### 3.48 $$\int \frac{1}{(-3-5 \sin (c+d x))^3} \, dx$$

Optimal. Leaf size=113 $-\frac{45 \cos (c+d x)}{512 d (5 \sin (c+d x)+3)}+\frac{5 \cos (c+d x)}{32 d (5 \sin (c+d x)+3)^2}+\frac{43 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+3 \cos \left (\frac{1}{2} (c+d x)\right )\right )}{2048 d}-\frac{43 \log \left (3 \sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{2048 d}$

[Out]

(43*Log[3*Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/(2048*d) - (43*Log[Cos[(c + d*x)/2] + 3*Sin[(c + d*x)/2]])/(20
48*d) + (5*Cos[c + d*x])/(32*d*(3 + 5*Sin[c + d*x])^2) - (45*Cos[c + d*x])/(512*d*(3 + 5*Sin[c + d*x]))

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Rubi [A]  time = 0.0770199, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 12, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.5, Rules used = {2664, 2754, 12, 2660, 616, 31} $-\frac{45 \cos (c+d x)}{512 d (5 \sin (c+d x)+3)}+\frac{5 \cos (c+d x)}{32 d (5 \sin (c+d x)+3)^2}+\frac{43 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+3 \cos \left (\frac{1}{2} (c+d x)\right )\right )}{2048 d}-\frac{43 \log \left (3 \sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{2048 d}$

Antiderivative was successfully veriﬁed.

[In]

Int[(-3 - 5*Sin[c + d*x])^(-3),x]

[Out]

(43*Log[3*Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/(2048*d) - (43*Log[Cos[(c + d*x)/2] + 3*Sin[(c + d*x)/2]])/(20
48*d) + (5*Cos[c + d*x])/(32*d*(3 + 5*Sin[c + d*x])^2) - (45*Cos[c + d*x])/(512*d*(3 + 5*Sin[c + d*x]))

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]

Rule 12

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

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(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]

Rule 616

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/Simp
[b/2 - q/2 + c*x, x], x], x] - Dist[c/q, Int[1/Simp[b/2 + q/2 + c*x, x], x], x]] /; FreeQ[{a, b, c}, x] && NeQ
[b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c] && PerfectSquareQ[b^2 - 4*a*c]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{1}{(-3-5 \sin (c+d x))^3} \, dx &=\frac{5 \cos (c+d x)}{32 d (3+5 \sin (c+d x))^2}+\frac{1}{32} \int \frac{6-5 \sin (c+d x)}{(-3-5 \sin (c+d x))^2} \, dx\\ &=\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))}+\frac{1}{512} \int \frac{43}{-3-5 \sin (c+d x)} \, dx\\ &=\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))}+\frac{43}{512} \int \frac{1}{-3-5 \sin (c+d x)} \, dx\\ &=\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))}+\frac{43 \operatorname{Subst}\left (\int \frac{1}{-3-10 x-3 x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{256 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))}-\frac{129 \operatorname{Subst}\left (\int \frac{1}{-9-3 x} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{2048 d}+\frac{129 \operatorname{Subst}\left (\int \frac{1}{-1-3 x} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{2048 d}\\ &=\frac{43 \log \left (3+\tan \left (\frac{1}{2} (c+d x)\right )\right )}{2048 d}-\frac{43 \log \left (1+3 \tan \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))}\\ \end{align*}

Mathematica [A]  time = 0.462492, size = 179, normalized size = 1.58 $\frac{60 \sin \left (\frac{1}{2} (c+d x)\right ) \left (\frac{3}{3 \sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}+\frac{1}{\sin \left (\frac{1}{2} (c+d x)\right )+3 \cos \left (\frac{1}{2} (c+d x)\right )}\right )-\frac{40}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+3 \cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{40}{\left (3 \sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+43 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+3 \cos \left (\frac{1}{2} (c+d x)\right )\right )-43 \log \left (3 \sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{2048 d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(43*Log[3*Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 43*Log[Cos[(c + d*x)/2] + 3*Sin[(c + d*x)/2]] - 40/(3*Cos[(c
+ d*x)/2] + Sin[(c + d*x)/2])^2 + 40/(Cos[(c + d*x)/2] + 3*Sin[(c + d*x)/2])^2 + 60*Sin[(c + d*x)/2]*((3*Cos[(
c + d*x)/2] + Sin[(c + d*x)/2])^(-1) + 3/(Cos[(c + d*x)/2] + 3*Sin[(c + d*x)/2])))/(2048*d)

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Maple [A]  time = 0.036, size = 114, normalized size = 1. \begin{align*}{\frac{25}{1152\,d} \left ( 3\,\tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-2}}-{\frac{155}{4608\,d} \left ( 3\,\tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-1}}-{\frac{43}{2048\,d}\ln \left ( 3\,\tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }-{\frac{25}{128\,d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +3 \right ) ^{-2}}+{\frac{15}{512\,d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +3 \right ) ^{-1}}+{\frac{43}{2048\,d}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +3 \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25/1152/d/(3*tan(1/2*d*x+1/2*c)+1)^2-155/4608/d/(3*tan(1/2*d*x+1/2*c)+1)-43/2048/d*ln(3*tan(1/2*d*x+1/2*c)+1)-
25/128/d/(tan(1/2*d*x+1/2*c)+3)^2+15/512/d/(tan(1/2*d*x+1/2*c)+3)+43/2048/d*ln(tan(1/2*d*x+1/2*c)+3)

