,d}+\frac {\frac {963\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{1280}+\frac {11739\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{6400}+\frac {2313\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{1280}+\frac {273}{256}}{d\,{\left (5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+5\right )}^2} \]

[In]

int(1/(3*sin(c + d*x) + 5)^3,x)

[Out]

(59*atan((5*tan(c/2 + (d*x)/2))/4 + 3/4))/(1024*d) - (59*(atan(tan(c/2 + (d*x)/2)) - (d*x)/2))/(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*(6*tan(c/2 + (d*x)/2) + 5*tan(c/2 + (d*x)/2)^2 + 5)^2)