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

Optimal. Leaf size=81 \[ -\frac {45 \cos (c+d x)}{512 d (3 \sin (c+d x)+5)}-\frac {3 \cos (c+d x)}{32 d (3 \sin (c+d x)+5)^2}-\frac {59 \tan ^{-1}\left (\frac {\cos (c+d x)}{\sin (c+d x)+3}\right )}{1024 d}-\frac {59 x}{2048} \]

[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))

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Rubi [A]  time = 0.06, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2664, 2754, 12, 2658} \[ -\frac {45 \cos (c+d x)}{512 d (3 \sin (c+d x)+5)}-\frac {3 \cos (c+d x)}{32 d (3 \sin (c+d x)+5)^2}-\frac {59 \tan ^{-1}\left (\frac {\cos (c+d x)}{\sin (c+d x)+3}\right )}{1024 d}-\frac {59 x}{2048} \]

Antiderivative was successfully verified.

[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 2658

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*ArcTan[(b*Cos[c + d*x])/(a - q + b*Sin[c + d*x])])/(d*q), x]] /; FreeQ[{a, b, c, d}, x] && GtQ[a^2 - b^2,
0] && NegQ[a]

Rule 2664

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] && Integer
Q[2*n]

Rule 2754

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*} \int \frac {1}{(-5-3 \sin (c+d x))^3} \, dx &=-\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 \tan ^{-1}\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*}

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Mathematica [A]  time = 0.36, size = 114, normalized size = 1.41 \[ \frac {\frac {3 (3 (60 \sin (c+d x)-15 \sin (2 (c+d x))-9 \cos (2 (c+d x))+59)-182 \cos (c+d x))}{(3 \sin (c+d x)+5)^2}-59 \tan ^{-1}\left (\frac {2 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}\right )}{1024 d} \]

Antiderivative was successfully verified.

[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])] + (3*(-182*Cos[c
+ d*x] + 3*(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
)

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fricas [A]  time = 0.47, size = 94, normalized size = 1.16 \[ -\frac {59 \, {\left (9 \, \cos \left (d x + c\right )^{2} - 30 \, \sin \left (d x + c\right ) - 34\right )} \arctan \left (\frac {5 \, \sin \left (d x + c\right ) + 3}{4 \, \cos \left (d x + c\right )}\right ) - 540 \, \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 1092 \, \cos \left (d x + c\right )}{2048 \, {\left (9 \, d \cos \left (d x + c\right )^{2} - 30 \, d \sin \left (d x + c\right ) - 34 \, d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 1.43, size = 121, normalized size = 1.49 \[ -\frac {1475 \, d x + 1475 \, c + \frac {24 \, {\left (1605 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3913 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3855 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2275\right )}}{{\left (5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5\right )}^{2}} + 2950 \, \arctan \left (-\frac {3 \, \cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 3}{\cos \left (d x + c\right ) - 3 \, \sin \left (d x + c\right ) - 9}\right )}{51200 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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) + s
in(d*x + c) + 3)/(cos(d*x + c) - 3*sin(d*x + c) - 9)))/d

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maple [B]  time = 0.09, size = 184, normalized size = 2.27 \[ -\frac {963 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{1280 d \left (5 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+5\right )^{2}}-\frac {11739 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6400 d \left (5 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+5\right )^{2}}-\frac {2313 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1280 d \left (5 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+5\right )^{2}}-\frac {273}{256 d \left (5 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+5\right )^{2}}-\frac {59 \arctan \left (\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4}+\frac {3}{4}\right )}{1024 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-963/1280/d/(5*tan(1/2*d*x+1/2*c)^2+6*tan(1/2*d*x+1/2*c)+5)^2*tan(1/2*d*x+1/2*c)^3-11739/6400/d/(5*tan(1/2*d*x
+1/2*c)^2+6*tan(1/2*d*x+1/2*c)+5)^2*tan(1/2*d*x+1/2*c)^2-2313/1280/d/(5*tan(1/2*d*x+1/2*c)^2+6*tan(1/2*d*x+1/2
*c)+5)^2*tan(1/2*d*x+1/2*c)-273/256/d/(5*tan(1/2*d*x+1/2*c)^2+6*tan(1/2*d*x+1/2*c)+5)^2-59/1024/d*arctan(5/4*t
an(1/2*d*x+1/2*c)+3/4)

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maxima [B]  time = 0.70, size = 173, normalized size = 2.14 \[ -\frac {\frac {12 \, {\left (\frac {3855 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3913 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {1605 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + 2275\right )}}{\frac {60 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {86 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {60 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {25 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 25} + 1475 \, \arctan \left (\frac {5 \, \sin \left (d x + c\right )}{4 \, {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {3}{4}\right )}{25600 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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*si
n(d*x + c)/(cos(d*x + c) + 1) + 3/4))/d

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mupad [B]  time = 4.65, size = 112, normalized size = 1.38 \[ \frac {59\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{1024\,d}-\frac {59\,\mathrm {atan}\left (\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+\frac {3}{4}\right )}{1024\,d}-\frac {\frac {963\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{1280}+\frac {11739\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{6400}+\frac {2313\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{1280}+\frac {273}{256}}{d\,{\left (5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+5\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[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*(6*tan(c/2 + (d*x)/2) + 5*tan(c/2 + (d*x)/2)^2 + 5)^2)

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sympy [A]  time = 4.75, size = 921, normalized size = 11.37 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[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 + 1
536000*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) + 6400
00*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/(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) - 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/(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) - 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
))

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