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Maxima [A]  time = 0.977954, size = 263, normalized size = 2.33 \begin{align*} -\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} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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 + 6
0*sin(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)/(c
os(d*x + c) + 1) + 1) - 387*log(sin(d*x + c)/(cos(d*x + c) + 1) + 3))/d

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Fricas [A]  time = 1.94951, size = 389, normalized size = 3.44 \begin{align*} \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 )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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) +
440*cos(d*x + c))/(25*d*cos(d*x + c)^2 - 30*d*sin(d*x + c) - 34*d)

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Sympy [A]  time = 6.38025, size = 1229, normalized size = 10.88 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)), (-387*log(tan(c/2 + d*x/2) + 1/3)*tan(c/2 + d*x/2)**4/(184
32*d*tan(c/2 + d*x/2)**4 + 122880*d*tan(c/2 + d*x/2)**3 + 241664*d*tan(c/2 + d*x/2)**2 + 122880*d*tan(c/2 + d*
x/2) + 18432*d) - 2580*log(tan(c/2 + d*x/2) + 1/3)*tan(c/2 + d*x/2)**3/(18432*d*tan(c/2 + d*x/2)**4 + 122880*d
*tan(c/2 + d*x/2)**3 + 241664*d*tan(c/2 + d*x/2)**2 + 122880*d*tan(c/2 + d*x/2) + 18432*d) - 5074*log(tan(c/2
+ d*x/2) + 1/3)*tan(c/2 + d*x/2)**2/(18432*d*tan(c/2 + d*x/2)**4 + 122880*d*tan(c/2 + d*x/2)**3 + 241664*d*tan
(c/2 + d*x/2)**2 + 122880*d*tan(c/2 + d*x/2) + 18432*d) - 2580*log(tan(c/2 + d*x/2) + 1/3)*tan(c/2 + d*x/2)/(1
8432*d*tan(c/2 + d*x/2)**4 + 122880*d*tan(c/2 + d*x/2)**3 + 241664*d*tan(c/2 + d*x/2)**2 + 122880*d*tan(c/2 +
d*x/2) + 18432*d) - 387*log(tan(c/2 + d*x/2) + 1/3)/(18432*d*tan(c/2 + d*x/2)**4 + 122880*d*tan(c/2 + d*x/2)**
3 + 241664*d*tan(c/2 + d*x/2)**2 + 122880*d*tan(c/2 + d*x/2) + 18432*d) + 387*log(tan(c/2 + d*x/2) + 3)*tan(c/
2 + d*x/2)**4/(18432*d*tan(c/2 + d*x/2)**4 + 122880*d*tan(c/2 + d*x/2)**3 + 241664*d*tan(c/2 + d*x/2)**2 + 122
880*d*tan(c/2 + d*x/2) + 18432*d) + 2580*log(tan(c/2 + d*x/2) + 3)*tan(c/2 + d*x/2)**3/(18432*d*tan(c/2 + d*x/
2)**4 + 122880*d*tan(c/2 + d*x/2)**3 + 241664*d*tan(c/2 + d*x/2)**2 + 122880*d*tan(c/2 + d*x/2) + 18432*d) + 5
074*log(tan(c/2 + d*x/2) + 3)*tan(c/2 + d*x/2)**2/(18432*d*tan(c/2 + d*x/2)**4 + 122880*d*tan(c/2 + d*x/2)**3
+ 241664*d*tan(c/2 + d*x/2)**2 + 122880*d*tan(c/2 + d*x/2) + 18432*d) + 2580*log(tan(c/2 + d*x/2) + 3)*tan(c/2
+ d*x/2)/(18432*d*tan(c/2 + d*x/2)**4 + 122880*d*tan(c/2 + d*x/2)**3 + 241664*d*tan(c/2 + d*x/2)**2 + 122880*
d*tan(c/2 + d*x/2) + 18432*d) + 387*log(tan(c/2 + d*x/2) + 3)/(18432*d*tan(c/2 + d*x/2)**4 + 122880*d*tan(c/2
+ d*x/2)**3 + 241664*d*tan(c/2 + d*x/2)**2 + 122880*d*tan(c/2 + d*x/2) + 18432*d) - 50*tan(c/2 + d*x/2)**4/(18
432*d*tan(c/2 + d*x/2)**4 + 122880*d*tan(c/2 + d*x/2)**3 + 241664*d*tan(c/2 + d*x/2)**2 + 122880*d*tan(c/2 + d
*x/2) + 18432*d) - 3540*tan(c/2 + d*x/2)**2/(18432*d*tan(c/2 + d*x/2)**4 + 122880*d*tan(c/2 + d*x/2)**3 + 2416
64*d*tan(c/2 + d*x/2)**2 + 122880*d*tan(c/2 + d*x/2) + 18432*d) - 3600*tan(c/2 + d*x/2)/(18432*d*tan(c/2 + d*x
/2)**4 + 122880*d*tan(c/2 + d*x/2)**3 + 241664*d*tan(c/2 + d*x/2)**2 + 122880*d*tan(c/2 + d*x/2) + 18432*d) -
490/(18432*d*tan(c/2 + d*x/2)**4 + 122880*d*tan(c/2 + d*x/2)**3 + 241664*d*tan(c/2 + d*x/2)**2 + 122880*d*tan(
c/2 + d*x/2) + 18432*d), True))

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Giac [A]  time = 1.16365, size = 144, normalized size = 1.27 \begin{align*} \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} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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